Question

Answer the following questions in a single digit integer.
What is the bond order of $N{e_2}$ ?

Answer 1

Hint : Bond order in a molecule can be defined as the chemical bonds that are formed between two atoms. It can also be defined as the number of overlapping electron pairs between a pair of atoms. Bond order can only be calculated between any two particular atoms in a molecule. Greater the bond order greater will be the strength of the bond.

Complete Step By Step Answer:
Bond order of a molecule can be defined as the chemical bonds that are formed between two atoms. It is basically the number of overlapping electron pairs between a pair of atoms.
Bond order was first introduced by Linus Pauling. Bond order can be calculated as the half the difference between the number of bonding electrons $({N_b})$ and the number of antibonding electrons $({N_a})$ . Therefore, the formula for bond order is:
Bond order $= \dfrac{1}{2}({N_b} - {N_a})$
Where, ${N_b}$ is the number of bonding electrons present between two molecules and ${N_a}$ is the number of antibonding electrons present.
Electronic configuration of $N{e_2}$ in molecular diagram is:
$$ \sigma 1{s^2};\sigma *1{s^2},;\sigma 2{s^2}\sigma 2{s^2},;\sigma p{x^2}\pi 2p{y^2}2{\pi ^}^2p{y^2}\pi *2p{y^2}2p{z^2}\sigma *2p{x^2} $$
The orbitals represented with a * are antibonding orbitals.
Hence, we can see that the number of bonding electrons are ten and antibonding electrons are 10. Put these values in the formula above:
Bond order $= \dfrac{1}{2}(10 - 10)$
Bond order of $N{e_2}$ $= 0$
Hence the answer is $0$ .

Note :
Bonding electrons and non-bonding electrons take part in determining the bond order of a molecule while the non-bonding electrons (which do not take part in bonding) are not included in determining the bond order of a molecule. If there are unpaired electrons in the molecule, then it is paramagnetic in nature and if there are paired electrons then it is diamagnetic in nature.