Question

The hybridization of the central atom in \[IC{l_2}^ + \] is:
A. \[ds{p^2}\]
B. \[sp\]
C. \[s{p^2}\]
D. \[s{p^3}\]

Answer 1

Hint: The hybridization of orbital hybridization is a term used to indicate the mixing of orbitals to generate new hybrid orbitals. The new orbitals formed are used for making bonds with other atoms.

Complete answer:
 The given compound is Iodine dichloride cation. The central atom is iodine in \[IC{l_2}^ + \]. Iodine is an element in the periodic table with atomic number \[53\] and electronic configuration is\[\left[ {Kr} \right]4{d^{10}}5{s^2}5{p^5}\] .
The iodine is attached to two chlorine atoms. Chlorine is an element in the periodic table with atomic number \[17\] and electronic configuration\[\left[ {Ne} \right]3{s^2}3{p^5}\] .
The hybridization of the central atom can be obtained using the VSEPR rules. At first the number of electrons around the central atom and VSEP number is calculated.
VSEP number = \[\frac{1}{2}\] (number of electrons around the central atom I + number of atoms bonded to central atom by single bond - unit of positive charge)
$\Rightarrow$ =\[\dfrac{1}{2}{\text{ }}\left[ {7 + 2 - 1} \right]\]
$\Rightarrow$ = \[4\]
The VSEP number equal to \[4\] belongs to \[s{p^3}\] hybridization
So, the correct answer is “Option D”.

Note:
Of the two bonded atoms the less electronegative atom is selected as the central atom. The total number of electrons which are only present on the outermost shell of the central atom is counted. The total number of electrons bonded with other atoms and also which are bonded with the central atom is counted. The charge on the molecules is added for negative ions and subtracted for positive ions. The summation of the values yields the valence shell electron pair number or the VSEP number.
The determination of the VSEP number is the key point to identify the hybridization. The hybrid orbital is of equal energy and used for bond formation.