Question

The correct formula for the complex compound, Na\(_2\) [Fe(CN)\(_4\)](III) is

• A. Na[Fe(CN)\(_4\)]
• B. Na\(_2\) [Fe(CN)\(_4\)]
• C. Na\(_3\) [Fe(CN)\(_4\)]
• D. Na\(_4\) [Fe(CN)\(_4\)]

Answer 1

The Correct Option is C

For Na\(_2\) [Fe(CN)\(_4\)](III), Fe is in +3 oxidation state (indicated by (III)). Each CN\(^-1\) ligand has a -1 charge.
Charge on complex ion: \( x + 4(-1) = -2 \Rightarrow x - 4 = -2 \Rightarrow x = +2 \).
But Fe(III) implies \( x = +3 \). Recalculate with correct formula:
For [Fe(CN)\(_4\)]\(^{n-}\), with Fe\(^{3+}\), charge: \( +3 + 4(-1) = -1 \).
To balance with Na\(^+\), formula is Na\(_3\) [Fe(CN)\(_4\)], as 3 Na\(^+\) neutralize -3 charge.
Answer: Na\(_3\) [Fe(CN)\(_4\)].