Question

The correct statement with regard to $H _{2}^{+}$and $H _{2}^{-}$is

• A. Both $H _{2}^{+}$and $H _{2}^{-}$do not exist
• B. $H _{2}{ }^{-}$is more stable than $H _{2}{ }^{+}$
• C. $H _{2}{ }^{+}$is more stable than $H _{2}{ }^{-}$
• D. Both $H _{2}^{+}$and $H _{2}^{-}$are equally stable

Answer 1

Answer: (C)

The electron presence in anti-bonding orbital result in repulsion and decrease of stability The stability order follows the molecular orbital theory The formula for bonding order can be regarded as follows
Bonding order $=\frac{1}{2}$ (Number of bonding electrons -Number of antibonding electrons)
The number of antibonding electrons in $H _{2}{ }^{+}$ is $0$
The number of antibonding electrons in $H _{2}^{-}$ is $1$
The bonding order of ${H}_{2}{ }^{+}$is $=\frac{1}{2}(1-0)=0.5$
The bonding order of ${H}_{2}{ }^{-}$is $=\frac{1}{2}(2-1)=0.5$
$H _{2}{ }^{+}$and $H _{2}{ }^{-}$molecule have the same bond order
Presence of anti-bonding electron decreases the stability of the molecule Therefore, $H _{2}{ }^{+}$is more stable than $H _{2}{ }^{-}$