A) bent, \[sp\]
B) trigonal, \[s{{p}^{2}}\]
C) octahedral, \[s{{p}^{3}}d\]
D) tetrahedral, \[s{{p}^{3}}\]
Correct Answer: D
Solution :
[d] Hybridisation in \[Si{{(C{{H}_{3}})}_{4}}({\scriptstyle{}^{1}/{}_{2}}\,\,[4+4+0-0]=4)\] is \[s{{p}^{3}}\] which is tetrahedral.
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