Arrange the following species in the correct | vedclass

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Question

(A) Stability is directly proportional to the bond order. The bond order ($B$.$O$.) is calculated as follows:$(i)$ $N_2^{-}$: Total electrons $= 7+7+1

Vedclass · Accepted Answer

$Ne_2 < O_2^{2-} < C_2 < N_2^{-}$

Vedclass · Answer

(A) Stability is directly proportional to the bond order. The bond order ($B$.$O$.) is calculated as follows:$(i)$ $N_2^{-}$: Total electrons $= 7+7+1 = 15$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1$. $B$.$O$. $= (10-5)/2 = 2.5$.$(ii)$ $C_2$: Total electrons $= 6+6 = 12$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. $B$.$O$. $= (8-4)/2 = 2.0$.$(iii)$ $Ne_2$: Total electrons $= 10+10 = 20$. Bonding and antibonding electrons are equal. $B$.$O$. $= 0$. Thus,$Ne_2$ does not exist.$(iv)$ $O_2^{2-}$: Total electrons $= 8+8+2 = 18$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. $B$.$O$. $= (10-8)/2 = 1.0$.Comparing the bond orders: $N_2^{-} (2.5) > C_2 (2.0) > O_2^{2-} (1.0) > Ne_2 (0)$.Therefore,the correct order of stability is $N_2^{-} > C_2 > O_2^{2-} > Ne_2$.

Arrange the following species in the correct order of their stabilities: $N_2^{-}, C_2, Ne_2, O_2^{2-}$

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