Match List - I with List - II.
List - I (Molecule) List - II (Bond order) (a) \(\mathrm{Ne_2}\) (i) 1 (b) \(\mathrm{N_2}\) (ii) 2 (c) \(\mathrm{F_2}\) (iii) 0 (d) \(\mathrm{O_2}\) (iv) 3
Choose the correct answer from the options given below :
| List - I (Molecule) | List - II (Bond order) | ||
| (a) | \(\mathrm{Ne_2}\) | (i) | 1 |
| (b) | \(\mathrm{N_2}\) | (ii) | 2 |
| (c) | \(\mathrm{F_2}\) | (iii) | 0 |
| (d) | \(\mathrm{O_2}\) | (iv) | 3 |
Choose the correct answer from the options given below :
Show Hint
- (a) \(\to\) (i), (b) \(\to\) (ii), (c) \(\to\) (iii), (d) \(\to\) (iv)
- (a) \(\to\) (iv), (b) \(\to\) (iii), (c) \(\to\) (ii), (d) \(\to\) (i)
- (a) \(\to\) (iii), (b) \(\to\) (iv), (c) \(\to\) (i), (d) \(\to\) (ii)
- (a) \(\to\) (ii), (b) \(\to\) (i), (c) \(\to\) (iv), (d) \(\to\) (iii)
The Correct Option is C
Solution and Explanation
\(\mathrm{Ne_2}\) (20 electrons): \(N_b = 10,\ N_a = 10\), bond order \(= 0\) (iii).
\(\mathrm{N_2}\) (14 electrons): \(N_b = 10,\ N_a = 4\), bond order \(= 3\) (iv).
\(\mathrm{F_2}\) (18 electrons): \(N_b = 10,\ N_a = 8\), bond order \(= 1\) (i).
\(\mathrm{O_2}\) (16 electrons): \(N_b = 10,\ N_a = 6\), bond order \(= 2\) (ii).
So (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii).
Answer: (3) (a) \(\to\) (iii), (b) \(\to\) (iv), (c) \(\to\) (i), (d) \(\to\) (ii)
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