Question:
What type of bond and type of overlapping present in \(\text{N}_2\) molecule ?
What type of bond and type of overlapping present in \(\text{N}_2\) molecule ?
Show Hint
N2 has a triple bond: one sigma from p-p axial overlap and two pi bonds from sidewise p-p overlap.
Updated On: Oct 1, 2026
- Single bond and s\(-\)s overlap
- Double bond and s\(-\)p overlap
- Triple bond and p\(-\)p overlap
- Double bond and p\(-\)p overlap
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Nitrogen has the configuration \(2s^2 2p^3\), so each atom has three unpaired p electrons. Each atom needs three more electrons to complete the octet, so two atoms share three pairs.
Step 2: Detailed Explanation
The molecule is \(\text{N} \equiv \text{N}\), a triple bond. Taking the molecular axis as z:
1. One \(\sigma\) bond forms by head-on (axial) overlap of \(2p_z\) orbitals.
2. Two \(\pi\) bonds form by sidewise overlap of \(2p_x\) and \(2p_y\) orbitals.
All three bonds are made from p orbitals, so the overlap is p-p.
Option (A) is the s-s overlap seen in \(\text{H}_2\). Options (B) and (D) call it a double bond, which is wrong because N2 has three shared pairs. Option (B) also mentions s-p overlap, as in HF.
Final Answer:
N2 has a triple bond formed by p-p overlap, option (C). \[ \boxed{\text{Triple bond, p-p overlap}} \]
Nitrogen has the configuration \(2s^2 2p^3\), so each atom has three unpaired p electrons. Each atom needs three more electrons to complete the octet, so two atoms share three pairs.
Step 2: Detailed Explanation
The molecule is \(\text{N} \equiv \text{N}\), a triple bond. Taking the molecular axis as z:
1. One \(\sigma\) bond forms by head-on (axial) overlap of \(2p_z\) orbitals.
2. Two \(\pi\) bonds form by sidewise overlap of \(2p_x\) and \(2p_y\) orbitals.
All three bonds are made from p orbitals, so the overlap is p-p.
Option (A) is the s-s overlap seen in \(\text{H}_2\). Options (B) and (D) call it a double bond, which is wrong because N2 has three shared pairs. Option (B) also mentions s-p overlap, as in HF.
Final Answer:
N2 has a triple bond formed by p-p overlap, option (C). \[ \boxed{\text{Triple bond, p-p overlap}} \]
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