A) \[\text{CH}=\overset{+}{\mathop{\text{C}}}\,\]
B) \[S-O\]
C) \[{{\text{C}}_{6}}{{H}_{5}}C{{H}_{2}}\overset{\bullet \bullet }{\mathop{C{{H}_{2}}}}\,\]
D) \[{{\text{C}}_{6}}{{H}_{5}}\overset{\bullet \bullet }{\mathop{C{{H}_{2}}}}\,\]
Correct Answer: C
Solution :
(c) Electro negativity values of \[C{{l}_{2}}\equiv {{I}_{2}}\equiv 2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\equiv 2{{S}_{2}}O_{3}^{2-}\] \[C{{l}_{2}}\equiv {{I}_{2}}\equiv 2{{S}_{2}}O_{3}^{2-}\equiv CaOC{{l}_{2}}\] The maximum difference in EN values wil \[2{{S}_{2}}O_{3}^{2-}=\frac{100\times 1}{1000}=0.1\] Thus, \[C{{l}_{2}}=0.1\]bond would be most polar.
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