Bond order of He_{2}, He_{2}^{+} and He_{2}^{ | vedclass

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Question

(B) The bond order is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$, where $N_b$ is the number of bonding electrons and $N_a$ is the

Vedclass · Accepted Answer

$0, 0.5, 1$

Vedclass · Answer

(B) The bond order is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $He_{2}$ $(4 \ e^{-})$: Configuration is $\sigma(1s)^{2}, \sigma^{*}(1s)^{2}$. $B.O. = \frac{2-2}{2} = 0$.For $He_{2}^{+}$ $(3 \ e^{-})$: Configuration is $\sigma(1s)^{2}, \sigma^{*}(1s)^{1}$. $B.O. = \frac{2-1}{2} = 0.5$.For $He_{2}^{2+}$ $(2 \ e^{-})$: Configuration is $\sigma(1s)^{2}$. $B.O. = \frac{2-0}{2} = 1$.Thus, the bond orders are $0, 0.5, 1$ respectively.

Bond order of $He_{2}$, $He_{2}^{+}$ and $He_{2}^{2+}$ are respectively:

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