Question

The bond lengths of all the $A-X$ bonds are identical in which of the following compounds?
[$A$ = central atom and $X$ = surrounding atom]

• A. $Xe{F}_4$
• B. $PC{l}_5$
• C. Both (a) and (b)
• D. $S{F}_4$

Answer 1

The correct option is A $Xe{F}_4$

In $Xe{F}_4$, all the four $Xe-F$ bonds are of the same length and are arranged in the square planar shape and the F atoms are present on the equatorial plane. The lone pairs are axially arranged for maximum stability.

In $PC{l}_5$, the shape is trigonal bipyramidal and the three $Cl$ atoms are arranged on the equatorial plane. The remaining two $Cl$ atoms are arranged axially and the axial bonds are longer than the equatorial bonds.

The shape of $S{F}_4$ is distorted see-saw and due to the lone pair-bond pair repulsion, the bond angle between the two S-F bonds become ${101.6}^o$. The bond length of these $S-F$ bonds is 154.5 pm. The other two S-F bonds make an angle of ${173.1}^o$ and have a bond length of 164.6 pm due to bond pair-bond pair and lone pair-bond pair repulsions.