Paper: CSIR-NET December 2019 · Subject: Inorganic Chemistry · Chapter: Chemical Bonding · Topic: Chemical Bonding – General · Marks: 2 · Difficulty: Medium
The magnitude of bond angles in gaseous $\mathrm{NF_3}$, $\mathrm{SbF_3}$ and $\mathrm{SbCl_3}$ follow the order
(a)$\mathrm{NF_3>SbF_3>SbCl_3}$
(b)$\mathrm{SbCl_3>SbF_3>NF_3}$
(c)$\mathrm{SbF_3>SbCl_3>NF_3}$
(d)$\mathrm{NF_3>SbCl_3>SbF_3}$
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Experimental bond angles: NF₃ 102°, SbCl₃ 97°, SbF₃ 87°. A larger central atom and a more electronegative halogen both pull the bond pairs away and close the angle, so NF₃ > SbCl₃ > SbF₃.