
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
161.1k+ views
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.
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.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Degree of Dissociation and Its Formula With Solved Example for JEE
