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JEE Main & Advanced Chemistry GOC Question Bank Bonding and hybridisation inorganic compounds

  • 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) \[C{{H}_{3}}-CH=CH-C{{H}_{2}}-CH=CH-C\equiv CH\]

    Correct Answer: D

    Solution :

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


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