Question

The bond dissociation energy $(E)$ and bond length $(R)$ of $O _{2}, N _{2}$ and $F _{2}$ follow the order as:

	(E)	(R)
(a)	$O _{2} > N _{2} > F _{2}$	$F _{2}> O _{2}> N _{2}$
(b)	$N _{2}> F _{2}> O _{2}$	$N _{2}> O _{2}> F _{2}$
(c)	$N _{2}> O _{2}> F _{2}$	$F _{2}> O _{2}> N _{2}$
(d)	$N _{2}> O _{2}> F _{2}$	$N _{2}> O _{2}> F _{2}$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (C)

Greater the bond order, shorter is the bond length and thus, stronger is the bond.

Since, for the given molecules, bond order follows the trend

$\underset{(B . O=3)}{N _{2}} > \underset{(B . O=2)}{O _{2}} > \underset{(B . O=1)}{F _{2}}$

Thus, the bond length decreases as $F _{2} > O _{2} > N _{2}$

and bond dissociation energy follows the trend $N _{2} >O _{2} > F _{2}$