JEE Main 2020 · 2 September · morning · Numerical answer · +4 / −0
The oxidation states of iron atoms in compounds (A), (B) and (C), respectively, are x, y and z. The sum of x, y and z is ________.
Na₄[Fe(CN)₅(NOS)]
(A)
Na₄[FeO₄]
(B)
[Fe₂(CO)₉]
(C)
Answer and solution
Answer 6
Solution
Chemistry ma’am · 4 steps
Read each ligand's charge first, then let the overall charge on the complex give you iron's number.
- Compound A. Four Na⁺ make the complex ion [Fe(CN)₅(NOS)]⁴⁻. This is the purple ion formed when sulfide attacks nitroprusside. Each CN⁻ is -1 and the thionitrosyl ligand NOS⁻ is -1, so x + 5(-1) + (-1) = -4, giving x = +2.
- Compound B. Four Na⁺ balance [FeO₄]⁴⁻. Each oxide is -2, so y + 4(-2) = -4, giving y = +4.
- Compound C. CO is a neutral ligand and [Fe₂(CO)₉] carries no charge. Both iron atoms are therefore in the zero state, so z = 0.
x + y + z = (+2) + (+4) + 0 = 6
- So. The sum of the three oxidation states is 6.
The common slip: 7. Treating NOS as a neutral ligand makes iron in A equal to +3, and then the sum comes out as 7.
The idea to keep. For any complex, first fix the charge of each ligand: CN⁻, NOS⁻ and oxide are anionic, CO is neutral. The metal's oxidation state is then whatever balances the overall charge.
Answer from the NTA final answer key. Solution by PrepWiser.