JEE Main · 2024 · 04 Apr 2024 · Shift 1 · d and f Block Elements

Consider the following reaction

MnO_{2} + KOH + O_{2} \to A + H_{2} O

.
Product 'A' in neutral or acidic medium disproportionate to give products ' B ' and ' C ' along with water. The sum of spin-only magnetic moment values of B and C is _____ BM. (nearest integer)
(Given atomic number of Mn is 25 )

Answer: 4

Solution

2Mn O_{2} + 4KOH + O_{2} \to 2 K_{2} Mn O_{4} + 2 H_{2} O (A = K_{2} Mn O_{4}, M n^{+ 6}) .

In neutral/acidic medium the manganate disproportionates:

3Mn O_{4}^{2 -} + 4 H^{+} \to 2Mn O_{4}^{-} (B, M n^{+ 7}) + Mn O_{2} (C, M n^{+ 4}) + 2 H_{2} O .

Spin-only moments:

Mn O_{4}^{-}

(

d^{0}

)

\Rightarrow μ = 0

;

Mn O_{2}

(

M n^{+ 4}, d^{3}

)

\Rightarrow μ = 3 (3 + 2) = 15 = 3.87

BM.
Sum

= 0 + 3.87 \approx 4

BM. Answer: 4.
(Note: this item is self-contained — no figure is required despite the export's figure flag.)

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on d and f Block Elements →

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · d and f Block Elements

Solution

More JEE Main questions on d and f Block Elements