The correct decreasing order of spin-only magnetic moment values (BM) of textCu^+$\text{Cu}^+$, textCu^2+$\text{Cu}^{2+}$, textCr^2+$\text{Cr}^{2+}$, and textCr^3+$\text{Cr}^{3+}$ ions is:

### 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$$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$ ### Step 1: Compare with Given Equation Comparing this with the given equation Cr_2O_7^2- + XH^+ + Ye^- rightarrow 2A + ZH_2O$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$: X = 14$X = 14$ Y = 6$Y = 6$ Z = 7$Z = 7$ A = Cr^3+$A = Cr^{3+}$ ### Pattern Recognition In acidic medium, dichromate (Cr_2O_7^2-$Cr_2O_7^{2-}$) always requires 14H^+$14H^+$ to balance 7O$7O$ atoms, forming 7H_2O$7H_2O$. Chromium reduces from +6 to +3 state, taking 6e^-$6e^-$ overall. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements Class 11 Chemistry: Redox Reactions

### Related Formula textElectronic configuration of mathrmEu = [mathrmXe] 4f^7 6s^2$$\text{Electronic configuration of } \mathrm{Eu} = [\mathrm{Xe}] 4f^7 6s^2$$ ### Core Logic The most common and stable oxidation state for lanthanoids is +3$+3$. In the case of Europium: mathrmEu^2+ = [mathrmXe] 4f^7$$\mathrm{Eu}^{2+} = [\mathrm{Xe}] 4f^7$$ This configuration possesses a highly stable half-filled f$f$-subshell. However, because the +3$+3$ state is universally favored by thermodynamics in solution, textEu^2+$\text{Eu}^{2+}$ readily undergoes oxidation to lose one more electron: mathrmEu^2+ rightarrow mathrmEu^3+ + 1e^-$$\mathrm{Eu}^{2+} \rightarrow \mathrm{Eu}^{3+} + 1e^-$$ By releasing an electron to stabilize into the +3$+3$ state, it behaves as a potent reducing agent. ### Step 1: Evaluation Conversely, textCe^4+$\text{Ce}^{4+}$ acts as a powerful oxidizing agent to return to +3$+3$, while textLu^3+$\text{Lu}^{3+}$ and textGd^3+$\text{Gd}^{3+}$ are already perfectly configured at their native stable limits. ### Pattern Recognition Europium(II$II$) has a stable half-filled f^7$f^7$ configuration, yet easily loses an electron to attain the highly stable +3$+3$ 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

### Related Formula textGeneral configuration of Group 12: (n-1)d^10 ns^2$$\text{General configuration of Group 12: } (n-1)d^{10} ns^2$$ ### Core Logic Analyzing each statement based on inorganic chemistry principles: * **Statement A is false**: Because their d$d$-subshell is completely full (d^10$d^{10}$), these elements do not form strong metallic bonds. As a result, they exhibit the *lowest* enthalpy of atomization in their respective periods. * **Statement B is true**: textZn$\text{Zn}$ and textCd$\text{Cd}$ show only a stable +2$+2$ oxidation state, whereas textHg$\text{Hg}$ exhibits variable states forming both +1$+1$ (as textHg_2^2+$\text{Hg}_2^{2+}$) and +2$+2$. * **Statement C is false**: With a fully paired d^10$d^{10}$ subshell, their compounds lack unpaired electrons and are explicitly diamagnetic. * **Statement D is true**: Due to weak metallic bonds, these elements have low melting points and are classified as soft metals. ### Step 1: Selection Verification Statements B and D are true, matching choice (1). ### Pattern Recognition Group 12 metals have a full d^10$d^{10}$ subshell, leading to exceptionally weak metallic bonding, low enthalpies of atomization, and diamagnetic characteristics. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements

### Related Formula Dissolution reaction pathway: textPbCrO_4 + 4textNaOH (hot excess) rightarrow textNa_2[textPb(OH)_4] + textNa_2textCrO_4$$\text{PbCrO}_4 + 4\text{NaOH (hot excess)} \rightarrow \text{Na}_2[\text{Pb(OH)}_4] + \text{Na}_2\text{CrO}_4$$ ### Core Logic The reaction yields sodium tetrahydroxoplumbate(II), [textPb(OH)_4]^2-$[\text{Pb(OH)}_4]^{2-}$. The charge of the complex species is -2$-2$ (dianionic), and it binds 4 hydroxo coordination ligands, matching a coordination number of four. ### Chapter Mix Class 12 Chemistry: d-and f-Block Elements Class 12 Chemistry: Coordination Compounds

### Step 1: Production of Reddish Fumes 4textNaCl + textK_2textCr_2textO_7 + 6textH_2textSO_4 rightarrow 2textCrO_2textCl_2uparrow + 2textKHSO_4 + 4textNaHSO_4 + 3textH_2textO$$4\text{NaCl} + \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{H}_2\text{SO}_4 \rightarrow 2\text{CrO}_2\text{Cl}_2\uparrow + 2\text{KHSO}_4 + 4\text{NaHSO}_4 + 3\text{H}_2\text{O}$$ Reddish brown vapors (B) are chromyl chloride (CrO_2Cl_2$CrO_2Cl_2$). ### Step 2: Conversion to Yellow Solution textCrO_2textCl_2 + 4textNaOH rightarrow textNa_2textCrO_4 + 2textNaCl + 2textH_2textO$$\text{CrO}_2\text{Cl}_2 + 4\text{NaOH} \rightarrow \text{Na}_2\text{CrO}_4 + 2\text{NaCl} + 2\text{H}_2\text{O}$$ Yellow solution (C) corresponds to sodium chromate (Na_2CrO_4$Na_2CrO_4$). ### Pattern Recognition Chloride detection signature: textCl^- rightarrow textCrO_2textCl_2text (red-brown) rightarrow textNa_2textCrO_4text (yellow chromate).$\text{Cl}^- \rightarrow \text{CrO}_2\text{Cl}_2\text{ (red-brown)} \rightarrow \text{Na}_2\text{CrO}_4\text{ (yellow chromate)}.$ ### Chapter Mix Class 12 Chemistry: d-and f-Block Elements