Question

Select the incorrect statement about \(\mathrm{N}_2\) molecule.

• A. It is more stable than \(\mathrm{O}_{2}\) molecule.
• B. It consists more electrons in bonding molecular orbitals than \(\mathrm{O}_{2}\)
• C. Its bond order is 3
• D. It is diamagnetic

Hint:

Compare electron counts in bonding and antibonding orbitals of N2 and O2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Bond order \( =\dfrac{N_b-N_a}{2} \). \( \text{N}_2 \) has 10 valence electrons and \( \text{O}_2 \) has 12.

Step 2: Electron Counts
\( \text{N}_2 \): \( N_b=10 \), \( N_a=4 \), bond order \( =3 \), diamagnetic.
\( \text{O}_2 \): \( N_b=10 \), \( N_a=6 \), bond order \( =2 \), paramagnetic.

Step 3: Check the Statements
(A) is true because the bond order is higher. (C) and (D) are true.
(B) is false because \( \text{N}_2 \) and \( \text{O}_2 \) have the same number of bonding electrons; \( \text{N}_2 \) has fewer antibonding electrons.

Step 4: Compare the options
The incorrect statement is option (B).

Final Answer:
Statement B is incorrect because N2 does not have more bonding electrons than O2. \[ \boxed{\text{(B) }\text{It consists more electrons in bonding molecular orbitals than }\mathrm{O}_{2}} \]