The metal ion in complex `underline(A ) ` has EAN identical to the atomic number of krypton . `underline (A) ` is
( Atomic no. Of `Cr =24,Fe=26 ,pd =46) `

Options

A. `[Pd(NH_(3))_(6)]Cl_(4)`
B. `[Cr(NH_(3))_(5)Cl]SO_(4) `
C. `Na_(4)[Fe(CN)_(6)]`
D. `K_(3)[Fe(CN)_(6)]`

Answer

C

Explanation

EAN =`Z-(o .N ) +2 (C .N) x` Electrons donated by `CN^(-) `
where ,ON = oxidation number
Z = atomic numebr
` [Fe(CN)_(6)]^(4-) :EAN = 26-(2) +2(6) - 2+12=36`