Q9 / 180 · COMEDK 2025 (10 May Shift 1) · Chemistry
Identify the correct statement.
- AThe central atoms in both $\mathrm{CH}_4$ and $\mathrm{SF}_4$ are in a state of $\mathrm{sp}^3$ hybridisation
- BThe $\mathrm{C}-\mathrm{O}$ bond length in $\mathrm{CO}_2$ molecule is 110 pm because of the Inductive effect $(+\mathrm{I})$ of oxygen
- CThe resultant dipole moment of $\mathrm{NF}_3$ is greater than that of $\mathrm{NH}_3$
- Correct answerDA molecule of the type $\mathrm{AB}_5 \mathrm{E}$ where the central atom has 5 bond pairs and 1 lone pair has a square pyramid geometry
Solution
(1) Incorrect: C in $CH_4$ is $sp^3$ hybridised (4 bond pairs, 0 lone pairs). S in $SF_4$ has 4 bond pairs and 1 lone pair (steric number 5), making it $sp^3d$ hybridised with a see-saw shape.
(2) Incorrect: Oxygen is highly electronegative and exerts a $-I$ effect (electron-withdrawing), not $+I$.
(3) Incorrect: Dipole moment of $NH_3$ is greater than that of $NF_3$. In $NH_3$, the lone pair dipole and N-H bond dipoles add up. In $NF_3$, the N-F bond dipoles oppose the lone pair dipole, partially canceling it.
(4) Correct: $AB_5E$ type has steric number 6 (octahedral electron geometry). With one lone pair occupying one position, the molecular geometry is square pyramidal (e.g., $BrF_5$, $IF_5$).
The correct statement is (4).
(2) Incorrect: Oxygen is highly electronegative and exerts a $-I$ effect (electron-withdrawing), not $+I$.
(3) Incorrect: Dipole moment of $NH_3$ is greater than that of $NF_3$. In $NH_3$, the lone pair dipole and N-H bond dipoles add up. In $NF_3$, the N-F bond dipoles oppose the lone pair dipole, partially canceling it.
(4) Correct: $AB_5E$ type has steric number 6 (octahedral electron geometry). With one lone pair occupying one position, the molecular geometry is square pyramidal (e.g., $BrF_5$, $IF_5$).
The correct statement is (4).
Hence, the correct answer is option: 'D'.