question_answer

The shape of sulphate ion is

A)  square planar  

B)  trigonal

C)  trigonal planar

D)  tetrahedral

Correct Answer: D

Solution :

                 Sulphate ion \[(SO_{4}^{2-})\] has tetrahedral geometry as in it S-atom undergoes \[s{{p}^{3}}\]-hybridisation. The valence shell configuration of S atom and oxygen atom are S-atom (G.S.) \[3{{s}^{2}}3p_{x}^{2}3p_{y}^{1}3p_{z}^{1}\] O-atom (G.S.) \[2{{s}^{2}}2p_{x}^{2}2p_{y}^{1}2p_{z}^{1}\] ion (G.S.) \[2{{s}^{2}}2p_{x}^{2}2p_{y}^{2}2p_{z}^{1}\] S-atom in II excited state \[3{{s}^{1}}3p_{x}^{1}3p_{y}^{1}3p_{z}^{1}3d_{{{z}^{2}}}^{1}3d_{{{x}^{2}}-{{y}^{2}}}^{\,1}\]                 formation of \[SO_{4}^{2-}\] ion Due to \[s{{p}^{3}}\] hybridisation four hybrid orbitals are directed towards four corners of tetrahedron. These half filled hybrid orbitals form four sigma bonds \[[s{{p}^{3}}(S)-p\,(O)]\] with four O-atoms and two pi-bonds are formed by \[{{d}_{\pi }}-{{p}_{\pi }}\] overlapping. Thus \[SO_{4}^{2-}\] ion has tetrahedral geometry.