Question

How do you write electronic configuration for $H$, ${H_2}$ , ${H_2}^ -$ and ${H_2}^{ - 2}$ . Calculate their bond order?

Answer 1

Electronic configuration is the process of distributing electrons of a particular atom or a molecule in their atomic orbitals or molecular orbitals respectively. Bond order is used to describe the presence of a total number of covalent bonds between two atoms.

Complete answer:
Hydrogen is the starting element of the periodic table which comes in groups $\left( I \right)$ and period $\left( I \right)$ . Atomic number of Hydrogen is $1$ , therefore it contains only electrons and protons but no neutrons.
According to the Aufbau rule, electrons start filling in an orbital in their increasing order of energy which means that the first electron filled in an orbital with lowest energy state.
Order of energy of orbitals is- $1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,7s$
Pauli’s exclusion principle explains the holding capacity of each orbital which extends maximum to two electrons and they must have opposite spin in an orbital.
Bond order of any molecule is calculated by using a formula which relates bonding electrons with nonbonding electrons and expressed mathematically as-
$BO = \dfrac{1}{2}\left[ {Nb - Na} \right]$
Where $BO =$ bond order
$Nb =$ Bonding electron
$Na =$ Nonbonding electron
In the case of hydrogen $H$, it has only one electron in its outermost shell so an electron must fill in $1s$ orbital. Hence, the electronic configuration of $H$ is $1{s^1}$ . As we know, a bond is formed between two atoms so the bond order is zero in the case of $H$.
In the case of dihydrogen ${H_2}$ , the total number of electrons present to the molecule is $\left( {1 + 1 = 2} \right)$ . Hence, electronic configuration of ${H_2}$ is $1{s^2}$ . In dihydrogen as we see from configuration there are two bonding electrons and zero nonbonding electrons so bond order of ${H_2}$ is
$BO = \dfrac{1}{2}\left[ {Nb - Na} \right]$
$BO = \dfrac{1}{2}\left[ {2 - 0} \right]$
After solving it, we get
$BO = 1$ .
In the case of ${H_2}^ -$ ion, total number of electrons present in ion is one electron from each hydrogen atom and one electron due to negative charge i.e., $\left( {1 + 1 + 1 = 3} \right)$ . Hence, electronic configuration of ${H_2}^ -$ is $1{s^2}2{s^1}$ . In ${H_2}^ -$ as we see from configuration there are two bonding electrons and one nonbonding electrons so bond order of ${H_2}^ -$ is -
$BO = \dfrac{1}{2}\left[ {Nb - Na} \right]$
$BO = \dfrac{1}{2}\left[ {2 - 1} \right]$
After solving it, we get
$BO = 0.5$ .
In the case of ${H_2}^{ - 2}$ ion, total number of electrons present in ion is one electron from each hydrogen atom and two electrons due to negative charge i.e., $\left( {1 + 1 + 2 = 4} \right)$ . Hence, the electronic configuration of ${H_2}^{ - 2}$ is $1{s^2}2{s^2}$ . In ${H_2}^{ - 2}$ as we see from configuration there are two bonding electrons and two nonbonding electrons so bond order of ${H_2}^{ - 2}$ is -
$BO = \dfrac{1}{2}\left[ {Nb - Na} \right]$
$BO = \dfrac{1}{2}\left[ {2 - 2} \right]$
After solving it, we get
$BO = 0$ .
$\Rightarrow$ electronic configuration of $H$is $1{s^1}$ and bond order is $0$ .
$\Rightarrow$ electronic configuration of ${H_2}$ is $1{s^2}$ and bond order is $1$ .
$\Rightarrow$ electronic configuration of ${H_2}^ -$ is $1{s^2}2{s^1}$ and bond order is $0.5$ .
$\Rightarrow$ electronic configuration of ${H_2}^{ - 2}$ is $1{s^2}2{s^2}$ and bond order is $0$ .

Note:
Bond order with value of zero indicates that molecules do not exist in nature in that particular form. Energy of different orbitals are calculated by using Bohr Burys’s rule which states that the energy of an orbital increases according to the $\left( {n + l} \right)$ rule. This rule justifies the filling of $4d$ orbital prior to $3d$ after filling of $3p$ orbital.