NCERT Solutions: Chemical Bonding & Molecular Structure

Q4.1: Explain the formation of a chemical bond.
Ans: 

•  A chemical bond is an attractive force that holds atoms, ions or molecules together in a chemical compound.
• The formation of chemical bonds is described by several complementary theories, including the Electronic Theory of Bonding, Valence Shell Electron Pair Repulsion (VSEPR) Theory, Valence Bond Theory (VBT) and Molecular Orbital Theory (MOT). Each theory emphasises different aspects: electronic arrangements, molecular shape, orbital overlap and molecular orbitals respectively.
• Chemical bond formation occurs because systems tend to attain greater stability. Noble gases are inert because their outer shells are full. Atoms with incomplete outer shells are reactive and combine to achieve a stable electronic configuration (often resembling the nearest noble gas) by either sharing electrons or by transferring electrons:
  i) When atoms share electrons, a covalent bond is formed.
  ii) When one atom transfers electrons to another, producing oppositely charged ions that attract each other, an ionic bond is formed.

Thus, chemical bonds stabilise atoms by allowing them to attain a more favourable valence-shell electron arrangement, lowering the total energy of the system.

Q4.2: Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Ans: Mg: Magnesium has two valence electrons. The Lewis symbol places two dots around Mg to show these valence electrons:

Na: Sodium has one valence electron. Its Lewis dot structure shows a single dot:

B: Boron has three valence electrons. Its Lewis symbol shows three dots:

O: Oxygen has six valence electrons. The Lewis symbol shows six dots arranged as three pairs and two singles:

N: Nitrogen has five valence electrons. The Lewis symbol for N shows five dots:

Br: Bromine has seven valence electrons. Its Lewis dot symbol shows seven dots:

To write Lewis symbols, determine the number of valence electrons from the group number and place that number of dots around the element symbol in pairs as far as possible.

Q4.3: Write Lewis symbols for the following atoms and ions:
S and S^2-; Al and Al³⁺; H and H^-
Ans: (i) S and S^2-
The number of valence electrons in sulphur is 6.
The Lewis dot symbol of sulphur (S) is

When sulphur gains two electrons its charge becomes 2- and the symbol is shown with eight electrons and the negative charge indicated: .

(ii) Al and Al³⁺
Aluminium has three valence electrons.
The Lewis dot symbol of aluminium (Al) is

When aluminium loses three electrons it becomes Al³⁺; the cation is usually shown in brackets with its charge: [Al]³⁺.

(iii) H and H^-
Hydrogen has one valence electron.
The Lewis dot symbol of hydrogen (H) is

When hydrogen gains one electron it becomes hydride (H^-) and is shown with two electrons: .

Q4.4: Draw the Lewis structures for the following molecules and ions:
H₂S, SiCl₄, BeF₂,  , HCOOH

Ans:

Q4.5: Define octet rule. Write its significance and limitations.
Ans: The octet rule (or electronic theory of bonding) was developed by Kossel and Lewis. It states that atoms tend to gain, lose or share electrons so as to attain a valence shell containing eight electrons, similar to the nearest noble gas.

Significance:
- It gives a simple explanation for the formation of many ionic and covalent compounds by predicting when atoms will gain, lose or share electrons.
- It helps to rationalise common formulas and valencies for many main-group elements.

Limitations of the octet rule:
(a) It does not predict molecular shapes or relative stabilities - for that, other theories (VSEPR, VBT, MOT) are needed.
(b) Some noble gases (for example xenon and krypton) form compounds (e.g., XeF₂, KrF₂), so the rule is not absolute.
(c) Elements in and beyond the third period can have expanded valence shells (more than eight electrons) because d-orbitals can be involved (examples: PF₅, SF₆).
(d) Molecules or ions with an odd number of electrons cannot satisfy the octet for all atoms (examples: NO, NO₂).

(e) Some compounds have fewer than eight electrons around the central atom (electron-deficient species) - examples: BeH₂, AlCl₃, BF₃.

(e) This rule cannot be applied to those compounds in which the number of electrons surrounding the central atom is less than eight. For example, LiCl, BeH₂, AlCl₃ do not obey the octet rule.

Q4.6: Write the favourable factors for the formation of ionic bond.
Ans: An ionic bond results from the transfer of one or more electrons from a metal to a non-metal. Factors that favour ionic bond formation are:

(i) Low ionization enthalpy of the metal (so it can lose electrons easily).
(ii) High (more negative) electron gain enthalpy of the non-metal (so it can gain electrons readily).
(iii) High lattice energy of the resulting ionic solid (large energy released on formation of the crystal lattice), which compensates for the energy required to form ions.

Additional factors include large difference in electronegativity between the atoms and suitable ionic sizes (small highly charged ions lead to strong lattice energies).

Q4.7: Discuss the shape of the following molecules using the VSEPR model:
BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃
Ans:
(i) BeCl_2:

The central Be atom has no lone pairs and two bond pairs (type AB₂). According to VSEPR, BeCl₂ is linear (bond angle 180°).

(ii) BCl_3:

Boron has no lone pair and three bond pairs (type AB₃), so BCl₃ is trigonal planar (bond angles 120°).

(iii) SiCl_4:

Silicon has four bond pairs and no lone pair (AB₄), giving a tetrahedral shape for SiCl₄ (bond angles ≈109.5°).

(iv) AsF_5:

The central atom has five bond pairs and no lone pairs (AB₅), so AsF₅ is trigonal bipyramidal. Equatorial positions are 120° apart, axial positions are 90° to the equatorial plane.

(v) H₂S:

The S atom has two lone pairs and two bond pairs (type AB₂E₂). The geometry is bent (similar to water) but with a smaller H-S-H angle due to larger size and weaker lp-bp repulsion compared with oxygen.

(vi) PH_3:

Phosphorus in PH₃ has one lone pair and three bond pairs (AB₃E), so the shape is trigonal pyramidal (not trigonal bipyramidal). The lone pair compresses the bond angles slightly below the ideal tetrahedral angle.

Q4.8: Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Ans: The central nitrogen in NH₃ has one lone pair and three bond pairs (AB₃E) - trigonal pyramidal. Oxygen in H₂O has two lone pairs and two bond pairs (AB₂E₂) - bent.

The lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion. Since oxygen has two lone pairs, the repulsion on the bond pairs is greater in H₂O than in NH₃, causing a larger compression of the bond angle. Thus, the H-O-H angle (~104.5°) is smaller than the H-N-H angle in ammonia (~107°).

Q4.9: How do you express the bond strength in terms of bond order?
Ans: Bond strength correlates with bond order: a higher bond order indicates more shared electron density between the two atoms, a larger bond energy and generally a shorter bond length. Thus, greater bond order → stronger bond.

Q4.10: Define the bond length.
Ans: Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule.

Bond lengths are expressed in Angstrom (Å = 10^-10 m) or picometre (pm = 10^-12 m) and are measured by spectroscopic, X-ray diffraction and electron-diffraction techniques.

In ionic compounds the bond length is approximately the sum of the ionic radii (d = r₊ + r_-), and in covalent compounds it is the sum of covalent radii (d = r_A + r_B).

Q4.11: Explain the important aspects of resonance concerning the ion.

Ans: Experimental evidence shows all C-O bonds in the carbonate ion are equivalent and of the same length. A single Lewis structure with one C=O and two C-O single bonds does not represent this equality.

Therefore carbonate ion is described as a resonance hybrid of three equivalent canonical structures in which the double bond is successively placed on each oxygen atom. The true structure is a delocalised average of these forms and shows equal bond orders (≈1⅓) for all C-O bonds.

Resonance stabilises the ion by delocalising electron density over the oxygens and the central carbon.

 Q4.12: 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.

Ans: The given two structures cannot be taken as canonical resonance forms because resonance requires the positions of atoms to remain the same while only the positions of electrons are delocalised. In the two structures shown the positions of atoms are changed (for example, which hydrogen is bonded to oxygen or phosphorus differs), so they are different structural isomers rather than resonance forms.

Q4.13: Write the resonance structures for SO₃, NO₂ and.

Ans: The resonance structures are:

(a) SO₃:

 (b) 

(c) :

Each case shows multiple canonical forms whose average (resonance hybrid) accounts for bond equivalence and partial bond orders.

Q4.14: 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.
Ans:  (a) K and S:

The electronic configurations of K and S are as follows:
K: 2, 8, 8, 1
S: 2, 8, 6

Sulphur requires two more electrons to complete its octet. Each potassium atom can donate one electron. Hence two K atoms transfer one electron each to S to give K⁺ ions and S^2- as:

(b) Ca and O:
The electronic configurations are:
Ca: 2, 8, 8, 2
O: 2, 6

Calcium can lose two electrons to become Ca²⁺, while oxygen needs two electrons to complete its octet. Thus, Ca transfers two electrons to O forming Ca²⁺ and O^2-:

(c) Al and N:

The electronic configurations are:

Al: 2, 8, 3

N: 2, 5

Aluminium can donate three electrons to become Al³⁺, and nitrogen requires three electrons to attain an octet (N^3-). Hence the electron transfer can be shown as:

Q4.15: 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.
Ans: Carbon dioxide has a zero dipole moment experimentally. This is possible only if the molecule is linear so that the two C-O bond dipoles are equal in magnitude and opposite in direction, cancelling each other (resultant μ = 0 D).

H₂O has a measured dipole moment of about 1.84 D. The bent geometry prevents cancellation of the O-H bond dipoles, so a net dipole exists. Thus the observed dipole moments explain the linear shape of CO₂ and the bent shape of H₂O.

Q4.16:Write the significance/applications of dipole moment.

Ans: In heteronuclear molecules a difference in electronegativity produces polar bonds and an uneven electron distribution; one end of the bond becomes slightly positive and the other slightly negative. The product of the magnitude of the partial charge (Q) and the distance (r) between the centres of positive and negative charge is the dipole moment (μ) of the molecule:

μ = Q × r

Dipole moment is a vector; an arrow is used with its tail at the positive end and head towards the negative end.

The SI unit of dipole moment is the coulomb metre (C·m). In practice chemists commonly use the Debye (D): 1 D = 3.33564 × 10^-30 C·m.

Applications and significance:
- Dipole moment measures molecular polarity and helps distinguish between polar and non-polar molecules (non-polar molecules have μ = 0).
- It aids estimation of bond polarity and the distribution of electron density in molecules.
- Dipole moments are used to estimate percentage ionic character of bonds and to predict intermolecular interactions (e.g., solubility, boiling points).

Q4.17: Define electronegativity. How does it differ from electron gain enthalpy?

Ans: Electronegativity is the tendency of an atom to attract the shared pair of electrons towards itself in a covalent bond; it is a relative, dimensionless scale (for example, Pauling scale) and not an absolute energy quantity.

Electron gain enthalpy (electron affinity) is the energy change when one mole of gaseous atoms gains one mole of electrons to form one mole of anions. It is a measurable thermodynamic quantity and is expressed in energy units (e.g., kJ mol^-1). Electronegativity is a qualitative measure of attraction for bonding electrons, whereas electron gain enthalpy is a specific energetic quantity measured for isolated gaseous atoms.

Q4.18: Explain with the help of suitable example polar covalent bond.
Ans: When two dissimilar atoms with different electronegativities form a covalent bond, the shared pair of electrons is not shared equally. The electron density shifts towards the more electronegative atom, producing partial charges at the ends and hence a polar covalent bond.

For example, in HCl the chlorine atom is more electronegative than hydrogen, so the bonding electron pair is displaced towards chlorine. Chlorine acquires a partial negative charge (δ-) and hydrogen a partial positive charge (δ+), creating a polar bond.

Q4.19:Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ & ClF₃

Ans: Ionic character depends largely on the electronegativity difference between the bonded atoms. Arranged in order of increasing ionic character:

N₂ < SO₂ < ClF₃ < K₂O < LiF.

Q4.20: 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.

Ans: The correct Lewis structure for acetic acid (showing all lone pairs and bond types) is:

Q4.21: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?

Ans: Ground state electronic configuration of carbon: 1s² 2s² 2p². In CH₄ carbon undergoes sp³ hybridisation (one 2s + three 2p) producing four equivalent sp³ orbitals arranged tetrahedrally. 

A square-planar arrangement would require dsp² hybridisation involving d-orbitals; carbon has no available d-orbitals in the valence shell, so dsp² is not possible. Moreover, a square-planar geometry would give 90° bond angles and greater bond-pair repulsions, lowering stability. VSEPR and hybridisation together therefore favour a tetrahedral CH₄.

Q4.22: Explain why BeH₂ molecule has a zero dipole moment although the Be-H bonds are polar.
Ans: The Lewis structure of BeH₂ shows no lone pairs on Be and two Be-H bond pairs (type AB₂), so the molecule is linear (H-Be-H, 180°):

Each Be-H bond dipole has equal magnitude but opposite direction along the molecular axis, so they cancel giving a net dipole moment of zero despite the individual bonds being polar.

Q4.23: Which out of NH₃ and NF₃ has a higher dipole moment and why?

Ans: Both NH₃ and NF₃ are pyramidal (N has one lone pair and three bond pairs). Electronegativity differences alone would suggest NF₃ might be more polar, but the directions of bond dipoles and the lone-pair dipole differ.

In NH₃ the bond dipoles (N-H) point from H to N and add to the lone-pair dipole, producing a larger net dipole (≈1.46 D). In NF₃ the N-F bond dipoles point from N to F and partially oppose the dipole due to N's lone pair, so the net dipole is smaller (≈0.24 D). Therefore NH₃ has the higher dipole moment.

Q4.24: What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.
Ans: Hybridisation is the intermixing of atomic orbitals of similar energies to form new equivalent hybrid orbitals with characteristic shapes and orientations. Hybridisation helps explain molecular geometry and equivalent bonding directions.

sp: Formed by mixing one s and one p orbital; two sp hybrids are arranged linearly (180°).

sp²: One s + two p orbitals give three sp² hybrids arranged trigonal planar (120°).

sp³: One s + three p orbitals give four sp³ hybrids arranged tetrahedrally (≈109.5°).

Q4.25: Describe the change in hybridisation (if any) of the Al atom in the following reaction.

Ans: The valence orbital picture of aluminium in the ground state can be represented as:

The orbital picture of aluminium in the excited state can be represented as:

Hence, it undergoes sp² hybridization to give a trigonal planar arrangement (in AlCl₃).

To form AlCl₄^-, the empty 3p_z orbital also gets involved and the hybridization changes from sp² to sp³. As a result, the shape gets changed to tetrahedral.

Q4.26: Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF₃  NH₃ → F₃B.NH₃
Ans: Boron atom in BF₃ is sp² hybridized. The orbital picture of boron in the excited state can be shown as:

Nitrogen atom in NH₃ is sp³ hybridized. The orbital picture of nitrogen can be represented as:

After the reaction has occurred, an adduct F₃B⋅NH₃ is formed and hybridization of 'B' changes to sp³. However, the hybridization of 'N' remains unchanged because N shares its lone pair with electron deficient B.

Q4.27: Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.
Ans: C₂H₄ :

The electronic configuration of C-atom in the excited state is:

In the formation of an ethene molecule (C₂H₄), one sp² hybrid orbital of carbon overlaps a sp² hybridized orbital of another carbon atom, thereby forming a C-C sigma bond.

The remaining two sp² orbitals of each carbon atom form a sp²-s sigma bond with two hydrogen atoms. The unhybridized orbital of one carbon atom undergoes sidewise overlap with the orbital of a similar kind present on another carbon atom to form a weak π-bond.

C₂H₂ :

In the formation of C₂H₂ molecule, each C-atom is sp hybridized with two 2p-orbitals in an unhybridized state.

One sp orbital of each carbon atom overlaps with the other along the internuclear axis forming a C-C sigma bond. The second sp orbital of each C-atom overlaps a half-filled 1s-orbital to form a σ bond.

The two unhybridized 2p-orbitals of the first carbon undergo sidewise overlap with the 2p orbital of another carbon atom, thereby forming two pi (π) bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two π-bonds.

Q4.28:What is the total number of sigma and pi bonds in the following molecules?

(a) C₂H₂

(b) C₂H₄

Ans: A single bond is a result of the axial overlap of bonding orbitals. Hence, it contributes a sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals. A pi-bond is always present in it. A triple bond is a combination of two pi-bonds and one sigma bond.

Structure of C₂H₂ can be represented as:

Hence, there are three sigma and two pi-bonds in C₂H₂.

The structure of C₂H₄ can be represented as:

Hence, there are five sigma bonds and one pi-bond in C₂H₄.

Q4.29: 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 2p_x (c) 2p_y and 2p_y (d) 1s and 2s.
Ans: (c)
Explanation: Sigma bonds are formed by axial (end-to-end) overlap of orbitals along the internuclear axis (x-axis). Overlaps such as s-s, s-p_x and s-s (1s-2s) occur along the axis and give sigma bonds. Two 2p_y orbitals are oriented perpendicular to the x-axis and overlap sideways (lateral overlap), producing a π bond rather than a σ bond. Therefore option (c) does not form a sigma bond.

Q4.30: Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH₃-CH₃ 
(b) CH₃-CH=CH₂ 
(c) CH₃-CH₂-OH 
(d) CH₃-CHO 
(e) CH₃COOH

Ans:- (a)

Both C₁ and C₂ in ethane are sp³ hybridised.

(b)

C₁ (the CH₃ end) is sp³ hybridised; C₂ (the =CH- carbon) and C₃ (the =CH₂ carbon) are sp² hybridised.

(c)

Both C₁ and C₂ are sp³ hybridised in ethanol.

(d)

In acetaldehyde, C₁ (methyl carbon) is sp³ hybridised and C₂ (carbonyl carbon) is sp² hybridised.

(e)

In acetic acid, C₁ (methyl carbon) is sp³ hybridised and C₂ (carboxyl carbon) is sp² hybridised.

Q4.31: What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Ans: When atoms share one or more valence electrons, a covalent bond forms. The shared electron pairs between bonded atoms are called bond pairs. Electron pairs that do not participate in bonding and remain localized on one atom are called lone pairs.

Example - In C₂H₆ (ethane) all valence electrons are used to form bond pairs (C-C and C-H); there are seven bond pairs and no lone pairs on the bonded atoms shown in the structure.

In H₂O the oxygen has two bond pairs (with H) and two lone pairs on the oxygen atom:

Q4.32: Distinguish between a sigma and a pi bond.

Ans: The following are the differences between sigma and pi-bonds:

 Q4.33: Explain the formation of H₂ molecule on the basis of valence bond theory.
Ans: 

Let us assume that two hydrogen atoms (A and B) with nuclei (N_A and N_B) and electrons (e_A and e_B) 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.

Attractive force arises between:
(a) Nucleus of one atom and its own electron i.e., N_A - e_A and N_B - e_B.
(b) Nucleus of one atom and electron of another atom i.e., N_A - e_B and N_B - e_A.

Repulsive force arises between:
(a) Electrons of two atoms i.e., e_A - e_B.
(b) Nuclei of two atoms i.e., N_A - N_B.

The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, 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.

Q4.34: Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Ans: Atomic orbitals combine to form molecular orbitals effectively when these conditions are met:

(a) The combining atomic orbitals must have the same or similar energies (so that constructive overlap is significant).
(b) The orbitals must have suitable orientations for effective overlap.
(c) The extent of overlap should be appreciable - greater overlap gives stronger bonding molecular orbitals.

Q4.35: Use molecular orbital theory to explain why the Be₂ molecule does not exist.
Ans: The electronic configuration of Beryllium is.

The molecular orbital electronic configuration for Be₂ molecule can be written as:

Hence, the bond order for Be₂ is

Where,

N_b = Number of electrons in bonding orbitals

N_a = Number of electrons in anti-bonding orbitals

Bond order of Be₂ = (4-4) / 2 = 0

A negative or zero bond order means that the molecule is unstable. Hence, Be₂ molecule does not exist.

Q4.36: Compare the relative stability of the following species and indicate their magnetic properties;
O₂,, (superoxide), (peroxide)

Ans: There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = N_b and the number of anti-bonding orbitals = 4 = N_a.

Bond order

(8 - 4) / 2 = 2

Similarly, the electronic configuration of can be written as:

N_b = 8
N_a = 3

Bond order of  

= 2.5

Electronic configuration of ion will be:

N_b = 8

N_a = 5

Bond order of  

= 1.5

Electronic configuration of the ion will be:

N_b = 8
N_a = 6
Bond order of  = 1

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. On this basis, the order of stability is

Q4.37: Write the significance of a plus and a minus sign shown in representing the orbitals.
Ans: Molecular orbitals are represented by wave functions that have phases. A plus sign denotes a positive phase of the wave function and a minus sign denotes a negative phase.

When atomic orbitals combine, regions of the same phase overlap constructively to give bonding molecular orbitals; regions of opposite phase overlap destructively to give antibonding molecular orbitals. The signs therefore indicate the relative phase used when forming constructive or destructive interference between orbitals.

Q4.38: Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?
Ans:

The ground state and excited state outer electronic configurations of phosphorus (Z = 15) are:

Phosphorus atom is sp³d hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as:

The five sp³d hybrid orbitals are directed towards the five corners of the trigonal bipyramidals. Hence, the geometry of PCl₅ can be represented as:

There are five P-Cl sigma bonds in PCl₅. Three P-Cl bonds lie in one plane and make an angle of 120° with each other. These bonds are called equatorial bonds.

The remaining two P-Cl bonds lie above and below the equatorial plane and make an angle of 90° with the plane. These bonds are called axial bonds.

As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.

Q4.39: Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Ans: A hydrogen bond is an attractive interaction in which a hydrogen atom covalently bonded to a strongly electronegative atom (such as N, O or F) interacts with a lone pair on an electronegative atom in another molecule (intermolecular) or elsewhere in the same molecule (intramolecular).

Hydrogen bonding is substantially stronger than van der Waals (dispersion) forces and is often considered a strong form of dipole-dipole interaction. It has a significant effect on physical properties such as boiling point, viscosity and solubility.

There are two types of hydrogen bonds:
(i) Intermolecular H-bond (e.g., HF, H₂O).
(ii) Intramolecular H-bond (e.g., o-nitrophenol).

Hydrogen bonds are stronger than van der Waals forces and are directional, which gives rise to specific molecular arrangements in liquids and solids.

Q4.40: What is meant by the term bond order? Calculate the bond order of: N₂, O₂, and.

Ans: 

Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.

If N_a is equal to the number of electrons in an anti-bonding orbital, then N_b is equal to the number of electrons in a bonding orbital.

Bond order =

If N_b > N_a, then the molecule is said be stable. However, if N_b ≤ N_a, then the molecule is considered to be unstable.

Bond order of N₂ can be calculated from its electronic configuration as:

Number of bonding electrons, N_b = 10

Number of anti-bonding electrons, N_a = 4

Bond order of nitrogen molecule  = 3

There are 16 electrons in a dioxygen molecule, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = N_b and the number of anti-bonding electrons = 4 = N_a.

Bond order

= 2

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration of can be written as:

N_b = 8

N_a = 3

Bond order of

= 2.5
Thus, the bond order of is 2.5.
The electronic configuration of ion will be:
N_b = 8
N_a = 5
Bond order of =
= 1.5

Thus, the bond order of ion is 1.5.