Oxidation State of Molybdenum in Ammonium Phosphomolybdate

Q: Can we directly find the oxidation state from the adduct formula without treating it as a complex ion?

Yes, in oxide adduct formulas like (NH_4)_3PO_4 · 12MoO_3, each component moiety is electrically neutral, so Mo in MoO_3 gives the exact same oxidation state as the complex ion representation [PMo_12O_40]^(3-).

In ammonium-phosphomolybdate, the oxidation state of Mo is ^(+)..................

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Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Can we directly find the oxidation state from the adduct formula without treating it as a complex ion?

Yes, in oxide adduct formulas like $({N H}_{4})_{3} {P O}_{4} \cdot 12 {M o O}_{3}$ , each component moiety is electrically neutral, so $M o$ in ${M o O}_{3}$ gives the exact same oxidation state as the complex ion representation $[{P M o}_{12} O_{40}]^{3 -}$ .

Answer

Accepted answer: 6

The oxidation state of molybdenum in ammonium phosphomolybdate, $(N H_{4})_{3} [P M o_{12} O_{40}]$ , is +6.

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Step-by-step solution

Given: Compound: Ammonium phosphomolybdate, formula $({N H}_{4})_{3} {P O}_{4} \cdot 12 {M o O}_{3}$ (or $({N H}_{4})_{3} [P ({M o}_{12} O_{40})]$ ).
Find: Find the oxidation state of Molybdenum ( $M o$ ) in the compound.
Strategise: In the adduct representation $({N H}_{4})_{3} {P O}_{4} \cdot 12 {M o O}_{3}$ , molybdenum is present as neutral ${M o O}_{3}$ units where oxygen is in the $- 2$ oxidation state. Setting the sum of oxidation numbers in ${M o O}_{3}$ to zero yields the oxidation state of $M o$ . Alternatively, in the polyoxometalate anion $[{P M o}_{12} O_{40}]^{3 -}$ , with $P$ as $+ 5$ and $O$ as $- 2$ : $5 + 12 x + 40 (- 2) = - 3 ⟹ 12 x = 72 ⟹ x = + 6$ .
Execute: Let the oxidation state of $M o$ be $x$ . In ${M o O}_{3}$ : $x + 3 (- 2) = 0 ⟹ x = 6$
Verify: In group 6 transition elements, $M o$ exhibits its maximum group oxidation state $+ 6$ (analogous to $C r (V I)$ in chromate/molybdate), matching the stability in oxide form ${M o O}_{3}$ .

Oxidation State of Molybdenum in Ammonium Phosphomolybdate

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The oxidation state of molybdenum in ammonium phosphomolybdate, $(N H_{4})_{3} [P M o_{12} O_{40}]$ , is +6.

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

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Oxidation State of Molybdenum in Ammonium Phosphomolybdate

The oxidation state of molybdenum in ammonium phosphomolybdate, (NH4)3[PMo12O40](NH4​)3​[PMo12​O40​], is +6.

01Given

02Find

03Strategise

04Execute

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The number of colloidal systems from the following, which will have 'liquid' as the dispe…

Formation of which complex, among the following, is not a confirmatory test of…

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The oxidation state of molybdenum in ammonium phosphomolybdate, $(N H_{4})_{3} [P M o_{12} O_{40}]$ , is +6.

A compound is formed by two elements $X$ and $Y$ . The element $Y$ fo…