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Determining the Sum of Transition Metal Oxidation States

The sum of oxidation state of the metals in Fe(CO)_5, VO^(2+) and WO_3 is:

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Correct answer

The oxidation states of the metals are 0 in $F e (C O)_{5}$ , +4 in ${V O}^{2 +}$ , and +6 in ${W O}_{3}$ , giving a total sum of 10.

Option analysis

Why each option works or fails

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Solution

StepWorking

01Given

Given species are $F e (C O)_{5}$ , $V O^{2 +}$ , and $W O_{3}$ .

02Find

We need to find the sum of the oxidation states of the transition metals ( $F e$ , $V$ , and $W$ ) in the given species.

03Strategise

Determine the oxidation state of each metal individually by assigning standard oxidation numbers to ligands/atoms (neutral $C O = 0$ , oxide $O^{2 -} = - 2$ ) and equating the algebraic sum of oxidation numbers to the net charge of each species.

04Execute

1. In $F e (C O)_{5}$ : $C O$ is a neutral ligand (charge 0). Thus, $O.S. of Fe + 5 (0) = 0 ⟹ O.S. (F e) = 0$ . 2. In $V O^{2 +}$ : Oxygen has an oxidation state of $- 2$ . Thus, $O.S. of V + (- 2) = + 2 ⟹ O.S. (V) = + 4$ . 3. In $W O_{3}$ : Oxygen has an oxidation state of $- 2$ . Thus, $O.S. of W + 3 (- 2) = 0 ⟹ O.S. (W) = + 6$ . Sum of oxidation states $= 0 + 4 + 6 = 10$ .

✓Verify

Check: $F e$ is in a homoleptic carbonyl complex (always 0), $V$ in vanadyl ion ${V O}^{2 +}$ is in $+ 4$ state, $W$ in tungsten trioxide is in group-maximum $+ 6$ state. Sum $= 0 + 4 + 6 = 10$ . Everything matches chemistry fundamentals.

Hints that build this answer step by step

What are the individual oxidation states of $F e$ in $F e (C O)_{5}$ , $V$ in ${V O}^{2 +}$ , and $W$ in ${W O}_{3}$ ?
$F e = 0$ , $V = + 4$ , $W = + 6$
What is the sum of these three oxidation states: $0 + (+ 4) + (+ 6)$ ?
10

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why is the oxidation state of Fe in Fe(CO)5 equal to 0?

Carbon monoxide (CO) is a neutral molecule with zero formal charge. Therefore, the net charge on the complex (0) belongs entirely to Fe. This makes its oxidation state 0.

Answer

The oxidation states of the metals are 0 in $F e (C O)_{5}$ , +4 in ${V O}^{2 +}$ , and +6 in ${W O}_{3}$ , giving a total sum of 10.

Why each option works or fails

Step-by-step solution

Given: Given species are $F e (C O)_{5}$ , $V O^{2 +}$ , and $W O_{3}$ .
Find: We need to find the sum of the oxidation states of the transition metals ( $F e$ , $V$ , and $W$ ) in the given species.
Strategise: Determine the oxidation state of each metal individually by assigning standard oxidation numbers to ligands/atoms (neutral $C O = 0$ , oxide $O^{2 -} = - 2$ ) and equating the algebraic sum of oxidation numbers to the net charge of each species.
Execute: 1. In $F e (C O)_{5}$ : $C O$ is a neutral ligand (charge 0). Thus, $O.S. of Fe + 5 (0) = 0 ⟹ O.S. (F e) = 0$ . 2. In $V O^{2 +}$ : Oxygen has an oxidation state of $- 2$ . Thus, $O.S. of V + (- 2) = + 2 ⟹ O.S. (V) = + 4$ . 3. In $W O_{3}$ : Oxygen has an oxidation state of $- 2$ . Thus, $O.S. of W + 3 (- 2) = 0 ⟹ O.S. (W) = + 6$ . Sum of oxidation states $= 0 + 4 + 6 = 10$ .
Verify: Check: $F e$ is in a homoleptic carbonyl complex (always 0), $V$ in vanadyl ion ${V O}^{2 +}$ is in $+ 4$ state, $W$ in tungsten trioxide is in group-maximum $+ 6$ state. Sum $= 0 + 4 + 6 = 10$ . Everything matches chemistry fundamentals.

Determining the Sum of Transition Metal Oxidation States

What feels right?

The oxidation states of the metals are 0 in $F e (C O)_{5}$ , +4 in ${V O}^{2 +}$ , and +6 in ${W O}_{3}$ , giving a total sum of 10.

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

✓Verify

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