Question

Which of the following has the highest value for bond order $-{\text{C}}_6{\text{H}}_6,\text{C}{\text{O}}_3^{2-},\text{N}{\text{O}}^{-}$ and ${\text{SO}}_4^{2-}?$

• A. ${\text{C}}_6{\text{H}}_6$
• B. $\text{C}{\text{O}}_3^{2-}$
• C. $\text{N}{\text{O}}^{-}$
• D. $\text{S}{\text{O}}_3$

Answer 1

Answer: (C)

Bond order (BO) can be calculated using the following methods:

(i) $\text{B}\text{O}=\frac{\text{T}\text{o}\text{t}\text{a}\text{l}\text{n}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{b}\text{o}\text{n}\text{d}\text{s}\text{a}\text{s}\text{s}\text{o}\text{c}\text{i}\text{a}\text{t}\text{e}\text{d}\text{w}\text{i}\text{t}\text{h}\text{t}\text{h}\text{e}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{l}\text{a}\text{t}\text{o}\text{m}}{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{σ}\text{b}\text{o}\text{n}\text{d}\text{s}\text{a}\text{s}\text{s}\text{o}\text{c}\text{i}\text{a}\text{t}\text{e}\text{d}\text{w}\text{i}\text{t}\text{h}\text{t}\text{h}\text{e}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{l}\text{a}\text{t}\text{o}\text{m}}$

(or)

$\text{B}\text{O}=\frac{\text{T}\text{o}\text{t}\text{a}\text{l}\text{n}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{b}\text{o}\text{n}\text{d}\text{s}\text{b}\text{e}\text{t}\text{w}\text{e}\text{e}\text{n}\text{t}\text{w}\text{o}\text{a}\text{t}\text{o}\text{m}\text{s}}{\text{T}\text{o}\text{t}\text{a}\text{l}\text{n}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{r}\text{e}\text{s}\text{o}\text{n}\text{a}\text{n}\text{c}\text{e}\text{s}\text{t}\text{r}\text{u}\text{c}\text{t}\text{u}\text{r}\text{e}\text{s}}$

(ii) $\text{B}\text{O}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{b}\text{o}\text{n}\text{d}\text{i}\text{n}\text{g}\text{e}\text{l}\text{e}\text{c}\text{t}\text{r}\text{o}\text{n}\text{s}-\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{a}\text{n}\text{t}\text{i}\text{b}\text{o}\text{n}\text{d}\text{i}\text{n}\text{g}\text{e}\text{l}\text{e}\text{c}\text{t}\text{r}\text{o}\text{n}\text{s}}{2}$

We use both the concepts as per situation.

Thus,

(I) We use concept (i) for the species ${\text{C}}_6{\text{H}}_6,\text{C}{\text{O}}_3^{2-}$ and ${\text{SO}}_4^{2-}$ and concept (ii) for $\text{N}{\text{O}}^{-}.$

On the basis of concept $\left(\text{i}\right)$

BO for ${\text{C}}_6{\text{H}}_6=\frac{2+1}{2}=1.5$

BO for $\text{C}{\text{O}}_3^{2-}=\frac{4}{3}=1.33$





BO for ${\text{SO}}_4^{2-}=\frac{6}{4}=1.5$

On the basis of concept (ii)

BO of $\text{N}{\text{O}}^{-}=7+8+1$

$=16$ electrons

$\therefore \text{B}\text{O}$ for $\text{N}{\text{O}}^{-}=\frac{10-6}{2}=2$

Hence highest value for $\text{B}\text{O}$ is of $\text{N}{\text{O}}^{-}.$