A) \[{{C}_{2}}{{H}_{6}}\]
B) \[N{{H}_{4}}Cl\]
C) \[HCl\]
D) \[AlC{{l}_{3}}\]
Correct Answer: B
Solution :
\[N{{H}_{4}}Cl\] has both types of bonds polar and non polar \[{{\left[ H-\underset{H}{\overset{H}{\mathop{\underset{|}{\overset{|}{\mathop{N}}}\,}}}\,\to H \right]}^{+}}C{{l}^{-}}\]
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