A) \[C{{H}_{4}}\]
B) \[CH\equiv CH\]
C) \[CH_{3}^{+}\]
D) \[CH_{3}^{-}\]
Correct Answer: B
Solution :
In sp type hybrid orbitals, s and p character are equal (.ie, 50% each). [a] In \[C{{H}_{4}}\to 4bp\Rightarrow s{{p}^{3}}\] hybridisation Thus, in it s-character is 25% and p-character is 75%. [b] In \[CH\equiv CH\Rightarrow 2bp\Rightarrow sp\] hybridisation Thus, s and p characters are equal. [c] In \[CH_{3}^{+}\Rightarrow 3bp\Rightarrow s{{p}^{2}}\] hybridisation. Thus, s-character in \[CH_{3}^{+}\] is\[~33.33\] and p-character is \[66.66%.\] [d] \[CH_{3}^{-}\Rightarrow 4bp\Rightarrow s{{p}^{3}}\] hybridisation Thus, .s-character is 25% and p-character 75%.
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