Question

The shape of $I{F}_7$ molecule is

• A. octahedral
• B. trigonal bipyramidal
• C. tetrahedral
• D. pentagonal bipyramidal

Answer 1

Answer: (D)

Number of hybrid orbitals,
$H=\frac{1}{2}[V+M+A-C]$
(where, $V=$ number of valence electrons of central atom)
$M=$ number of monovalent atoms
$A=$ negative charge
$C=$ positive charge]
$\therefore H=\frac{1}{2}[7+7+0-0]=7$
Thus, the hybridisation of $I{F}_7$ is $s{p}^3{d}^3$ and its geometry is pentagonal bipyramidal.