Question

Match List-I with List-II.

List-I	List-II
(Molecule)	(Bond order)
a. $N{e}_2$	i. 1
b. $N_2$	ii. 2
c. $F_2$	iii. 0
d. $O_2$	iv. 3

Choose the correct answer from the options given below:

• A. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
• B. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
• C. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
• D. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Answer 1

The correct option is A (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

As per the molecular orbital theory,
$N{e}_2$ does not exist. Thus,
$N{e}_2\to \text{Bond order}=0$

$N\equiv N$,
$N_2\to \text{Bond order}=3$

$F-F$,
$F_2\to \text{Bond order}=1$

$O=O$,
$O_2\to \text{Bond order}=2$

Hence, option (a) is correct.