Question

MO configuration of He is

• A. $(\sigma 1s{)}^2,(\stackrel{\ast }{\sigma }1s{)}^2,(\sigma 2s{)}^1$
• B. $(\sigma 1s{)}^2,(\stackrel{\ast }{\sigma }1s{)}^2,(\stackrel{\ast }{\sigma }2s{)}^1$
• C. $(\sigma 1s{)}^2,(\stackrel{\ast }{\sigma }1s{)}^2,(\sigma 2s{)}^2$
• D. $(\sigma 1s{)}^2,(\stackrel{\ast }{\sigma }1s{)}^2,(\sigma 2s{)}^1$

Answer 1

Answer: (A)

MO configuration of $H{e}_2^{-}$
This molecule contains five electrons, two from one helium atom, and three from the other ($H{e}^{-}$ ion). The molecular orbital configuration of the molecule is as
$H{e}_2^{-}(\sigma 1s{)}^2{\left(\stackrel{\ast }{\sigma }1s\right)}^2,(\sigma 2s{)}^1$
Bond order of $H{e}_2^{-}=\frac{N_b-N_a}{2}=\frac{3-2}{2}=\frac{1}{2}$
and molecule is expected to be paramagnetic due to the presence of unpaired electron.