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When N(2) goes to N(2)^(+), the N-N bond...

When `N_(2)` goes to `N_(2)^(+)`, the `N-N` bond distance …………, and when `O_(2)` goes to `O_(2)^(+)` the `O-O` bond distance…….

A

Decrease, increase

B

Increase, decrease

C

Increase, increase

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`N_2 = KK. (sigma2s)^2(sigma^(**)2s)^2(pi2p_x)^2(pir2p_y)^2(sigma2p_z)^2`
`N_b=8,N_a=2`
For `N_2^(+),N_b=7,N_a=2`
So when `N_2` goes to `N_2^(+)` Bond order decreases or bond length increases.
`O_2 = KK. (sigma2s)^2(sigma*2s)^2 (sigma2p_x )^2 (pi2p_x)^2 (pi2py)^2 (pi^(**) 2p_x)^(1) (pi^(**) 2p_y)^1`
`N_b=8,N_a=4`
For `O_2^(+),N_b=8,N_a=3`
So when `O_2` goes to `O_2^(+)` : Bond order increases or bond length decreases.
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