A) \[~s{{p}^{2}},sp,\text{ }s{{p}^{3}}\]
B) \[sp,\text{ }s{{p}^{3}},\text{ }s{{p}^{2}}\]
C) \[sp,\text{ }s{{p}^{2}},\text{ }s{{p}^{3}}\]
D) \[s{{p}^{3}},s{{p}^{2}},\text{ }sp\]
Correct Answer: B
Solution :
Key Idea: \[-C-C-s{{p}^{3}}\] \[-C=C-s{{p}^{2}}\] \[-C\equiv C-sp\] \[=C\equiv C=sp\] \[\underset{6}{\overset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}}\,-\underset{5}{\overset{s{{p}^{3}}}{\mathop{CH}}}\,=\underset{4}{\overset{s{{p}^{2}}}{\mathop{CH}}}\,-\underset{3}{\overset{s{{p}^{3}}}{\mathop{C{{H}_{2}}}}}\,-\underset{2}{\overset{sp}{\mathop{C}}}\,\equiv \underset{1}{\overset{sp}{\mathop{C}}}\,H\] Hence, the state of hybridisation of carbons 1, 3 and 5 are \[sp,\,\,s{{p}^{3}}\] and \[s{{p}^{2}}\] respectively.
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