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 $K_2MnO_4$ $K_2MnO_4$ 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:

Q: 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 K2

CrO42- (chromate ion) is tetrahedral, not square planar. Statement A is incorrect. 2Na2CrO4 + 2H^+ arrow Na2Cr2O7 + 2Na^+ + H2O. Dichromates are indeed prepared from chromates. Statement B is correct. The green manganate ion (MnO42-) has manganese in the +6 oxidation state (3d1). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect. Dark green coloured K2MnO4 undergoes disproportionation in neutral or acidic media to yield permanganate (MnO4^-) and

A. $textB, C, D only$ $\text{B, C, D only}$
B. $textA, D, E only$ $\text{A, D, E only}$
C. $textA, B, C only$ $\text{A, B, C only}$
D. $textB, D, E only$ $\text{B, D, E only}$

Solution & Explanation

### Step 1: Statement A Analysis

CrO_4^2-

(chromate ion) is tetrahedral, not square planar. Statement A is incorrect. ### Step 2: Statement B Analysis

2Na_2CrO_4 + 2H^+ rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O

. Dichromates are indeed prepared from chromates. Statement B is correct. ### Step 3: Statement C Analysis The green manganate ion (

MnO_4^2-

) has manganese in the +6 oxidation state (

3d^1

). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect. ### Step 4: Statement D Analysis Dark green coloured

K_2MnO_4

undergoes disproportionation in neutral or acidic media to yield permanganate (

MnO_4^-

) and manganese dioxide (

MnO_2

). Statement D is correct. ### Step 5: Statement E Analysis Fajans' rule dictates that as the oxidation state increases, polarizing power increases, leading to a decrease in ionic character (increase in covalent character). Statement E is correct. ### Final Conclusion The correct statements are B, D, and E. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

Reference Study Guides

The D And F Block Elements

More The d- and f-Block Elements Previous-Year Questions — Page 3

Q47 jee_main_2025_04_april_morning Chemical Properties of KMnO4

KMnO_4

acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of

X + Y

is ______.

Numerical Answer. Answer: 10 to 10

Solution

### Core Logic Let's resolve both components step by step: 1. **Finding X:** In an acidic medium, the permanganate ion (

KMnO_4

, where

Mn

is in the

+7

state) is reduced to the divalent manganese cation (

Mn^2+

, state

+2

X = 7 - 2 = 5

2. **Finding Y:** During qualitative salt analysis, the acetate ion reacts with neutral ferric chloride to produce a characteristic blood-red coordination solution. Boiling this solution throws down a **brown-red precipitate** of basic ferric acetate,

[Fe(OH)_2(CH_3COO)]

. In this complex, Iron retains its

+3

oxidation state:

Fe^3+ implies [Ar] 3d^5 4s^0 implies textNumber of d-electrons (Y) = 5

Summing the values yields:

X + Y = 5 + 5 = 10

### Pattern Recognition This problem elegantly links standard redox transitions with qualitative inorganic salt tests. Remember that throughout the basic ferric acetate precipitation test, Iron remains steadily in its ferric

+3

(

d^5

) core configuration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements Inorganic Qualitative Analysis

Q35 jee_main_2025_24_jan_evening Magnetic Properties of Transition Metals

Match List-I with List-II.

List-I (Transition metal ion)	List-II (Spin only magnetic moment (B.M.))
(A) $mathrmTi^3+$ $\mathrm{Ti}^{3+}$	(I) $3.87$ $3.87$
(B) $mathrmV^2+$ $\mathrm{V}^{2+}$	(II) $0.00$ $0.00$
(C) $mathrmNi^2+$ $\mathrm{Ni}^{2+}$	(III) $1.73$ $1.73$
(D) $mathrmSc^3+$ $\mathrm{Sc}^{3+}$	(IV) $2.84$ $2.84$

Choose the correct answer from the options given below :

A. \text{(A)-(III), (B)-(I), (C)-(II), (D)-(IV)}
B. \text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}
C. \text{(A)-(IV), (B)-(II), (C)-(III), (D)-(I)}
D. \text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}

Solution

### Related Formula

mu = sqrtn(n+2) text B.M.

where

n

represents the number of unpaired electrons. ### Core Logic Let's calculate the number of unpaired d-electrons (

n

) and the resulting spin-only magnetic moment for each transition metal ion: * (A)

mathrmTi^3+

: Electronic configuration

= [Ar] 3d^1 ightarrow n = 1

mu = sqrt1(1+2) = sqrt3 approx 1.73text B.M. ightarrow text(III)

* (B)

mathrmV^2+

: Electronic configuration

= [Ar] 3d^3 ightarrow n = 3

mu = sqrt3(3+2) = sqrt15 approx 3.87text B.M. ightarrow text(I)

* (C)

mathrmNi^2+

: Electronic configuration

= [Ar] 3d^8

. The

3d

subshell has 3 paired orbitals and 2 unpaired orbitals

ightarrow n = 2

mu = sqrt2(2+2) = sqrt8 approx 2.84text B.M. ightarrow text(IV)

* (D)

mathrmSc^3+

: Electronic configuration

= [Ar] 3d^0 ightarrow n = 0

mu = 0.00text B.M. ightarrow text(II)

Matching these values yields the sequence: (A)-(III), (B)-(I), (C)-(IV), (D)-(II). ### Pattern Recognition Shortcut: The digit before the decimal point in a spin-only magnetic moment matches the number of unpaired electrons (

n

). For example, a value of

3.87text B.M.

means there are exactly

3

unpaired electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

Q31 jee_main_2025_24_jan_morning Lanthanoids Oxidation States

Which of the following ions is the strongest oxidizing agent? [Atomic Number of

Ce=58

Eu=63

Tb=65,

Lu=71

]

A. $Lu^3+$ $Lu^{3+}$
B. $Eu^2+$ $Eu^{2+}$
C. $Tb^4+$ $Tb^{4+}$
D. $Ce^3+$ $Ce^{3+}$

Solution

### Core Logic The most common and chemically robust oxidation state for lanthanoid elements is

+3

. Consequently, ions existing in unstable

+4

oxidation states exhibit a pronounced thermodynamic driving force to capture electrons and revert to the

+3

form. Among the options,

Tb^4+

acts as a potent oxidizing agent due to this stability drive. ### Pattern Recognition

Ln^4+

forms naturally act as electron grabbers to sink back into the thermodynamic sweet spot of

+3

states. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements

Q37 jee_main_2025_24_jan_morning Preparation and Properties of Potassium Permanganate

Preparation of potassium permanganate from

mathrmMnO_2

involves two step process in which the

1^textst

step is a reaction with KOH and

mathrmKNO_3

to produce

A. $mathrmK_4[mathrmMn(mathrmOH)_6]$
B. $mathrmK_3mathrmMnO_4$ $\mathrm{K}_3\mathrm{MnO}_4$
C. $mathrmKMnO_4$ $\mathrm{KMnO}_4$
D. $mathrmK_2mathrmMnO_4$ $\mathrm{K}_2\mathrm{MnO}_4$

Solution

### Related Formula

2MnO_2 + 4KOH + O_2 xrightarrowKNO_3 2K_2MnO_4 + 2H_2O

### Core Logic The standard preparation of potassium permanganate begins with the oxidative fusion of pyrolusite ore (

MnO_2

). Fusing the solid reactant directly along an alkaline base payload (

KOH

) combined explicitly with an oxidizing carrier (

KNO_3

) yields the intermediate green product, **potassium manganate** (

K_2MnO_4

). ### Pattern Recognition Step 1 yields the

+6

green compound (

K_2MnO_4

); the subsequent Step 2 steps oxidize this intermediate to synthesize the target deep purple

+7

agent (

KMnO_4

). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements

Q27 jee_main_2025_28_jan_evening Oxides and Oxoanions of Transition Metals

The amphoteric oxide among

V_2O_3

V_2O_4

and

V_2O_5

upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :

A. $+3$ $+3$
B. $+7$ $+7$
C. $+5$ $+5$
D. $+4$ $+4$

Solution

### Related Formula Oxidation state equation for an oxoanion

VO_4^3-

x + 4(-2) = -3

### Core Logic Among the given oxides of Vanadium: -

V_2O_3

is basic. -

V_2O_4

is less basic / amphoteric. -

V_2O_5

is predominantly amphoteric (reacts with both acids and alkalies). When

V_2O_5

reacts with an alkali, it forms the orthovanadate ion (

VO_4^3-

). ### Step 1: Finding the Oxidation State In

VO_4^3-

ion:

x - 8 = -3 implies x = +5

Thus, the oxidation state of Vanadium in the resulting oxide anion is

+5

. ### Pattern Recognition As the oxidation state of a transition metal increases, its oxide shifts from basic to amphoteric to acidic.

V_2O_5

has the highest oxidation state (

+5

) here and dissolves in alkali to retain its

+5

oxidation state in

VO_4^3-

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements

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