Question

In which of the following species the central atom has the type of hybridization which is not the same as that present in the other three?
A) $PC{l_5}$
B) $S{F_4}$
C) $I_3^ -$
D) $SbCl_5^{2 - }$

Answer 1

The total number of electrons around the central atom must be the same for those which have the same hybridization. So we have to find the number of valence electrons of the central atom and number of electrons shared by the atom which is attached to the central atom.

Complete step-by-step answer: “The hybridization is the intermixing of atomic orbitals of similar energy to form new hybrid orbitals”. For this we have to find valence electrons of the central atom. In the compound the atom whose electronegativity is lower is considered as a central atom. Those which have the same number of valence electrons have the same hybridization.
$Option\left( 1 \right)PC{l_5}$, In this compound the Phosphorous $\left( P \right)$ has lower electronegativity than the Chlorine $\left( {Cl} \right)$, So P is the central atom. Phosphorus belongs to the ${15^{th}}$ group in periodic table, hence it has five valence electrons. In this compound phosphorus is surrounded by five chlorine atoms and that will contribute one-one electrons each. That means in the given compound total 10 electrons are present around the central atom and that means 5 electron pairs are there, hence its hybridization is $s{p^3}d$.
$Option\left( 2 \right)S{F_4}$, In this compound the Sulfur $\left( S \right)$ has lower electronegativity than the Fluorine $\left( F \right)$, So Sulfur is the central atom. Sulfur belongs to the ${16^{th}}$group in periodic table, hence it has six valence electrons. In this compound sulfur is surrounded by four fluorine atoms and that will contribute one-one electrons each. That means in the given compound total 10 electrons are present around the central atom and that means 5 electron pairs are there, hence its hybridization is $s{p^3}d$.
$Option\left( 3 \right)I_3^ -$, In this compound iodine $\left( I \right)$ itself is the central atom. Iodine belongs to the ${17^{th}}$group in periodic table, hence it has seven valence electrons. In this compound Iodine is surrounded by two other iodine will contribute one-one electron each, so negative charge is present means one more electron. That means in the given compound total 10 electrons are present around the central atom and that means 5 electron pairs are there, hence its hybridization is $s{p^3}d$.
$Option\left( 4 \right)SbCl_5^{2 - }$, In this compound Bismuth $\left( {Sb} \right)$has lower electronegativity than Chlorine $\left( {Cl} \right)$, So bismuth is central atom. Bismuth belongs to the ${15^{th}}$group in periodic table, hence it has five valence electrons. In this compound bismuth is surrounded by five chlorine atoms and that will contribute one-one electron each and here, two negative charges are present means two more electrons. That means in the given compound total 12 electrons are present around the central atom and that means 6 electron pairs are there, hence hybridization is $s{p^3}{d^2}$.

Correct option is $\left( 4 \right)SbCl_5^{ - 2}$. Because it has different hybridization than the other three.

Note: As, explanation is in above solution it is concluded that number of surrounded electrons of central atom is key factor to determine the hybridization of the given compounds for the comparison it is an effective way to determine hybridization in less time.