Among, mathrmSc, mathrmMn, mathrmCo and mathrmCu, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is ________ BM (in nearest integer).

Numerical Answer Type:
Enter a numerical value Answer: 4 to 4 +4 marks

Solution & Explanation

### Related Formula Spin-only magnetic moment (mu) is given by: mu = sqrtn(n+2)mathrm~BM where n is the number of unpaired d-electrons. ### Core Logic Enthalpies of atomization of the given 3d transition elements (in mathrmkJ/mol): - Scandium (mathrmSc): 326 - Manganese (mathrmMn): 281 - Cobalt (mathrmCo): 425 - Copper (mathrmCu): 339 Thus, Cobalt (mathrmCo) has the highest enthalpy of atomization. ### Step 1: Determine unpaired electrons in mathrmCo^2+ Electronic configuration of Cobalt (Z=27): mathrmCo: [mathrmAr] 3d^7 4s^2 For divalent Cobalt ion (mathrmCo^2+): mathrmCo^2+: [mathrmAr] 3d^7 In the d-subshell (five orbitals): - Three orbitals are paired, and three are unpaired (n=3). ### Step 2: Calculate spin-only magnetic moment mu = sqrt3(3+2) = sqrt15 approx 3.87mathrm~BM Rounding to the nearest integer gives 4. ### Pattern Recognition Enthalpy of atomization generally peaks near the middle of transition series due to maximum metallic bonding. However, mathrmMn (3d^5 4s^2) is an anomaly with an exceptionally low value (281mathrm~kJ/mol) due to its highly stable half-filled d^5 subshell configuration which reduces electron delocalization in metallic bonding. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements

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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
  • C. mathrmKMnO_4
  • D. mathrmK_2mathrmMnO_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
  • B. +7
  • C. +5
  • D. +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
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

### Formulas Used Melting point trends in 3d, 4d, and 5d series transition metals depend on the extent of metallic bonding and d-electron participation. ### Core Logic According to NCERT transition element periodic trends: * **3d Series (mathrmMn vs mathrmFe)**: Manganese (mathrmMn, 3d^5 4s^2) has an abnormally low melting point compared to Iron (mathrmFe, 3d^6 4s^2) because its stable, half-filled d^5 configuration holds d-electrons more tightly, reducing their participation in metallic bonding rightarrow mathbfmathrmMn < mathrmFe. * **4d Series (mathrmTc vs mathrmRu)**: Technetium (mathrmTc, 4d^5 5s^2) similarly shows a dip in melting point compared to Ruthenium (mathrmRu, 4d^7 5s^1) due to the stable 4d^5 configuration rightarrow mathbfmathrmTc < mathrmRu. * **5d Series (mathrmRe vs mathrmOs)**: Rhenium (mathrmRe, 5d^5 6s^2) has optimal interatomic interaction and a higher melting point than Osmium (mathrmOs, 5d^6 6s^2) rightarrow mathbfmathrmOs < mathrmRe. Combining these trends yields: **mathrmMn < mathrmFe, mathrmTc < mathrmRu, and mathrmOs < mathrmRe** ### Pattern Recognition Stable half-filled d^5 configurations in 3d (mathrmMn) and 4d (mathrmTc) restrict d-electron delocalization, creating characteristic dips in melting point curves compared to adjacent metals. **Correct Option:** **(C)**
Q 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 rightarrow 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 mathrm~g/mol ### Pattern Recognition Transition metal oxides in high oxidation states with minimal ionic breakdown parameters reliably act as insoluble precipitates in water. ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements

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