Question

The correct order of F–M–F bond angle of compounds given in the options is (where M represents the central element)

• A. XeF$_2$ $>$ BF$_3$ $>$ OF$_2$ $>$ [NF$_4$]$^+$
• B. XeF$_2$ $>$ BF$_3$ $>$ [NF$_4$]$^+$ $>$ OF$_2$
• C. BF$_3$ $>$ XeF$_2$ $>$ [NF$_4$]$^+$ $>$ OF$_2$
• D. XeF$_2$ $>$ OF$_2$ $>$ BF$_3$ $>$ [NF$_4$]$^+$

Hint:

Bond angle decreases with increase in lone pairs due to stronger lone pair repulsion compared to bond pair repulsion

Answer 1

The Correct Option is A

Step 1: Find each shape by VSEPR.
XeF$_2$ is linear with a 180 degree angle. BF$_3$ is trigonal planar at 120 degrees. [NF$_4$]$^+$ is tetrahedral at about 109.5 degrees. OF$_2$ is bent at about 103 degrees.

Step 2: Order the angles.
\[ 180 > 120 > 109.5 > 103 \]

Step 3: Answer.
\[ \boxed{\text{XeF}_2 > \text{BF}_3 > [\text{NF}_4]^+ > \text{OF}_2} \]