Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

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 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:

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 4

Q46 jee_main_2025_28_jan_evening Magnetic Properties and Oxidation States
The spin only magnetic moment (mu) value (B.M.) of the compound with strongest oxidising power among Mn_2O_3, TiO and VO is ______ B.M. (Nearest integer).
Numerical Answer. Answer: 5 to 5

Solution

### Related Formula Spin-only magnetic moment expression: mu = sqrtn(n+2)mathrm\ B.M. ### Core Logic Evaluating the oxidation states and stability profiles: - In TiO: Ti^2+ - In VO: V^2+ - In Mn_2O_3: Mn^3+ Mn^3+ possesses a very high reduction potential (E^circ_Mn^3+/Mn^2+ = +1.57mathrm\ V), making it an exceptionally strong oxidizing agent because it easily gains an electron to form stable Mn^2+ (d^5 configuration). ### Step 1: Calculate the Magnetic Moment of Mn(III) Electronic configuration of Mn^3+: Mn^3+ = [Ar]3d^4 implies n = 4text unpaired electrons Calculating the spin-only magnetic moment: mu = sqrt4(4+2) = sqrt24 approx 4.89mathrm\ B.M. ### Step 2: Rounding to Nearest Integer Rounding 4.89mathrm\ B.M. to the nearest integer gives 5. ### Pattern Recognition High reduction potentials are strongly tied to manganese in its +3 oxidation state. To quickly estimate magnetic moments, remember that a system with n unpaired electrons always results in a value of 'n.textsomething' B.M. Thus, 4 unpaired electrons rightarrow 4.89mathrm\ B.M., which rounds up to 5. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements
Q jee_main_2025_29_jan_morning Melting Points of Transition Elements
The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is :
  • A. mathrmFe < mathrmMn , mathrmRu < mathrmTc and mathrmRe < mathrmOs
  • B. mathrmMn < mathrmFe, mathrmTc < mathrmRu and mathrmRe < mathrmOs
  • C. mathrmMn < mathrmFe, mathrmTc < mathrmRu and mathrmOs < mathrmRe
  • D. mathrmFe < mathrmMn , mathrmRu < mathrmTc and mathrmOs < mathrmRe

Solution

### Related Formula textMelting Point trends in 3d, 4d, and 5d series transition metals ### Core Logic According to the official NCERT transition elements structural profile trends : * In the 3d series, Manganese (mathrmMn) has an unexpectedly low melting point relative to Iron (mathrmFe) due to its stable half-filled d^5 configuration, which restricts metallic bonding options ightarrow mathrmMn < mathrmFe . * In the 4d series, Technetium (mathrmTc) similarly exhibits a lower melting point drop compared to Ruthenium (mathrmRu) ightarrow mathrmTc < mathrmRu . * In the 5d series, Rhenium (mathrmRe) displays a higher melting point relative to Osmium (mathrmOs) due to optimal effective nuclear charge constraints ightarrow mathrmOs < mathrmRe . Combining these standard physical periodic entries yields option (3). ### Pattern Recognition The half-filled d^5 configuration introduces localized stability spikes that lower metallic bond cohesion dramatically for mathrmMn and mathrmTc, creating distinct periodic deep dips in their melting point curves. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements
Q47 jee_main_2025_29_jan_morning Preparation and Properties of Potassium Dichromate
The molar mass of the water insoluble product formed from the fusion of chromite ore mathrm(FeCr_2O_4) with mathrmNa_2mathrmCO_3 in presence of mathrmO_2 is ________ mathrmg \, mol^-1.
Numerical Answer. Answer: 160 to 160

Solution

### Related Formula textBalanced fusion reaction process description ### Core Logic Write the balanced chemical equation for the industrial preparation stage of chromate salts : 4mathrmFeCr_2O_4 + 8mathrmNa_2CO_3 + 7mathrmO_2 ightarrow 8mathrmNa_2CrO_4 + 2mathrmFe_2O_3 + 8mathrmCO_2 Evaluating the solubilities of the products: * mathrmNa_2CrO_4 is highly soluble in water. * mathrmFe_2O_3 (Iron(III) oxide) is water-insoluble . Molar Mass of mathrmFe_2O_3 : M = (2 cdot 55.85) + (3 cdot 16.0) simeq (2 cdot 56) + (3 cdot 16) = 112 + 48 = 160 \, mathrmg/mol ### Pattern Recognition Transition metal oxides in high oxidation states with minimal ionic breakdown parameters reliably act as insoluble precipitates in water. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements
Q63 jee_main_2024_01_february_morning Oxidising Properties
In acidic medium, K_2Cr_2O_7 shows oxidising action as represented in the half reaction Cr_2O_7^2- + XH^+ + Ye^- rightarrow 2A + ZH_2O X, Y, Z and A are respectively are:
  • A. 8, 6, 4 text and Cr_2O_3
  • B. 14, 7, 6 text and Cr^3+
  • C. 8, 4, 6 text and Cr_2O_3
  • D. 14, 6, 7 text and Cr^3+

Solution

### Core Logic The balanced half-reaction for the dichromate ion acting as an oxidising agent in an acidic medium is: Cr_2O_7^2- + 14H^+ + 6e^- rightarrow 2Cr^3+ + 7H_2O ### Step 1: Compare with Given Equation Comparing this with the given equation Cr_2O_7^2- + XH^+ + Ye^- rightarrow 2A + ZH_2O: X = 14 Y = 6 Z = 7 A = Cr^3+ ### Pattern Recognition In acidic medium, dichromate (Cr_2O_7^2-) always requires 14H^+ to balance 7O atoms, forming 7H_2O. Chromium reduces from +6 to +3 state, taking 6e^- overall. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements Class 11 Chemistry: Redox Reactions
Q73 jee_main_2024_29_january_evening Lanthanoid Oxidation States
Which of the following acts as a strong reducing agent? (Atomic number : Ce = 58, Eu = 63, Gd = 64, Lu = 71)
  • A. mathrmLu^3+
  • B. mathrmGd^3+
  • C. mathrmEu^2+
  • D. mathrmCe^4+

Solution

### Related Formula textElectronic configuration of mathrmEu = [mathrmXe] 4f^7 6s^2 ### Core Logic The most common and stable oxidation state for lanthanoids is +3. In the case of Europium: mathrmEu^2+ = [mathrmXe] 4f^7 This configuration possesses a highly stable half-filled f-subshell. However, because the +3 state is universally favored by thermodynamics in solution, textEu^2+ readily undergoes oxidation to lose one more electron: mathrmEu^2+ rightarrow mathrmEu^3+ + 1e^- By releasing an electron to stabilize into the +3 state, it behaves as a potent reducing agent. ### Step 1: Evaluation Conversely, textCe^4+ acts as a powerful oxidizing agent to return to +3, while textLu^3+ and textGd^3+ are already perfectly configured at their native stable limits. ### Pattern Recognition Europium(II) has a stable half-filled f^7 configuration, yet easily loses an electron to attain the highly stable +3 state typical of lanthanoids, making it a strong reducing agent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements

More The d- and f-Block Elements Questions — jee_main_2024_31_jan_morning

Practice all The d- and f-Block Elements previous-year questions →

YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Under Development
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics
Latest from the Blog
View all →

Loading articles...