A) \[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{{{a}^{x}}+{{b}^{x}}+{{c}^{x}}}{3} \right)}^{2/x}}\]
B) \[ab{{c}^{3}}\]
C) \[{{(abc)}^{1/3}}\]
D) \[f(x)=[x]\cos \left[ \frac{2x-1}{2} \right]\pi ,\]
Correct Answer: A
Solution :
The molecular orbital electronic configuration of\[NaHC{{O}_{3}}\]is\[Ca{{({{H}_{2}}P{{O}_{4}})}_{2}}\] \[NaAl{{(S{{O}_{4}})}_{2}}\] Bond order (BO) \[Cl{{O}^{-}}\] In \[ClO_{2}^{-}\] there is one electron less in anti bonding MO therefore, \[ClO_{3}^{-}\] In \[ClO_{4}^{-}\]there is one electron more in anti bonding MO therefore, \[N{{H}_{4}}N{{O}_{3}}\xrightarrow[{}]{\Delta }{{N}_{2}}O+2{{H}_{2}}O\] Bond length \[N{{H}_{4}}N{{O}_{2}}\xrightarrow[{}]{\Delta }{{N}_{2}}+2{{H}_{2}}O\] Hence, the order of bond length in given series is \[PC{{l}_{5}}\xrightarrow[{}]{\Delta }PC{{l}_{3}}+C{{l}_{2}}\]
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