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

Bond order is defined as the difference between the number of electrons present in bonding of the molecular orbitals and the anti bonding molecular orbitals. If the attraction present between the electrons is more then it will have high bond order. We can determine the bond order with the help of molecular orbital theory.

Complete answer: The bond order helps to determine the number of chemical bonds present in the pair of atoms. If a covalent bond is considered between the two atoms then the single bond is said to be having a bond order of one. If double bond is present then the bond order will be of two and if triple bond is present then the bond order will be of three. To calculate the bond order we need to know the electronic configuration of $H{{e}_{2}}^{+}$which is $1{{s}^{2}},2{{s}^{1}}$. This means it has three electrons in it. The molecular orbital electronic configuration of $H{{e}_{2}}^{+}$will be $\sigma 1{{s}^{2}},\sigma *1{{s}^{1}}$. So the bonding electrons present in it is two and the anti bonding electrons present in it is 1.
The formula of bond order is $\dfrac{1}{2}[{{N}_{b}}-{{N}_{a}}]$.
So now substituting the values in the above formula, we get,
Bond order= $\dfrac{1}{2}[2-1]=0.5$
Since the electrons in it are odd that is 3 so it will have only one pair of electrons and the other remaining electron will remain unpaired. So as one electron is unpaired so it will be paramagnetic in nature.

So the correct answer is option (A).

Note: The paramagnetic substances contain some unpaired electrons with them due to which their net magnetic moment of all electrons is not equal to zero. The paramagnetics substances have low and positive magnetic susceptibility and relative permeability is greater than 1. The intensity of magnetization is small and positive.