A) \[s{{p}^{2}}>s{{p}^{3}}>sp\]
B) \[s{{p}^{3}}<s{{p}^{2}}>sp\]
C) \[s{{p}^{3}}>s{{p}^{2}}>sp\]
D) \[sp>s{{p}^{2}}>s{{p}^{3}}\]
Correct Answer: D
Solution :
As the s-character increases, electronegativity increases and thus, the bond angle increases. The order of s-character in different hybrid orbitals is\[\underset{50%}{\mathop{sp\,\,>}}\,\,\,\,\,\,\,\,\underset{33.3%}{\mathop{s{{p}^{2}}}}\,>\,\,\,\,\underset{25%}{\mathop{s{{p}^{3}}}}\,\] s-character Thus, order of bond angle is \[\underset{({{180}^{o}})}{\mathop{sp\,\,}}\,\,\,\,>\,\,\underset{({{120}^{o}})}{\mathop{s{{p}^{2}}}}\,>\,\,\underset{({{109.5}^{o}})}{\mathop{s{{p}^{3}}}}\,\]
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