A) \[C{{H}_{3}}CN\xrightarrow[HCl]{SnC{{l}_{2}}}\]\[C{{H}_{3}}CH=NH\xrightarrow{{{H}_{3}}{{O}^{+}}}C{{H}_{3}}CHO\]
B) \[C{{H}_{3}}COCl+{{H}_{2}}\xrightarrow[BaS{{O}_{4}}]{Pd}C{{H}_{3}}CHO+HCl\]
C) \[C{{H}_{2}}=C{{H}_{2}}+{{H}_{2}}O\xrightarrow{P{{d}^{2+}}}C{{H}_{3}}CHO\]
D) None of the above
Correct Answer: C
Solution :
Among isoelectronic ions (ions having same number of electrons but different nuclear charge) the greater the nuclear charge, the greater is the attraction for electrons and smaller is the ionic radius. Hence, the increasing order of ionic size is as\[A{{l}^{3+}}<N{{a}^{+}}<{{F}^{-}}<{{O}^{2-}}\].
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