Question

FeSO4

reacts with excess KCN solution to form a complex then no. of unpaired ‘d’ electrons

in central atom of that complex.

1) 4 2) 5 3) 3 4) 0​

Answer 1

FeSO4+KCN-->K4[Fe(CN)6] +K2SO4

K4[Fe(CN)6]-->4K+. + [Fe(CN)6]4-

[Fe(CN)6]4-

Fe+6(CN) =-4

Fe -6= -4

Fe=+2

pls refer the attachment for the diagram

no of unpaired e- is 0

as CN is a strong field ligands

I hope this helps ( ╹▽╹ )

Answer 2

Hence the central metal atom has 0 unpaired electrons.

• Ferrous sulfate reacts with KCN to form potassium hexacyanoiron(II) complex as follows-
• $FeSO_4+KCN\rightarrow K_4[Fe(CN)_6]$.
• In this complex, the oxidation state of iron is +2.
• Hence iron contains 26-2=24 electrons.
• There are 6 electrons residing at the 3d orbital.
• As cyanide is a strong field ligands so, it makes pairing and there are no unpaired electrons present in Fe.
• Hence the central metal atom has 0 unpaired electrons.
• Thus, the correct option is 4.