Chapter 21, Problem 126SCQ

(a)

Interpretation Introduction

Interpretation:

To predict why the $\text{N}-\text{O}$ bonds are shorter than the $\text{N}-\text{OH}$ bonds in nitric acid.

Concept introduction:

The formula for nitric acid is ${\text{HNO}}_{\text{3}}$. It is present in the aqueous form. It is a very strong acid and on decomposition produces nitrogen and oxygen gas. The central element in the molecule is nitrogen. Nitrogen belongs to group $\text{3A}$ of the periodic table. The molecule is planar in shape.

Bond length is the distance between the nuclei in a bond and it is related to the sum of the covalent radii at the bonded atoms.

(b)

Interpretation Introduction

Interpretation:

To rationilize the bond angles in $HN{O}_3$.

Concept introduction:

The formula for nitric acid is ${\text{HNO}}_{\text{3}}$. It is present in the aqueous form. It is a very strong acid and on decomposition produces nitrogen and oxygen gas. The central element in the molecule is nitrogen. Nitrogen belongs to group $\text{3A}$ of the periodic table. The molecule is planar in shape.

Bond angle is the angle defined by lines joining the centres of two atoms to a third atom to which they are covalently bonded.

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& Geometry& Bondangle\\ A{X}_2& sp& Linear& {180}^{°}\\ A{X}_3,A{X}_{2}B& s{p}^2& Trigonalplanar& {120}^{°}\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& Tetrahedral& {109.5}^{°}\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& Trigonalbipyramidal& {120}^{°},{90}^{°}\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& Octahedral& {90}^{°}\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

(c)

Interpretation Introduction

Interpretation:

The hybridization in the central nitrogen atom in nitric acid and the orbitals overlap to form $N-O$ $\pi$ bond  should be determined

Concept introduction:

The formula for nitric acid is ${\text{HNO}}_{\text{3}}$. It is present in the aqueous form. It is a very strong acid and on decomposition produces nitrogen and oxygen gas. The central element in the molecule is nitrogen. Nitrogen belongs to group $\text{3A}$ of the periodic table. The molecule is planar in shape.

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$