Which of the following has the least bond ene | vedclass

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Question

(A) The bond energy is directly proportional to the bond order. Let us calculate the bond order for each species using Molecular Orbital Theory:$(A)$

Vedclass · Accepted Answer

$N_{2}^{2-}$

Vedclass · Answer

(A) The bond energy is directly proportional to the bond order. Let us calculate the bond order for each species using Molecular Orbital Theory:$(A)$ $N_{2}^{2-}$: Total electrons = $14 + 2 = 16$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. Bond order = $(8 - 6) / 2 = 1$.$(B)$ $N_{2}^{-}$: Total electrons = $14 + 1 = 15$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. Bond order = $(8 - 5) / 2 = 1.5$.$(C)$ $N_{2}^{+}$: Total electrons = $14 - 1 = 13$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^1 (\pi 2p_x)^2 (\pi 2p_y)^2$. Bond order = $(7 - 2) / 2 = 2.5$.$(D)$ $N_{2}$: Total electrons = $14$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. Bond order = $(8 - 2) / 2 = 3$.Since $N_{2}^{2-}$ has the lowest bond order $(1)$,it has the least bond energy.

Which of the following has the least bond energy?

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