Question

The bond order in $H{e}_2^{+}$ ion is

• A. 0.5
• B. 1.0
• C. 1.5
• D. 2.0

Answer 1

Answer: (A)

$H{e}_2^{+}(2+2-1=3)$
$=\sigma 1s^2,{\sigma }^{\ast }1s^1$
Bond order $=\frac{N_b-N_a}{2}$
$(N_b,=$ bonding electrons, $N_a,=$ anti bonding electrons $)$
$=\frac{2-1}{2}=\frac{1}{2}=0.5$