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Bond order of $\mathrm{He}_{2}, \mathrm{He}_{2}^{+}$ and $\mathrm{He}_{2}^{2+}$ are respectively:
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Verified Answer
The correct answer is:
$0, \frac{1}{2}, 1$
Hint:
$\mathrm{He}_{2}\left(4 \mathrm{e}^{-}\right)=\sigma(1 \mathrm{~s})^{2}, \sigma^{*}(1 \mathrm{~s})^{2} ; \mathrm{B} . \mathrm{O}=\frac{2-2}{2}=0$
$\mathrm{He}_{2}^{+}\left(3 \mathrm{e}^{-}\right)=\sigma(1 \mathrm{~s})^{2} \sigma^{*}(1 \mathrm{~s})^{1} ; \mathrm{B} . \mathrm{O}=\frac{2-1}{2}=\frac{1}{2}$
$\mathrm{He}_{2}^{+2}\left(2 \mathrm{e}^{-}\right)=\sigma(\mathrm{1s})^{2} ; \quad$ B.O. $=\frac{2-0}{2}=1$
$\mathrm{He}_{2}\left(4 \mathrm{e}^{-}\right)=\sigma(1 \mathrm{~s})^{2}, \sigma^{*}(1 \mathrm{~s})^{2} ; \mathrm{B} . \mathrm{O}=\frac{2-2}{2}=0$
$\mathrm{He}_{2}^{+}\left(3 \mathrm{e}^{-}\right)=\sigma(1 \mathrm{~s})^{2} \sigma^{*}(1 \mathrm{~s})^{1} ; \mathrm{B} . \mathrm{O}=\frac{2-1}{2}=\frac{1}{2}$
$\mathrm{He}_{2}^{+2}\left(2 \mathrm{e}^{-}\right)=\sigma(\mathrm{1s})^{2} ; \quad$ B.O. $=\frac{2-0}{2}=1$
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