Question

The number of dπ - pπ bonds present respectively in $$S{O}_2$$, $$S{O}_3$$, $$Cl{O_4}^{-}$$ are:
(A) 0,1,2
(B) 1,2,3
(C) 2,3,4
(D) 2,3,3

Answer 1

Hint: pπ-dπ bonds will be formed when π bonds are more than the number of unhybridized p orbitals left. First pπ-pπ bonds are formed, after that pπ-dπ if there are not enough p orbitals left for multiple bond formation.

Complete step by step solution:
Now in the given question, in $$S{O}_2$$, sulphur has six valence electrons ($$3s^{2}3p^4$$) and it forms bond with two oxygen atoms, the sulphur atom forms one sigma and one pi bond with each oxygen atom and has one lone pair and its hybridization is $$s{p}^2$$. There are two paired electrons in the 3s orbital and four electrons in 3p orbital. In order to form four bonds, it needs four unpaired electrons. Therefore, one 3p electron shifts to an empty 3d orbital. Now, there are four unpaired electrons i.e. three unpaired electrons in three 3p orbitals and one unpaired electron in one 3d orbital. One 3s and two 3p orbitals get hybridized to form three equal $$s{p}^2$$ hybrid orbitals. The remaining 3p and 3d orbitals remain unhybridized. The unpaired 3p electron forms pπ - pπ bond whereas the unpaired in d orbital forms dπ - pπ bond. Thus, there will be one dπ - pπ bond.
Similarly, in $$S{O}_3$$, the sulphur atom forms one sigma and one pi bond with each oxygen atom and has no lone pair, it also has the hybridization of $$s{p}^2$$. There are two paired electrons in the 3s orbital and four electrons in 3p orbital. In order to form four bonds, it needs six unpaired electrons. Therefore, one s electron and two 3p electrons shifts to an empty 3d orbital. Now, there are six unpaired electrons i.e. one in s orbital, three unpaired electrons in three 3p orbitals and two unpaired electrons in one 3d orbital. One 3s and two 3p orbitals get hybridized to form three equal $$s{p}^2$$ hybrid orbitals. The remaining 3p and 3d orbitals remain unhybridized. The unpaired 3p electron forms pπ - pπ bond whereas the two unpaired in d orbital forms dπ - pπ bond. Thus, there will be two dπ - pπ bonds.
In $$Cl{O_4}^{-}$$, the chlorine atom has 7 valence electrons, and it forms a bond with four oxygen atoms, the chlorine atom forms one sigma and one pi bond with three of the oxygen atom and a single bond with the left one, its hybridization is $$s{p}^3$$. There are two paired electrons in the 3s orbital and 5 electrons in 3p orbital. In order to form four bonds, it needs seven unpaired electrons. Therefore, two 3p electrons and a 3s shift to an empty 3d orbital. Now, there are seven unpaired electrons i.e. three unpaired electrons in three 3p orbitals and three unpaired electrons in one 3d orbital and one in 3s. One 3s and three 3p orbitals get hybridized to form three equal $$s{p}^3$$ hybrid orbitals. The remaining 3p and 3d orbitals remain unhybridized. The unpaired 3p electron forms pπ - pπ bond whereas the unpaired in d orbital forms dπ - pπ bond. Thus, there are three dπ - pπ bonds.


So, the correct option is (b).

Note: You can also find the hybridization by adding the number of sigma bonds and the lone pairs. dπ - pπ bonding is formed due to the sideways overlap of p and d orbitals. pπ-dπ bonds will be formed when π bonds are more than the no. of unhybridized p orbitals left.