Name: Hybridisation in SF₆ Molecule
Uploaded: 2019-07-19
Duration: 2 min 31 s
Description: Understand the bonding present in the case of sulphur hexafluoride. This video illustrates that the hybridisation present in the case of sulphur hexafluoride...

Question

Use the energy level diagram to show that

N_{2}

would be expected to have a triple bond,

F_{2}

a single bond and

{Ne}_{2}

no bond.

Accepted Answer

Energy level diagram for $N_{2}$ .

The electronic configuration of $N_{2}$ molecule may be written as : $[σ (1 s)^{2}] [σ^{*} (1 s)^{2}] [σ (2 s)^{2}] [σ^{*} (2 s)^{2}] [π (2 p_{x})^{2} = π (2 p_{y})^{2}] [σ (2 p_{z})^{2}]$
Here, there are four completely filled bonding molecular orbitals(outer shell) and one completely filled antibonding molecular orbital(outer shell). So that, $N_{b}$ = 8, $N_{a}$ = 2
Bond order = $\frac{1}{2} (N_{b} - N_{a})$
= $\frac{1}{2} (8 - 2)$ = 3.
Bond order value of 3 means that ${Ne}_{2}$ contains triple bond.

Energy level diagram for $F_{2}$ .

Total number of electrons in $F_{2}$ molecule is 18, 9 from each atom. Thus, it has two electrons more than that of $O_{2}$ molecule. These two additional electrons will enter into $π^{*} (2 p_{x})$ and $π^{*} (2 p_{y})$ orbital one in each of them.
Thus, the electronic configuration of $F_{2}$ will be : $KK {[σ (2 s)]}^{2} {[σ^{*} (2 s)]}^{2} {[σ (2 p_{z})]}^{2} {[π (2 p_{x})]}^{2} = [π (2 p_{y})]^{2} {[π * (2 p_{x})]}^{2} [π * (2 p_{y})]^{2}$
$N_{b}$ = 8, $N_{a}$ = 6(outer shell)
Therefore,
BO = $\frac{1}{2} (N_{b} - N_{a})$ = $\frac{1}{2} (8 - 6)$ = 1.
It justifies the presence of single bond.

The total number of electron in ${Ne}_{2}$ = 20. Thus, four bonding and four anti-bonding orbitals are completely filled.

$N_{a} = 8, N_{b} = 8$ (outer shell)
So, BO = $\frac{1}{2} (8 - 8) = 0$
This means that no bond is found between two Ne atoms or in other words, ${Ne}_{2}$ does not exist. According to Molecular Orbital theory, only those molecule can exists which have net positive bond order while the molecules with negative or zero bond order will not exists.

Use the energy level diagram to show that $N_{2}$ would be expected to have a triple bond, $F_{2}$ a single bond and ${Ne}_{2}$ no bond.

Important Questions on Chemical Bonding and Molecular Structure

Learn and Prepare for any exam you want !

Use the energy level diagram to show that N2 would be expected to have a triple bond, F2 a single bond and Ne2 no bond.

Important Questions on Chemical Bonding and Molecular Structure

Learn and Prepare for any exam you want !

Use the energy level diagram to show that $N_{2}$ would be expected to have a triple bond, $F_{2}$ a single bond and ${Ne}_{2}$ no bond.