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Q. The bond order in $ He_{2}^{+} $ ion is

Chhattisgarh PMTChhattisgarh PMT 2010

Solution:

$He _{2}^{+}(2+2-1=3)$
$=\sigma 1 s^{2}, \sigma ^* 1 s^{1}$
Bond order $=\frac{N_{b}-N_{a}}{2}$
$\left(N_{b},=\right.$ bonding electrons, $N_{a},=$ anti bonding electrons $)$
$=\frac{2-1}{2}=\frac{1}{2}=0.5$