A) \[s{{p}^{2}},\text{ }s{{p}^{3}}\]and\[s{{p}^{2}}\]respectively
B) \[sp,\text{ }s{{p}^{3}}\]and\[s{{p}^{2}}\]respectively
C) \[sp,\text{ }s{{p}^{3}}\]and\[sp\]respectively
D) \[s{{p}^{2}},\text{ }s{{p}^{3}}\]and\[sp\]respectively
Correct Answer: B
Solution :
Number of hybrid orbitals, \[H=\frac{1}{2}[V+M-C+A]\] (Here, V = valence electrons, M = monovalent atom, C = positive charge, A = negative charge). In\[C{{O}_{2}},\] \[H=\frac{1}{2}[4+0-0+0]=2\] Thus, hybridisation is sp. In\[C{{H}_{4}},\] \[H=\frac{1}{2}[4+4-0+0]=4\] Thus, hybridisation is\[s{{p}^{3}}\]. In \[CH_{3}^{+},\] \[H=\frac{1}{2}[4+3-1+0]=3\] Thus, hybridisation is\[s{{p}^{2}}\].
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