The bond orders of He_2^{+} and He_2 are resp | vedclass

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Question

(A) The molecular orbital electronic configuration of $He_2$ is $\sigma 1s^2, \sigma^* 1s^2$. Bond order $= \frac{(	ext{electrons in bonding MO}) - (

Vedclass · Accepted Answer

$1/2, 0$

Vedclass · Answer

(A) The molecular orbital electronic configuration of $He_2$ is $\sigma 1s^2, \sigma^* 1s^2$. Bond order $= \frac{(	ext{electrons in bonding MO}) - (	ext{electrons in antibonding MO})}{2}$. For $He_2$,bond order $= \frac{2-2}{2} = 0$. Since the bond order is $0$,the $He_2$ molecule does not exist. For $He_2^{+}$,the molecular orbital electronic configuration is $\sigma 1s^2, \sigma^* 1s^1$. Bond order $= \frac{2-1}{2} = 1/2$.

The bond orders of $He_2^{+}$ and $He_2$ are respectively

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