On heating a mixture of common salt and textK_2textCr_2textO_7$\text{K}_2\text{Cr}_2\text{O}_7$ in equal amount along with concentrated textH_2textSO_4$\text{H}_2\text{SO}_4$ in a test tube, a gas is evolved. Formula of the gas evolved and oxidation state of the central metal atom in the gas respectively are: (1) textCrO_2textCl_2$\text{CrO}_2\text{Cl}_2$ and +5$+5$ (2) textCrO_2textCl_2$\text{CrO}_2\text{Cl}_2$ and +6$+6$ (3) textCr_2textO_2textCl_2$\text{Cr}_2\text{O}_2\text{Cl}_2$ and +6$+6$ (4) textCr_2textO_2textCl_2$\text{Cr}_2\text{O}_2\text{Cl}_2$ and +3$+3$

### 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 textBond Strength propto frac1textSize difference$$\text{Bond Strength} \propto \frac{1}{\text{Size difference}}$$ textAcidic Strength propto frac1textM-H Bond Dissociation Energy$$\text{Acidic Strength} \propto \frac{1}{\text{M-H Bond Dissociation Energy}}$$ ### Core Logic Let's analyze the properties of Tellurium compounds: 1. **Tellurium Dioxide** (mathrmTeO_2$\mathrm{TeO}_2$): - Due to the **inert pair effect**, the +6$+6$ oxidation state of Tellurium is less stable, whereas its +4$+4$ state is relatively stable. However, in comparison to sulphur dioxide (which is a strong reducing agent), mathrmTeO_2$\mathrm{TeO}_2$ is oxidising because the lower oxidation states (like element Tellurium or +2$+2$) are chemically accessible. Thus, mathrmTeO_2$\mathrm{TeO}_2$ acts as an **oxidising agent**. 2. **Tellurium Hydride** (mathrmTeH_2$\mathrm{TeH}_2$): - Tellurium is a very large atom. The orbital overlap between Tellurium and Hydrogen is extremely poor. Hence, the mathrmTe-H$\mathrm{Te-H}$ bond is very long and has very **low bond dissociation energy**. - This allows mathrmTeH_2$\mathrm{TeH}_2$ to easily release mathrmH^+$\mathrm{H^+}$ in solution, making it highly **acidic**. ### Step 1: Final Verification Therefore, the nature of mathrmTeO_2$\mathrm{TeO}_2$ is oxidising, and the nature of mathrmTeH_2$\mathrm{TeH}_2$ is acidic. ### Pattern Recognition Periodic Trend: As we go down Group 16: - Acidic strength of hydrides increases: mathrmH_2O < H_2S < H_2Se < H_2Te$\mathrm{H_2O < H_2S < H_2Se < H_2Te}$. - Reducing character of hydrides also increases. - Reducing power of dioxides decreases: mathrmSO_2$\mathrm{SO_2}$ (reducing) rightarrow mathrmTeO_2$\rightarrow \mathrm{TeO_2}$ (oxidising). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements