Question

Bond-order of $ H{e_2}^ + $ is ____ but it is paramagnetic due to ____
(A) $ 0.5 $ , one unpaired electron
(B) 1, two unpaired electron
(C) $ 0.5 $ , no unpaired electron
(D) 1, one unpaired electron

Answer 1

Hint: Molecular orbital theory states that atoms combine together to form molecular orbitals. The total number of molecular orbitals formed will be equal to the atomic orbitals forming molecular orbitals. We solve this sum by applying the relation between molecular orbital theory and bond order. Paramagnetism is caused due to unpaired electrons and the presence of an odd number of unpaired electrons give rise to fractional bond order.
 $ {\text{Bond Order}} = \dfrac{{{\text{bonding electrons}} - {\text{antibonding electrons}}}}{{{\text{number of molecules}}}} $

Complete Step by step solution
In molecular orbital theory, there are bonding molecular orbitals and antibonding molecular orbitals. In bonding molecular orbitals, the probability of finding electrons is high. In antibonding molecular orbitals, the probability of finding electrons is low. Also, in antibonding molecular orbitals, there is a node where electron density is zero. The bonding molecular orbitals are formed by the additive effect of the atomic orbitals. The antibonding molecular orbitals are formed by the subtractive effect of the atomic orbitals.
The number of the chemical bonds between the pair of atoms is indicated by the bond order. Bond order indicates the stability of a bond.
Bond order can be calculated by the subtracting bonding electrons with antibonding electrons and dividing by 2.
 $ H{e_2}^ + $ has 3 electrons. The molecular orbital electronic configuration is as follows,
 $ H{e_2}^ + = \sigma 1{s^2} * ,1{s^1} $
Here, the bonding electrons are 2 and the antibonding electron is 1.
Bond order of $ H{e_2}^ + $ molecule = $ \dfrac{{2 - 1}}{2} $ = $ 0.5 $
Therefore, the bond order of $ H{e_2}^ + $ is $ 0.5 $ .
Seeing the electronic configuration, there is one unpaired electron present in the antibonding orbital. Hence $ H{e_2}^ + $ is paramagnetic.
The correct answer is option A.

Note
The bond order of a single bond is 1 and the bond order of a double bond is 2. More is the bond order, more is the strength of the bond. This also means that more energy is required to break that bond. As a result, nitrogen in our air is inert as it is a triple bonded diatomic molecule, while hydrogen is highly reactive as it is a single bonded diatomic molecule.