Question

In which of the following molecule, all the atoms lie in same plane?

• A. $Xe{F}_4$
• B. $B{F}_3$
• C. $P{F}_5$
• D. $N{H}_3$

Answer 1

Answer: (A), (B)

The correct options are
A $Xe{F}_4$
B $B{F}_3$

From structure it can be seen that,
In $B{F}_3$, Boron is sp${}^2$-hybridized with bond angle 120${}^{\circ }$ and triangular planar.
And $Xe{F}_4$ is square planer molecule.
$N{H}_3$ has pyramidal structure and $PC{l}_5$ has traingular bipyramidal structure.