jee main chemistry important questions | d And f-Block Elements jee main questions FREE

Q 11 :
Consider the following reaction

${MnO}_{2} + KOH + O_{2} \to A + H_{2} O .$

Product ‘A’ in neutral or acidic medium disproportionates to give products ‘B’ and ‘C’ along with water. The sum of spin-only magnetic moment values of B and C is ______ B.M. (nearest integer) (Given atomic number of Mn is 25). [2024]

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Potassium permanganate is prepared by fusion of ${MnO}_{2}$ with an alkali metal hydroxide and an oxidising agent. This produces the dark green $K_{2} {MnO}_{4}$ which disproportionates in a neutral or acidic solution to give permanganate.

$2 {MnO}_{2} + 4 KOH + O_{2} \to 2 K_{2} {MnO}_{4} (A) + 2 H_{2} O$

$3 {MnO}_{4}^{2 -} + 4 H^{+} \to 2 {MnO}_{4}^{-} (B) + {MnO}_{2} (C) + 2 H_{2} O$

Compound	Electronic configuration of central metal ion	Number of unpaired electron in central metal ion (n)	Spin only magnetic moment ( $μ$ ) = $\sqrt{n (n + 2)}$ BM
${MnO}_{4}^{-} (B)$	${Mn}^{7 +} = {}_{18}{[Ar]} 3 d^{0}$	0	0 BM
${MnO}_{2} (C)$	${Mn}^{4 +} = {}_{18}{[Ar]} 3 d^{3}$	3	$\sqrt{3 (3 + 2)}$ BM = 3.87 BM

Sum = (0 + 3.87) BM = 3.87 BM

Nearest integer is 4

Q 12 :
A first row transition metal with highest enthalpy of atomisation, upon reaction with oxygen at high temperature forms oxides of formula $M_{2} O_{n}$ (where n = 3,4,5). The ‘spin-only’ magnetic moment value of the amphoteric oxide from the above oxides is ___ B.M. (near integer)
(Given atomic number : Sc : 21, Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26, Co : 27, Ni : 28, Cu : 29, Zn : 30) [2024]

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V has highest value of enthalpy of atomization for first transition series.

Element	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn
$∆ H_{a} (kJ {mol}^{- 1})$	326	473	515	397	281	416	425	430	339	126

$V_{2} O_{3}$ is basic, $V_{2} O_{4}$ is less basic and $V_{2} O_{5}$ is amphoteric.

Oxidation number of V in $V_{2} O_{5}$ is + 5

$V = {}_{18}{[Ar]} 3 d^{3} 4 s^{2},$ $V^{+ 5} = {}_{18}{[Ar]}$

Number of unpaired electrons $(n)$ in $(V^{+ 5}) = 0$

Spin only magnetic moment $(μ) = \sqrt{n (n + 2)} BM = 0$

Q 13 :
The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products A and B along with the evolution of ${CO}_{2}$ . The sum of spin-only magnetic moment values of A and B is ______ B.M. (Nearest integer)

(Given atomic number: C : 6, Na : 11, O : 8, Fe : 26, Cr : 24) [2024]

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Dichromates are generally prepared from chromate, which in turn are obtained by the fusion of chromite ore $({FeCr}_{2} O_{4})$ with sodium or potassium carbonate in free access of air.

The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of sodium chromate $({Na}_{2} {CrO}_{4})$ and ferric oxide $({Fe}_{2} O_{3})$ .

$4 {FeCr}_{2} O_{4} + 8 {Na}_{2} {CO}_{3} + 7 O_{2} \to 8 {Na}_{2} {CrO}_{4} (A) + 2 {Fe}_{2} O_{3} (B) + 8 {CO}_{2}$

Compound	Electronic configuration of central atom	Spin only magnetic moment ( $μ = \sqrt{n (n + 2)} BM$ )
${Na}_{2} {CrO}_{4} (A)$	${Cr}^{6 +} = {}_{18}{[Ar]} 3 d^{0} 4 s^{0}$	$\sqrt{0 (0 + 2)} BM = 0 BM$
${Fe}_{2} O_{3} (B)$	${Fe}^{3 +} = {}_{18}{[Ar]} 3 d^{5} 4 s^{0}$	$\sqrt{5 (5 + 2)} BM = 5.92 BM$

Sum = 0 + 5.92 $≃$ 6.

Q 14 :
The difference in the ‘spin-only’ magnetic moment values of ${KMnO}_{4}$ and the manganese product formed during titration of ${KMnO}_{4}$ against oxalic acid in acidic medium is _____ B.M. (nearest integer) [2024]

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In acidic medium while acting as an oxidizing agent, ${KMnO}_{4}$ is reduced to ${Mn}^{2 +}$ .

${KMnO}_{4} + H_{2} C_{2} O_{4} \to {Mn}^{2 +} + {CO}_{2}$ (unbalanced)

Substance	Electronic configuration of metal ion	Number of unpaired electrons (n)	Spin only magnetic moment ( $μ$ ) = $\sqrt{n (n + 2)}$ BM
${KMnO}_{4}$	${Mn}^{7 +} = {}_{18}{[Ar]} 3 d^{0} 4 s^{0}$	0	0 BM
${Mn}^{2 +}$	${Mn}^{2 +} = {}_{18}{[Ar]} 3 d^{5} 4 s^{0}$	5	$\sqrt{5 (5 + 2)} BM$ = 5.92 BM

Difference = 5.92 - 0 $≃$ 6

Q 15 :
Among $CrO$ , ${Cr}_{2} O_{3}$ and ${CrO}_{3}$ , the sum of spin-only magnetic moment values of basic and amphoteric oxides is _______ $10^{- 2}$ BM (nearest integer).

(Given atomic number of Cr is 24) [2024]

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Oxide	Nature	Electronic configuration of central metal ion	Number of unpaired electrons (n)	Spin only magnetic moment ( $μ$ ) = $\sqrt{n (n + 2)}$ BM
$CrO$	Basic	${Cr}^{+ 2} = {}_{18}{[Ar]} 3 d^{4} 4 s^{0}$	4	$\sqrt{4 (4 + 2)} BM = 4.9 BM$
${Cr}_{2} O_{3}$	Amphoteric	${Cr}^{+ 3} = {}_{18}{[Ar]} 3 d^{3} 4 s^{0}$	3	$\sqrt{3 (3 + 2)} BM = 3.87 BM$
${CrO}_{3}$	Acidic	${Cr}^{+ 6} = {}_{18}{[Ar]} 3 d^{0} 4 s^{0}$	0	0

Sum = 4.9 + 3.87 = 8.77 = 877 $\times 10^{- 2}$

Q 16 :
Among ${VO}_{2}^{+}, {MnO}_{4}^{-}$ and ${Cr}_{2} O_{7}^{2 -}$ , the spin-only magnetic moment value of the species with least oxidizing ability is .......... B.M. (Nearest integer). (Given atomic number V = 23, Mn = 25, Cr = 24). [2024]

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Oxidizing ability of given oxoanions is

${VO}_{2}^{+} < {Cr}_{2} O_{7}^{2 -} < {MnO}_{4}^{-}$

This is due to increasing stability of lower species to which they are reduced.

Compound	Electronic configuration of central metal ion	Number of unpaired electrons (n)	Spin only magnetic moment ( $μ$ ) = $\sqrt{n (n + 2)}$ BM
${VO}_{2}^{+}$	$V^{5 +} = {}_{18}{[Ar]} 3 d^{0} 4 s^{0}$	0	0 BM

Q 17 :
The 'spin only' magnetic moment value of ${MO}_{4}^{2 -}$ is ______ B.M. (Where M is a metal having least metallic radii. among Sc, Ti, V, Cr, Mn, and Zn).

(Given atomic number: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25, and Zn = 30) [2024]

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Element	Sc	Ti	V	Cr	Mn	Zn
Metallic radius (pm)	164	147	135	129	137	137

Least metallic radius is of Cr among given elements.

Compound	Electronic configuration of central atom	Number of unpaired electrons (n)	Spin only magnetic moment ( $μ$ )= $\sqrt{n (n + 2)}$ BM
${CrO}_{4}^{2 -}$	${Cr}^{+ 6} = {}_{18}{[Ar]} 3 d^{0} 4 s^{0}$	0	0

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Q 14 :
The difference in the ‘spin-only’ magnetic moment values of ${KMnO}_{4}$ and the manganese product formed during titration of ${KMnO}_{4}$ against oxalic acid in acidic medium is _____ B.M. (nearest integer) [2024]

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Q 15 :
Among $CrO$ , ${Cr}_{2} O_{3}$ and ${CrO}_{3}$ , the sum of spin-only magnetic moment values of basic and amphoteric oxides is _______ $10^{- 2}$ BM (nearest integer).

(Given atomic number of Cr is 24) [2024]

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Q 16 :
Among ${VO}_{2}^{+}, {MnO}_{4}^{-}$ and ${Cr}_{2} O_{7}^{2 -}$ , the spin-only magnetic moment value of the species with least oxidizing ability is .......... B.M. (Nearest integer). (Given atomic number V = 23, Mn = 25, Cr = 24). [2024]

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Q 17 :
The 'spin only' magnetic moment value of ${MO}_{4}^{2 -}$ is ______ B.M. (Where M is a metal having least metallic radii. among Sc, Ti, V, Cr, Mn, and Zn).

(Given atomic number: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25, and Zn = 30) [2024]

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Topic Question Set

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Q 14 : The difference in the ‘spin-only’ magnetic moment values of KMnO4 and the manganese product formed during titration of KMnO4 against oxalic acid in acidic medium is _____ B.M. (nearest integer) [2024]

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Q 15 : Among CrO, Cr2O3 and CrO3, the sum of spin-only magnetic moment values of basic and amphoteric oxides is _______ 10-2 BM (nearest integer). (Given atomic number of Cr is 24) [2024]

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Q 16 : Among VO2+,MnO4- and Cr2O72-, the spin-only magnetic moment value of the species with least oxidizing ability is .......... B.M. (Nearest integer). (Given atomic number V = 23, Mn = 25, Cr = 24). [2024]

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Q 17 : The 'spin only' magnetic moment value of MO42- is ______ B.M. (Where M is a metal having least metallic radii. among Sc, Ti, V, Cr, Mn, and Zn). (Given atomic number: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25, and Zn = 30) [2024]

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Q 14 :
The difference in the ‘spin-only’ magnetic moment values of ${KMnO}_{4}$ and the manganese product formed during titration of ${KMnO}_{4}$ against oxalic acid in acidic medium is _____ B.M. (nearest integer) [2024]

Q 15 :
Among $CrO$ , ${Cr}_{2} O_{3}$ and ${CrO}_{3}$ , the sum of spin-only magnetic moment values of basic and amphoteric oxides is _______ $10^{- 2}$ BM (nearest integer).

(Given atomic number of Cr is 24) [2024]

Q 16 :
Among ${VO}_{2}^{+}, {MnO}_{4}^{-}$ and ${Cr}_{2} O_{7}^{2 -}$ , the spin-only magnetic moment value of the species with least oxidizing ability is .......... B.M. (Nearest integer). (Given atomic number V = 23, Mn = 25, Cr = 24). [2024]

Q 17 :
The 'spin only' magnetic moment value of ${MO}_{4}^{2 -}$ is ______ B.M. (Where M is a metal having least metallic radii. among Sc, Ti, V, Cr, Mn, and Zn).

(Given atomic number: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25, and Zn = 30) [2024]