### Core Logic Let's resolve both components step by step: 1. **Finding X:** In an acidic medium, the permanganate ion (KMnO_4$KMnO_4$, where Mn$Mn$ is in the +7$+7$ state) is reduced to the divalent manganese cation (Mn^2+$Mn^{2+}$, state +2$+2$): X = 7 - 2 = 5$X = 7 - 2 = 5$ 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)]$[Fe(OH)_2(CH_3COO)]$. In this complex, Iron retains its +3$+3$ oxidation state: Fe^3+ implies [Ar] 3d^5 4s^0 implies textNumber of d-electrons (Y) = 5$Fe^{3+} \implies [Ar] 3d^5 4s^0 \implies \text{Number of d-electrons } (Y) = 5$ Summing the values yields: X + Y = 5 + 5 = 10$X + Y = 5 + 5 = 10$ ### 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$+3$ (d^5$d^5$) core configuration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements Inorganic Qualitative Analysis

### Related Formula mu = sqrtn(n+2) text B.M.$\mu = \sqrt{n(n+2)} \text{ B.M.}$ where n$n$ represents the number of unpaired electrons. ### Core Logic Let's calculate the number of unpaired d-electrons (n$n$) and the resulting spin-only magnetic moment for each transition metal ion: * (A) mathrmTi^3+$\mathrm{Ti}^{3+}$: Electronic configuration = [Ar] 3d^1 ightarrow n = 1$= [Ar] 3d^1 ightarrow n = 1$ mu = sqrt1(1+2) = sqrt3 approx 1.73text B.M. ightarrow text(III)$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\text{ B.M.} ightarrow \text{(III)}$ * (B) mathrmV^2+$\mathrm{V}^{2+}$: Electronic configuration = [Ar] 3d^3 ightarrow n = 3$= [Ar] 3d^3 ightarrow n = 3$ mu = sqrt3(3+2) = sqrt15 approx 3.87text B.M. ightarrow text(I)$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\text{ B.M.} ightarrow \text{(I)}$ * (C) mathrmNi^2+$\mathrm{Ni}^{2+}$: Electronic configuration = [Ar] 3d^8$= [Ar] 3d^8$. The 3d$3d$ subshell has 3 paired orbitals and 2 unpaired orbitals ightarrow n = 2$ ightarrow n = 2$ mu = sqrt2(2+2) = sqrt8 approx 2.84text B.M. ightarrow text(IV)$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84\text{ B.M.} ightarrow \text{(IV)}$ * (D) mathrmSc^3+$\mathrm{Sc}^{3+}$: Electronic configuration = [Ar] 3d^0 ightarrow n = 0$= [Ar] 3d^0 ightarrow n = 0$ mu = 0.00text B.M. ightarrow text(II)$\mu = 0.00\text{ B.M.} ightarrow \text{(II)}$ 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$n$). For example, a value of 3.87text B.M.$3.87\text{ B.M.}$ means there are exactly 3$3$ unpaired electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

### Core Logic The most common and chemically robust oxidation state for lanthanoid elements is +3$+3$. Consequently, ions existing in unstable +4$+4$ oxidation states exhibit a pronounced thermodynamic driving force to capture electrons and revert to the +3$+3$ form. Among the options, Tb^4+$Tb^{4+}$ acts as a potent oxidizing agent due to this stability drive. ### Pattern Recognition Ln^4+$Ln^{4+}$ forms naturally act as electron grabbers to sink back into the thermodynamic sweet spot of +3$+3$ states. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements

### Related Formula 2MnO_2 + 4KOH + O_2 xrightarrowKNO_3 2K_2MnO_4 + 2H_2O$2MnO_2 + 4KOH + O_2 \xrightarrow{KNO_3} 2K_2MnO_4 + 2H_2O$ ### Core Logic The standard preparation of potassium permanganate begins with the oxidative fusion of pyrolusite ore (MnO_2$MnO_2$). Fusing the solid reactant directly along an alkaline base payload (KOH$KOH$) combined explicitly with an oxidizing carrier (KNO_3$KNO_3$) yields the intermediate green product, **potassium manganate** (K_2MnO_4$K_2MnO_4$). ### Pattern Recognition Step 1 yields the +6$+6$ green compound (K_2MnO_4$K_2MnO_4$); the subsequent Step 2 steps oxidize this intermediate to synthesize the target deep purple +7$+7$ agent (KMnO_4$KMnO_4$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements

### Related Formula textMelting Point trends in 3d, 4d, and 5d series transition metals$\text{Melting Point trends in 3d, 4d, and 5d series transition metals}$ ### Core Logic According to the official NCERT transition elements structural profile trends : * In the 3d$3d$ series, Manganese (mathrmMn$\mathrm{Mn}$) has an unexpectedly low melting point relative to Iron (mathrmFe$\mathrm{Fe}$) due to its stable half-filled d^5$d^5$ configuration, which restricts metallic bonding options ightarrow mathrmMn < mathrmFe$ ightarrow \mathrm{Mn} < \mathrm{Fe}$ . * In the 4d$4d$ series, Technetium (mathrmTc$\mathrm{Tc}$) similarly exhibits a lower melting point drop compared to Ruthenium (mathrmRu$\mathrm{Ru}$) ightarrow mathrmTc < mathrmRu$ ightarrow \mathrm{Tc} < \mathrm{Ru}$ . * In the 5d$5d$ series, Rhenium (mathrmRe$\mathrm{Re}$) displays a higher melting point relative to Osmium (mathrmOs$\mathrm{Os}$) due to optimal effective nuclear charge constraints ightarrow mathrmOs < mathrmRe$ ightarrow \mathrm{Os} < \mathrm{Re}$ . Combining these standard physical periodic entries yields option (3). ### Pattern Recognition The half-filled d^5$d^5$ configuration introduces localized stability spikes that lower metallic bond cohesion dramatically for mathrmMn$\mathrm{Mn}$ and mathrmTc$\mathrm{Tc}$, 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