### 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

### Core Logic The noble gas configuration of Xenon (Z=54$Z=54$) provides the primary core layout. For Neodymium (Z=60$Z=60$), the 6 remaining valence electrons distribute into the inner 4textf$4\text{f}$ orbital subshell rather than filling the 5textd$5\text{d}$ subshell due to shielding effects. This results in an absolute atomic ground state electronic configuration of text[Xe] 4textf^4 6texts^2$\text{[Xe]} 4\text{f}^4 6\text{s}^2$. ### Pattern Recognition Lanthanide filling sequences generally bypass 5d$5d$ progression except for specific exceptions (La, Gd, Lu). ### Chapter Mix Class 12 Chemistry: d-and f-Block Elements

### Core Logic The reaction sequence for the chromyl chloride test is: Cl^- + K_2Cr_2O_7 + H_2SO_4 rightarrow CrO_2Cl_2$Cl^- + K_2Cr_2O_7 + H_2SO_4 \rightarrow CrO_2Cl_2$ The chromyl chloride gas is then passed through a basic medium (like NaOH$NaOH$) to form a yellow solution of chromate ions: CrO_2Cl_2 xrightarrowtextBasic medium CrO_4^2- + Cl^-$CrO_2Cl_2 \xrightarrow{\text{Basic medium}} CrO_4^{2-} + Cl^-$ Acidification of the yellow CrO_4^2-$CrO_4^{2-}$ solution followed by the addition of H_2O_2$H_2O_2$ and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO_5$CrO_5$). CrO_4^2- xrightarrow[textyellow solution, 1. textAcidification CrO_5 text (blue compound)$CrO_4^{2-} \xrightarrow[\text{yellow solution}, {1. \text{Acidification}} CrO_5 \text{ (blue compound)}$ ### Step 1: Oxidation State Calculation

The structure of chromium pentoxide (CrO_5$CrO_5$) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O^2-$O^{2-}$) and four peroxide oxygens (O_2^2-$O_2^{2-}$). Therefore, there are 2 peroxo linkages. Let the oxidation state of Chromium be x$x$. x + 1(-2) + 4(-1) = 0$x + 1(-2) + 4(-1) = 0$ x - 2 - 4 = 0$x - 2 - 4 = 0$ x = +6$x = +6$ Thus, the oxidation state of Cr in CrO_5$CrO_5$ is +6$+6$. ### Pattern Recognition A classic oxidation state trap. Calculating simply via formula CrO_5$CrO_5$ yields x - 10 = 0 implies x = +10$x - 10 = 0 \implies x = +10$, which is impossible for Chromium (max +6). Whenever calculation exceeds the maximum group valency, peroxide bonds are present. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements Class 11 Chemistry: Redox Reactions