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 $\text{N}\text{O}^{-}.$