chemical bond.
The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond.
A few important questions.
Why do atoms combine?
Why are only certain combinations possible?
Why do some atoms combine while certain others do not?
Why do molecules possess definite shapes?
To answer such questions different theories and concepts have been put forward from time to time.
These are Kössel-Lewis approach,
Valence Shell
Electron Pair Repulsion (VSEPR) Theory,
Valence Bond (VB)
Theory and Molecular Orbital (MO) Theory.
KÖssel-Lewis Approach to Chemical Bonding (1916)
Worked independently.
Lewis pictured the atom in terms of a positively charged ‘Kernel’ (the nucleus plus
the inner electrons) and the outer shell that
could accommodate a maximum of eight
electrons.
Kernel- Lewis pictured the atom in terms of a positively charged ‘Kernel’ (the nucleus plus the inner electrons) and the outer shell that could accommodate a maximum of eight electrons. He, further assumed that these eight electrons occupy the corners of a cube which surround the ‘Kernel’.
The octet of electrons, represents a particularly stable electronic arrangement.
Lewis postulated that atoms achieve the stable octet when they are linked by
chemical bonds.
Write atomic number of first twenty elements of periodic table, Also give their electronic configuration, also show their Lewis Structures.
Lewis Symbols:
In the formation of a molecule, only the outer shell electrons take part in chemical combination and they are known as valence electrons. The inner shell electrons are well protected and are generally not involved in the combination process.
G.N. Lewis, an American chemist introduced simple notations to represent valence electrons in an atom. These notations are called Lewis symbols.
Significance of Lewis Symbols :
a. The number of dots around the symbol represents the number of valence electrons.
b. This number of valence electrons helps to calculate the common or group valence of the element. The group valence of the elements is generally either equal to the number of dots in Lewis symbols or 8 minus the number of dots or valence electrons.
In relation to chemical bonding, Kössel gave the following postulates known as kössel’s theory of chemical bonding :
The following example can elaborate on it.
Ca has a configuration [Ar]4s2. It is a group 2 element and can easily lose 2 electrons to attain the configuration of argon.
F has a configuration of [He] 2s2.
It is a group 17 element and can gain an electron to attain the stable electron configuration of neon. So, the 2 electrons lost by Ca can be gained by 2 F atoms.
Ca→ Ca2++2e
F+e- → F
Ca2++2F-→Ca F2+
The bond formed due to electrostatic attraction between the positive cation and negative anion is termed an electrovalent bond. We can say that the number of unit charge(s) on the ion is equal to the electrovalence and that Ca has an electrovalence of +2, while F has electrovalence of -1.
The importance of Kössel Postulates can be understood from the fact that it provides the abscess for modern concepts concerning the formation of ions by transfer of electrons and the formation of ionic crystalline compounds.
His views also helped in the systematisation and understanding of ionic compounds.
It is also to be noted that he was aware that there are still a large number of compounds that did not fit in this concept.
Except for the noble gases, which are capable of independent existence, most of the elements combine to form compounds. Kössel and Lewis were the first scientists to provide an explanation as to why chemical bonds are formed. They explained chemical bonds in terms of electrons.
Lewis gave a notation of writing symbols of elements. He imagined atoms to be made up of a kernel and outer shell with a maximum of 8 electrons. The term kernel was used to describe the nucleus along with inner electrons.He presumed that the outer 8 electrons will occupy the corner of the cube surrounding the kernel. All the eight corners will be occupied in noble gases. This stable arrangement was called octet of electrons.
In his postulates. Kössel explained that an electropositive element loses electrons and an electronegative element gains electrons to achieve a stable electronic configuration of noble gases.
The chemical bond formed between the negative anion and positive cation is called electrovalent bond because of electrostatic attraction present between them. So we can say that the number of unit charge(s) on the ion is equal to the electrovalence.
The importance of Kössel Postulates can be understood from the fact that it provides the abscess for modern concepts concerning the formation of ions by transfer of electrons and the formation of ionic crystalline compounds.
For example, the formation of NaCl and CaF2from can be explained as: The bond formed, as a result of the electrostatic attraction between the positive and negative ions was termed as the electrovalent bond. The electrovalence is thus equal to the number of unit charge(s) on the ion. Thus, calcium is assigned a
positive electrovalence of two, while chlorine a negative electrovalence of one.
What is the Kossel theory?
Kössel and Lewis explained chemical bonding in terms of electrons. They explained that electronegative elements gain electrons and electropositive elements lose electrons to form bonds and acquire a stable noble gas configuration.
Who were kossel and Lewis?
Kossel and Lewis are two scientists who made significant contributions to our understanding of chemical bonding. Walther Kossel was a German physicist who postulated that atoms gain, lose, or share electrons to achieve a noble gas electron configuration.
Lewis postulated that atoms achieve
the stable octet when they are linked by
chemical bonds. In the case of sodium and
chlorine, this can happen by the transfer of
an electron from sodium to chlorine thereby
giving the Na+ and Cl- ions.
In the case of
other molecules like Cl2, H2, F2, etc., the bond is formed by the sharing of a pair of electrons between the atoms. In the process each atom attains a stable outer octet of electrons.
What are kernels in chemistry?
Kernel refers to the nucleus and all of the electrons, except those in the valance (outer) shell. Positive ions surrounded and held togerher by sea of free electrons in metallic solids, these are called kernels. Kernel electrons are those which are not present in valence shell.
Kernel = Nucleus + inner electrons.
Octet Rule
According to octet rule that atoms are most stable when their valence shells are filled with eight electrons.
It is based on the observation that the atoms of the main group elements have a tendency to participate in chemical bonding in such a way that each atom of the resulting molecule has eight electrons in the valence shell.
The octet rule is only applicable to the main group elements.
The elements that obey this rule include the s-block elements and the p-block elements (except hydrogen, helium, and lithium).
4.1.2 Covalent Bond
Langmuir (1919) refined t h e Lewis postulations by abandoning the idea of
the stationary cubical arrangement of the octet, and by introducing the term covalent bond.
The Lewis-Langmuir theory can be
understood by considering the formation of
the chlorine molecule, Cl2.
Q. What is atomic number of Chlorine? Write its electronic configuration, also mention valence electrons.
The Cl atom with electronic configuration, [Ne]3s2 3p 5, is one electron short of the argon configuration.
The formation of the Cl2 molecule can be understood in terms of the sharing of a pair of electrons between the two chlorine atoms, each chlorine atom contributing one electron to the shared pair. In the process both chlorine atoms attain the outer shell octet of the nearest noble gas (i.e., argon).
How Noble gas electronic configuration can be ?
i. Transference of electrons
ii. Mutual sharing of electrons
iii. Donation of lone pair of electrons
Covalent Bond
A force which binds atoms of same or different elements by mutual sharing of electrons is called a covalent bond, If the combining atoms are same the covalent molecule is known as homoatomic. If they are different, they are known as heteroatomic molecule.
Covalency:
The number of electrons which an atom contributes towards mutual sharing during the formation of a chemical bond called its covalency in that compound.
Single covalent bond:
A covalent bond formed by the mutual sharing of one pair of electrons is called a single covalent bond, or simply a single bond. A single covalent bond is represented by a small line (−) between the two atoms.
Double covalent bond:
A covalent bond formed by the mutual sharing of
two pair of electrons is called a double covalent bond, or simply a double bond. A
double covalent bond is represented by two small horizontal lines (=) between the
two atoms. E.g. O=O, O=C=O etc.
Triple covalent bond:
A covalent bond formed by the mutual sharing of
three pair of electrons is called a triple covalent bond, or simply a triple bond.
A
triple covalent bond is represented by three small horizontal lines (≡) between the
two atoms. E.g. N≡N, H-C≡C-H etc.
Formation of a covalent bond:
Formation of a covalent bond is favoured by
(i) High ionisation enthalpy of the combining elements.
(ii) Nearly equal electron gain enthalpy and equal electro-negativities of
combining elements.
(iii) High nuclear charge and small atomic size of the combining elements.
Polar covalent bond: The bond between two unlike atoms which differ in their affinities for electrons is said to be polar covalent bond. E.g. H-Cl
Coordinate bond:
The bond formed when one sided sharing of electrons
take place is called a coordinate bond. Such a bond is also known as dative bond.
It
is represented by an arrow (→) pointing towards the acceptor atom. E.g. H3N, BF3.
Bond Length:
Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
4.1.3
Lewis Representation of Simple Molecules (the Lewis Structures
The Lewis dot structures provide a picture of bonding in molecules and ions in terms of the shared pairs of electrons and the octet rule.
Steps for writing Lewis dot structures)
a. the total number of valence electrons present in the molecule is calculated by adding the individual valencies of each atom. For example, in the CH4. molecule there are eight valence electrons available for bonding (4 from carbon and 4 from the four hydrogen atoms).
b. For anions, each negative charge would mean addition of one electron.
c. For cations, each positive charge would result by subtraction of one electron from the total count.
d. In general the least electronegative atom occupies the central position in the
molecule/ion.
e. Now, the lone pairs of electrons are assigned to each atom belonging to the molecule.
Generally, the lone pairs are assigned to the most electronegative atoms first.
f. Once the lone pairs are assigned, if every atom does not have an octet configuration, a double or triple bond must be drawn to satisfy the octet valency of each atom.
g. If required, a lone pair can be converted into a bond pair in order to satisfy the octet rule for two atoms.
The Lewis Representation of Some Molecules
Determine the total number of valence electrons in chlorine molecule.
Formal charge on an atom is the difference between the number of valence electrons is an isolated atom and the number of electrons assigned to that atom in a Lewis structure.
4.1.5 Limitations of the Octet Rule
The octet rule, though useful, is not universal.
Other drawbacks of the octet theory
Noble gas electronic configuration can be achieved by
In order to explain the formation of a chemical bond in terms of electrons, Lewis postulated that atoms achieve stable octet when they are linked by a chemical bond. On the basis of this chemical bonds are following type
The formation of a positive ion involves ionization, i.e., removal of electron(s) from the neutral atom and Ionization enthalpy is defined as the minimum amount of energy that is required to remove the most loosely bounded electrons that is electron present in the outermost shell from an isolated gaseous atom. Electron gain enthalpy is defined as the amount of energy released when an electron is added to an isolated gaseous atom.
Ionic bonds will be formed more easily between elements with
comparatively low ionization enthalpies and elements with comparatively high
negative value of electron gain enthalpy.
Covalent bonds are formed between two atoms or ions in which the electron pairs are shared between them; they are also known as molecular bonds. The forces of attraction or repulsion between two atoms (when they share an electron pair or bonding pair) are called Covalent Bonding. The pair of electrons which are shared by the two atoms now extend around the nuclei of atoms leading to the creation of a molecule.
1. They are hard, brittle and crystalline. They have high melting and boiling points.
The chemical formula of sodium chloride is NaCl.
1. It is distributed abundantly in nature. Salt is a major ingredient of the dissolved materials in seawater.
1. It is easily soluble in water and partially soluble or insoluble in other liquids.
1. It is widely used in food industries as a food preservative and as a flavour enhancer. The Lattice Enthalpy of an ionic solid is defined as the energy required to
completely separate one mole of a solid ionic compound into gaseous constituent ions. For example, the lattice enthalpy of NaCl is 788 kJ mol–1. This means that 788 kJ of energy is required to separate one mole of solid NaCl into one mole of Na+ (g) and one mole of Cl– (g) to an infinite distance.
Bond Enthalpy: It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.
4.3.3 Bond Enthalpy Structure of Methane, Ammonia and Water and resonance in O3 molecule
4.3.4 Bond Order
Structure of Methane, Ammonia and Water and resonance in O3 molecule
Properties of carbon dioxide
Significance –
Few important points-
4.3.6 Polarity of Bonds
Electronegativity of few elements
Dipole moment (μ) :
Sidgwick and Powell in 1940, proposed a simple theory based on the repulsive interactions of the electron pairs in the valence shell of the atoms. Valence bond theory was introduced by Heitler and London (1927) and developed by Pauling and others.
It is based on the concept of atomic orbitals and the electronic configuration of the atoms. Let two hydrogen atoms A and B having their nuclei NA
and NB
and electrons present in them are eA
and eB.
Postulates of Valence Bond Theory
Orbital overlap concept-
If we refer to the minimum energy state in the formation of hydrogen molecule the two H-atoms are enough near so as to allow their atomic orbitals to undergo partial interpenetration. This partial interpenetration of atomic orbitals is called overlapping of atomic orbitals. ⇒ The Schrodinger wave equation is a mathematical expression describing the energy and position of the electron in space and time, taking into account the matter wave nature of the electron inside an atom.
However, since it cannot be solved for any system containing more than one electron, molecular orbitals which are one electron wave functions for molecules are difficult to obtain directly from the solution of Schrödinger wave equation. To overcome this problem, an approximate method known as linear combination of atomic orbitals (LCAO) has been adopted e.g. ⇒
Conditions for the Combination of Atomic Orbitals ⇒
Why are spectroscopic techniques used? ⇒ ⇔ The energy levels of these molecular orbitals have been determined experimentally
from spectroscopic data for homonuclear diatomic molecules of second row elements of the periodic table. The increasing order of
energies of various molecular orbitals for O2
and F2 is given below: ⇒
⇔
Electronic Configuration and Molecular Behaviour ⇒
Hydrogen Bonding Nitrogen, oxygen and fluorine are the highly electronegative elements. (3, 3.44, 4.0) When they are attached to a hydrogen atom to form covalent
bond, the electrons of the covalent bond are
shifted towards the more electronegative
atom. The Linear combination of atomic orbitals which is also known as LCAO is an approximate method for representing molecular orbitals. ⇒ O. Q What is Electron affinity?
SVL ⇒
BF4– - ⇒ ⇔
OL5.a. Give lewis representation of PO43–ion, CO32–ion, HNO3, O3 and NF3.
The chlorine molecule contains two chlorine atoms. In the periodic table, chlorine is a group VIIA element with seven electrons in its last shell.
Therefore, the total number of valence electrons= 7(2)= 14.
Total electron pairs exist in the form of lone pairs and bonds.
Total electron pairs are calculated by dividing the total valence electron count by two. For the Cl2 molecule, the total number of electron pairs in their valence shells is seven.
Determine the central atom in Cl2
There are only two atoms and they both belong to the same element, therefore, the central atom will be chlorine only.
4.1.4 Formal Charge
Def. - The formal charge over an atom of a polyatomic molecule or ion is the difference between the valence electron of that atom in the elemental state and the number of electrons assigned to that atom in Lewis structure.
A polyatomic molecule is a molecule made up of more than three atoms of an element. Example, Phosphorus (P4) and sulphur are examples (S8).
Formal charge (F.C.) on an atom in a Lewis structure =
[total number of valence electrons in the free atom]-[ total number of non bonding (lone pair) Electrons] – 1/2 [ total number ofbonding(shared) electrons]
Importance of Formal charge:
a. The formal charge being a theoretical charge doesn’t indicate any real charge separation in the molecule.
b. Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.
c. Knowledge of the lowest energy structure helps in predicting the major product of a reaction and also describes a lot of phenomena.
d. Generally, the lowest energy structure is the one with the smallest formal charges on the atoms and the most distributed charge.
Generally the lowest energy structure is the one with the smallest formal charges on the atoms. The formal charge is a factor based on a pure covalent view of bonding in which electron pairs are shared equally by neighbouring atoms.
Ozone molecule (O3).
There are three types of exceptions to the octet rule.
a. The incomplete octet of the central atom – LiCl, BeH2, BCl3.
b. Odd-electron molecules- NO ans NO2.
c. The expanded octet- PF5, SF6, H2SO4.
-It is clear that octet rule is based upon
the chemical inertness of noble gases.
However, some noble gases (for example
xenon and krypton) also combine with
oxygen and fluorine to form a number of
compounds like XeF2, KrF, XeOF etc.
- This theory does not account for the shape of molecules.
-It does not explain the relative stability of the molecules being totally silent about
the energy of a molecule.
4.2 Ionic or Electrovalent Bond
An ionic bond is formed by complete transference of one or more electrons from the valence shell of one atom to the valence shell of another atom. In this way both the atoms acquire stable electronic configurations of noble gases. The atom which loses electron becomes a positive ion and the atom which gains electron becomes negative ion.
Electrovalency is the number of electrons lost or gained during the formation of an ionic bond or electrovalent bond.
Formation of ionic compounds depends upon
) or zero (i.e., Nb = Na) bond order means an unstable molecule.
that of the negative ion involves the addition of electron(s) to the neutral atom.
M(g) → M+(g)+ e– ;
Ionization enthalpy
X(g) + e– → X-(g);
Electron gain enthalpy
M+(g)+ X-(g)
→ MX(s)
The electron gain process may be exothermic
or endothermic. The ionization, on the other
hand, is always endothermic.
Covalent Bond
Most ionic compounds have cations derived from metallic elements and anions from non-metallic elements.
Write electronic configuration of the following -
a. Na; b. Na+;
c. Ca; d. Ca++;
e. Mg; f. Mg ++;
g. Cl; h. Cl-;
h. S; i. S--;
The ammonium
ion, NH4 + (made up of two non-metallic elements) is an exception.
Characteristics of Ionic Compounds
2. They are polar in nature.
3. The linkage between oppositely charged ions is non rigid and non directional.
4.They are soluble in polar solvents such as water and insoluble in non polar solvents such as CCl4, Benzene, ether etc.
They are good conductors of electricity in fused state and in solution due to mobility of the ions. They are bad conductors of electricity in solid state because ions are unable to move.
For example: the ionization enthalpy for Na+(g) formation from Na(g) is 495.8 kJ mol–1;
while the electron gain enthalpy for the change
Cl(g) + e– → Cl-(g);
is, – 348.7 kJ mol–1 only.
The sum of the two, 147.1 kJ mol–1 is more than compensated for by the enthalpy of lattice formation of NaCl(s) (–788 kJ mol-1).
Therefore, the energy released in the processes is more than the energy absorbed.
Thus a qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state.
Sodium Chloride, NaCl Structure.
Sodium Chloride and Calcium fluoride
⇒ ⇒
The Occurrence of Sodium Chloride
2. Pure salt can be obtained from mineral halite. Sodium chloride is obtained by mining the deposits and brine solution is obtained by passing water into the deposits. Hence the salts get dissolved then the solution is pumped out.
Evaporation of the seawater is one of the major processes used to obtain salt and is most widely followed in countries like India. The crystals obtained usually consist of impurities such as calcium sulphate, sodium sulphate etc. Pure crystals are obtained by dissolving the salts with little water and filtering the solution.
Properties of Sodium Chloride
2. They are white crystals which do not have an odour but possess a taste.
3. In its aqueous state NaCl acts as a good conductor of electricity due to the free movement of the ions.
4. It has a melting point of 801°C and a boiling point of 1,413°C.
Uses of Sodium Chloride
2. It is a major raw material in the industrial manufacturing of various chemicals such as sodium carbonate, sodium hydrogen carbonate etc.
3. This salt is used in glass production.
4. In cold countries, it is used to prevent the build-up of ice on roads, bridges etc which is important for safe driving conditions.
What is sodium chloride used for?
The basic compound used by our body to digest and transport nutrients is sodium chloride ( NaCl), also known as salt. Preservation of blood pressure. Keeping the correct fluid balance.
What are the uses of saline solution?
Saline solution is commonly referred to as regular saline, although it is sometimes referred to as isotonic or physiological saline. In medicine, saline has many applications. It’s used for wound washing, sinus clearance, and dehydration care. It may be topically added or intravenously used.
Why the formula of sodium chloride is NaCl?
If sodium atoms interact with chlorine atoms, sodium chloride is formed. Sodium will donate an electron (which is a negative-charged particle) to chlorine as this happens. The chemical formula for sodium chloride is NaCl, indicating that there is precisely one chloride atom for every sodium atom present.
Does sodium chloride kill bacteria?
Sodium chloride is not only used for a number of different thing, but is a good antibacterial agent as well. An antibacterial agent is one that prevents bacteria from developing and multiplying.
What is the primary composition of NaCl?
Formula and structure: NaCl is the molecular formula of sodium chloride and 58.44 g / mol is its molar mass. It is an ionic compound which consists of a chloride anion (Cl-) and a sodium cation (Na+).
⇔ ⇔
4.2.1 Lattice Enthalpy
- The molar enthalpy change accompanying the complete separation of the constituent particles that compose of the solids (such as ions for ionic solid, molecules for molecular solids) under standard conditions is called lattice enthalpy (∆1⁰). The lattice enthalpy is a positive quantity.
Formation of an ionic bond:
It is favoured by,
(i) the low ionisation enthalpy of a metallic element which forms the cations,
(ii) High electron gain enthalpy of non- metallic element which forms the anions,
(iii) Large lattice enthalpy i.e; the smaller size and the higher charge of the atoms.
Bond Order:
In the Lewis description of covalent bond, the Bond Order is given by the number of bonds between the two atoms in a molecule
Resonance:
whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately
Electro valency: The number of electrons lost or gain by an atom of an element is called as electrovalency. The element which give up electrons to form positive ions are said to have positive valency, while the elements which accept electrons to form negative ions are said to have negative valency.
Conditions
a.The ease of formation of the positive and negative ions from the respective neutral atoms;
b. The arrangement of the positive and negative ions in the solid, that is, the
lattice of the crystalline compound.
The formation of a positive ion involves ionization, i.e., removal of electron(s) from the neutral atom and that of the negative ion involves the addition of electron(s) to the neutral atom.
M(g) → M+(g) + e– ;
Ionization enthalpy
X(g) + e– → X – (g) ;
Electron gain enthalpy ΔegH
M+(g) + X –(g) → MX(s)
The electron gain process may be exothermic or endothermic. The ionization, on the other hand, is always endothermic. Electron affinity, is the
negative of the energy change accompanying
electron gain.
How to determine Electron Pair Geometry?
Bond Length
Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
How to measure bond length?
Bond lengths are measured by spectroscopic, X-ray diffraction and electron-diffraction techniques Each atom of the bonded pair contributes to the bond length (Fig. 4.1). In the case of a covalent bond, the contribution from each atom is called the covalent radius of that atom.
The covalent radius is measured approximately as the radius of an atom’s core which is in contact with the core of an adjacent atom in a bonded situation.
The van der Waals radius represents the overall size of the atom which includes its valence shell in a nonbonded situation. Further, the van der Waals radius is half of the distance between two similar atoms in separate molecules in
a solid. Covalent and van der Waals radii of chlorine are depicted in Fig. 4.2.
Covalent and van der Waals
radius
Difernces between Covalent and van der Waals radius
Bond and Bond Length(pm)
O–H 96
C–H 107
N–O 136
C–O 143
C–N 143
C–C 154
C=O 121
N=O 122
C=C 133
C=N 138
C≡N 116
C≡C 120
Molecules and Bond Length (pm)
H2 (H – H) 74
F2 (F – F) 144
Cl2 (Cl – Cl) 199
Br2 (Br – Br) 228
I2 (I – I) 267
N2 (N≡N) 109
O2 (O = O) 121
HF (H – F) 92
HCl (H – Cl) 127
HBr (H – Br) 141
HI (H – I) 160
4.3.2 Bond Angle
Bond Angle is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule/complex ion. Bond angle is expressed in degree which can be experimentally determined by spectroscopic methods.
It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.
The unit of bond enthalpy is kJ mol–1.
When a bond is formed between atoms energy is released. When bonds dissociates energy required for dissociation.
The energy released when a bond is formed or the energy required to dissociate the bonds is called Bond enthalpy.
Definition:
The amount of energy released when 1 mole of bonds are formed from isolated atoms in the gaseous state or the amount of energy required to dissociate 1 mole of bonds present between the atoms in the gaseous molecules.
Bond dissociation enthalpy depends upon -
a. Size of the bonded atom and
b. Bond length.
The unit of bond enthalpy is kJ mol-1. For
example, the H – H bond enthalpy in hydrogen
molecule is 435.8 kJ mol-1.
H2(g) → H(g) + H(g);
ΔaH= 435.8 kJ mol-1.
Similarly the bond enthalpy for molecules
containing multiple bonds, for example O2 and
N2 will be as under :
O2 (O = O)(g)→ O(g) + O(g);
ΔaH = 498 kJ mol-1
N2 (N ≡N) (g) → N(g) + N(g);
ΔaH = 946.0 kJ mol-1
It is important that larger the bond
dissociation enthalpy, stronger will be the
bond in the molecule. For a heteronuclear
diatomic molecules like HCl, we have
HCl (g) → H(g) + Cl (g); ΔaH= 431.0 kJ mol-1
In case of polyatomic molecules, the
measurement of bond strength is more
complicated. For example in case of H2O
molecule, the enthalpy needed to break the
two O – H bonds is not the same.
H2O(g) → H(g) + OH(g);
ΔaH1⁰
= 502 kJ mol-1
OH(g) → H(g) + O(g);
ΔaH2= 427 kJ mol–1
The difference in the ΔaH that the second O–H bond undergoes
some change because of changed chemical environment. This is the reason for some difference in energy of the same O–H bond in different molecules like C2H5OH (ethanol) and water.
Therefore in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule,
Average bond enthalpy =(502 +427)/2
= 464.5 kJ mol–1
In the Lewis description of covalent bond, the Bond Order is given by the number
of bonds between the two atoms in a molecule. The bond order, for example in
H2 (with a single shared electron pair), in O2 (with two shared electron pairs) and in N2 (with three shared electron pairs) is 1,2,3 respectively. Similarly in CO (three shared electron pairs between C and O) the bond order is 3.
For N2, bond order is 3 and its ΔaH is 946 kJ mol-1; being one of the highest for a diatomic molecule.
Isoelectronic molecules and ions have identical bond orders; for example, F2 and
O22-have bond order 1.
N2, CO and NO+ have bond order 3.
A general correlation useful for understanding the stablities of molecules
is that: with increase in bond order, bond enthalpy increases and bond length
decreases.
Resonance Structures
According to the concept of resonance, whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately.
Thus for O3, the two structures shown above constitute the canonical
structures or resonance structures and their hybrid i.e., the III structure represents the structure of O3 more accurately. This is also called resonance hybrid. Resonance is represented by a double headed arrow.
Resonance in O3molecule
Explain the structure of CO32- ion in terms of resonance.
The carbon atom is connected to two oxygen atoms in the first resonance structure. A dashed line indicates the double bond between the carbon and oxygen atoms. The carbon atom is connected to a single oxygen atom and a second carbon atom in the second resonance structure.
Explain the structure of carbon dioxide.
Carbon dioxide is colourless and odourless gas in nature. It has the chemical formula CO2. It plays a vital role in the occurrence of life on the earth. The gas is vital for the process of photosynthesis.
Structure of carbon dioxide
The carbon dioxide molecule has two double bonds between the carbon and oxygen atoms. Each double bond is made up of one sigma and one pi bond, so a carbon dioxide molecule contains two sigma and two pi bonds. Electron dot structures or Lewis dot formula can be drawn if the molecular formula of the compound is known.
The bond angle is 180° between both the carbon and the two oxygen atoms, carbon dioxide has a linear structure.
The carbon atom is sp hybridization and both oxygen atoms are hybridized by sp2.
The two individual bonds formed are polar, because oxygen is more electronegative than carbon, but because of the cancellation of the dipole moment, the molecule as a whole is not polar.
The chemical formula of carbon dioxide is CO2. Carbon dioxide is colourless and odourless gas in nature.
It plays a vital role in the existence of life on the earth.
This gas is found in traces in the atmosphere, but it is of great importance.
The experimentally determined carbon to oxygen bond length in CO2 is 115 pm.
The lengths of a normal carbon to oxygen double bond (C=O)
and carbon to oxygen triple bond (C≡O) are 121 pm and 110 pm respectively. The carbon-oxygen bond lengths in CO2 (115 pm) lie between the values
for C=O and C≡O.
Obviously, a single Lewis structure cannot depict this position and it becomes necessary to write more than one Lewis structures and to consider that the structure of CO2 is best described as a hybrid of the canonical or resonance forms I, II and III.
Resonance in CO2 molecule, I, II and III represent the three canonical forms.
Resonance stabilizes the molecule as the energy of the resonance hybrid is less than the energy of any single cannonical structure;
and, Resonance averages the bond characteristics as a whole.
Thus the energy of the O3 resonance hybrid is lower than either of the two cannonical froms I and II (Fig).
In general, it may be stated that
• Resonance stabilizes the molecule as the energy of the resonance hybrid is less than the energy of any single cannonical structure; and,
• Resonance averages the bond characteristics as a whole.
Thus the energy of the O3 resonance hybrid is lower than either of the two cannonical froms I and II.
• The cannonical forms have no real existence.
• The molecule does not exist for a certain fraction of time in one
cannonical form and for other fractions of time in other cannonical
forms.
• There is no such equilibrium between the cannonical forms as we have
between tautomeric forms (keto and enol) in tautomerism.
• The molecule as such has a single structure which is the resonance
hybrid of the cannonical forms and which cannot as such be depicted by
a single Lewis structure.
The existence of a hundred percent ionic or covalent bond represents an ideal situation. In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond between two hydrogen atoms, there is some ionic character.
When covalent bond is formed between two similar atoms, for example in H2, O2, Cl2, N2 or F2, the shared pair of electrons is equally attracted by the two atoms.
As a result electron pair is situated exactly between the two identical nuclei. The bond so formed is called nonpolar covalent bond.
Contrary to this in case of a heteronuclear molecule like HF, the shared electron pair between the two atoms gets displaced more towards fluorine since the electronegativity of fluorine (4) is far greater than that of hydrogen(2.2). The
resultant covalent bond is a polar covalent bond.
1H 2.20
6C 2.55
7N 3.04
8O 3.44
9F 3.98
11Na 0.93
12Mg 1.31
16S 2.58
17Cl 3.16
20Ca 1.00
35Br 2.96
53I 2.66
The product of the magnitude of the charge and the distance between the centres of positive and negative charge. It is a vector quantity and is represented by an arrow with its tail at the positive centre and head pointing towards a negative centre.
Dipole moment (μ) = charge (Q) × distance of separation (r)
It can be noted that the symbols 𝛿+ and 𝛿– represent the two electric charges that arise in a molecule which are equal in magnitude but are of opposite signs.
They are separated by a set distance, which is commonly denoted by ‘r’.
Important Points
1. The dipole moment of a single bond in a polyatomic molecule is known as the bond dipole moment and it is different from the dipole moment of the molecule as a whole.
2. It is a vector quantity, i.e. it has magnitude as well as definite directions.
Being a vector quantity, it can also be zero as the two oppositely acting bond dipoles can cancel each other.
3. By convention, it is denoted by a small arrow with its tail on the negative center and its head on the positive center.
In chemistry, the dipole moment is represented by a slight variation of the arrow symbol. It is denoted by a cross on the positive center and arrowhead on the negative center. This arrow symbolizes the shift of electron density in the molecule.
4. In the case of a polyatomic molecule, the dipole moment of the molecule is the vector sum of the all present bond dipoles in the molecule.
Polyatomic molecules are electrically neutral groups of three or more atoms held together by covalent bonds.
Mathematically,
Dipole Moment (µ) = Charge (Q) X distance of separation (r)
It is measured in Debye units denoted by ‘D’.
1 D = 3.33564 × 10-30C. m, where C is Coulomb and m denotes a metre.
The bond dipole moment that arises in a chemical bond between two atoms of different electronegativities can be expressed as follows:
μ = Q.r
Where: μ is the bond dipole moment,
𝛿 is the magnitude of the partial charges 𝛿+ and 𝛿–,
And d is the distance between 𝛿+ and 𝛿–.
The bond dipole moment (μ) is also a vector quantity, whose direction is parallel to the bond axis. In chemistry, the arrows that are drawn in order to represent dipole moments begin at the positive charge and end at the negative charge.
When two atoms of varying electronegativities interact, the electrons tend to move from their initial positions to come closer to the more electronegative atom. This movement of electrons can be represented via the bond dipole moment.
(Peter Debye, the Dutch chemist received Nobel prize in 1936 for his work on X-ray diffraction and dipole moments. The magnitude of the dipole moment is given in Debye units in order to honour him.)
Dipole moment of BeF2
In a beryllium fluoride molecule, the bond angle between the two beryllium-fluorine bonds is 180⁰.
Fluorine, being the more electronegative atom, shifts the electron density towards itself. The individual bond dipole moments in a BeF2 molecule are illustrated below.
In BeF2 the two individual bond dipole moments cancel each other out because they are equal in magnitude but are opposite in direction. Therefore, the net dipole moment of a BeF2 molecule is zero. Fig.
Dipole moment of H2O (Water)
In a water molecule, the electrons are localised around the oxygen atom since it is much more electronegative than the hydrogen atom. However, the presence of a lone pair of electrons in the oxygen atom causes the water molecule to have a bent shape (as per the VSEPR theory).
For a free water monomer, its dipole moment has been experimentally determined to be 1.855 D.
(1) In water clusters and condensed phases, the dipole moment of an individual water molecule is known to be enhanced significantly due to polarization and delocalization effects.
Net Dipole moment, μ =1.85 D=1.85×3.33564×10-30Cm =
6.17 ×10-30Cm.
In water Therefore, the individual bond dipole moments do not cancel each other out as is the case in the BeF2 molecule.
Dipole moment of BF3
In tetra-atomic molecule, for example in BF3, the dipole moment is zero although the
B – F bonds are oriented at an angle of 120o to one another, the three bond moments give a net sum of zero as the resultant of any two is
equal and opposite to the third.
Dipole moment of NH3
and NF3 molecule.
Both the molecules have pyramidal shape with a lone pair of electrons on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment of NH3 (4.90 ×10-30C. m) is greater than that of NF3 (0.8 × 10-30 C.m). This
is because, in case of NH3 the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N – H bonds, whereas in NF3 the orbital dipole is in the direction opposite to the resultant dipole
moment of the three N–F bonds. The orbital dipole because of lone pair decreases the effect of the resultant N – F bond moments, which
results in the low dipole moment of NF3.
Dipole moments of some molecules are shown below
Molecule (AB)
HF 1.78 linear
HCl 1.07 linear
HBr 0.79 linear
Hl 0.38 linear
H2 0 linear
Molecule (AB2)
H2O 1.85 bent
H2S 0.95 bent
CO2 0 linear
Molecule (AB3)
NH31.47 trigonal-pyramidal
NF3 0.23 trigonal-pyramidal
BF3 0 trigonal-planar
Molecule (AB4)
CH40 tetrahedral
CHCl31.04 tetrahedral
CCl4 0 tetrahedral
4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory
It was further developed and redefined by Nyholm and Gillespie (1957).
The main postulates of VSEPR theory -
1. The shape of a molecule depends upon the number of valence shell electron pairs (bonded or nonbonded) around the central atom.
2. Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.
3. These pairs of electrons tend to occupy such positions in space that minimize repulsion and thus maximise distance between them.
4. The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at maximum distance from one another.
5. A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair.
6. Where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structure.
The repulsive interaction of electron pairs decrease in the order:
Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp)
For the prediction of geometrical shapes of molecules with the help of VSEPR theory,
it is convenient to divide molecules into two categories as
(i) molecules in which the central atom has no lone pair and
(ii) molecules in which the central atom has one or more lone pairs.
Table
Shapes of molecules -
linear, trigonal planar, tetrahedral, trigonalbipyramidal and octahedral,
Importance of VSEPR Models
1. Lewis structures only tell the number and types of bonds between atoms, as they are limited to two dimensions. The VSEPR model predicts the 3-D shape of molecules and ions but is ineffective in providing any specific information regarding the bond length or the bond itself.
2. VSEPR models are based on the concept that electrons around a central atom will configure themselves to minimize repulsion, and that dictates the geometry of the molecule.
3. It can predict the shape of nearly all compounds that have a central atom, as long as the central atom is not a metal. Each shape has a name and an idealized bond angle associated with it.
Steps to Using VSEPR
1. Draw a Lewis structure for the ion or molecule in question.
2. Determine the number of electron groups around the central atom. Each lone pair of electrons counts as a single group. Each bond counts as a single group, even if it is a double or triple bond. Find the corresponding electron geometry from the table.
3. Determine the number of lone pairs and the number of bonding pairs around the central atom, and use that to find the molecular geometry.
VSEPR Notation
VSEPR notation gives a general formula for classifying chemical species based on the number of electron pairs around a central atom.
In general, A is used to represent the central atom.
B or X is used to represent the number of atoms bonded to the central atom.
E represents the number of lone pairs on the central atom (ignore lone pairs on bonded atoms).
Again, this theory is also not void of any limitations.
Limitations of the VSEPR theory
1. The VSEPR model is not a theory. It does not explain or attempt to explain any observations or predictions. Rather, it is an algorithm that accurately predicts the structures of a large number of compounds.
2. VSEPR is simple and useful but does not work for all chemical species.
The idealized bond angles do not always match the measured values.
Quantum mechanics and atomic orbitals can give more sophisticated predictions when VSEPR is inadequate.
Valence Bond Theory
As these two atoms come closer new attractive and repulsive forces begin to operate.
The nucleus of one atom is attracted towards its own electron and the electron of the other and vice versa. Repulsive forces arise between the electrons of two atoms and nuclei of two atoms. Attractive forces tend to bring the two atoms closer whereas repulsive forces tend to push them apart.
Electrons in a molecule occupy atomic orbitals rather than molecular orbitals. The overlapping of atomic orbitals results in the formation of a chemical bond and the electrons are localized in the bond region due to overlapping.
The Lewis approach to chemical bonding failed to shed light on the formation of chemical bonds. Also, valence shell electron pair repulsion theory (or VSEPR theory) had limited applications (and also failed in predicting the geometry corresponding to complex molecules).
In order to address these issues, the valence bond theory was put forth by the German physicists Walter Heinrich Heitler and Fritz Wolfgang London. The Schrodinger wave equation was also used to explain the formation of a covalent bond between two hydrogen atoms. The chemical bonding of two hydrogen atoms as per the valence bond theory is illustrated below.
1. Covalent bonds are formed when two valence orbitals (half-filled) belonging to two different atoms overlap on each other. The electron density in the area between the two bonding atoms increases as a result of this overlapping, there by increasing the stability of the resulting molecule.
2. The presence of many unpaired electrons in the valence shell of an atom enables it to form multiple bonds with other atoms. The paired electrons present in the valence shell do not take participate in the formation of chemical bonds as per the valence bond theory.
3. Covalent chemical bonds are directional and are also parallel to the region corresponding to the atomic orbitals that are overlapping.
4. Sigma bonds and pi bonds differ in the pattern that the atomic orbitals overlap in, i.e. pi bonds are formed from sidewise overlapping whereas the overlapping along the axis containing the nuclei of the two atoms leads to the formation of sigma bonds.
Applications of Valence Bond Theory
The maximum overlap condition which is described by the valence bond theory can explain the formation of covalent bonds in several molecules.
This is one of its most important applications. For example, the difference in the length and strength of the chemical bonds in H2 and F2 molecules can be explained by the difference in the overlapping orbitals in these molecules.
The covalent bond in an HF molecule is formed from the overlap of the 1s orbital of the hydrogen atom and a 2p orbital belonging to the fluorine atom, which is explained by the valence bond theory.
e.g. Consider two hydrogen atoms A and B
approaching each other having nuclei NA
and NB and electrons present in them are
represented by eA and eB. When the two atoms
are at large distance from each other, there is no interaction between them. As these two
atoms approach each other, new attractive
and repulsive forces begin to operate.
Attractive forces arise between:
(i) nucleus of one atom and its own electron
that is NA–eA and NB–eB
(ii) nucleus of one atom and electron of
other atom i.e., NA – eB, NB– eA.
Similarly repulsive forces arise between
(i) electrons of two atoms like eA–eB
,
(ii) nuclei of two atoms NA–NB.
Attractive forces tend to bring the two atoms close to each other whereas repulsive
forces tend to push them apart (Fig. 4.7).
Experimentally it has been found that the magnitude of new attractive force is
more than the new repulsive forces.
As a
result, two atoms approach each other and
potential energy decreases.
Ultimately a stage
is reached where the net force of attraction
balances the force of repulsion and system
acquires minimum energy.
At this stage
two hydrogen atoms are said to be bonded
together to form a stable molecule having the
bond length of 74 pm.
Since the energy gets released when the
bond is formed between two hydrogen atoms,
the hydrogen molecule is more stable than
that of isolated hydrogen atoms. The energy
so released is called as bond enthalpy, which
is corresponding to minimum in the curve
depicted in Fig. 4.8. Conversely, 435.8 kJ of
energy is required to dissociate one mole of
H2 molecule.
H2 (g) + 435.8 kJ mol-1→ H(g) + H(g)
Limitations of Valence Bond Theory
1. Could not explain the tetravalency exhibited by carbon.
2. No explanation on the energies of the electrons.
3. It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds.
4. The theory assumes that electrons are localized in specific areas.
5. No distinction between weak and strong ligands and failed to offer any explanation for the colour exhibited by coordination compounds.
The overlap between the atomic orbitals can be positive, negative or zero depending upon the characteristics of the orbitals participating to overlap.
Directional Properties of Bonds
As we have already seen, the covalent bond
is formed by overlapping of atomic orbitals.
The molecule of hydrogen is formed due to the
overlap of 1s-orbitals of two H atoms.
In case of polyatomic molecules like CH4,
NH3 and H2O, the geometry of the molecules
is also important in addition to the bond
formation. For example
Why is it so that CH4 molecule has tetrahedral shape and HCH
bond angles are 109.5°?
Carbon has four electrons in its valence shell. In methane molecule, the carbon atom forms covalent bonds with four different hydrogen atoms. There are no lone pairs on carbon atom. Therefore, the four hydrogen atoms are arranged at four corners of tetrahedron and carbon at the centre of the tetrahedron.
Why is the shape of
NH3 molecule pyramidal ?
It is trigonal pyramidal due to yhe presence of alone pair of electrons on the central nitrogen atom. The nonbonding pair of electrons pushes away from the bonding pairs producing a trigonal pyramidal shape. If the central atom with no lone pair is bonded to three other atoms the molecule will have a trigonal planar shape.
The valence bond theory explains the
shape, the formation and directional properties
of bonds in polyatomic molecules like CH4,
NH3 and H2O, etc. in terms of overlap and
hybridisation of atomic orbitals.
Overlapping of Atomic Orbitals
When orbitals of two atoms come close to form
bond, their overlap may be positive, negative
or zero depending upon the sign (phase) and
direction of orientation of amplitude of orbital
wave function in space (Fig. 4.9).
Positive and negative sign on boundary surface diagrams
in the Fig. 4.9 show the sign (phase) of orbital
wave function and are not related to charge.
Orbitals forming bond should have same sign
(phase) and orientation in space. This is called
positive overlap. Various overlaps of s and p
orbitals are depicted in Fig. 4.9.
The criterion of overlap, as the main factor
for the formation of covalent bonds applies
uniformly to the homonuclear/heteronuclear
diatomic molecules and polyatomic molecules.
We know that the shapes of CH4
, NH3, and H
O molecules are tetrahedral, pyramidal
and bent respectively.
Use of VB theory to find out geometrical shapes in terms
of the orbital overlaps.
Let us first consider the CH4 (methane) molecule. The electronic configuration of carbon in its ground state is
[He]2s22p2
which in the excited state becomes [He] 2s1, 2px1 2py1 2pz1.
The energy required for this excitation is compensated by the release of energy due to overlap between the orbitals of carbon and the hydrogen.
The four atomic orbitals of carbon, each with an unpaired electron can overlap
with the 1s orbitals of the four H atoms which are also singly occupied. This will result in the formation of four C-H bonds. It will, however, be observed that while the three p orbitals of carbon are at 90° to one another, the HCH angle for these will also be 90°. That is three C-H bonds will be oriented at 90° to one another. The 2s orbital of carbon and the 1s orbital of H are spherically symmetrical and
they can overlap in any direction.
Therefore the direction of the fourth C-H bond cannot be ascertained. This description does not fit in with the tetrahedral HCH angles of 109.5°.
Clearly, it follows that simple atomic orbital overlap does not account for the directional characteristics of bonds in CH4.
Using similar
procedure and arguments, it can be seen that in
the case of NH3 and H2O molecules, the HNH
and HOH angles should be 90°. This is in disagreement with the actual bond angles of 107° and 104.5° in the NH3 and H2O molecules respectively.
What is positive overlap?
If the two interacting orbitals have the same phase, then the overlap is positive and a bond is formed. The phase of the two interacting orbitals (+ or -) is the sign of orbital wave function and is not related to the charge.
What do you mean by zero overlaps?
A zero overlap is a result of orbitals do not overlap at all or do not overlap efficiently. A zero overlap does not lead to bond formation between the atomic orbitals.
What are orbital phases?
Orbitals are essentially mathematical functions that define complex patterns of standing waves that can be graphed on a graph but do not have any physical reality. An orbital process is a direct result of electrons’ wave-like properties.
types of Overlapping and Nature of Covalent Bonds
(i) Sigma(σ) bond, and (ii) pi(π) bond
Sigma(σ) bond
This type of covalent bond is formed by the end to end (headon) overlap of bonding orbitals along the internuclear axis. This is called as head on overlap or axial overlap. This can be formed by any one of the following types of combinations of atomic orbitals.
s-s overlapping
In this case, there is overlap of two half filled s-orbitals along the internuclear axis as shown.
s-p overlapping
This type of overlap occurs between half filled s-orbitals of one atom and half filled p-orbitals of another atom.
p–p overlappingp–p overlapping : This type of overlap takes place between half filled p-orbitals of the two approaching atoms.
pi(π) bond
In the formation of π bond the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the
internuclear axis. The orbitals formed due to sidewise overlapping consists
of two saucer type charged clouds above and below the plane of the participating atoms.
Strength of Sigma and pi Bonds
Basically the strength of a bond depends upon
the extent of overlapping. In case of sigma bond,
the overlapping of orbitals takes place to a
larger extent. Hence, it is stronger as compared
to the pi bond where the extent of overlapping
occurs to a smaller extent.
Also to note that in the formation of
multiple bonds between two atoms of a
molecule, pi bond(s) is formed in addition to a
sigma bond.
Hybridisation
In order to explain the characteristic
geometrical shapes of polyatomic molecules
like CH4, NH3 and H2O etc., Pauling introduced the concept of hybridisation.
According to him the atomic orbitals combine to form new set of equivalent orbitals known as hybrid orbitals.
Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridisation which can be defined
as the process of intermixing of the orbitals of
slightly different energies so as to redistribute
their energies, resulting in the formation of
new set of orbitals of equivalent energies and
shape.
For example when one 2s and three 2p-orbitals of carbon hybridise, there is the
formation of four new sp3 hybrid orbitals.
Salient features of hybridisation:
1. The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised.
2. The hybridised orbitals are always equivalent in energy and shape.
3. The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
4. These hybrid orbitals are directed in space in some preferred direction to have
minimum repulsion between electron pairs and thus a stable arrangement.
Important conditions for hybridisation
(i) The orbitals present in the valence shell of the atom are hybridised.
(ii) The orbitals undergoing hybridization should have almost equal energy.
(iii) Promotion of electron is not essential condition prior to hybridisation.
(iv) It is not necessary that only half filled orbitals participate in hybridisation. In
some cases, even filled orbitals of valence shell take part in hybridisation.
Types of Hybridisation
Sp hybridization, sp 2 Hybridisation, sp3 Hybridisation, sp3d Hybridisation, sp3d2 Hybridisation etc.
There are various types of hybridisation
involving s, p and d orbitals. The different
types of hybridisation are as under:
(I) sp hybridisation: This type of hybridisation
involves the mixing of one s and one p orbital
resulting in the formation of two equivalent
sp hybrid orbitals. The suitable orbitals for
sp hybridisation are s and pz, if the hybrid
orbitals are to lie along the z-axis. Each sp
hybrid orbitals has 50% s-character and
50% p-character. Such a molecule in which
the central atom is sp-hybridised and linked
directly to two other central atoms possesses
linear geometry. This type of hybridisation is
also known as diagonal hybridisation.
The two sp hybrids point in the opposite
direction along the z-axis with projecting
positive lobes and very small negative lobes,
which provides more effective overlapping
resulting in the formation of stronger bonds.
Example of molecule having sp
hybridisation
BeCl2: .
Atomic number of Be is 4. The ground state electronic configuration of Be is 1s2, 2s2.
In the exited state one of the 2s-electrons is promoted to vacant 2p orbital to account for its bivalency.
One 2s and one 2p-orbital gets hybridised to
form two sp hybridised orbitals. These two
sp hybrid orbitals are oriented in opposite
direction forming an angle of 180°.
Each of the sp hybridised orbital overlaps with the 2p-orbital of chlorine axially and form two Be- Cl sigma bonds. This is shown in Fig. 4.10.
(II) sp2 hybridisation:
In this hybridisation
there is involvement of one s and two
p-orbitals in order to form three equivalent
sp2 hybridised orbitals. For example, in
BCl3 molecule, the ground state electronic
configuration of central boron atom is
1s22s22p1. In the excited state, one of the 2s
electrons is promoted to vacant 2p orbital as
These three orbitals (one 2s and two 2p)
hybridise to form three sp2 hybrid orbitals.
The three hybrid orbitals so formed are
oriented in a trigonal planar arrangement
and overlap with 2p orbitals of chlorine to
form three B-Cl bonds. Therefore, in BCl3
(Fig. 4.11), the geometry is trigonal planar
with Cl B Cl bond angle of 120°.
(III) sp3 hybridisation: This type of
hybridisation can be explained by taking the
example of CH4 molecule in which there is
mixing of one s-orbital and three p-orbitals
of the valence shell to form four sp3 hybrid
orbital of equivalent energies and shape. There
is 25% s-character and 75% p-character in
each sp3 hybrid orbital. The four sp3 hybrid
orbitals so formed are directed towards the
four corners of the tetrahedron. The angle
between sp3 hybrid orbital is 109.5° as shown
in Fig. 4.12.
The structure of NH3 and H2O molecules
can also be explained with the help of sp3
hybridisation. In NH3, the valence shell (outer)
electronic configuration of nitrogen in the ground state is 2S2 2px1 2py1 2pz1 having three y
unpaired electrons in the sp3 hybrid orbitals
and a lone pair of electrons is present in the
fourth one. These three hybrid orbitals overlap
with 1s orbitals of hydrogen atoms to form
three N–H sigma bonds. We know that the
force of repulsion between a lone pair and a
bond pair is more than the force of repulsion
between two bond pairs of electrons. The
molecule thus gets distorted and the bond
angle is reduced to 107° from 109.5°. The
geometry of such a molecule will be pyramidal
as shown in Fig. 4.13.
In case of H2O molecule, the four oxygen
orbitals (one 2s and three 2p) undergo sp3
hybridisation forming four sp3 hybrid orbitals
out of which two contain one electron each and
the other two contain a pair of electrons. These
four sp3 hybrid orbitals acquire a tetrahedral
geometry, with two corners occupied by
hydrogen atoms while the other two by the
lone pairs. The bond angle in this case is
reduced to 104.5° from 109.5° (Fig. 4.14)
and the molecule thus acquires a V-shape or
angular geometry.
Other Examples of sp3, sp2 and sp
Hybridisation
sp3 Hybridisation in C2H6 molecule: In
ethane molecule both the carbon atoms
assume sp3 hybrid state. One of the four
sp3 hybrid orbitals of carbon atom overlaps
axially with similar orbitals of other atom to
form sp3-sp3 sigma bond while the other three
hybrid orbitals of each carbon atom are used
in forming sp3–s sigma bonds with hydrogen
atoms as discussed in section 4.6.1(iii).
Therefore in ethane C–C bond length is 154
pm and each C–H bond length is 109 pm.
sp2 Hybridisation in C2H4: In the formation
of ethene molecule, one of the sp2 hybrid
orbitals of carbon atom overlaps axially with
sp2 hybridised orbital of another carbon atom
to form C–C sigma bond. While the other two
sp2 hybrid orbitals of each carbon atom are
used for making sp2–s sigma bond with two
hydrogen atoms. The unhybridised orbital (2px
or 2py) of one carbon atom overlaps sidewise
with the similar orbital of the other carbon
atom to form weak π bond, which consists of
two equal electron clouds distributed above
and below the plane of carbon and hydrogen
atoms.
Thus, in ethene molecule, the carboncarbon
bond consists of one sp2–sp2 sigma
bond and one pi (π ) bond between p orbitals
which are not used in the hybridisation and
are perpendicular to the plane of molecule; the
bond length 134 pm. The C–H bond is sp2–s
sigma with bond length 108 pm. The H–C–H
bond angle is 117.6° while the H–C–C angle
is 121°. The formation of sigma and pi bonds
in ethene is shown in Fig. 4.15.
sp Hybridisation in C2H2 : In the formation
of ethyne molecule, both the carbon atoms
undergo sp-hybridisation having two
unhybridised orbital i.e., 2py and 2px.
One sp hybrid orbital of one carbon atom
overlaps axially with sp hybrid orbital of the
other carbon atom to form C–C sigma bond,
while the other hybridised orbital of each
carbon atom overlaps axially with the half
filled s orbital of hydrogen atoms forming
σ bonds. Each of the two unhybridised p
orbitals of both the carbon atoms overlaps
sidewise to form two π bonds between the
carbon atoms. So the triple bond between the
two carbon atoms is made up of one sigma
and two pi bonds as shown in Fig. 4.16.
Hybridisation of Elements
involving d Orbitals
The elements present in the third period
contain d orbitals in addition to s and p
orbitals. The energy of the 3d orbitals are
comparable to the energy of the 3s and 3p
orbitals. The energy of 3d orbitals are also
comparable to those of 4s and 4p orbitals.
As a consequence the hybridisation involving
either 3s, 3p and 3d or 3d, 4s and 4p is
possible. However, since the difference in
energies of 3p and 4s orbitals is significant, no
hybridisation involving 3p, 3d and 4s orbitals
is possible.
The important hybridisation schemes
involving s, p and d orbitals are summarised
below:
(i) Formation of PCl5 (sp3d hybridisation):
The ground state and the excited state outer
electronic configurations of phosphorus
(Z=15) are represented below.
Now the five orbitals (i.e., one s, three
p and one d orbitals) are available for
hybridisation to yield a set of five sp3d hybrid
orbitals which are directed towards the five
corners of a trigonal bipyramidal as depicted
in the Fig. 4.17.
It should be noted that all the bond angles
in trigonal bipyramidal geometry are not
equivalent. In PCl5 the five sp3d orbitals of
phosphorus overlap with the singly occupied
p orbitals of chlorine atoms to form five P–Cl
sigma bonds. Three P–Cl bond lie in one
plane and make an angle of 120° with each
other; these bonds are termed as equatorial
bonds. The remaining two P–Cl bonds–one
lying above and the other lying below the
equatorial plane, make an angle of 90° with
the plane. These bonds are called axial bonds.
As the axial bond pairs suffer more repulsive
interaction from the equatorial bond pairs,
therefore axial bonds have been found to
be slightly longer and hence slightly weaker
than the equatorial bonds; which makes PCl5
molecule more reactive.
(ii) Formation of SF6 (sp3d2 hybridisation):
In SF6 the central sulphur atom has the
ground state outer electronic configuration
3s23p4. In the exited state the available six
orbitals i.e., one s, three p and two d are
singly occupied by electrons. These orbitals
hybridise to form six new sp3d2 hybrid
orbitals, which are projected towards the six
corners of a regular octahedron in SF6. These six sp3d2 hybrid orbitals overlap with singly
occupied orbitals of fluorine atoms to form
six S–F sigma bonds. Thus SF6 molecule has
a regular octahedral geometry as shown in
Fig. 4.18.
Molecular Orbital Theory
(031104.2)
Molecular orbital (MO) theory was developed
by F. Hund and R.S. Mulliken in 1932. The
salient features of this theory are :
(i) The electrons in a molecule are present
in the various molecular orbitals as the
electrons of atoms are present in the
various atomic orbitals.
(ii) The atomic orbitals of comparable
energies and proper symmetry combine
to form molecular orbitals.
(iii) While an electron in an atomic orbital
is influenced by one nucleus, in a
molecular orbital it is influenced by
two or more nuclei depending upon the
number of atoms in the molecule. Thus,
an atomic orbital is monocentric while
a molecular orbital is polycentric.
(iv) The number of molecular orbital formed
is equal to the number of combining
atomic orbitals. When two atomic
orbitals combine, two molecular orbitals
are formed. One is known as bonding
molecular orbital while the other is
called antibonding molecular orbital.
(v) The bonding molecular orbital has
lower energy and hence greater stability
than the corresponding antibonding
molecular orbital.
(vi) Just as the electron probability
distribution around a nucleus in an
atom is given by an atomic orbital, the
electron probability distribution around
a group of nuclei in a molecule is given
by a molecular orbital.
(vii) The molecular orbitals like atomic
orbitals are filled in accordance with
the aufbau principle obeying the Pauli’s
exclusion principle and the Hund’s rule.
Formation of Molecular Orbitals Linear Combination of Atomic
Orbitals (LCAO)
⇉
According to wave mechanics, the atomic orbitals can be expressed by wave functions (ψ ’s) which represent the amplitude of the electron waves. These are obtained from the solution of Schrödinger wave equation.
It is based on three considerations.
a. Classical plane wave equation
b. Broglie’s hypothesis of matter-wave
c. Conservation of energy
What is a wave function?
A wave function is used to describe ‘matter waves’. Matter waves are very small particles in motion having a wave nature – dual nature of particle and wave. Any variable property that makes up the matter waves is a wave function of the matter wave. A wave function is denoted by the symbol ‘Ψ’.
Amplitude, a property of a wave, is measured by following the movement of the particle with its Cartesian coordinates with respect to time. The amplitude of a wave is a wave function. The wave nature and the amplitudes are a function of coordinates and time, such that,
Wave function amplitude = Ψ = Ψ(r,t), where ‘r’ is the position of the particle in terms of x, y and z directions.
What is meant by stationary state, and what is its relevance to the atom?
Stationary state is a state of a system whose probability density given by | Ψ2 | (psi) is invariant with time. In an atom, the electron is a matter wave with quantised angular momentum, energy, etc. The movement of the electrons in their orbit is such that probability density varies only with respect to the radius and angles.
The movement is akin to a stationary wave between two fixed ends and is independent of time. The wave function concept of matter waves is applied to the electrons of an atom to determine its variable properties.
What is the physical significance of the Schrodinger wave function?
The Bohr concept of an atom is simple. But it cannot explain the presence of multiple orbitals and the fine spectrum arising out of them. It is applicable only to the one-electron system.
The Schrodinger wave function has multiple unique solutions representing characteristic radius, energy and amplitude.
The probability density of the electron calculated from the wave function shows multiple orbitals with unique energy and distribution in space.
The Schrodinger equation could explain the presence of multiple orbitals and the fine spectrum arising out of all atoms, not necessarily hydrogen-like atoms.
What is the Hamilton operator used in the Schrodinger equation?
Answer: In Mathematics, the operator is a rule that converts observed properties into another property. For example, ‘A’ will be an operator if it can change a property f(x) into another f(y).
f(x)= f(y)
Hamiltonian operator is the sum of potential and kinetic energies of particles calculated over three coordinates and time.
Hamiltonian operator = Ȟ = T + V = Kinetic energy + Potential energy
The electrons are more likely to be found:
⇔
Hydrogen molecule-
Consider the hydrogen molecule consisting of two atoms A and B. Each hydrogen atom in the ground state has one electron in 1s orbital. The atomic orbitals of these atoms may be represented by the wave functions ψA and ψB. Mathematically, the formation of molecular orbitals may be described by the linear combination of atomic orbitals that can take place by addition and by subtraction of wave functions of individual atomic orbitals as shown below :
Ψ MO ± ψA + ψB
Therefore, the two molecular orbitals
σ and σ* are formed as :
σ = ψA + ψB
σ* = ψA – ψB
The molecular orbital σ formed by the addition of atomic orbitals is called the bonding molecular orbital while the molecular orbital σ* formed by the subtraction of atomic orbital is called antibonding molecular orbital as depicted in Fig. 4.19.
(031104.2_1a)
Qualitatively, the formation of molecular orbitals can be understood in terms of the
constructive or destructive interference of the electron waves of the combining atoms. In the formation of bonding molecular orbital, the two electron waves of the bonding atoms reinforce each other due to constructive interference while in the formation of waves cancel each other due to destructive interference. As a result, the electron density in a bonding molecular orbital is located between the nuclei of the bonded atoms because of which the repulsion between the nuclei is very less while in case of an antibonding molecular orbital, most of the electron density is located away from the space between the nuclei.
Infact, there is a nodal plane (on which the electron density is zero) between the nuclei and hence the repulsion between the nuclei is high. Electrons placed in a bonding molecular orbital tend to hold the nuclei together and stabilise a molecule. Therefore, a bonding molecular orbital always possesses lower energy than either of the atomic orbitals that have combined to form it. In contrast, the electrons placed in the antibonding molecular orbital destabilise the molecule. This is
because the mutual repulsion of the electrons in this orbital is more than the attraction between the electrons and the nuclei, which causes a net increase in energy. It may be noted that the energy of the antibonding orbital is raised above the energy of the parent atomic orbitals that have combined and the energy of the bonding orbital has been lowered than the parent orbitals. The total energy of two molecular orbitals, however, remains the same as that of two original atomic orbitals.
⇔
i. The combining atomic orbitals must have the same or nearly the same energy.
This means that 1s orbital can combine with another 1s orbital but not with 2s orbital because the energy of 2s orbital is appreciably higher than that of 1s orbital. This is not true if the atoms are very different.
ii. The combining atomic orbitals must have the same symmetry about the molecular axis. By convention z-axis is taken as the molecular axis. It is important to note that atomic orbitals having same or nearly the same energy will not combine if they do not have the same symmetry.
For example, 2pz orbital of one atom can combine with 2pz orbital of the other atom but not with the 2px or 2py orbitals because of their different symmetries.
iii. The combining atomic orbitals must overlap to the maximum extent. Greater
the extent of overlap, the greater will be the electron-density between the nuclei of a molecular orbital.
Types of Molecular Orbitals
Molecular orbitals of diatomic molecules are designated as σ (sigma), π (pi), δ (delta), etc. In this nomenclature, the sigma (σ ) molecular orbitals are symmetrical around the bond-axis while pi (π ) molecular orbitals are not symmetrical. For example, the linear combination of 1s orbitals centered on two nuclei produces two molecular orbitals which are symmetrical around the bond-axis. Such molecular orbitals are of the σ type and are designated as σ1s and σ*1s.Fig.
(031104.2_1a)
If internuclear axis is taken to be in the z-direction, it can be seen that a linear
combination of 2pz- orbitals of two atoms also produces two sigma molecular orbitals designated as 2pz and *2pz. Fig.
Molecular orbitals obtained from 2px and 2py orbitals are not symmetrical around the bond axis because of the presence of positive lobes above and negative lobes below the molecular plane.
Such molecular orbitals,
are labelled as π and =π * . Fig.
A π bonding MO has larger electron density above and below the inter-nuclear axis. The π* antibonding MO has a node between the nuclei.
Energy Level Diagram for Molecular Orbitals
We have seen that 1s atomic orbitals on two atoms form two molecular orbitals designated as σ1s and σ*1s. In the same manner, the 2s and 2p atomic orbitals (eight atomic orbitals on two atoms) give rise to the following eight molecular orbitals:
Antibonding MOs - σ∗2s σ∗2pz π∗2px π∗2py
Bonding MOs - σ2s σ2pz π2px π2py
Spectroscopy also provides a precise analytical method for finding the constituents in material having unknown chemical composition. In a typical spectroscopic analysis, a concentration of a few parts per million of a trace element in a material can be detected through its emission spectrum.
There are three basic types of spectroscopy: atomic absorption spectroscopy (AAS), atomic emission spectroscopy (AES), and atomic fluorescence spectroscopy (AFS). AAS includes methods such as infrared (IR) spectroscopy and ultraviolet-visible (UV-Vis) spectroscopy.
Which spectroscopic technique is best?
Because the energy of the absorbed radiation depends on the environment around the absorbing nucleus in a molecule, NMR (Nuclear magnetic resonance) spectroscopy provides the most structural information of all the spectroscopic techniques used in chemistry.
What is the principle of spectrophotometry?
The fundamental theory is that light is absorbed or emitted over a certain wavelength spectrum by each compound.
While spectroscopy studies how radiated matter interacts, spectrometry is the study of the color spectra produced by that radiated matter. Spectrophotometry requires special instruments that are capable of both observing the change in radiated matter and the colors it produces.
⇔
σ1s < σ ∗1s < σ 2s < σ ∗2s < σ 2pz < (π 2px=π 2py) < (π ∗2px= π∗ 2py) < σ ∗2pz
However, this sequence of energy levels of molecular orbitals is not correct for the
remaining molecules Li2, Be2, B2, C2, N2.
For
instance, it has been observed experimentally
that for molecules such as B2, C2, N2etc.
the increasing order of energies of various molecular orbitals is
σ 1s < σ ∗1s < σ 2s < σ ∗2s < (π 2 px = π 2 py)
< σ 2pz < (π ∗2px =π∗2py) < σ ∗2pz
The important characteristic feature of this order is that the energy of 2pz molecular orbital is higher than that of 2px and 2py molecular orbitals.
The distribution of electrons among various molecular orbitals is called the electronic configuration of the molecule. From the electronic configuration of the molecule, it is possible to get important information about the molecule as discussed below.
Stability of Molecules: If Nb is the number of electrons occupying bonding orbitals and Na the number occupying the antibonding
orbitals, then
(i) the molecule is stable if Nb is greater than Na, and
(ii) the molecule is unstable if Nb is less than Na.
In (i) more bonding orbitals are occupied and so the bonding influence is stronger and a stable molecule results.
In (ii) the antibonding influence is stronger and therefore the
molecule is unstable.
Bond order
Bond order (b.o.) is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals i.e.,
Bond order (b.o.) = ½ (Nb–Na)
In bond order -
i. A positive bond order (i.e., Nb > Na) means a stable molecule while
ii. a negative (i.e., Nb
Nature of the bond
Integral bond order values of 1, 2 or 3 correspond to single, double or triple bonds
respectively.
Bond-length
The bond order between two atoms in a molecule may be taken as an approximate
measure of the bond length. The bond length decreases as bond order increases.
Magnetic nature
If all the molecular orbitals in a molecule are
doubly occupied, the substance is diamagnetic
(repelled by magnetic field).
However if one or
more molecular orbitals are singly occupied it
is paramagnetic (attracted by magnetic field), e.g., O2 molecule.
Bonding in some homonuclear diatomic molecules
1. Hydrogen molecule (H2)
2. Helium molecule (He2)
3. Lithium molecule (Li2)
4. Carbon molecule (C2 ):
5. Oxygen molecule (O2 ):
1. Hydrogen molecule (H2 ): /
It is formed by the combination of two hydrogen atoms. Each hydrogen atom has one electron in 1s orbital. Therefore, in all there are two electrons in hydrogen molecule which are present in σ1s molecular orbital. So electronic configuration
of hydrogen molecule is
H2 : (σ1s)2
The bond order of H2 molecule can be calculated as given below:
Bond order =Nb -Na 2
This means that the two hydrogen atoms
are bonded together by a single covalent bond.
The bond dissociation energy of hydrogen
molecule has been found to be 438 kJ mol –1 and
bond length equal to 74 pm.
Since no unpaired electron is present in hydrogen molecule, therefore, it is diamagnetic.
2. Helium molecule (He2 ): The electronic
configuration of helium atom is 1s 2. Each
helium atom contains 2 electrons, therefore,
in He2 molecule there would be 4 electrons.
These electrons will be accommodated in
σ1s and σ*1s molecular orbitals leading to
electronic configuration:
He2
: (σ1s) 2 (σ*1s).
Bond order of He2
is ½(2 – 2) = 0
He2 molecule is therefore unstable and does not exist.
Similarly, it can be shown that Be2 molecule
(σ1s) 2 (σ*1s) (σ2s) 2 (σ*2s) 2 also does not exist.
3. Lithium molecule (Li2 ): The electronic
configuration of lithium is 1s2, 2s1.
There are six electrons in Li2. The electronic
configuration of Li2 molecule, therefore, is
Li2 : (σ1s) 2 (σ*1s) 2 (σ2s) 2
The above configuration is also written
as KK(σ2s) 2 where KK represents the closed
K shell structure (σ1s) 2 (σ*1s) 2.
From the electronic configuration of Li2
molecule it is clear that there are four electrons
present in bonding molecular orbitals and two
electrons present in antibonding molecular
orbitals.
Its bond order, therefore, is ½ (4 –2) = 1.
It means that Li2 molecule is stable
and since it has no unpaired electrons it
should be diamagnetic.
Diamagnetic
Li2 molecules are known to exist in the
vapour phase.
4. Carbon molecule (C2 ): The electronic
configuration of carbon is 1s2 2s2 2p2.
There are twelve electrons in C2.
The electronic configuration of C2 molecule, therefore, is
C2 : (σ1s) 2 (σ ∗1s) 2 (σ ∗ 2s) 2 (π2p2x = π2p2y)
or KK (σ2s) 2 (σ ∗ 2s) 2 (π2p2x
= π2p2y)
The bond order of C2 is ½ (8 – 4) = 2
and C2 should be diamagnetic. Diamagnetic
C2 molecules have indeed been detected in
vapour phase.
It is important to note that
double bond in C2 consists of both pi bonds
because of the presence of four electrons in
two pi molecular orbitals. In most of the other
molecules a double bond is made up of a
sigma bond and a pi bond. In a similar fashion
the bonding in N2 molecule can be discussed.
5. Oxygen molecule (O2 ): The electronic
configuration of oxygen atom is 1s2 2s2 2p4.
Each oxygen atom has 8 electrons, hence,
in O2 molecule there are 16 electrons. The
electronic configuration of O2 molecule,
therefore, is
O2 : (σ1s) 2 (σ ∗1s)2 (σ 2s) 2 (σ ∗ 2s) 2σ (σ2pz) 2
(π2px2 ≡ π2p y 2) (π∗2p1x ≡ π ∗2py1).
O2 : [KK (σ 2s) 2 (σ ∗ 2s) 2σ (σ2pz) 2
(π2px2 ≡ π2p y 2) (π∗2p1x ≡ π ∗2py1)].
From the electronic configuration of O2
molecule it is clear that ten electrons are
present in bonding molecular orbitals and six
electrons are present in antibonding molecular
orbitals. Its bond order, therefore, is
Bond order = [Nb – Na] = [10 – 6] =2
So in oxygen molecule, atoms are held
by a double bond.
Moreover, it may be noted
that it contains two unpaired electrons in
π ∗2px and π ∗2py molecular orbitals, therefore,
O2 molecule should be paramagnetic,
a prediction that corresponds to
experimental observation.
In Fig are given the molecular
orbital occupancy and molecular properties
for B2 through Ne2.
This partially positively charged
hydrogen atom forms a bond with the other
more electronegative atom.
This bond is
known as hydrogen bond and is weaker
than the covalent bond.
For example, in HF
molecule, the hydrogen bond exists between
hydrogen atom of one molecule and fluorine
atom of another molecule as depicted below :
– – – Hδ+–Fδ– – – –Hδ+–Fδ– –– – – Hδ+–Fδ– –
Here, hydrogen bond acts as a bridge between
two atoms which holds one atom by covalent
bond and the other by hydrogen bond.
Hydrogen bond is represented by a dotted
line (– – –) while a solid line represents the
covalent bond.
Thus, hydrogen bond can be
defined as the attractive force which binds
hydrogen atom of one molecule with the
electronegative atom (F, O or N) of another
molecule.
Cause of Formation of Hydrogen
Bond
When hydrogen is bonded to strongly electronegative element ‘X’, the electron pair shared between the two atoms moves far away from hydrogen atom.
As a result the hydrogen atom becomes highly electropositive with respect to the other atom ‘X’.
Since there is displacement of electrons towards X, the hydrogen acquires fractional positive charge (δ+) while ‘X’ attain fractional negative charge (δ–).
This results in the formation of a polar molecule having electrostatic force of attraction which can be represented as:
Hδ+–Xδ–– – –δ+–Xδ––– – – Hδ+–Xδ––
The magnitude of H-bonding depends on the physical state of the compound.
It is maximum in the solid state and minimum in the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds.
Types of H-Bonds
There are two types of H-bonds
(i) Intermolecular hydrogen bond
(ii) Intramolecular hydrogen bond
(a) Intermolecular hydrogen bond: It is
formed between two different molecules of the
same or different compounds. For example,
H-bond in case of HF molecule, alcohol or
water molecules, etc.
(b) Intramolecular hydrogen bond : It is
formed when hydrogen atom is in between
the two highly electronegative (F, O, N)
atoms present within the same molecule. For
example, in o-nitrophenol the hydrogen is in
between the two oxygen atoms.
Fig. ⇔
It’s more of a superimposition method where constructive interference of two atomic wave functions produces a bonding molecular orbital whereas destructive interference produces non-bonding molecular orbital.
A nonbonding molecular orbital (NMBO) is a molecular orbital wherein the introduction or withdrawal of electrons has no effect on the bond sequence among atoms. This orbital is frequently denoted by the letter “n.” These orbitals are analogous to the lone electron pairs seen in Lewis structures.
What causes nonbonding orbitals to form?
The linear combination of atomic orbitals (LCAO) yields all molecular orbitals, such as a nonbonding molecular orbital. Let’s take the example of HF (Hydrogen fluoride) to understand better. Here, F contains additional electrons than H in a basic diatomic compound like HF. H’s s orbital can coincide with fluorine’s 2pz orbital to generate a bonding and an antibonding σ* orbital. The F’s px and py orbitals have no additional orbitals with which to mix. Instead, they’ve evolved into nonbonding molecular orbitals. Hence, they have the appearance of px and py orbitals. However, they will be considered molecular orbitals.
The energy of these orbitals in the molecule is similar to those in an isolated F atom. As a result, injecting one electron into them has no effect on the molecule’s durability.
Two Important Factors That Leads to Form Nonbonding Orbitals are
a. Atomic Orbital’s Uneven Number
If the proportion of atomic orbitals with appropriate symmetry is unequal, orbitals having nonbonding nature will develop. For instance, when three atomic orbitals merge, the most typical consequence is the development of a low-energy bonding molecular orbital, an intermediate-energy nonbonding molecular orbital, and a high-energy antibonding molecular orbital.
b. Energy Differences
Combining orbitals with various energies can result in orbitals with nonbonding molecular properties. For example, atomic orbitals with comparable energies will actively interact, resulting in bonded molecule orbitals with significantly lower energies than those of the constituent atomic orbitals. Conversely, atomic orbitals with extremely uneven energies have weaker interactions.
The conditions that are required for a linear combination of atomic orbitals are as follows:
a. Same Energy of combining orbitals – The combining atomic orbitals must have the same or nearly the same energy. This means that the 2p orbital of an atom can combine with another 2p orbital of another atom but 1s and 2p cannot combine as they have appreciable energy difference.
Same symmetry about the molecular axis – The combining atoms should have the same symmetry around the molecular axis for proper combination, otherwise, the electron density will be sparse.
For e.g. all the sub-orbitals of 2p have the same energy but still, the 2pz orbital of an atom can only combine with a 2pz orbital of another atom but cannot combine with 2px and 2py orbital as they have a different axis of symmetry.
In general, the z-axis is considered as the molecular axis of symmetry.
Proper Overlap between the atomic orbitals – The two atomic orbitals will combine to form molecular orbital if the overlap is proper. Greater the extent of overlap of orbitals, greater will be the nuclear density between the nuclei of the two atoms.
Types of Molecular Orbitals
According to the molecular orbital theory, there exist three primary types of molecular orbitals that are formed from the linear combination of atomic orbitals.
a. Anti Bonding Molecular Orbitals
The electron density is concentrated behind the nuclei of the two bonding atoms in anti-bonding molecular orbitals. This results in the nuclei of the two atoms being pulled away from each other. These kinds of orbitals weaken the bond between two atoms.
b. Non-Bonding Molecular Orbitals
In the case of non-bonding molecular orbitals, due to a complete lack of symmetry in the compatibility of two bonding atomic orbitals, the molecular orbitals formed have no positive or negative interactions with each other. These types of orbitals do not affect the bond between the two atoms.
c. Formation of Molecular Orbitals
An atomic orbital is an electron wave; the waves of the two atomic orbitals may be in phase or out of phase. Suppose ΨA and ΨB represent the amplitude of the electron wave of the atomic orbitals of the two atoms A and B.
Case 1: When the two waves are in phase so that they add up and the amplitude of the wave is Φ = ΨA + ΨB.
Case 2: When the two waves are out of phase, the waves are subtracted from each other so that the amplitude of the new wave is Φ´ = ΨA – ΨB
Characteristics of Bonding Molecular Orbitals
i. The probability of finding the electron in the internuclear region of the bonding molecular orbital is greater than that of combining atomic orbitals.
ii. The electrons present in the bonding molecular orbital result in the attraction between the two atoms.
iii. The bonding molecular orbital has lower energy as a result of attraction and, hence, has greater stability than that of the combining atomic orbitals.
They are formed by the additive effect of the atomic orbitals. The amplitude of the new wave is given by Φ= ΨA + ΨB
They are represented by σ, π, and δ.
Characteristics of Anti-bonding Molecular Orbitals
i. The probability of finding the electron in the internuclear region decreases in the anti-bonding molecular orbitals.
ii. The electrons present in the anti-bonding molecular orbital result in the repulsion between the two atoms.
iii. The anti-bonding molecular orbitals have higher energy because of the repulsive forces and lower stability.
iv. They are formed by the subtractive effect of the atomic orbitals. The amplitude of the new wave is given by Φ ´= ΨA – ΨB
They are represented by σ∗, π∗, δ∗
Why Are Antibonding Orbitals Higher in Energy?
The energy levels of bonding molecular orbitals are always lower than those of anti-bonding molecular orbitals. This is because the electrons in the orbital are attracted by the nuclei in the case of bonding molecular orbitals, whereas the nuclei repel each other in the case of anti-bonding molecular orbitals.
(031104.2)
Electronic Configuration and Molecular Behaviour
Hydrogen Bonding
The established order of hydrogen bonds, which can be a type of appealing intermolecular force, is known as hydrogen bonding.
Hydrogen bonding is a special type of dipole-dipole attraction between molecules, not a covalent bond to a hydrogen atom.
In water molecules, for example, hydrogen is covalent hydrogen bonded to the greater electronegative oxygen atom.
Hydrogen Bonding Conditions:
a. In order for the molecule to function, an incredibly electronegative atom ought to be linked to the hydrogen atom. The molecule will become more and more polarized as its electronegativity increases.
b. The electronegative atom needs to be small. The smaller the size, the more potent the electrostatic attraction
Types of hydrogen bonding
Intermolecular hydrogen bonding:-It is formed between two different molecules of the same or different compounds. For example, H-bond in case of HF molecule, alcohol or water molecules, etc.
Intramolecular hydrogen bonding:-It is formed when a hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule.
E1. Write the Lewis dot structure of CO molecule.
Step 1. Count the total number of valence electrons of carbon and oxygen atoms.
The outer (valence) shell configurations of carbon and oxygen atoms are: 2s2 2p2
and 2s2 2p4, respectively. The valence electrons available are 4 + 6 =10.
Step 2. The skeletal structure of CO is written as: CO
Step 3. Draw a single bond (one shared electron pair) between C and O and
complete the octet on O, the remaining two electrons are the lone pair on C.
This does not complete the octet on carbon and hence we have to resort to
multiple bonding (in this case a triple bond) between C and O atoms. This
satisfies the octet rule condition for both atoms.
E2. Write the Lewis structure of the nitrite ion, NO2– .
Step 1. Count the total number of valence electrons of the nitrogen atom,
the oxygen atoms and the additional one negative charge (equal to one electron).
N(2s22p3), O(2s22p4)5 + (2 × 6) +1 = 18 electrons
Step 2. The skeletal structure of NO2–is written as : O N O
Step 3. Draw a single bond (one shared electron pair) between the nitrogen and
each of the oxygen atoms completing the octets on oxygen atoms. This, however,
does not complete the octet on nitrogen if the remaining two electrons constitute
lone pair on it.
Hence we have to resort to multiple bonding between nitrogen and one of
the oxygen atoms (in this case a double bond). This leads to the following Lewis
dot structures.
E3. Explain the structure of CO32– ion in terms of resonance.
The single Lewis structure based on the presence of two single bonds and
one double bond between carbon and oxygen atoms is inadequate to
represent the molecule accurately as it represents unequal bonds. According
to the experimental findings, all carbon to oxygen bonds in CO32– are equivalent.
Therefore the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below.
E4. Explain the structure of CO2 molecule.
The experimentally determined carbon to oxygen bond length in CO2 is 115 pm. The lengths of a normal carbon to oxygen double bond (C=O) and carbon to oxygen triple bond (C≡O) are 121 pm and 110 pm respectively. The carbon-oxygen bond lengths in CO2 (115 pm) lie between the values for C=O and C≡O. Obviously, a single
Lewis structure cannot depict this position and it becomes necessary to write more than one Lewis structures and to consider that the structure of CO2 is best described as a hybrid of the canonical or resonance forms I, II and III.
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B1. Explain the formation of a chemical bond.
1. The attractive force uhich holds together the constituent particles (atoms, ions or molecules) in a chemical species is known as chemical bond.
Various theories have been suggested for the formation of chemical bonds such as the electronic theory, valence shell electron pair repulsion theory, valence bond theory, and molecular orbital theory.
A chemical bond formation is attributed to the tendency of a system to attain stability. Atoms combine with each other and complete their respective octets or duplets to attain the stable configuration of the nearest noble gases. This combination can occur either by sharing of electrons or by transferring one or more electrons from one atom to another. The chemical bond formed as a result of sharing of electrons between atoms is called a covalent bond. An ionic bond is formed as a result of the transference of electrons from one atom to another.
B2. Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br.
2. Mg- There are two valence electrons in Mg atom.
Na- There is only one valence electron in an atom of sodium.
B- There are 3 valence electrons in Boron atom.
O- There are six valence electrons in an atom of oxygen.
N- There are five valence electrons in an atom of nitrogen.
Br-There are seven valence electrons in bromine.
B3. Write Lewis symbols for the following atoms and ions:
S and S2-; Al and Al3+; H and H–
3. (i) S and S2-
The number of valence electrons in sulphur (16 S)is 6.
The Lewis dot symbol of sulphur (S) is
The 2 negative charge infers that there will be two electrons more in addition to the six valence electrons. Hence, the Lewis dot symbol of S2- is
(ii) Al and Al3+
The number of valence electrons in aluminium (13Al) isThe tripositive charge on a species infers that it has donated its three electrons.
(031104.1b2)
(iii) H and H-
The number of valence electrons in hydrogen is 1.
The Lewis dot symbol of hydrogen (H) is:
The unit negative charge infers that there will be one electron more in addition to the one valence electron. Hence, the Lewis dot symbol is:
B4. Draw the Lewis structures for the following molecules and ions :
H2S, SiCl4, BeF2, CO3
2-, HCOOH
4.
B5. Define octet rule. Write its significance and limitations.
5. Atoms combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to complete their octets in their valence shell: This is called the octet rule.
Significance : The octet rule helps to explain as to why different atoms combine with each other to form ionic or covalent compounds.
- Electron affinity is defined as. The amount of energy released when an electron is added to a neutral atom to form an anion. The electron affinity is the potential energy change of the atom when an electron is added to a neutral gaseous atom to form a negative ion. So the more negative the electron affinity the more favourable the electron addition process is. Not all elements form stable negative ions in which case the electron affinity is zero or even positive.
The amount of energy released when an electron is added to a neutral atom to form an anion is called electron affinity. Electron affinities are difficult to measure.
a. Electron affinity increases going left to right across a period because of increased nuclear attraction.
b. Going down the group the electron affinity should decrease since the electron is being added increasingly further away from the nucleus. Electron becomes less tightly bound and can be easily removed.
0. Q What is the trend for electron affinity?
Electron Affinity increases across a period from left to right because of increasing effective nuclear charge and decreasing size of atoms. Electron Affinity decreases down the group due to increased size of atoms.
0.Q
Is electron affinity positive or negative?
When an electron is added to a neutral atom, energy is released and Electron Affinity is generally exothermic. Due to electronic repulsions, the second Electron Affinity can be positive.
0.Q Who discovered the concept of electron affinity?
Electron Affinity was a concept that was discovered in 1901 in view of the discovery of electron negativity by Linus Carl Pauling. Electron Affinity is the amount of change in energy when an electron in a gaseous state is applied to a neutral atom.
0.Q
Why is the electron affinity of noble gases positive?
The enthalpy of electron gain for halogens is highly negative because by accepting an extra electron they can acquire the nearest stable noble gas configuration. Noble gases have significant positive enthalpy in the gain of electrons.
0.Q
Why do halogens have high electron affinity?
The halogens’ high electron affinities are due to their small size, high effective nuclear charge and an almost full outer shell of electrons. When an electron is added to halogens with very high electron affinity, high energy is released.
⇔
B6. Write the favourable factors for the formation of ionic bond.
6. The favourable factors for the formation of ionic bonds are-
i. Low ionization enthalpy of the metal atom.
ii. High electron gain enthalpy of the non-metal atom.
iii. High lattice enthalpy of the compound. formed.
iv. The electronegativity between two ions should be greater than 1.7
B7. Discuss the shape of the following molecules using the VSEPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
(031104.1b5)
7. a. BeCl2 – It is of the type EX2
The central atom has no lone pair and there are two bond pairs. i.e., BeCl2.
Shape is linear.
b. BCl3- it is of the type EX3
The central atom has no lone pair and there are three bond pairs.
Shape is trigonal planar.
c. SiCl4- It is of the type of EX4 type molecule.
The central atom has no lone pair and there are four bond pairs.
Shape of SiCl4 is tetrahedral.
d. AsF5 – It is EX5 type of molecule.
The central atom has no lone pair and there are five bond pairs.
Shape is trigonal bipyramidal.
e. H2S is of the type AB2E2.
The central atom has one lone pair and there are two bond pairs.
The shape is Bent.
f. PH3: - PH3 is of the EX5 type
The central atom has one lone pair and there are three bond pairs.
Therefore, the shape is trigonal bipyramidal.
B8. Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
8. NH3 molecule has one lone pair while H2O has two lone pairs ofelectrons. Due to the presence of lone pairs the geometries of NH3 and H2O are distorted.
Due to the presence of stronger lp-bp repulsion than bp-bp repulsion, the bond angle in NH3 is reduced from normal tetrahedral bond angle ( l09.5⁰) to l07⁰). In case of H2O, two lp of electrons force the O-H bonds more closely than N-H bonds in NH3. So the bond angle decreases to a larger extent to 104.5⁰.
B9. How do you express the bond strength in terms of bond order ?
9. The bond order shows the number of chemical bonds present between a pair of atoms.Greater the bond order, smaller is the bond length and greater bond strength.
B10. Define the bond length.
10. Bond length is defined as the average distance belween the centres of the nuclei of two bonded atoms in a molecule. In other words, it represents the equilibrium internuclear separation distance of the bonded atoms in a molecule.
B11. Explain the important aspects of resonance with reference to the CO32-ion.
(031104.1c2)
11. In some molecules single Lewis structure fails to explain all the characteristics of the molecule. As a result a number of structures can be drawn to explain all the characteristics. Such structures are called resonating structures.
Resonance is defined as the phenomenon as a result of which a molecule can be expressed in different forms none of which can explain all the properties of the molecule.
As in case of ion all the C- O bonds are equivalent. Therefore it is not correct to represent ion through one Lewis strucuture.
Instead following three resonating structures can be drawn-
In this case, each atom has an octet of electrclns. According to this structure, there are single bonds between two carbon-oxygen atoms and one double bond between carbon and oxygen atoms. Therefore, the two C-O bonds should he different than the third C=O bond. However, experimentaily, it is observed that all the three bond lengths are equal
and the bonds are intermediate between single ancl double bonds. To solve the problem, the CO32- may be represented as resonance hybrid of the structures given below.
(031104.1c2)
B12. H3PO3 can be represented by structures 1 and 2 shown below. Can these two
structures be taken as the canonical forms of the resonance hybrid representing
H3PO3 ? If not, give reasons for the same.
12. No, these cannotbe taken as canonicai forms because the position of atoms have changed.
Oz. What are Canonical forms?
Canonical form, any of a set of representations of the resonance structure of a molecule each of which contributes to the real structure.
B13. Write the resonance structures for SO3, NO2 and NO3-.
13. Resonance is defined as the phenomenon as a result of which a molecule can be expressed in different forms, none of which can explain all the properties of the molecules. The actual structure of the molecule is called resonance hybrid.
The resonating structures must have same position of the atoms, they must have same number of paired & unpaired electrons, they should have nearly same energy. The resonance structures are: (031104.1c2)
B14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N.
14. K and S:
a. The electronic configurations of K and S are as follows:
K: 2, 8, 8, 1
S: 2, 8, 6
Sulphur (S) requires 2 more electrons to complete its octet. Potassium (K) requires one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as: (031104.1c3)
b. Ca and O: The electronic configurations of Ca and O are as follows:
Ca: 2, 8, 8, 2
O: 2, 6
Oxygen requires two electrons more to complete its octet, whereas calcium has two electrons more than the nearest noble gas i.e., Argon. Hence, the electron transfer takes place as: (031104.1c3)
c. Al and N: The electronic configurations of Al and N are as follows:
Al: 2, 8, 3
N: 2, 5
Nitrogen is three electrons short of the nearest noble gas (Neon), whereas aluminium has three electrons more than Neon. Hence, the electron transference can be shown as:
(031104.1c3)
B15. Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
15. Dipole moment is the product of the magnitude of the positive or negative charge (q) and the distance (d) between the charges, i.e
μ = q × d.
(031104.1c3)
CO2 has zero dipole moment. This means that CO2 molecule is linear so that the two C = O bond moments get cancelled giving zero resultant dipole moment. However, H2O molecule has resultant dipole moment showing that it cannot be linear.The two O-H bonds are aranged in angular shape and the bond moment of two O-H bonds give resultant dipole moment.
Dipole moment is the product of the magnitude of the positive or negative charge (q) and the distance (d) between the charges, i.e
μ = q × d.
The SI unit is coulomb metre.
The molecule with symmetrical and linear geometries have zero dipole moment because they are vector in nature & the dipole of different bonds cancel with one another. CO2is symmetrical and has dipole moment 0.Therefore it will have a linear shape with bond angle 180⁰.
Resultant μ = 0 D
H2O is a unsymmetrical molecule & bent geometries, the geometries have specific dipole moment (1.84D) because the bond polarities do not cancel each other.
Therefore H2O will have a bent structure with bond angle 104⁰.
B16. Write the significance/applications of dipole moment.
16. The important significane of dipole moment are as follows:
i. For predicting the nature of the molecules- molecules with zero dipole moment are non polar while the molecule with specific dipole moment are polar in nature.
ii. For determining the shape of the molecule-
If a molecule has a specific dipole moment then its shape will not be symmetrical,they may be bent or angular & the molecule with zero dipole moment will be symmetrical & has linear shape.
iii. For comparing the polarities of the molecules- greater the dipole moment value, more will be the polarity and vice versa.
B17. Define electronegativity. How does it differ from electron gain enthalpy?
17. Electronegativity is defined as the tendency of an element to attract the shared pair of electrons towards itself in a covalent bond.There is no specific units for electronegativity.It is only a relative tendency.
Electron gain enthalpy is the energy released when one mole of electron are added to gaseous atoms of an element. It can be negative or positive depending upon whether the electron is added or removed. An element has a constant value of the electron gain enthalpy that can be measured experimentally.
B18. Explain with the help of suitable example polar covalent bond.
18. When two dissimilar atoms having different electronegativity combine together to form a covalent bond, the shared pair of electrons does not lie at equal distances from the nuclei of both the bonded atoms but shifts towards the atom having greater electronegativity. Since the more electronegative atom attracts the electrons more strongly, the distribution of electrons gets distorted. As a result, one end of the molecule, having more electronegative atom becomes slightly negatively charged while the other acquires slightly positive charge. Thus, positive and negative poles are developed and this type of bond is called polar covalent bond.
For example in HCl, chlorine being more electronegative than hydrogen will pull bonded electrons towards itself and develop a negative charge while hydrogen will have a partial positive charge.
B19. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3.
19. Dipole moment helps in calculating the percentage ionic character of polar bonds.It is the ratio of observed dipole moment to the dipole moment for the complete transfer of electrons. Greater the difference in electronegativity of bonding atoms ,the greater will be the ionic character.
On this basis, the order of increasing ionic character in the given molecules is
N2 < SO2 < ClF3 < K2O < LiF.
B20. The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
20. Since all the atoms are in either period 1 or 2,this molecule will adhere to the octet rule.The exception,of course being the hydrogen’s.they follow the duet rule (2 electrons). In the given structure, the H atom is bonded to carbon with a double bond, which is not possible because H has only one electron to share with carbon. Also the second carbon atom is not having its valency completed ,which is four , so it has to make one more bond with Oxygen.Therefore the correct Lewis structure for acetic acid is as follows:
(031104.1c5)
B21. Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar ?
Electronic configuration of carbon atom:
6C:1s22s22p2
In the excited state, the orbital picture of carbon can be represented as:
6C:1s22s22px1
2py1 2pz0
For a square planar shape, the hybridization of the central atom has to be dsp2. However, an atom of carbon does not have d-orbitals to undergo dsp2 hybridization. Hence, the structure of CH4 cannot be square planar.
B22. Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are
polar.
B23. Which out of NH3 and NF3 has higher dipole moment and why ?
B24. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
(031104.1c10)
B25. Describe the change in hybridisation (if any) of the Al atom in the following reaction.
AlCl Cl AlCl 3 4 ¬
B26. Is there any change in the hybridisation of B and N atoms as a result of the following
reaction?
B27. Draw diagrams showing the formation of a double bond and a triple bond between
carbon atoms in C2H4 and C2H2 molecules.
B28. What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2 (b) C2H4
B29. Considering x-axis as the internuclear axis which out of the following will
not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px; (c) 2py and 2py
(d) 1s and 2s.
B30. Which hybrid orbitals are used by carbon atoms in the following molecules?
CH3 –CH3; (b) CH3CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH
(031104.1_3k)
(a) CH3 –CH3; Both Carbon atoms are sp3 hybridized.
(b) CH3CH=CH3; C1 is sp3 hybridized, while C2 and C3 are sp2 hybridized.
(c) CH3-CH2-OH; Both C1 and C2 are sp3 hybridized.
(d) CH3-CHO; C1 is sp3 hybridized and C2 is sp2 hybridized.
(e) CH3COOH; C1 is sp3 hybridized and C2 is sp2 hybridized.
(031104.1_3k)
B31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type.
B32. Distinguish between a sigma and a pi bond.
Sigma bond is a chemical bond formed by the linear or co-axial overlapping of the atomic orbitals of two atoms. A pi bond is a type of covalent bond that exists between atoms where the electrons are on top and bottom of the axis connecting the nuclei of the joined atoms, also called lateral overlap of two atomic orbitals
Properties of σ bond-
i. It is made up of atomic orbitals that overlap head-to-head.
ii. Around the bond axis, it is cylindrically symmetrical.
iii. Because of the maximal overlap, it is a significantly stronger bond.
iv. It is created by the linear overlap of the s-s, s-p, and p-p orbitals
Properties of π bond-
i. It is formed by the atomic orbitals overlapping side by side.
ii. Above and below the two nuclei, there are two sections of the electronic cloud.
ii. Because there is less overlap, it is a weaker tie.
iv. It's made up of parallel p-p orbital overlap that's coplanar
B33. Explain the formation of H2 molecule on the basis of valence bond theory.
Let us assume that two hydrogen atoms (A and B) with nuclei (N A and NB ) and electrons (eA and eB) are taken to undergo a reaction to form a hydrogen molecule.
When A and B are at a large distance, there is no interaction between them. As they begin to approach each other, the attractive and repulsive forces start operating.
Following attractive forces arise-
(a) Nucleus of one atom and its own electron i.e., NA–(eA and NB – eB.
(b) Nucleus of one atom and electron of another atom i.e., NA– eB and Nb– eA.
Following repulsive forces arise-
(a) Electrons of two atoms i.e., eA–eB.
(b) Nuclei of two atoms i.e., NA–NB.
The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.
(01104.1_3j)
As the magnitude of the attractive forces is more than that of the repulsive forces, the the two atoms approach each other.
As a result, the potential energy decreases. Finally, a state is reached when the attractive forces balance the repulsive forces and the system acquires minimum energy.
This leads to the formation of a dihydrogen molecule.
B34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
The conditions that are required for a linear combination of atomic orbitals are as follows:
a. Same Energy of combining orbitals – The combining atomic orbitals must have the same or nearly the same energy. This means that the 2p orbital of an atom can combine with another 2p orbital of another atom but 1s and 2p cannot combine as they have appreciable energy difference.
Same symmetry about the molecular axis – The combining atoms should have the same symmetry around the molecular axis for proper combination, otherwise, the electron density will be sparse.
For e.g. all the sub-orbitals of 2p have the same energy but still, the 2pz orbital of an atom can only combine with a 2pz orbital of another atom but cannot combine with 2px and 2py orbital as they have a different axis of symmetry.
In general, the z-axis is considered as the molecular axis of symmetry.
Proper Overlap between the atomic orbitals – The two atomic orbitals will combine to form molecular orbital if the overlap is proper. Greater the extent of overlap of orbitals, greater will be the nuclear density between the nuclei of the two atoms.
B35. Use molecular orbital theory to explain why the Be2 molecule does not exist.
The electronic configuration of Beryllium is 1s22s2.
The molecular orbital electronic configuration for Be2 molecule can be written as:
(031104.1_3j)
Hence, the bond order for Be2 is 1/2(Nb−Na)
Where,
Nb = Number of electrons in bonding orbitals
Na = Number of electrons in anti-bonding orbitals
∴ Bond order of Be2 =1/2(4−4) = 0
A negative or zero bond order means that the molecule is unstable.
Hence, Be2 molecule does not exist.
B36. Compare the relative stability of the following species and indicate their magnetic properties; O2, O2+, O2−(superoxide),O2 2− (peroxide).
O. Give examples of superoxides and peroxides.
(031104.1_3k)
Alkali metals (which have a +1 oxidation state) form oxides, M2O,
peroxides, M2O2,
and superoxides, MO2.
dioxygenyl hexafluoroplatinate (O2PtF6),
B37. Write the significance of a plus and a minus sign shown in representing the orbitals.
B38. Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared
to equatorial bonds?
B39. Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
B40. What is meant by the term bond order? Calculate the bond order of : N2, O2, O2
+
and O2
–.
Go back - Top
L1. Isostructural species are those which have the same shape and hybridisation.
Among the given species identify the isostructural pairs.
(a) [NF3 and BF3]
(b) [BF4– and NH4+ ]
(c) [BCl3 and BrCl3]
(d) [NH3 and NO3– ]
B
OL1.a. Draw Lewis structures of the following -
(031104.1_2a)
NF3, BF3, BF4–, NH4+, BCl3, BrCl3, NH3, NO3–
The atomic number of B is 5. Its ground state electronic configuration: 1s2 2s2 2px1. When B forms BF3, one of 2s electrons becomes excited and is promoted to 2py orbital. Thus B has three unpaired electrons in its valence shell. To get the most stable structure of BF3, one 2s orbital and two 2p orbitals get hybridized and form three sp2 hybridized orbitals. These three sp2 hybridized orbitals (each having one electron) are in one plane and overlap with p orbital (or sp3, each having one unpaired electron) of three F atoms and hence BF3 is trigonal planar in shape with three covalent bonds and one pz orbital is still vacant that remains perpendicularly to the plane of the molecule. When a F- ion approaches towards the vacant pz orbital of B in BF3, to attain the possible most stable symmetrical shape B changes its hybridization to sp3 and overlapping occurs between vacant orbital of B and one filled orbital of F- ion and thus gets the tetrahedral geometry.
⇔
L2. Polarity in a molecule and hence the dipole moment depends primarily on
electronegativity of the constituent atoms and shape of a molecule. Which of
the following has the highest dipole moment?
(a) CO2
(b) HI
(c) H2O
(d) SO 2
C
OL2.a. Give electronegativity of H, O, C, S and I.
OL2.b. Draw dipole structures along with their dipole moments of the following.
CO2, HI, H2O, SO 2
L3. The types of hybrid orbitals of nitrogen in NO2+, NO3– and NH4+ respectively are
expected to be
(a) sp, sp3 and sp2
(b) sp, sp2 and sp3
(c) sp2, sp and sp3
(d) sp2, sp3 and sp
B
OL3.a Explain the structures of NO2+, NO3– and NH4+. Also give their orbital picture.
L4. Hydrogen bonds are formed in many compounds e.g., H2O, HF, NH3. The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing
order of the boiling points of above compounds is :
(a) HF > H2O > NH3
(b) H2O > HF > NH3
(c) NH3 > HF > H2O
(d) NH3 > H2O > HF
B
OL4.a. Write boiling points of H2O, HF, NH3
OL4.b. Draw structures of H2O, HF and NH3 showing hydrogen bonding in dotted lines.
L5. In PO43–ion the formal charge on the oxygen atom of P–O bond is
(a) + 1
(b) – 1
(c) – 0.75
(d) + 0.75
B
OL5.b. What is polyatomic molecule and polyatomic ion?
Also define formal charge.
A polyatomic molecule is a molecule made up of more than three atoms of an element. Example, Phosphorus (P4) and sulphur are examples (S8).
A polyatomic ion, also known as a molecular ion, is a covalently linked set of two or more atoms, or a metal complex, that behaves as a single unit and has a net charge greater than zero. This chemical species is an ion, as opposed to a molecule, which has a net charge of zero.
The formal charge over an atom of a polyatomic molecule or ion is the difference between the valence electron of that atom in the elemental state and the number of electrons assigned to that atom in Lewis structure.
L6. In NO3–ion, the number of bond pairs and lone pairs of electrons on nitrogen atom are
(a) 2, 2
(b) 3, 1
(c) 1, 3
(d) 4, 0
D
OL6.a. Give lewis representation of NO3–. (031104.1_2) (OL3a).
L7. Which of the following species has tetrahedral geometry?
(a) BH4–
(b) NH2–
(c) CO32–
(d) H3O+
A
OL7.a. Give structure of the following with shape -
(031104.1_2) (OL3a).
BH4–; - The central atom B (boron) in BH4− undergoes sp3 hybridization which results in tetrahedral geometry with 4 bond pairs and 0 lone pairs of electrons.
NH2–; Hybridisation of NH2–is sp3. Because there - charge, instead of 5 there will be 6 e two with hydragen and two lone pairs. The electronic geometry of NH2– is tetrahedral and the molecular shape is Bent shape.
CO32–; molecular geometry is trigonal planar with bond angles of 120°.
H3O+;
The shape is pyramidal as the 3 hydrogen atoms lie at three corners of oxygen as a triangle, and one lone pair of electron on oxygen attains pyramidal shape. to H3O+.
L8. Number of ¬ bonds and bonds in the following structure is–
(i) 6, 19
(ii) 4, 20
(iii) 5, 19
(iv) 5, 20
L9. Which molecule/ion out of the following does not contain unpaired electrons?
(a) N2+
(b) O2
(c) O22–
(d) B2
C
OL9.a. Give molecular structure and bond order of the following-
N2+
Total no. of electrons = 13
Its electronic configuration is : σ1s² σ*1s² σ2s² σ*2s² π 2py² [π2pz² σ2px2 ]
Bond order = 1/2[Nb-Na],
So bond order = (9–4)/2 = 5/2 = 2.5
O2
Total no. of electrons = 16
increasing order of energies of various molecular orbitals is σ 1s2 < σ ∗1s2 < σ 2s2 < σ ∗2s2 < (π 2 px2 = π 2 py2) < σ 2pz2 < (π ∗2px1 =π∗2py1)
The number of bonding electrons = 10
The number of anti-bonding electrons = 6
Bond order = 1/2[Nb-Na],
Bond order= (10-6)/2=2
O22–
Total no. of electrons = 18
increasing order of energies of various molecular orbitals is σ 1s2 < σ ∗1s2 < σ 2s2 < σ ∗2s2 < (π 2 px2 = π 2 py2) < σ 2pz2 < (π ∗2px2 =π∗2py2)
The number of bonding electrons = 10
The number of anti-bonding electrons = 8
Bond order = 1/2[Nb-Na],
Bond order= (10-8)/2=1
B2
Total no. of electrons = 10
increasing order of energies of various molecular orbitals is σ 1s2 < σ ∗1s2 < σ 2s2 < σ ∗2s2 < (π 2 px1 = π 2 py1)
The number of bonding electrons = 10
The number of anti-bonding electrons = 8
Bond order = 1/2[Nb-Na],
Bond order= (6-4)/2=1
L10. In which of the following molecule/ion all the bonds are not equal?
(i) XeF4
(ii) BF4–
(iii) C2H4
(iv) SiF4
C
(031104.1_2d)
In XeF4, how many 90-degree angles are there? Ans. Around Xenon, there are six electron pairs (four bonding and two lone pairs).
How many bonds does BF4 have?
The boron has a -1 charge since it has four bonds and it is in group 3: The central atom has 4 atoms and no lone pairs, therefore, both the electron and molecular geometries are tetrahedral. Steric number 4 corresponds to sp3-hybridization where the idealized bond angles are 109.5⁰.
There are 2 covalent bonds between the two carbon and 4 carbon-hydrogen covalent bonds.
These four hybrid orbitals are then utilized by silicon to create covalent bonds with the four fluorine atoms. Therefore, the hybridization of the silicon atom in SiF4 is sp³, which allows it to form the four covalent bonds with the four fluorine atoms in a tetrahedral arrangement.
L11. In which of the following substances will hydrogen bond be strongest?
(i) HCl
(ii) H2O
(iii) HI
(iv) H2S
L12. If the electronic configuration of an element is 1s2 2s2 2p6 3s2 3p6 3d2 4s2, the
four electrons involved in chemical bond formation will be_____.
(i) 3p6
(ii) 3p6, 4s2
(iii) 3p6, 3d2
(iv) 3d2, 4s2
L13. Which of the following angle corresponds to sp2 hybridisation?
(i) 90°
(ii) 120°
(iii) 180°
(iv) 109°
L13. The electronic configurations of three elements, A, B and C are given below.
Answer the questions 14 to 17 on the basis of these configurations.
A 1s2 2s2 2p6
B 1s2 2s2 2p6 3s2 3p3
C 1s2 2s2 2p6 3s2 3p5
L14. Stable form of A may be represented by the formula :
(i) A
(ii) A2
(iii) A3
(iv) A4
L15. Stable form of C may be represented by the formula :
(i) C
(ii) C2
(iii) C3
(iv) C4
L16. The molecular formula of the compound formed from B and C will be
(i) BC
(ii) B2C
(iii) BC2
(iv) BC3
L17. The bond between B and C will be
(i) Ionic
(ii) Covalent
(iii) Hydrogen
(iv) Coordinate
L18. Which of the following order of energies of molecular orbitals of N2 is correct?
(i) (¬2py ) < ( 2pz) < (¬*2px) (-*2py)
(ii) (¬2py ) > ( 2pz) > (¬*2px) (¬*2py )
(iii) (¬2py ) < ( 2pz) > (¬*2px) (¬*2py )
(iv) (¬2py ) > ( 2pz) < (¬*2px) (¬*2py )
L19. Which of the following statement is not correct from the view point of molecular
orbital theory?
(i) Be2 is not a stable molecule.
(ii) He2 is not stable but He2
+ is expected to exist.
(iii) Bond strength of N2 is maximum amongst the homonuclear diatomic
molecules belonging to the second period.
(iv) The order of energies of molecular orbitals in N2 molecule is
2s < * 2s < 2pz < (¬2px = ¬2py ) < (¬* 2px = ¬* 2py) < *2pz
L20. Which of the following options represents the correct bond order :
(i) O2– > O2 > O2+
(ii) O2– < O2 < O2+
(iii) O2– > O2 < O2+
(iv) O2– < O2 > O2+
L21. The electronic configuration of the outer most shell of the most electronegative
element is
(i) 2s22p5
(ii) 3s23p5
(iii) 4s24p5
(iv) 5s25p2
L22. Amongst the following elements whose electronic configurations are given
below, the one having the highest ionisation enthalpy is
(i) [Ne]3s23p1
(ii) [Ne]3s23p2
(iii) [Ne]3s23p2
(iv) [Ar]3d104s24p3
In the following questions two or more options may be correct.
L23. Which of the following have identical bond order?
(i) CN–
(ii) NO+
(iii) O2–
(iv) O22–
L24. Which of the following attain the linear structure:
(i) BeCl2
(ii) NCO+
(iii) NO2
(iv) CS2
L25. CO is isoelectronic with
(i) NO+
(ii) N2
(iii) SnCl2
(iv) NO2 –
L26. Which of the following species have the same shape?
(i) CO2
(ii) CCl4
(iii) O3
(iv) NO2–
L27. Which of the following statements are correct about CO32– ?
(i) The hybridisation of central atom is sp3.
(ii) Its resonance structure has one C–O single bond and two C=O double
bonds.
(iii) The average formal charge on each oxygen atom is 0.67 units.
(iv) All C–O bond lengths are equal.
L28. Dimagnetic species are those which contain no unpaired electrons. Which
among the following are dimagnetic?
(i) N2
(ii) N22–
(iii) O2
(iv) O22–
L29. Species having same bond order are :
(i) N2
(ii) N2 –
(iii) F2+
(iv) O2–
L30. Which of the following statements are not correct?
(i) NaCl being an ionic compound is a good conductor of electricity in the
solid state.
(ii) In canonical structures there is a difference in the arrangement of atoms.
(iii) Hybrid orbitals form stronger bonds than pure orbitals.
(iv) VSEPR Theory can explain the square planar geometry of XeF4.
III. Short Answer Type
L31. Explain the non linear shape of H2S and non planar shape of PCl3 using valence
shell electron pair repulsion theory.
L32. Using molecular orbital theory, compare the bond energy and magnetic
character of O2+ and O2– species.
L33. Explain the shape of BrF2.
L34. Structures of molecules of two compounds are given below :
(a) Which of the two compounds will have intermolecular hydrogen bonding
and which compound is expected to show intramolecular hydrogen
bonding.
(b) The melting point of a compound depends on, among other things, the
extent of hydrogen bonding. On this basis explain which of the above
two compounds will show higher melting point.
(c) Solubility of compounds in water depends on power to form hydrogen
bonds with water. Which of the above compounds will form hydrogen
bond with water easily and be more soluble in it.
L35. Why does type of overlap given in the following figure not result in bond
formation?
L36. Explain why PCl5 is trigonal bipyramidal whereas IF2 is square pyramidal.
L37. In both water and dimethyl ether (CH3—O—CH3), oxygen atom is central atom, and has the same hybridisation, yet they have different bond angles.
Which one has greater bond angle? Give reason.
L38. Write Lewis structure of the following compounds and show formal charge
on each atom. HNO3, NO2, H2SO4
L39. The energy of σ2pz molecular orbital is greater than -π2px and π2py molecular
orbitals in nitrogen molecule. Write the complete sequence of energy levels in
the increasing order of energy in the molecule. Compare the relative stability
and the magnetic behaviour of the following species :
N2, N2+ , N2– , N22+
L40. What is the effect of the following processes on the bond order in N2 and O2?
(i) N2 N2+ e–
(ii) O2 O2+ e–
L41. Give reasons for the following :
(i) Covalent bonds are directional bonds while ionic bonds are nondirectional.
(ii) Water molecule has bent structure whereas carbon dioxide molecule is
linear.
(iii) Ethyne molecule is linear.
L42. What is an ionic bond? With two suitable examples explain the difference
between an ionic and a covalent bond?
L43. Arrange the following bonds in order of increasing ionic character giving
reason.
N—H, F—H, C—H and O—H
L44. Explain why CO3
2– ion cannot be represented by a single Lewis structure. How
can it be best represented?
L45. Predict the hybridisation of each carbon in the molecule of organic compound
given below. Also indicate the total number of sigma and pi bonds in this
molecule.
L46. Group the following as linear and non-linear molecules :
H2O, HOCl, BeCl2, Cl2O
L47. Elements X, Y and Z have 4, 5 and 7 valence electrons respectively. (i) Write
the molecular formula of the compounds formed by these elements individually
with hydrogen. (ii) Which of these compounds will have the highest dipole
moment?
L48. Draw the resonating structure of
(i) Ozone molecule
(ii) Nitrate ion
L49. Predict the shapes of the following molecules on the basis of hybridisation.
BCl3, CH4, CO2, NH3
L50. All the C—O bonds in carbonate ion (CO3
2– ) are equal in length. Explain.
L51. What is meant by the term average bond enthalpy? Why is there difference in
bond enthalpy of O—H bond in ethanol (C2H5OH) and water?
IV. Matching Type
L52. Match the species in Column I with the type of hybrid orbitals in Column II.
Column I Column II
(i) SF4 (a) sp3d2
(ii) IF5 (b) d2sp3
(iii) NO2
+ (c) sp3d
(iv) NH4
+ (d) sp3
(e) sp
16
L53. Match the species in Column I with the geometry/shape in Column II.
Column I Column II
(i) H3O+ (a) Linear
(ii) HC CH (b) Angular
(iii) ClO2
– (c) Tetrahedral
(iv) NH4
+ (d) Trigonal bipyramidal
(e) Pyramidal
L54. Match the species in Column I with the bond order in Column II.
Column I Column II
(i) NO (a) 1.5
(ii) CO (b) 2.0
(iii) O2
– (c) 2.5
(iv) O2 (d) 3.0
L55. Match the items given in Column I with examples given in Column II.
Column I Column II
(i) Hydrogen bond (a) C
(ii) Resonance (b) LiF
(iii) Ionic solid (c) H2
(iv) Covalent solid (d) HF
(e) O3
L56. Match the shape of molecules in Column I with the type of hybridisation in
Column II.
Column I Column II
(i) Tetrahedral (a) sp2
(ii) Trigonal (b) sp
(iii) Linear (c) sp3
V. Assertion and Reason Type
In the following questions a statement of Assertion (A) followed by a statement of
Reason (R) is given. Choose the correct option out of the choices given below each
question.
L57. Assertion (A) : Sodium chloride formed by the action of chlorine gas on
sodium metal is a stable compound.
Reason (R) : This is because sodium and chloride ions acquire octet in
sodium chloride formation.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
L58. Assertion (A) : Though the central atom of both NH3 and H2O molecules are
sp3 hybridised, yet H–N–H bond angle is greater than that of
H–O–H.
Reason (R) : This is because nitrogen atom has one lone pair and oxygen
atom has two lone pairs.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
L59. Assertion (A): Among the two O–H bonds in H2O molecule, the energy
required to break the first O–H bond and the other O–H bond
is the same.
Reason (R) : This is because the electronic environment around oxygen is
the same even after breakage of one O–H bond.
(i) A and R both are correct, and R is correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
VI. Long Answer Type
L60. (i) Discuss the significance/ applications of dipole moment.
(ii) Represent diagrammatically the bond moments and the resultant dipole
moment in CO2 , NF3 and CHCl3.
L61. Use the molecular orbital energy level diagram to show that N2 would be
expected to have a triple bond, F2, a single bond and Ne2, no bond.
L62. Briefly describe the valence bond theory of covalent bond formation by taking
an example of hydrogen. How can you interpret energy changes taking place
in the formation of dihydrogen?
L63. Describe hybridisation in the case of PCl5 and SF6. The axial bonds are longer
as compared to equatorial bonds in PCl5 whereas in SF6 both axial bonds and
equatorial bonds have the same bond length. Explain.
L64. (i) Discuss the concept of hybridisation. What are its different types in a
carbon atom.
(ii) What is the type of hybridisation of carbon atoms marked with star. Comprehension given below is followed by some multiple choice questions.
Each question has one correct option. Choose the correct option.
Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals
combine to form two molecular orbitals called bonding molecular orbital (BMO)
and anti bonding molecular orbital (ABMO). Energy of anti bonding orbital is raised
above the parent atomic orbitals that have combined and the energy of the bonding
orbital is lowered than the parent atomic orbitals. Energies of various molecular
orbitals for elements hydrogen to nitrogen increase in the order :
x y z x y z 1s < * 1s < 2s < * 2s < (¬2p ¬2p ) < 2p < (¬ * 2p ¬ * 2p ) < * 2p and
for oxygen and fluorine order of energy of molecular orbitals is given below :
z x y x y z 1s < * 1s < 2s < * 2s < 2p < (¬2p ¬2p ) < (¬ * 2p ¬ * 2p ) < * 2p
Different atomic orbitals of one atom combine with those atomic orbitals of the
second atom which have comparable energies and proper orientation. Further, if
the overlapping is head on, the molecular orbital is called ‘Sigma’, ( ) and if the
overlap is lateral, the molecular orbital is called ‘pi’, (¬). The molecular orbitals
are filled with electrons according to the same rules as followed for filling of atomic
orbitals. However, the order for filling is not the same for all molecules or their
ions. Bond order is one of the most important parameters to compare the strength
of bonds.
L65. Which of the following statements is correct?
(i) In the formation of dioxygen from oxygen atoms 10 molecular orbitals
will be formed.
(ii) All the molecular orbitals in the dioxygen will be completely filled.
(iii) Total number of bonding molecular orbitals will not be same as total
number of anti bonding orbitals in dioxygen.
(iv) Number of filled bonding orbitals will be same as number of filled anti
bonding orbitals.
MCQs
Which one of them is the weakest?
(a) Ionic bond
(b) Covalent bond
(c) Metallic Bond
(d) van der Waals force
D
Among them van der Waals force is the weakest force.
The bond angle around atom which uses sp2 hybridization is ———–
(a) 120⁰
(b) 180⁰
(c) 107⁰
(d) 109⁰. 28’
A
The bond angle around an atom which uses sp2 hybridization is 120⁰.
The correct order of hybridisation of the central atom in the following species: NH3, [PtCl4]2-, PCl5 and BCl3 is
(a) dsp2, dsp3, sp2 and sp3
(b) sp3, dsp2, dsp3, sp2
(c) dsp2, sp2, sp3, dsp3
(d) dsp2, sp3, sp2, dsp3.
B
sp3, dsp2, dsp3, sp2
Amongst H2O, H2S, H2Se and H2Te the one with the highest boiling point is
(a) H2O because of hydrogen bonding
(b) H2Te because of higher molecular weight
(c) H2S because of hydrogen bonding
(d) H2Se because of lower molecular weight.
A
H2O because of hydrogen bonding
Number of π bonds in Naphthalene is
(a) 6
(b) 3
(c) 4
(d) 5
D
5
Hydrogen bonding is not present in
(a) Glycerine
(b) Water
(c) Hydrogen sulphide
(d) Hydrogen fluoride
C
Hydrogen sulphide
In a double bond connecting two atoms there is a sharing of
(a) 2 electrons
(b) 4 electrons
(d) 1 electron
(d) all electrons
B
4 electrons
Atomic orbitals of carbon in Carbon dioxide are
(a) sp hybridized
(b) sp³d hybridized
(c) sp² hybridized
(d) sp³ hybridized
A
sp hybridized
The shape and hybridisation in BF3 is
(a) sp2, linear
(b) sp3d, plannar
(c) sp2, planar
(d) sp3planar
C
sp2, planar
Bond dissociation energies of HF, HCl, HBr follow the order
(a) HCl > HBr > HF
(b) HF > HBr > HCl
(c) HF > HCl > HBr
(d) HBr > HCl > HF
C
HF > HCl > HBr
In the resonating structures of benzene, the number of sigma and pi bonds are
(a) 3π and 12σ
(b) 3σ and 3π
(c) 6σ and 6π
(d) 12σ and 12π
A
In Benzene (C6H6) ring contain 3 alternative double bond = 3π bonds and (six C-C sigma bond + six C-H sigma bond) = 12 σ bonds.
Which of the following substances has a dipole moment more than zero?
(a) Water
(b) Methane
(c) Carbon dioxide
(d) Nitrogen
A
The dipole moment (μ) of H2O = 1.84 D.
The ion which is iso-electronic with CO is —————
(a) CN-
(b) O2-
(c) N2+
(d) O2+
A
Both CO (6 + 8=14) and CN– (6 + 7+1= 14) have the same electrons. So they are iso-electronic with each other.
The correct bond order in the following species is —————.
(a) O2+< O2-< O22+
(b) O2-< O2+< O22+
(c) O22+< O2+ < O2-
(d) O22+< O2-< O2+
B
The correct one is O2-< O2+< O22+,
bond order is 1.5 < 2.5 < 3.0 respectively.
Which of the following molecules have trigonal planar geometry?
(a) BF3
(b) NH3
(c) PCl3
(d) IF3
A
BF3 has trigonal planar geometry. The hybridization of BF3 is sp2 hybridization.
During change of O2 to O22-ion, the electron adds on which of the following orbitals?
(a) σ* orbital
(b) π orbital
(c) σ orbital
(d) π* orbital
D
The incoming electrons added to π* orbital in the change of O2 to O22- ion.
In which of the following pairs, the two molecules have identical bond orders:
(a) N2 , O22-
(b) N2 , O2-
(c) N2-, O2
(d) O22-, N2
A
Both N2 and O22- have bond order equal to 3.0.
Ionic bonds will be formed more easily between elements with comparatively:
(a) low ionization enthalpy and high electron affinity
(b) high ionization enthalpy and high electron affinity
(c) low ionization enthalpy and low electron affinity
(d) high ionization enthalpy and low electron affinity
A
Ionic bonds will be formed more easily between elements with low ionization enthalpy and high electron affinity.
Which one of the following molecules will form a linear polymeric structure due to hydrogen bonding?
(a) HCl
(b) HF
(c) H2O
(d) NH3
D
NH3
Which among the following has the largest dipole moment?
(a) NH3
(b) H2O
(c) HI
(d) SO3
B
H2O
Maximum bond angle is present in case of
(a) BBr3
(b) BCl3
(c) BF3
(d) Same in all
D
Same in all
Which of the following pairs of molecules will have permanent dipole moment for both members?
(a) NO2 and CO2
(b) NO2 and O3
(c) SiF4and CO2
(d) SiF4 and NO2
B
NO2 and O3
Which one of the following does not contain coordinate bond?
(a) BH4-
(b) NH4+
(c) CO32-
(d) H3O+
C
CO32-
Which of the following are iso-structural?
(a) XeF2, IF~2
(b) NH3, BF3
(c) CO32-, SO32-
(d) PCl5, ICl5
A
XeF2, IF~2
Which of the following are isoelectronic and iso-structural NO3-, CO32-, CO3-, SO 3
(a) NO3-, CO32-
(b) SO3, NO3-
(c) ClO3-, CO32-
A
NO3-, CO32-
Which of the following hydrocarbons has the lowest dipole moment?
(a) CH2 = CH – C ≡ CH
(b) CH3C ≡ C CH3
(c) CH3CH2C ≡ CH
B
CH3C ≡ C CH3
Which of the following has zero dipole moment?
(a) CIF
(b) PCl3
(c) SiF4
(d) Cl3CF
Answer
Answer: (c) SiF4
Question 21.
Among the following the molecule with, highest dipole moment is
(a) CH3Cl
(b) CH2Cl2
(c) CHCl3
(d) CCl4
Answer
Answer: (a) CH3Cl
Question 22.
In which of the following species is the underlined carbon having sp3 hybridisation?
(a) CH3COOH
(b) CH3CH2OH
(c) CH3-CO-CH3
(d) CH2=CH-CH3
Answer
Answer: (b) CH3CH2OH
Question 23.
Which one of the following arrangements of molecules is correct on the basis of their dipole moments?
(a) BF3 > NF3 > NH3
(b) NF3 > BF3 > NH3
(c) NH3 > BF3 > NF3
(d) NH3 > NF3 > BF3
(e) NH3 = NF3 > BF3
Answer
Answer: (d) NH3> NF3 > BF3
Question 24.
In OF2, number of bond pairs and lone pairs of electrons are respectively
(a) 2, 0
(b) 2, 8
(c) 2, 10
(d) 2, 9
Answer
Answer: (b) 2, 8
Question 26.
In which of the following pairs the two species have identical bond order?
(a) N−2, O2−2
(b) N+2, O−2
(c) N−2, O+2
(d) O+2, N2−2
Answer
Answer: (c) N−2, O+2
Question 29.
The number of antibonding electron pairs in O2−2 molecular ion on the basis of M.O. theory (At. No. of O = 8) is
(a) 2
(b) 3
(c) 4
(d) 5
Answer
Answer: (c) 4
Main axis of a diatomic molecule is Z. Atomic orbitals px and py overlap to form which of the following orbital?
(a) π-molecular orbital
(b) σ-molecular orbital
(c) δ-molecular orbital
(d) no bond will form
D
no bond will form
Among KO2, AlO−2, BaO2 and NO+2 unpaired electron is present in
(a) NO+2 and BaO2
(b) KO2 and AIO−2
(c) KG2 only
(d) BaO2 only
C
KG2 only
In which of the following bond angle is maximum?
(a) NH3
(b) NH+4
(c) PCl5
(d) SCl2
B
NH+4
In NO−3 ion, the number of bond pairs and lone pairs of electrons on N atoms are
(a) 2, 2
(b) 3, 1
(c) 1, 3
(d) 4, 0
D
4, 0
The bond order of O−2 is
(a) 0.5
(b) 1.5
(c) 2.5
(d) 3.5.
B
1.5
Which of the following is not paramagnetic?
(a) N+2
(b) CO
(c) O−2
(d) NO.
B
CO
Fill in the Blanks
Atomic orbitals of carbon in carbon dioxide are ……………….. hybridized.
Answer: sp
MCQ Questions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure with Answers 2 molecule contains ……………….. σ bonds and ……………….. π bonds.
Answer: 19, 4
H2O shows ……………….. hybridization, whereas BeF2 shows ……..hybridization. .
sp³, sp
O2 has ………………. unpaired electron, whereas O2 has unpaired electrons. .
one, two
Out of C2H6, C2H4, CH4, C2H2 it is only ……………….. which shows sp hybridization.
C2H2
CaC2 molecule has ……………….. bonds in it.
Ionic
XeF2 and CO2 molecules have ……………….. shapes.
identical or linear
PF5 has ……………….. geometry whereas IF5 has geometry.
Trigonal bipyramid, square planar
The lone pair-lone pair repulsion are ……………….. than lone pair-bond pair repulsions.
More
A bond formed between two atoms by the overlap of their atomic orbitals along the internuclear axis is called a ……………….. bond.
sigma (σ)
True/False Statements
Bond order of O2 is greater than that of O+2
False
Both benzene and carbon tetrachloride are non-polar.
True
The presence of polar bonds in a polyatomic molecule suggests that the molecule has non-zero dipole moment.
False
Whereas NH3 has one lone pair of electrons, NH+4 ion does not have any.
True
The dipole moment of NF3 is more than NH3.
True
Bond length in N2 is greater than that of O2.
False
O2 is paramagnetic whereas O2−2 is diamagnetic.
True
BF3 is triangular planar, whereas NH3 is pyramidal.
True
HF shows hydrogen bonding but HBr does not.
True
Out of sp³, sp², sp hybrid orbitals, sp³ has maximum s-character.
False
