A) sp, linear
B) \[s{{p}^{2}}\] , trigonal planar
C) \[s{{p}^{3}}\] , tetrahedral
D) \[s{{p}^{3}}d\], octahedral
Correct Answer: C
Solution :
Si (atomic number = 14) \[=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{6}},4{{s}^{2}}\] (ground state) Excited state. \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{1}}3p_{x}^{1}3p_{y}^{1}3p_{z}^{1}\] \[s{{p}^{3}}\] hybridization Shape = tetrahedral
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