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Chemical Bonding and Molecular Structure - Bond Parameters

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Bond Length: It is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond length decreases with the increase in bond multiplicity. For example, the bond length order is C−C>C=C>C≡CC-C > C=C > C \equiv C.

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Bond Angle: It is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule. It helps in determining the shape of the molecule. For example, in H2OH_2O, the bond angle is 104.5∘104.5^\circ due to lp−lplp-lp repulsions.

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Bond Enthalpy: It is the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state. Higher bond enthalpy indicates a stronger and more stable bond. Bond enthalpy ∝\propto Bond order ∝1Bond length\propto \frac{1}{\text{Bond length}}.

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Bond Order: In the Lewis description of covalent bonds, the Bond Order is given by the number of bonds between the two atoms in a molecule. For N2N_2, the bond order is 33; for O2O_2, it is 22.

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Resonance: When a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy and positions of nuclei are used. This stabilizes the molecule and averages the bond parameters. For example, in O3O_3, both O−OO-O bonds are of equal length.

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Dipole Moment (μ\mu): It is a measure of the polarity of a bond. It is a vector quantity, defined as the product of the magnitude of the charge (qq) and the distance between the centers of positive and negative charge (rr).

📐Formulae

μ=q×r\mu = q \times r

Bond Order=12(Nb−Na)\text{Bond Order} = \frac{1}{2}(N_b - N_a) (where NbN_b is number of bonding electrons and NaN_a is number of anti-bonding electrons)

Bond Order (Resonance)=Total number of bonds between two atoms in all structuresTotal number of resonating structures\text{Bond Order (Resonance)} = \frac{\text{Total number of bonds between two atoms in all structures}}{\text{Total number of resonating structures}}

Resultant Dipole Moment μnet=μ12+μ22+2μ1μ2cos⁡θ\text{Resultant Dipole Moment } \mu_{net} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta}

💡Examples

Problem 1:

Explain why the bond angle in NH3NH_3 (107∘107^\circ) is less than the tetrahedral angle (109.5∘109.5^\circ).

Solution:

In NH3NH_3, the central nitrogen atom is sp3sp^3 hybridized. However, it contains one lone pair and three bond pairs. According to VSEPR theory, lone pair-bond pair (lp−bplp-bp) repulsions are stronger than bond pair-bond pair (bp−bpbp-bp) repulsions. This causes the N−HN-H bonds to be pushed closer together, reducing the angle from 109.5∘109.5^\circ to 107∘107^\circ.

Explanation:

The deviation from ideal geometry occurs due to the presence of non-bonding electrons occupying more space.

Problem 2:

Calculate the bond order of the carbonate ion, CO32−CO_3^{2-}.

Solution:

The carbonate ion has three resonating structures. In these structures, there are a total of 44 bonds shared between the Carbon and the three Oxygen atoms (one double bond and two single bonds). Using the formula: Bond Order=43=1.33\text{Bond Order} = \frac{4}{3} = 1.33

Explanation:

Due to resonance, the double bond character is delocalized over all three C−OC-O positions, resulting in identical bond lengths.

Problem 3:

Why is the dipole moment of BeF2BeF_2 zero, even though the Be−FBe-F bonds are polar?

Solution:

BeF2BeF_2 is a linear molecule with a bond angle of 180∘180^\circ. The dipole moment is a vector quantity. The two Be−FBe-F bond dipoles are equal in magnitude but point in exactly opposite directions: μ⃗net=μ⃗1+μ⃗2=0\vec{\mu}_{net} = \vec{\mu}_1 + \vec{\mu}_2 = 0.

Explanation:

The individual bond dipoles cancel each other out due to the symmetrical linear geometry.

Bond Parameters - Revision Notes & Key Formulas | CBSE Class 11 Chemistry