Question
Which of the following has $$p\pi - d\pi $$ bonding?
A.
$$NO_3^ - $$
B.
$$SO_3^{2 - }$$
C.
$$BO_3^{3 - }$$
D.
$$CO_3^{2 - }$$
Answer :
$$SO_3^{2 - }$$
Solution :
In $$SO_3^{2 - },$$ $$S$$ is $$s{p^3}$$ hybridised, so
\[\underset{\begin{smallmatrix} \text{(Sulphur atom in} \\ \text{excited state)} \end{smallmatrix}}{\mathop{_{16}S=1{{s}^{2}},2\,{{s}^{2}}2{{p}^{6}},}}\,\] \[\underbrace{3{{s}^{2}}3p_{x}^{1}\,\,3p_{y}^{1}3p_{z}^{1}}_{s{{p}^{3}}\,\,\text{hybridisation}}\,\,\underset{\text{Unhybridised}}{\mathop{3d_{xy}^{1}}}\,\]
In $$'S'$$ the three $$p - {\text{orbitals}}$$ forms $$\sigma - {\text{bonds}}$$ with three oxygen atoms and unhybridised $$d - {\text{orbitals}}$$ is involved in $$\pi - {\text{bond}}$$ formation.
$${{\text{O}}_8} = 1{s^2},2{s^2}2p_x^22p_y^12p_z^1$$
In oxygen two unpaired $$p - {\text{orbitals}}$$ are present, one is involved in $$\sigma - {\text{bonds}}$$ formation while other is used in $$\pi - {\text{bond}}$$ formation.
Thus in $$SO_3^{2 - },$$ $$p$$ and $$d - {\text{orbitals}}$$ are involved for $$p\pi - d\pi $$ bonding.
In $$SO_3^{2 - },$$ $$S$$ is $$s{p^3}$$ hybridised, so
\[\underset{\begin{smallmatrix} \text{(Sulphur atom in} \\ \text{excited state)} \end{smallmatrix}}{\mathop{_{16}S=1{{s}^{2}},2\,{{s}^{2}}2{{p}^{6}},}}\,\] \[\underbrace{3{{s}^{2}}3p_{x}^{1}\,\,3p_{y}^{1}3p_{z}^{1}}_{s{{p}^{3}}\,\,\text{hybridisation}}\,\,\underset{\text{Unhybridised}}{\mathop{3d_{xy}^{1}}}\,\]
In $$'S'$$ the three $$p - {\text{orbitals}}$$ forms $$\sigma - {\text{bonds}}$$ with three oxygen atoms and unhybridised $$d - {\text{orbitals}}$$ is involved in $$\pi - {\text{bond}}$$ formation.
$${{\text{O}}_8} = 1{s^2},2{s^2}2p_x^22p_y^12p_z^1$$
In oxygen two unpaired $$p - {\text{orbitals}}$$ are present, one is involved in $$\sigma - {\text{bonds}}$$ formation while other is used in $$\pi - {\text{bond}}$$ formation.
Thus in $$SO_3^{2 - },$$ $$p$$ and $$d - {\text{orbitals}}$$ are involved for $$p\pi - d\pi $$ bonding.