Question

"$H{e}_2$ does not exist since its bond-order is zero."
Answer whether the above statement is true or false.

Answer 1

$H{e}_2$ has MO electronic configuration as

$\underset{bonding}{\sigma 1s^2}\underset{Anti-bonding}{{\sigma }^{\ast }1s^2}$

Bond order $=\frac{N_B-N_A}{2}=(2-2)/2=0$

Thus, electronandBond−

But $H{e}^{+}$ ion exists, as it has a configuration of $\underset{bonding}{\sigma 1s^2}\underset{Anti-bonding}{{\sigma }^{\ast }1s^1}$

Bond order $=\frac{N_B-N_A}{2}=\frac{2-1}{2}$

As bond order is positive $H{e}^{+}$ exists. Due to one unpaired electron $H{e}_2^{+}$ is paramagnetic.
The statement is true.