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NEET Sample Paper NEET Sample Test Paper-86

  • question_answer
    Which of the two ions from the list given below that have the geometry that is explained by the same hybridization of orbitals, \[N{{O}_{2}}^{-},\] \[N{{O}_{3}}^{-},\]\[N{{H}_{2}},\] \[N{{H}_{4}}^{+},\] \[SC{{N}^{-}}?\]

    A) \[N{{O}_{2}}^{-}\]and\[N{{O}_{3}}^{-}\]     

    B) \[N{{H}_{4}}^{+}\]and\[N{{O}_{3}}^{-}\]

    C) \[SC{{N}^{-}}\] and \[N{{H}_{2}}^{-}\]        

    D) \[N{{O}_{2}}^{-}\]and\[N{{H}_{2}}^{-}\]

    Correct Answer: A

    Solution :

    [a] Hybridisation\[=\frac{1}{2}\] [No. of valence electrons of central atom\[+\] no. of monovalent atoms attached to it \[+.\] Negative charge if any \[-\] positive charge if any] \[N{{O}_{2}}^{-}\,\,H=\frac{1}{2}[5+0+1-0]=3=s{{p}^{2}}\] \[N{{O}_{3}}^{-}\,\,H=\frac{1}{2}[5+0+1-0]=3=s{{p}^{2}}\] \[N{{H}_{2}}^{-}\,\,H=\frac{1}{2}[5+2+1+0]=4=s{{p}^{3}}\] \[N{{H}_{4}}^{+},\,\,H=\frac{1}{2}[5+4+0-1]=4=s{{p}^{3}}\] i.e., \[N{{O}_{2}}^{-}\] and \[N{{O}_{3}}^{-}\] have same hybridisation.


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