Which of the following statements is not corr | vedclass

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(D) According to molecular orbital theory: $1$. $B_2$ has $10$ electrons,configuration is $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}

Vedclass · Accepted Answer

In $C_2$ molecule,one sigma and one pi bond is present.

Vedclass · Answer

(D) According to molecular orbital theory: $1$. $B_2$ has $10$ electrons,configuration is $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \pi_{2p_x}^1 = \pi_{2p_y}^1$. It is paramagnetic due to unpaired electrons. $2$. Bond order of $O_2^{+}$ is $2.5$ and $O_2^{-}$ is $1.5$. Higher bond order implies higher stability,so $O_2^{+}$ is more stable than $O_2^{-}$. $3$. In $N_2$ ($14$ electrons),the energy of $\pi_{2p_x} = \pi_{2p_y}$ is lower than $\sigma_{2p_z}$ due to $s-p$ mixing. Thus,the statement that $\sigma_{2p_z}$ is higher than $\pi_{2p_x}$ is correct. $4$. In $C_2$ ($12$ electrons),the configuration is $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \pi_{2p_x}^2 = \pi_{2p_y}^2$. Both bonds are pi bonds; there is no sigma bond. Thus,the statement in option $D$ is incorrect.

Which of the following statements is not correct according to molecular orbital theory $:-$

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