Question

If a molecule\[{{X}_{2}}\] has a triple bond, then X will have the electronic configuration
A.\[1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}\]
B. \[1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}\]
C.\[1{{s}^{2}}2{{s}^{1}}\]
D.\[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\]

Answer 1

Hint: \[{{X}_{2}}\] possesses a triple bond. So, this molecule can hold or accept 3 electrons.
This molecule possesses three bonds which are formed by the sharing of three electrons from each X atom.
We can also say that the outermost shell is less by three electron which can be shown by
Nitrogen.

Complete step by step solution:Here in this question, we are provided a molecule X2 possesses a triple bond.
We have to find out the electronic configuration of X.
Molecules have covalent bonding. So, the given molecule also possesses covalent bonds.
We know that one bond is formed by two electrons.
The molecule possesses three bonds which are formed by the sharing of three electrons from each X atom.
It implies that the valency of X is 3. It possesses three electrons in the outermost shell.

A. \[1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}\]
Five electrons exist in the outermost shell. So, A is incorrect.

B. \[1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}\]
Three electrons exist in the outermost shell.
So, B is correct.

C. \[1{{s}^{2}}2{{s}^{1}}\]
One electron in the outermost shell.
So, C is incorrect.

D. \[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\]
One electron in the outermost shell.
So, D is incorrect.

So, option B is correct.

Note: Nitrogen carries four valence orbitals (one 2s and three 2p), so it can take part in most four electron-pair bonds by forming\[s{{p}^{3}}\]hybrid orbitals.
It does not constitute long chains due to the repulsion between lone pairs of electrons on neighboring atoms.
It is the only group-15 element that generally forms multiple bonds with itself and other elements of period-2, utilizing \[\pi \] overlap of adjacent np orbitals.