A) \[{{K}_{4}}[Fe{{(CN)}_{6}}]\]
B) \[{{K}_{3}}[Fe{{(CN)}_{6}}]\]
C) \[[Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}\]
D) \[[Cu{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}\]
Correct Answer: A
Solution :
For \[{{K}_{4}}[Fe{{(CN)}_{6}}],\], the BAN of \[F{{e}^{2+}}\] ion \[=(26-2+12)=36.\] Hence it follows BAN rule, as its BAN is equal to number of electrons of Kr (inert gas), i.e., 36.
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