Question

The ion that has \[s{p^3}{d^2}\] hybridization for the central atom is:
A.\[{\left[ {IC{l_2}} \right]^ - }\]
B.\[{\left[ {I{F_6}} \right]^ - }\]
C.\[{\left[ {IC{l_4}} \right]^ - }\]
D.\[{\left[ {Br{F_3}} \right]^ - }\]

Answer 1

Hint: Hybridization is a process that is used to determine the geometry and shape of a molecule. The formula used to find out the hybridization of a molecule is given as:$X = \dfrac{1}{2}[V + M - C + A]$
Where V = Number of valence electrons of the central atom.
M = number of monovalent groups.
C = number of positive charge
A = number of negative charges.

Complete step by step answer:
Let us consider the first option here: the central atom is I. It has 7 valence electrons, 2 monovalent groups, and the number of negative charges is 1. By using the above formula we will find the hybridization.
$
  X = \dfrac{1}{2}[7 + 2 + 1] \\
   \Rightarrow X = \dfrac{1}{2}[10] \\
  \therefore X = 5 \\
$
Hence the hybridization is $s{p^3}d$. Hence the geometry of the compound is trigonal bipyramidal
Now in the case of the second option, again central atom is I hence, the valence electron on the central atom is 7 and monovalent groups are 6, and the number of positive charges is 1,
$
  X = \dfrac{1}{2}[7 + 6 + 1] \\
   \Rightarrow X = \dfrac{1}{2}[14] \\
  \therefore X = 7 \\
$
Hence the hybridization is $s{p^3}{d^3}$. The geometry of the compound is pentagonal bipyramidal.
For the third option, the valence electron on central atom I is 7 and monovalent groups is 4 and the negative charge is 1, so the hybridization is calculated as,
$
  X = \dfrac{1}{2}[7 + 4 + 1] \\
  X = \dfrac{{12}}{2} \\
  X = 6 \\
$
Hence the hybridization is $s{p^3}{d^2}$. The geometry of the compound is octahedral.
In the fourth case, the valence electron is 7, and the number of monovalent groups is 4 and the negative charge is 1. So, Hybridization can be calculated as:
$
  X = \dfrac{1}{2}[7 + 2 + 1] \\
   \Rightarrow X = \dfrac{1}{2}[10] \\
  \therefore X = 5 \\
$
Hence the hybridization is $s{p^3}d$. The geometry of the compound is trigonal bipyramidal.

Hence the correct answer is (C).

Note:
Hybridization is defined as the concept of intermixing of the orbitals of an atom having nearly the same energy to give exactly equivalent orbitals with the same energy, identical shape, and symmetrical orientation in space. The concept of hybridization is also used to explain the observed values of bond angles.