Consider three metal chlorides x, y and z, where x is water soluble at room temperature, y is sparingly soluble in water at room temperature and z is soluble in hot water. x, y and z are respectively
A.MgCl₂, AgCl and AlCl₃$MgCl_2, AgCl \text{ and } AlCl_3$
B.AgCl, Hg₂Cl₂ and PbCl₂$AgCl, Hg_2Cl_2 \text{ and } PbCl_2$
C.AlCl₃, PbCl₂ and BaCl₂$AlCl_3, PbCl_2 \text{ and } BaCl_2$
D.CuCl₂, AgCl and PbCl₂$CuCl_2, AgCl \text{ and } PbCl_2$
Solution & Explanation
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.
Solubility data
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
A. Among the isotopes of carbon, ¹³C${}^{13}C$ is a radioactive isotope.
B. Carbon exhibits negative oxidation states along with +4 and +2.
C. Carbon cannot exceed its covalency more than four.
D.CO₂$CO_{2}$ is the most acidic oxide among the dioxides of group of 14 elements.
Solution
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
Q58jee_main_2026_24_january_eveningChlorine and its Compounds
One mole of Cl₂$Cl_{2}$ (g) was passed into 2 L of cold 2M KOH solution. After the reaction, the concentrations of Cl⁻$Cl^{-}$, ClO⁻$ClO^{-}$ and OH⁻$OH^{-}$ are respectively (assume volume remains constant)
A. 0.75 M, 0.75 M, 1 M
B. 0.5 M, 0.5 M, 0.5 M
C. 0.5 M, 0.5 M, 1 M
D. 1 M, 1 M, 1 M
Solution
Core Logic
Reaction of chlorine with cold and dilute alkali forms chloride and hypochlorite:
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$[\mathrm{ClO}^-] = \frac{\text{Moles of KClO}}{\text{Volume}} = \frac{1}{2} = 0.5\mathrm{M}$
Regarding the hydrides of group 15 elements EH₃$EH_{3}$ (E = N, P, As, Sb), select the correct statement from the following:
A. The stability of hydrides decreases down the group.
B. The basicity of hydrides decreases down the group.
C. The reducing character increases down the group.
D. The boiling point increases down the group.
Choose the correct answer from the options given below:
A. A, B & C only
B. A & D only
C. A, B, C & D
D. B & C only
Solution
Core Logic
Analyzing Group 15 hydrides (NH₃, PH₃, AsH₃, SbH₃, BiH₃$NH_3, PH_3, AsH_3, SbH_3, BiH_3$):
$\bullet$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.
$\bullet$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.
$\bullet$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.
$\bullet$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
Q52jee_main_2026_28_january_eveningProperties Of Hydrogen Halides
Given below are two statements:
Statement I: The increasing order of boiling point of hydrogen halides is HCl < HBr < HI < HF$HCl < HBr < HI < HF$
Statement II: The increasing order of melting point of hydrogen halides is HCl < HBr < HF < HI$HCl < HBr < HF < HI$
In the light of the above statements, choose the correct answer from the options given below:
A.(1) Both Statement I and Statement II are true$(1)\text{ Both Statement I and Statement II are true}$
B.(2) Statement I is true but Statement II is false$(2)\text{ Statement I is true but Statement II is false}$
C.(3) Both Statement I and Statement II are false$(3)\text{ Both Statement I and Statement II are false}$
D.(4) Statement I is false but Statement II is true$(4)\text{ Statement I is false but Statement II is true}$
Solution
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
More The p-Block Elements Questions — jee_main_2026_24_january_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
