A white solid X dissolves in water. With ammonium molybdate in warm conc. HNO3 it gives a canary yellow precipitate (NH4)3[PMo12O40], and the anion in the precipitate is [PMo12O40]^3- with P in the +5 state. The oxidation state of Mo in it and the number of NH4+ per formula unit are

JEE Advanced levelSingle correct optionDifficulty 3 of 3qa196

A white solid X dissolves in water. With ammonium molybdate in warm conc. $HNO X_{3}$ it gives a canary yellow precipitate $(NH X_{4}) X_{3} [PMo X_{12} O X_{40}]$ , and the anion in the precipitate is $[PMo X_{12} O X_{40}] X^{3 -}$ with P in the $+ 5$ state. The oxidation state of Mo in it and the number of $NH X_{4} X^{+}$ per formula unit are

For

[PMo X_{12} O X_{40}] X^{3 -}

+ 5 + 12 x + 40 (- 2) = - 3

gives

12 x = 72

and

x = + 6

. The anion carries a charge of

3 -

, so three

NH X_{4} X^{+}

are needed for neutrality. Molybdenum stays in its highest oxidation state, +6.