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

Q 1 :

In alkaline medium, ${MnO}_{4}^{-}$ oxidises $I^{-}$ to [2024]

${IO}^{-}$
${IO}_{4}^{-}$
$I_{2}$
${IO}_{3}^{-}$

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(4)

In neutral or faintly alkaline medium, permanganate $({MnO}_{4}^{-})$ oxidizes iodide $(I^{-})$ to iodate $({IO}_{3}^{-})$

$2 {MnO}_{4}^{-} + H_{2} O + I^{-} \to 2 {MnO}_{2} + 2 {OH}^{-} + {IO}_{3}^{-}$

Q 2 :

${KMnO}_{4}$ decomposes on heating at 513K to form $O_{2}$ along with [2024]

${Mn and KO}_{2}$
$K_{2} {MnO}_{4} {and MnO}_{2}$
$K_{2} {MnO}_{4} and Mn$
${MnO}_{2} {and K}_{2} O_{2}$

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(2)

$2 {KMnO}_{4} (s) \overset{Δ}{\to} {MnO}_{2} (s) + K_{2} {MnO}_{4} (s) + O_{2} (g)$

Q 3 :

The orange colour of $K_{2} {Cr}_{2} O_{7}$ and purple colour of ${KMnO}_{4}$ is due to [2024]

Charge transfer transition in both.
d → d transitions in ${KMnO}_{4}$ and charge transfer transitions in $K_{2} {Cr}_{2} O_{7}$ .
d → d transitions in $K_{2} {Cr}_{2} O_{7}$ and charge transfer transitions in ${KMnO}_{4}$ .
d → d transitions in both

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(1)

A compound is coloured if it has electronic energy levels where electrons can undergo transition by absorbing light in visible range. For most d-block elements, energy difference in d-orbitals lies in visible range. Hence due to d-d electronic transitions, these are coloured. But in $K_{2} {Cr}_{2} O_{7}$ (orange) and ${KMnO}_{4}$ (purple), central atom has no d-electron. Colour of these compounds is due to charge transfer from oxide to metal ion.

Q 4 :

Alkaline oxidative fusion of ${MnO}_{2}$ gives ''A'' which on electrolytic oxidation in alkaline solution produces B. A and B respectively are [2024]

${Mn}_{2} O_{7}$ and ${MnO}_{4}^{-}$
${MnO}_{4}^{2 -}$ and ${MnO}_{4}^{-}$
${Mn}_{2} O_{3}$ and ${MnO}_{4}^{2 -}$
${MnO}_{4}^{2 -}$ and ${Mn}_{2} O_{7}$

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(2)

Alkaline oxidative fusion of ${MnO}_{2}$ gives manganate $({MnO}_{4}^{2 -})$ , which upon electrolytic oxidation gives permanganate.

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

${MnO}_{4}^{2 -} \overset{Electrolytic oxidation}{\to} {MnO}_{4}^{-}$

Manganate Permanganate

This is commercial method of preparation of Permanganate

Q 5 :

Identify correct statements from below:

A. The chromate ion is square planar.
B. Dichromates are generally prepared from chromates.
C. The green manganate ion is diamagnetic.
D. Dark green coloured $K_{2} {MnO}_{4}$ disproportionates in a neutral or acidic medium to give permanganate.
E. With increasing oxidation number of transition metal, ionic character of the oxides decreases.

Choose the correct answer from the options given below: [2024]

B, C, D only
A, D, E only
A, B, C only
B, D, E only

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(4)

(A) The chromate ion is tetrahedral in shape.

(B) Dichromates are prepared from chromates.

(C) In manganate ion, Mn is in +6 oxidation state, hence has one unpaired electron and is paramagnetic.

(D) The dark green $K_{2} {MnO}_{4}$ disproportionates in a neutral or acidic solution to give permanganate.

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

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

(E) With increase in oxidation number of central metal atom, its polarising power increases and covalent character increase and therefore, ionic character decreases.

Q 6 :

Choose the correct statements from the following

A. ${Mn}_{2} O_{7}$ is an oil at room temperature
B. $V_{2} O_{4}$ reacts with acid to give ${VO}_{2}^{2 +}$
C. $CrO$ is a basic oxide
D. $V_{2} O_{5}$ does not react with acid

Choose the correct answer from the options given below: [2024]

A, B and D only
A and C only
A, B and C only
B and C only

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(2)

(A) ${Mn}_{2} O_{7}$ is a covalent green oil.

(B) $V_{2} O_{4}$ is a basic oxide. It dissolves in acids to give ${VO}^{2 +}$ salts (not ${VO}_{2}^{2 +}$ ).

(C) $CrO$ is basic and ${Cr}_{2} O_{3}$ is amphoteric.

(D) $V_{2} O_{5}$ is amphoteric. It reacts with alkalies as well as acids to give ${VO}_{4}^{3 -}$ respectively.

Q 7 :

Which of the following compounds show colour due to $d - d$ transition? [2024]

$K_{2} {Cr}_{2} O_{7}$
${CuSO}_{4} . 5 H_{2} O$
${KMnO}_{4}$
$K_{2} {CrO}_{4}$

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(2)

A compound is coloured if it has electronic energy levels where electrons can undergo transition by absorbing light in visible range. For most d block elements energy difference in d orbitals lies in visible range. Hence, due to d-d electronic transitions, These are coloured. In ${CuSO}_{4} . 5 H_{2} {O, Cu}^{2 +} (3 d^{9})$ splits into $t_{2 g}^{6}, e_{g}^{3}$ . $d - d$ transition from $t_{2 g}$ to $e_{g}$ makes it coloured. But in $K_{2} {Cr}_{2} O_{7}$ (orange), ${KMnO}_{4}$ (purple) and $K_{2} {CrO}_{4}$ (yellow), central atom has no d electron. Colour of these compounds is due to charge transfer from oxide to metal ion.

Q 8 :

Iron (III) catalyses the reaction between iodide and persulphate ions, in which

A. ${Fe}^{3 +}$ oxidises the iodide ion
B. ${Fe}^{3 +}$ oxidises the persulphate ion
C. ${Fe}^{2 +}$ reduces the iodide ion
D. ${Fe}^{2 +}$ reduces the persulphate ion

Choose the most appropriate answer from the options given below: [2024]

B only
A only
B and C only
A and D only

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(4)

$2 I^{-} + S_{2} O_{8}^{2 -} \to I_{2} + 2 {SO}_{4}^{2 -}$

In this reaction iron acts as catalyst as explained below:

${Fe}^{3 +}$ oxidizes the iodide ion

$2 {Fe}^{3 +} + 2 I^{-} \to 2 {Fe}^{2 +} + I_{2}$

${Fe}^{2 +}$ reduces the persulphate ion (oxygen is reduced from -1 to -2).

$2 {Fe}^{2 +} + S_{2} O_{8}^{2 -} \to 2 {Fe}^{3 +} + 2 {SO}_{4}^{2 -}$

Q 9 :

Given below are two statements:

Statement (I): Fusion of ${MnO}_{2}$ with KOH and an oxidising agent gives dark green $K_{2} {MnO}_{4}$ .

Statement (II): Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion.

In the light of the above statements, choose the correct answer from the options given below. [2024]

Statement I is true but Statement II is false
Both Statement I and Statement II are false
Statement I is false but Statement II is true
Both Statement I and Statement II are true

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(4)

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.

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

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

Q 10 :

Number of moles of $H^{+}$ ions required by 1 mole of ${MnO}_{4}^{-}$ to oxidise oxalate ion to ${CO}_{2}$ is _____________ . [2024]

(8)

${MnO}_{4}^{-}$ $+$ $8 H^{+}$ $+$ $5 e^{-}$ $\to$ ${Mn}^{2 +}$ $+$ $4 H_{2} O]$ $\times$ $2$

${C_{2} O}_{4}^{2 -}$ $\to$ $2 {CO}_{2}$ $+$ $2 e^{-}$ $]$ $\times$ $5$

_____________________________________________

$2 {MnO}_{4}^{-}$ $+$ $5 C_{2} O_{4}^{2 -}$ $+$ $16 H^{+}$ $\to$ $2 {Mn}^{2 +}$ $+$ $10 {CO}_{2}$ $+$ $8 H_{2} O$

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]

(4)

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]

(0)

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]

(6)

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]

(6)

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]

(877)

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]

(0)

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]

(0)

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

Q 18 :

The equilibrium $C r_{2} O_{7}^{2 -} ⇌ 2 C r O_{4}^{2 -}$ is shifted to the right in: [2024]

an acidic medium
a basic medium
a neutral medium
a weakly acidic medium

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(2)

Q 19 :

The correct set of ions (aqueous solution) with same colour from the following is: [2025]

${Zn}^{2 +}, V^{3 +}, {Fe}^{3 +}$
${Sc}^{3 +}, {Ti}^{3 +}, {Cr}^{2 +}$
${Ti}^{4 +}, V^{4 +}, {Mn}^{2 +}$
$V^{2 +}, {Cr}^{3 +}, {Mn}^{3 +}$

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(4)

Ions	Colour
${Sc}^{3 +}$	colourless
${Ti}^{4 +}$	colourless
${Ti}^{3 +}$	purple
$V^{4 +}$	blue
$V^{3 +}$	green
$V^{2 +}$	violet
${Cr}^{3 +}$	violet
${Mn}^{3 +}$	violet
${Cr}^{2 +}$	blue
${Mn}^{2 +}$	pink
${Fe}^{3 +}$	yellow
${Zn}^{2 +}$	colourless

Q 20 :

Match List–I with List–II. [2025]

List–I List–II

A. Bronze I. Cu, Ni

B. Brass II. Fe, Cr, Ni, C

C. UK silver coin III. Cu, Zn

D. Stainless steel IV. Cu, Sn

Choose the correct answer from the options given below:

A–III, B–IV, C–II, D–I
A–IV, B–II, C–III, D–I
A–IV, B–III, C–I, D–II
A–III, B–I, C–IV, D–II

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(3)

Q 21 :

Consider the following reactions

$K_{2} {Cr}_{2} O_{7} \to_{- H_{2} O}^{KOH} [A] \to_{- H_{2} O}^{H_{2} {SO}_{4}} [B] + K_{2} {SO}_{4}$

The products [A] and [B], respectively are: [2025]

$K_{2} {CrO}_{4} and K_{2} {Cr}_{2} O_{7}$
$K_{2} {CrO}_{4} and CrO$
$K_{2} {CrO}_{4} and {Cr}_{2} O_{3}$
$K_{2} Cr {(OH)}_{6} and {Cr}_{2} O_{3}$

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(1)

Chromate $({CrO}_{4}^{2 -})$ and dichromate $({Cr}_{2} O_{7}^{2 -})$ are inter-convertible into each other depending upon the pH of the solution. In acidic medium, ${Cr}_{2} O_{7}^{2 -}$ is more stable and in basic medium, ${CrO}_{4}^{2 -}$ is more stable.

$K_{2} {Cr}_{2} O_{7} \to_{- H_{2} O}^{KOH} \underset{[A]}{K_{2} {CrO}_{4}} \to_{- H_{2} O}^{H_{2} {SO}_{4}} \underset{[B]}{K_{2} {Cr}_{2} O_{7}} + K_{2} {SO}_{4}$

Q 22 :

Preparation of potassium permanganate from ${MnO}_{2}$ involves a two-step process in which the first step is a reaction with $KOH$ and ${KNO}_{3}$ to produce [2025]

$K_{2} {MnO}_{4}$
$K_{3} {MnO}_{4}$
${KMnO}_{4}$
$K_{4} [Mn {(OH)}_{6}]$

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(1)

Potassium permanganate is prepared by fusion of ${MnO}_{2}$ (pyrolusite) 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.

${MnO}_{2} + 2 KOH + {KNO}_{3} \to K_{2} {MnO}_{4} + {KNO}_{2} + H_{2} O$

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

Q 23 :

Which of the following oxidation reactions are carried out by both $K_{2} {Cr}_{2} O_{7}$ and ${KMnO}_{4}$ in acidic medium? [2025]

(A)   $I^{-} \to I_{2}$

(B)    $S^{2 -} \to S$

(C)   ${Fe}^{2 +} \to {Fe}^{3 +}$

(D)   $I^{-} \to {IO}_{3}^{-}$

(E)   $S_{2} O_{3}^{2 -} \to {SO}_{4}^{2 -}$

Choose the correct answer from the options given below:

(C), (D) and (E) only
(A), (B) and (C) only
(B), (C) and (D) only
(A), (D) and (E) only

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(2)

Reactant	Product with acidic ${KMnO}_{4}$	Product with faintly alkaline ${KMnO}_{4}$	Product with acidic $K_{2} {Cr}_{2} O_{7}$
$I^{-}$	$I_{2}$	${IO}_{3}^{-}$	$I_{2}$
$S^{2 -}$	$S$	$S$	$S$
${Fe}^{2 +}$	${Fe}^{3 +}$	${Fe}^{3 +}$	${Fe}^{3 +}$
$S_{2} O_{3}^{2 -}$	${SO}_{4}^{2 -} + S$	${SO}_{4}^{2 -}$	${SO}_{4}^{2 -} + S$

Q 24 :

The amphoteric oxide among $V_{2} O_{3}, V_{2} O_{4}$ and $V_{2} O_{5}$ , upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is: [2025]

+4
+7
+3
+5

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(4)

$V_{2} O_{3}$ – basic, $V_{2} O_{4}$ – less basic, $V_{2} O_{5}$ – amphoteric. $V_{2} O_{5}$ reacts with alkalies as well as acids to give ${VO}_{4}^{3 -}$ and ${VO}_{4}^{+}$ respectively. Oxidation number of V in ${VO}_{4}^{3 -}$ is +5.

Q 25 :

The molar mass of the water insoluble product formed from the fusion of chromite ore $({FeCr}_{2} O_{4})$ with ${Na}_{2} {CO}_{3}$ in presence of $O_{2}$ is ______ g ${mol}^{- 1}$ . [2025]

(160)

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

${Fe}_{2} O_{3}$ is insoluble in water. Its molar mass $= [2 \times 56 + 3 \times 16]$ $g/mol$

$= 160 g/mol$

Q 26 :

$K_{2} {Cr}_{2} O_{7}$ paper acidified with dilute $H_{2} {SO}_{4}$ turns green when exposed to [2023]

Hydrogen sulphide
Sulphur dioxide
Carbon dioxide
Sulphur trioxide

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(2)

${Cr}_{2} O_{7}^{2 -} + 2 H^{+} + 3 {SO}_{2} \to 2 {Cr}^{3 +} (green) + 3 {SO}_{4}^{2 -} + H_{2} O$

Q 27 :

Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from [2023]

+6 to +3
+6 to +2
+3 to +1
+2 to +1

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(1)

${\overset{+ 6}{Cr}}_{2} O_{7}^{- 2} \overset{H^{+}}{\to} {Cr}^{+ 3}$

Q 28 :

${KMnO}_{4}$ oxidises $I^{-}$ in acidic and neutral/faintly alkaline solution, respectively, to [2023]

${IO}_{3}^{-}$ and ${IO}_{3}^{-}$
$I_{2}$ and ${IO}_{3}^{-}$
$I_{2}$ and $I_{2}$
${IO}_{3}^{-}$ and $I_{2}$

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(2)

In acidic medium

$2 {MnO}_{4}^{-} + 10 I^{-} + 16 H^{+} ⟶ 2 {Mn}^{2 +} + 8 H_{2} O + 5 I_{2}$

In neutral or faintly alkaline medium

$2 {MnO}_{4}^{-} + I^{-} + H_{2} O ⟶ 2 {MnO}_{2} + 2 {OH}^{-} + {IO}_{3}^{-}$

Q 29 :

When ${Cu}^{2 +}$ ion is treated with $KI$ , a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are: [2023]

$X = {CuI}_{2}$ $Y = {Na}_{2} S_{2} O_{3}$
$X = {CuI}_{2}$ $Y = {Na}_{2} S_{4} O_{6}$
$X = {Cu}_{2} I_{2}$ $Y = {Na}_{2} S_{4} O_{6}$
$X = {Cu}_{2} I_{2}$ $Y = {Na}_{2} S_{4} O_{5}$

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(3)

${Cu}^{2 +} + 4 I^{-} ⟶ \underset{\underset{' X'}{White}}{{Cu}_{2} I_{2}} + I_{2}$

$I_{2} + 2 {Na}_{2} S_{2} O_{3} ⟶ \underset{' Y'}{2 {Na}_{2} S_{4} O_{6}} + 2 NaI$

$X = {Cu}_{2} I_{2}$ and $Y = {Na}_{2} S_{4} O_{6}$

Q 30 :

A solution of ${FeCl}_{3}$ , when treated with $K_{4} [Fe {(CN)}_{6}]$ gives a Prussian blue precipitate due to the formation of [2023]

$Fe [Fe {(CN)}_{6}]$
$K [{Fe}_{2} {(CN)}_{6}]$
${Fe}_{4} {[Fe {(CN)}_{6}]}_{3}$
${Fe}_{3} {[Fe {(CN)}_{6}]}_{2}$

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(3)

$4 {Fe}^{3 +} + 3 {[Fe {(CN)}_{6}]}^{4 -} ⟶ {Fe}_{4} {[Fe {(CN)}_{6}]}_{3} ↓$ Prussian blue precipitate.

Topic Question Set

Q 1 :

In alkaline medium, ${MnO}_{4}^{-}$ oxidises $I^{-}$ to [2024]

Q 2 :

${KMnO}_{4}$ decomposes on heating at 513K to form $O_{2}$ along with [2024]

Q 3 :

The orange colour of $K_{2} {Cr}_{2} O_{7}$ and purple colour of ${KMnO}_{4}$ is due to [2024]

Q 4 :

Alkaline oxidative fusion of ${MnO}_{2}$ gives ''A'' which on electrolytic oxidation in alkaline solution produces B. A and B respectively are [2024]

Q 6 :

Choose the correct statements from the following

A. ${Mn}_{2} O_{7}$ is an oil at room temperature
B. $V_{2} O_{4}$ reacts with acid to give ${VO}_{2}^{2 +}$
C. $CrO$ is a basic oxide
D. $V_{2} O_{5}$ does not react with acid

Choose the correct answer from the options given below: [2024]

Q 7 :

Which of the following compounds show colour due to $d - d$ transition? [2024]

Q 10 :

Number of moles of $H^{+}$ ions required by 1 mole of ${MnO}_{4}^{-}$ to oxidise oxalate ion to ${CO}_{2}$ is _____________ . [2024]

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]

Q 18 :

The equilibrium $C r_{2} O_{7}^{2 -} ⇌ 2 C r O_{4}^{2 -}$ is shifted to the right in: [2024]

Q 19 :

The correct set of ions (aqueous solution) with same colour from the following is: [2025]

Q 20 :

Match List–I with List–II. [2025]

List–I List–II

A. Bronze I. Cu, Ni

B. Brass II. Fe, Cr, Ni, C

C. UK silver coin III. Cu, Zn

D. Stainless steel IV. Cu, Sn

Choose the correct answer from the options given below:

(3)

Q 21 :

Consider the following reactions

$K_{2} {Cr}_{2} O_{7} \to_{- H_{2} O}^{KOH} [A] \to_{- H_{2} O}^{H_{2} {SO}_{4}} [B] + K_{2} {SO}_{4}$

The products [A] and [B], respectively are: [2025]

Q 22 :

Preparation of potassium permanganate from ${MnO}_{2}$ involves a two-step process in which the first step is a reaction with $KOH$ and ${KNO}_{3}$ to produce [2025]

Q 24 :

The amphoteric oxide among $V_{2} O_{3}, V_{2} O_{4}$ and $V_{2} O_{5}$ , upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is: [2025]

Q 25 :

The molar mass of the water insoluble product formed from the fusion of chromite ore $({FeCr}_{2} O_{4})$ with ${Na}_{2} {CO}_{3}$ in presence of $O_{2}$ is ______ g ${mol}^{- 1}$ . [2025]

Q 26 :

$K_{2} {Cr}_{2} O_{7}$ paper acidified with dilute $H_{2} {SO}_{4}$ turns green when exposed to [2023]

Q 27 :

Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from [2023]

Q 28 :

${KMnO}_{4}$ oxidises $I^{-}$ in acidic and neutral/faintly alkaline solution, respectively, to [2023]

Q 29 :

When ${Cu}^{2 +}$ ion is treated with $KI$ , a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are: [2023]

Q 30 :

A solution of ${FeCl}_{3}$ , when treated with $K_{4} [Fe {(CN)}_{6}]$ gives a Prussian blue precipitate due to the formation of [2023]

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	List–I		List–II
A.	Bronze	I.	Cu, Ni
B.	Brass	II.	Fe, Cr, Ni, C
C.	UK silver coin	III.	Cu, Zn
D.	Stainless steel	IV.	Cu, Sn

Topic Question Set

Q 1 : In alkaline medium, MnO4- oxidises I- to [2024]

Q 2 : KMnO4 decomposes on heating at 513K to form O2 along with [2024]

Q 3 : The orange colour of K2Cr2O7 and purple colour of KMnO4 is due to [2024]

Q 4 : Alkaline oxidative fusion of MnO2 gives ''A'' which on electrolytic oxidation in alkaline solution produces B. A and B respectively are [2024]

Q 6 : Choose the correct statements from the following A. Mn2O7 is an oil at room temperature B. V2O4 reacts with acid to give VO22+ C. CrO is a basic oxide D. V2O5 does not react with acid Choose the correct answer from the options given below: [2024]

Q 7 : Which of the following compounds show colour due to d-d transition? [2024]

Q 8 : Iron (III) catalyses the reaction between iodide and persulphate ions, in which A. Fe3+ oxidises the iodide ion B. Fe3+ oxidises the persulphate ion C. Fe2+ reduces the iodide ion D. Fe2+ reduces the persulphate ion Choose the most appropriate answer from the options given below: [2024]

Q 10 : Number of moles of H+ ions required by 1 mole of MnO4- to oxidise oxalate ion to CO2 is _____________ . [2024]

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]

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]

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]

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]

Q 18 : The equilibrium Cr2O72-⇌2CrO42- is shifted to the right in: [2024]

Q 19 : The correct set of ions (aqueous solution) with same colour from the following is: [2025]

Q 20 : Match List–I with List–II. [2025] List–I List–II A. Bronze I. Cu, Ni B. Brass II. Fe, Cr, Ni, C C. UK silver coin III. Cu, Zn D. Stainless steel IV. Cu, Sn Choose the correct answer from the options given below:

(3)

Q 21 : Consider the following reactions K2Cr2O7→-H2OKOH[A]→-H2OH2SO4[B]+K2SO4 The products [A] and [B], respectively are: [2025]

Q 22 : Preparation of potassium permanganate from MnO2 involves a two-step process in which the first step is a reaction with KOH and KNO3 to produce [2025]

Q 23 : Which of the following oxidation reactions are carried out by both K2Cr2O7 and KMnO4 in acidic medium? [2025] (A) I-→I2 (B) S2-→S (C) Fe2+→Fe3+ (D) I-→IO3- (E) S2O32-→SO42- Choose the correct answer from the options given below:

Q 24 : The amphoteric oxide among V2O3, V2O4 and V2O5, upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is: [2025]

Q 25 : The molar mass of the water insoluble product formed from the fusion of chromite ore (FeCr2O4) with Na2CO3 in presence of O2 is ______ g mol-1. [2025]

Q 26 : K2Cr2O7 paper acidified with dilute H2SO4 turns green when exposed to [2023]

Q 27 : Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from [2023]

Q 28 : KMnO4 oxidises I- in acidic and neutral/faintly alkaline solution, respectively, to [2023]

Q 29 : When Cu2+ ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are: [2023]

Q 30 : A solution of FeCl3, when treated with K4[Fe(CN)6] gives a Prussian blue precipitate due to the formation of [2023]

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Q 1 :

In alkaline medium, ${MnO}_{4}^{-}$ oxidises $I^{-}$ to [2024]

Q 2 :

${KMnO}_{4}$ decomposes on heating at 513K to form $O_{2}$ along with [2024]

Q 3 :

The orange colour of $K_{2} {Cr}_{2} O_{7}$ and purple colour of ${KMnO}_{4}$ is due to [2024]

Q 4 :

Alkaline oxidative fusion of ${MnO}_{2}$ gives ''A'' which on electrolytic oxidation in alkaline solution produces B. A and B respectively are [2024]

Q 6 :

Choose the correct statements from the following

A. ${Mn}_{2} O_{7}$ is an oil at room temperature
B. $V_{2} O_{4}$ reacts with acid to give ${VO}_{2}^{2 +}$
C. $CrO$ is a basic oxide
D. $V_{2} O_{5}$ does not react with acid

Choose the correct answer from the options given below: [2024]

Q 7 :

Which of the following compounds show colour due to $d - d$ transition? [2024]

Q 10 :

Number of moles of $H^{+}$ ions required by 1 mole of ${MnO}_{4}^{-}$ to oxidise oxalate ion to ${CO}_{2}$ is _____________ . [2024]

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]

Q 18 :

The equilibrium $C r_{2} O_{7}^{2 -} ⇌ 2 C r O_{4}^{2 -}$ is shifted to the right in: [2024]

Q 19 :

The correct set of ions (aqueous solution) with same colour from the following is: [2025]

Q 20 :

Match List–I with List–II. [2025]

List–I List–II

A. Bronze I. Cu, Ni

B. Brass II. Fe, Cr, Ni, C

C. UK silver coin III. Cu, Zn

D. Stainless steel IV. Cu, Sn

Choose the correct answer from the options given below:

Q 21 :

Consider the following reactions

$K_{2} {Cr}_{2} O_{7} \to_{- H_{2} O}^{KOH} [A] \to_{- H_{2} O}^{H_{2} {SO}_{4}} [B] + K_{2} {SO}_{4}$

The products [A] and [B], respectively are: [2025]

Q 22 :

Preparation of potassium permanganate from ${MnO}_{2}$ involves a two-step process in which the first step is a reaction with $KOH$ and ${KNO}_{3}$ to produce [2025]

Q 24 :

The amphoteric oxide among $V_{2} O_{3}, V_{2} O_{4}$ and $V_{2} O_{5}$ , upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is: [2025]

Q 25 :

The molar mass of the water insoluble product formed from the fusion of chromite ore $({FeCr}_{2} O_{4})$ with ${Na}_{2} {CO}_{3}$ in presence of $O_{2}$ is ______ g ${mol}^{- 1}$ . [2025]

Q 26 :

$K_{2} {Cr}_{2} O_{7}$ paper acidified with dilute $H_{2} {SO}_{4}$ turns green when exposed to [2023]

Q 27 :

Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from [2023]

Q 28 :

${KMnO}_{4}$ oxidises $I^{-}$ in acidic and neutral/faintly alkaline solution, respectively, to [2023]

Q 29 :

When ${Cu}^{2 +}$ ion is treated with $KI$ , a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are: [2023]

Q 30 :

A solution of ${FeCl}_{3}$ , when treated with $K_{4} [Fe {(CN)}_{6}]$ gives a Prussian blue precipitate due to the formation of [2023]