Given below are two statements: Statement I: The halogen that makes longest bond with hydrogen in HX, has the smallest covalent radius in its group. Statement II: A group 15 element's hydride EH_3$EH_{3}$ has the lowest boiling point among corresponding hydrides of other group 15 elements. The maximum covalency of that element E is 4. In the light of the above statements, choose the correct answer from the options given below.

### Core Logic The precipitation and complex formation reactions of Lead are: mathrmPbCl_2 + mathrmK_2mathrmCrO_4 rightarrow mathrmPbCrO_4 + 2mathrmKCl quad (textHot solution) quad textso, A is mathrmPbCrO_4$$\mathrm{PbCl}_2 + \mathrm{K}_2\mathrm{CrO}_4 \rightarrow \mathrm{PbCrO}_4 + 2\mathrm{KCl} \quad (\text{Hot solution}) \quad \text{so, A is } \mathrm{PbCrO}_4$$ mathrmPbCrO_4 + 4mathrmNaOH \ (excess) rightarrow mathrmNa_2[mathrmPb(OH)_4] + mathrmNa_2mathrmCrO_4 quad textso, B is mathrmNa_2[mathrmPb(OH)_4]$$\mathrm{PbCrO}_4 + 4\mathrm{NaOH \ (excess)} \rightarrow \mathrm{Na}_2[\mathrm{Pb(OH)}_4] + \mathrm{Na}_2\mathrm{CrO}_4 \quad \text{so, B is } \mathrm{Na}_2[\mathrm{Pb(OH)}_4]$$ mathrmPbSO_4 + 4mathrmCH_3mathrmCOONH_4 rightarrow (mathrmNH_4)_2 [mathrmPb(CH_3COO)_4] + (mathrmNH_4)_2mathrmSO_4 quad textso, X is (mathrmNH_4)_2[mathrmPb(CH_3COO)_4]$$\mathrm{PbSO}_4 + 4\mathrm{CH}_3\mathrm{COONH}_4 \rightarrow (\mathrm{NH}_4)_2 [\mathrm{Pb(CH_3COO)}_4] + (\mathrm{NH}_4)_2\mathrm{SO}_4 \quad \text{so, X is } (\mathrm{NH}_4)_2[\mathrm{Pb(CH_3COO)}_4]$$ ### Pattern Recognition Lead forms a yellow precipitate of lead chromate (A), which is amphoteric and dissolves in excess NaOH to form soluble plumbate(II) complex (B). It also forms a stable soluble complex with ammonium acetate (X). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: p-Block Elements Class 12 Chemistry: d and f Block Elements

### Core Logic Evaluating Statement I: - SiO_2$SiO_2$, CO_2$CO_2$, GeO$GeO$, GeO_2$GeO_2$ are acidic in nature. - SnO$SnO$, SnO_2$SnO_2$, PbO$PbO$, PbO_2$PbO_2$ are amphoteric in nature. Therefore, the pairs [SnO, SnO_2]$[SnO, SnO_2]$ and [PbO, PbO_2]$[PbO, PbO_2]$ contain oxides that are both amphoteric. Number of such pairs = 2. Statement I is True. Evaluating Statement II: - BF_3$BF_3$ has 6 electrons in the outermost shell of the central Boron atom. It is electron-deficient and acts as a Lewis acid. - It accepts a lone pair from Lewis bases like NH_3$NH_3$ to form an adduct. - In BF_3$BF_3$, Boron is sp^2$sp^2$ hybridized, resulting in a trigonal planar geometry. Statement II is True. ### Step 1: Conclusion Both statements are factually correct. ### Pattern Recognition Oxides of heavier Group 14 elements (Sn, Pb) are typically amphoteric in both their +2 and +4 oxidation states. BF_3$BF_3$ is the quintessential Lewis acid. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: p-Block Elements Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic This is the classic Chromyl Chloride test: 4textNaCl + textK_2textCr_2textO_7 + 6textH_2textSO_4 longrightarrow 2textKHSO_4 + 2textCrO_2textCl_2 + 4textNaHSO_4 + 3textH_2textO$$4\text{NaCl} + \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{H}_2\text{SO}_4 \longrightarrow 2\text{KHSO}_4 + 2\text{CrO}_2\text{Cl}_2 + 4\text{NaHSO}_4 + 3\text{H}_2\text{O}$$ In chromyl chloride (textCrO_2textCl_2$\text{CrO}_2\text{Cl}_2$), chromium is in the +6$+6$ oxidation state. ### Step 1: Final Conclusion The gas is textCrO_2textCl_2$\text{CrO}_2\text{Cl}_2$ and the oxidation state of Cr is +6$+6$, corresponding to option (2). ### Pattern Recognition Sees: qualitative analysis test for chloride ions (chromyl chloride test). Trap: Confusing oxidation state of chromium in dichromate versus chromyl chloride. ### Chapter Mix Class 12 Chemistry: p-Block Elements

### Related Formula NH_3 + K_2HgI_4 + KOH rightarrow HgOcdot Hg(NH_2)I downarrow + KI + H_2O$$NH_3 + K_2HgI_4 + KOH \rightarrow HgO\cdot Hg(NH_2)I \downarrow + KI + H_2O$$ ### Core Logic The reagent K_2HgI_4$K_2HgI_4$ in alkaline medium is Nessler's reagent. It gives a brown precipitate (iodide of Millon's base) with ammonia (NH_3$NH_3$). Therefore, the compound EH_3$EH_3$ is NH_3$NH_3$, and the element (E) is Nitrogen (N). The binary cation is the ammonium ion (NH_4^+)$(NH_4^+)$. We need to find the first ionization enthalpy of Nitrogen among the first elements of groups 13 (B), 14 (C), 15 (N), and 16 (O). Due to its stable half-filled 2p^3$2p^3$ configuration, Nitrogen has a remarkably high first ionization energy, higher than Oxygen. The order is: B < C < O < N$B < C < O < N$. Among the given values (801, 1086, 1312, 1402), 1402 kJtext mol^-1$kJ\text{ mol}^{-1}$ is the highest and corresponds to Nitrogen. ### Step 1: Final Conclusion Element E is Nitrogen, and its correct first ionization enthalpy is 1402 kJtext mol^-1$kJ\text{ mol}^{-1}$. ### Pattern Recognition Always remember the ionization energy exception across period 2: N > O and Be > B due to stable half-filled and fully-filled orbitals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: p-Block Elements