Question

In certain polar solvents $PC{l}_5$ undergoes an ionisation reaction in which $C{l}^{-}$ ion leaves one $PC{l}_5$ molecule and attaches itself to another.
$2PC{l}_5\rightleftharpoons PC{l}_4{}^{+}+PC{l}_6{}^{-}$
What are the changes in the geometrical shapes that occur in this ionisation?

• A. Trigonal Bipyramidal to tetrahedral and octahedral
• B. Trigonal Bipyramidal to pentagonal planar and square planar
• C. Pentagonal planar to tetrahedral and octahedral
• D. None of these

Answer 1

The correct option is A Trigonal Bipyramidal to tetrahedral and octahedral

Steps	$PC{l}_5$	$PC{l}_4^{+}$	$PC{l}_5^{-}$
(1) Total number of electron pairs	$\frac{5+35}{2}=20$	$\frac{5+28-1}{2}=16$	$\frac{5+42+1}{2}=24$
(2) Number of bonded electron pairs	$5$	$4$	$6$
(3) Number of electron pairs around the central atom	$20-15=5$	$16-12=4$	$24-18=6$
(4) Number of lone pairs	$0$	$0$	$0$
(5) VSEPR notation	$A{X}_5$	$A{X}_4$	$A{X}_6$
(6) Structure	trigonal bipyramidal	tetrahedral	octahedral

If you manage to determine the type of hybridisation of $P$ atom in each, then the structure can be decided, but this will also require whether a lone pair is present or not.
$PC{l}_{5}s{p}^{3}d$ trigonal bipyramidal
$PC{l}_4^{+}s{p}^3$ tetrahedral
$PC{l}_6^{-}s{p}^3{d}^2$ octahedral