The correct statements from the following are: (A) Tl³⁺$\text{Tl}^{3+}$ is a powerful oxidising agent (B) Al³⁺$\text{Al}^{3+}$ does not get reduced easily (C) Both Al³⁺$\text{Al}^{3+}$ and Tl³⁺$\text{Tl}^{3+}$ are very stable in solution (D) Tl⁺$\text{Tl}^{+}$ is more stable than Tl³⁺$\text{Tl}^{3+}$ (E) Al³⁺$\text{Al}^{3+}$ and Tl⁺$\text{Tl}^{+}$ are highly stable Choose the correct answer from the options given below:

Core Logic

Analyzing the standard solubility rules for metal chlorides: Most metal chlorides are soluble in water at room temperature (e.g., MgCl₂$MgCl_2$, AlCl₃$AlCl_3$, CuCl₂$CuCl_2$, BaCl₂$BaCl_2$). This represents 'x'. However, there are notable exceptions that are sparingly soluble: AgCl$AgCl$, Hg₂Cl₂$Hg_2Cl_2$, and PbCl₂$PbCl_2$. These represent 'y'. Among these exceptions, PbCl₂$PbCl_2$ has a unique property: it is sparingly soluble in cold water but highly soluble in hot water. This represents 'z'.

Step 1: Option Elimination

Option (1): AlCl₃$AlCl_3$ (z) is highly soluble in cold water, not specifically requiring hot water. Option (2): AgCl$AgCl$ (x) is not water soluble at room temperature. Option (3): BaCl₂$BaCl_2$ (z) is soluble in cold water. Option (4): CuCl₂$CuCl_2$ (x) is soluble at RT. AgCl$AgCl$ (y) is sparingly soluble. PbCl₂$PbCl_2$ (z) is soluble in hot water.

Step 2: Final Conclusion

The set that perfectly maps to x, y, and z is CuCl₂, AgCl and PbCl₂$CuCl_2, AgCl \text{ and } PbCl_2$.

Pattern Recognition

The keyword "soluble in hot water" uniquely flags PbCl₂$PbCl_2$ in qualitative analysis (Group I cations).

Chapter Mix

Class 11 Chemistry: The p-Block Elements Class 12 Chemistry: Qualitative Analysis

Core Logic

Isotopes of carbon include C¹²$\mathbf{C}^{12}$, C¹³$\mathbf{C}^{13}$, and C¹⁴$\mathbf{C}^{14}$. C¹³$\mathbf{C}^{13}$ is a stable, non-radioactive isotope. C¹⁴$\mathbf{C}^{14}$ is the radioactive isotope of carbon used in radiocarbon dating. Therefore, the statement that ¹³C${}^{13}C$ is radioactive is incorrect.

Pattern Recognition

Carbon has three main naturally occurring isotopes. Always remember that 12 and 13 are stable, while 14 is the famous radioactive tracer.

Chapter Mix

Class 11 Chemistry: The p-Block Elements

Core Logic

Reaction of chlorine with cold and dilute alkali forms chloride and hypochlorite:

Initial moles: Moles of Cl₂$\mathrm{Cl}_2$ = 1 mole$1 \text{ mole}$ Moles of KOH = M × V = 2M × 2L = 4 moles$\mathrm{KOH} = M \times V = 2\mathrm{M} \times 2\mathrm{L} = 4 \text{ moles}$

Step 1: Stoichiometry

t = 0$t = 0$: 1 mole$1 \text{ mole}$ Cl₂$\mathrm{Cl}_2$, 4 moles$4 \text{ moles}$ KOH$\mathrm{KOH}$ Since 1 mole of Cl₂$\mathrm{Cl}_2$ reacts with 2 moles of KOH$\mathrm{KOH}$, Cl₂$\mathrm{Cl}_2$ is the limiting reagent.

At tf$t_f$ (final): Moles of Cl₂$\mathrm{Cl}_2$ remaining = 0$0$ Moles of KOH$\mathrm{KOH}$ remaining = 4 - 2(1) = 2 moles$4 - 2(1) = 2 \text{ moles}$ Moles of KCl$\mathrm{KCl}$ formed = 1 mole$1 \text{ mole}$ Moles of KClO$\mathrm{KClO}$ formed = 1 mole$1 \text{ mole}$

Step 2: Final Concentrations

Total volume = 2 L. [OH^-] = Moles of unreacted KOHVolume = (2)/(2) = 1M$[\mathrm{OH}^-] = \frac{\text{Moles of unreacted KOH}}{\text{Volume}} = \frac{2}{2} = 1\mathrm{M}$ [Cl^-] = Moles of KClVolume = (1)/(2) = 0.5M$[\mathrm{Cl}^-] = \frac{\text{Moles of KCl}}{\text{Volume}} = \frac{1}{2} = 0.5\mathrm{M}$ [ClO^-] = Moles of KClOVolume = (1)/(2) = 0.5M$[\mathrm{ClO}^-] = \frac{\text{Moles of KClO}}{\text{Volume}} = \frac{1}{2} = 0.5\mathrm{M}$

Pattern Recognition

Cold & dilute alkali + Cl₂$Cl_2$ arrow$\rightarrow$ Chloride + Hypochlorite (Cl^-$Cl^-$ & ClO^-$ClO^-$). Hot & conc. alkali + Cl₂$Cl_2$ arrow$\rightarrow$ Chloride + Chlorate (Cl^-$Cl^-$ & ClO₃^-$ClO_3^-$).

Chapter Mix

Class 12 Chemistry: The p-Block Elements Class 11 Chemistry: Some Basic Concepts of Chemistry

Core Logic

Analyzing Group 15 hydrides (NH₃, PH₃, AsH₃, SbH₃, BiH₃$NH_3, PH_3, AsH_3, SbH_3, BiH_3$):\n$\bullet$ Stability: Decreases down the group (NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃$NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$) because E-H bond length increases and bond dissociation energy decreases.\n$\bullet$ Basicity: Decreases down the group (NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃$NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$) due to the increasing size of the central atom causing a decrease in electron density on the lone pair.\n$\bullet$ Reducing Character: Increases down the group (NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃$NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$) as E-H bonds become easier to break.\n$\bullet$ Boiling Point: Order is PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃$PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$. Ammonia (NH₃$NH_3$) has an anomalously high boiling point due to strong intermolecular hydrogen bonding. Thus, it does NOT simply increase steadily down the group.

Final Conclusion

Statements A, B, and C are correct.

Pattern Recognition

Hydrogen bonding in periods 2 (NH₃, H₂O, HF$NH_3, H_2O, HF$) breaks the monotonic trend for boiling points down their respective groups.

Chapter Mix

Class 12 Chemistry: The p-Block Elements

Core Logic

The boiling points and melting points of hydrogen halides depend on the interplay between van der Waals forces (which increase with molar mass) and hydrogen bonding.

Boiling point order: HF > HI > HBr > HCl$HF > HI > HBr > HCl$ HF$HF$ has the highest boiling point due to strong intermolecular hydrogen bonding.

Melting point order: HI > HF > HBr > HCl$HI > HF > HBr > HCl$ For melting point, the crystal lattice energy of HI$HI$ dominates the hydrogen bonding of HF$HF$.

Step 1: Final Conclusion

Both Statement I and Statement II represent the correct orders respectively.

Pattern Recognition

For Boiling Point, Hydrogen bonding wins (HF$HF$ is max). For Melting Point, massive size and dispersion forces win in solid lattice (HI > HF$HI > HF$).

Chapter Mix

Class 12 Chemistry: The p-Block Elements