Question

The bond order of $$H{e}_2^{+}$$ molecule ion is:
(A) $$1$$
(B) $$2$$
(C) $${\displaystyle \frac{1}{2}}$$
(D) $${\displaystyle \frac{1}{4}}$$

Answer 1

Hint: As we know that the total number of electrons in a helium molecule is four. In the given question one electron is lost and therefore three electrons are left in the molecule. The electrons in the molecules want to stay in the ground state but maximum of two electrons. The bond order is the strength of bonding in the molecule and measured by $${\displaystyle \frac{Bondingelectrons\left(n_b\right)-Nonbondingelectrons\left(n_{ab}\right)}{2}}$$.

Complete step by step answer:
The helium atom has electronic configuration $$1s^2$$ and therefore the helium molecule contains 4 electrons. In the given question one electron is lost in a helium molecule and the total number of left electrons in the molecule is three. The orbitals in which the electrons are placed is known as an atomic orbital.
Molecular orbitals are formed by the combination of atomic orbitals. The combining atomic orbitals must have the same or nearly the same energies and the extent of overlapping must between the atomic orbitals of two atoms should be large. In this helium molecule, the combining orbitals are $$1s^2$$ and $$1s^2$$. These atomic orbitals are very close to the helium nucleus.
Now, $$({\sigma }_{1s}{)}^2$$ atomic orbitals of two helium atoms form two molecular orbitals designated as $${\sigma }_{1s}$$ and $${{\sigma }^{\ast }}_{1s}$$. $${\sigma }_{1s}$$ is known as bonding orbital and $${{\sigma }^{\ast }}_{1s}$$ is known as an anti-bonding orbital.
The electrons first occupy the minimum energy and are located in a bonding orbital. According to Pauli’s exclusion principle “no more than two electrons can occupy the same orbitals and must have opposite spin.” so the electronic configuration in molecular orbital is:
$$({\sigma }_{1s}{)}^2,({{\sigma }^{\ast }}_{1s}{)}^1$$
The bonding electrons = 2
The non- bonding electrons = 1
The bond order is the number of covalent bonds in a molecule. It is calculated as
$Bondorder={\displaystyle \frac{Bondingelectrons\left(n_b\right)-Nonbondingelectrons\left(n_{ab}\right)}{2}}$
$Bondorder={\displaystyle \frac{2-1}{2}}={\displaystyle \frac{1}{2}}$

Therefore, the correct option is (C).

Note: The bond order of $$H{e}_2^{+}$$ is half and is very stable. More the bonding electrons in the molecular orbital and lesser the anti- bonding electrons, smaller will be the bond order and stable will be the molecule.