Question

Arrange the following in order of decreasing N – O bond length : NO2+ , NO2– , NO3–

• A. $NO_{3}^{-} > NO_{2}^{+} > NO_{2}^{-}$
• B. $NO_{3}^{-} > NO_{2}^{-} > NO_{2}^{+}$
• C. $NO_{2}^{+} > NO_{3}^{-} > NO_{2}^{-}$
• D. $NO_{2}^{-} > NO_{3}^{-} > NO_{2}^{+}$

Answer 1

Answer: (B)

$NO_{3}^{-} > NO_{2}^{-} > NO_{2}^{+}$

Answer 2

N atom in NO2+ is sp hybridised while in NO2– and NO3–, it is sp2 hybridised.

Bond order = 2

$\longleftrightarrow$Bond order = = 1.5

$\longleftrightarrow$$\longleftrightarrow$Bond order = $\frac{2 + 1 + 1}{3}$ = 1.33

Bond order $\propto$ $\frac{1}{Bondlength}$

So, bond length order is. $NO_{3}^{-} > NO_{2}^{-} > NO_{2}^{+}$