Subject: Inorganic Chemistry · Chapter: Chemical Bonding · Topic: Chemical Bonding – General · Exam: CSIR-NET DEC 2019 · 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
SELF-PRACTICE — the source book printed no answer.
Nothing is invented here, so this question has no answer on record.
Explanation
According to VSEPR theory as the size of central atom increases bond angle decreases.Hence, Bond angle of $\mathrm{NF_3}$ > $\mathrm{SbCl_3}$, $\mathrm{SbF_3}$Now, in between $\mathrm{SbCl_3}$ and $\mathrm{SbF_3}$, 'F' is more electronegative and pull the bond pair towards itself due to which bond angle FSbF decreases.The experimental bond angles are