In oxygen (O_2) and carbon (C_2) molecules,th | vedclass

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Question

(B) According to Molecular Orbital Theory $(MOT)$: For $O_2$ ($16$ electrons),the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2

Vedclass · Accepted Answer

$O_2: 1 \sigma, 1 \pi ; C_2: 0 \sigma, 2 \pi$

Vedclass · Answer

(B) According to Molecular Orbital Theory $(MOT)$: For $O_2$ ($16$ electrons),the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. The bond order is $2$,consisting of $1 \sigma$ and $1 \pi$ bond. For $C_2$ ($12$ electrons),the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. The bond order is $2$,consisting of $2 \pi$ bonds (no $\sigma$ bond). Thus,option $(B)$ is correct.

In oxygen $(O_2)$ and carbon $(C_2)$ molecules,the bonding is:

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