• # question_answer A straight chain hydrocarbon has the molecular formula ${{C}_{8}}{{H}_{10}}$. The hybridisation for the carbon atoms from one end of the chain to the other are respectively $s{{p}^{3}},\text{ }s{{p}^{2}},\text{ }s{{p}^{2}}\text{, }s{{p}^{3}},\text{ }s{{p}^{2}},\text{ }s{{p}^{2}},\text{ }sp$ and $sp$. The structural formula of the hydrocarbon would be [CBSE PMT 1992] A) $C{{H}_{3}}-C\equiv C-C{{H}_{2}}-CH=CH-CH=C{{H}_{2}}$ B) $C{{H}_{3}}-C{{H}_{2}}-CH=CH-C{{H}_{2}}-C\equiv C-CH=C{{H}_{2}}$ C) $C{{H}_{3}}-CH=CH-C{{H}_{2}}-C\equiv C-CH=C{{H}_{2}}$ D) $C{{H}_{3}}-CH=CH-C{{H}_{2}}-CH=CH-C\equiv CH$

$\underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,-\overset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,-\overset{s{{p}^{3}}}{\mathop{C{{H}_{2}}}}\,-\underset{s{{p}^{2}}}{\mathop{CH}}\,=\overset{s{{p}^{2}}}{\mathop{CH}}\,-\underset{sp}{\mathop{C}}\,\equiv \underset{sp}{\mathop{CH}}\,$