Consider the following elements In, Tl, Al, Pb, Sn and Ge. The most stable oxidation states of elements with highest…

Q: Consider the following elements In, Tl, Al, Pb, Sn and Ge. The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are

Let us check the trends for the provided main group elements (Al, In, Tl from Group 13 and Ge, Sn, Pb from Group 14): - **Highest First Ionization Enthalpy (IE1):** Out of these options, Germanium (Ge) sits highest and further right along its period layout, demonstrating the highest IE1 value among this set. Its most stable oxidation state is **+4**. - **Lowest First Ionization Enthalpy (IE1):** Indium (In) lies lowest leftward among these relative coordinates, maintaining the lowest IE1. Its mo

Year	2026	2025	2024	Total
Questions	16	13	13	42

Q60 jee_main_2026_24_january_morning Solubility of Chlorides

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

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

Q51 jee_main_2026_24_january_evening Group 14 Elements

Choose the INCORRECT statement

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

Q58 jee_main_2026_24_january_evening Chlorine and its Compounds

One mole of

Cl₂

(g) was passed into 2 L of cold 2M KOH solution. After the reaction, the concentrations of

Cl⁻

,

ClO⁻

and

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:

Cl₂ + 2KOH arrow KCl + KClO + H₂O

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

Step 1: Stoichiometry

Cl₂ + 2KOH arrow KCl + KClO + H₂O

$t = 0$ $t = 0$: $1 mole$ $1 \text{ mole}$ $Cl₂$ $\mathrm{Cl}_2$, $4 moles$ $4 \text{ moles}$ $KOH$ $\mathrm{KOH}$ 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. $[OH^-] = Moles of unreacted KOHVolume = (2)/(2) = 1M$ $[Cl^-] = Moles of KClVolume = (1)/(2) = 0.5M$ $[ClO^-] = Moles of KClOVolume = (1)/(2) = 0.5M$

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

Q61 jee_main_2026_28_january_morning Group 15 Hydrides

Regarding the hydrides of group 15 elements

EH₃

(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 ( $NH₃, PH₃, AsH₃, SbH₃, BiH₃$ $NH_3, PH_3, AsH_3, SbH_3, BiH_3$):\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

Q52 jee_main_2026_28_january_evening Properties Of Hydrogen Halides

Given below are two statements: Statement I: The increasing order of boiling point of hydrogen halides is

HCl < HBr < HI < HF

Statement II: The increasing order of melting point of hydrogen halides is

HCl < HBr < HF < HI

In the light of the above statements, choose the correct answer from the options given below:

A. $(1) Both Statement I and Statement II are true$
B. $(2) Statement I is true but Statement II is false$
C. $(3) Both Statement I and Statement II are false$
D. $(4) Statement I is false but Statement II is true$

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

Consider the following elements In, Tl, Al, Pb, Sn and Ge. The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More The p-Block Elements Previous-Year Questions — Page 3

Solution

Core Logic

Step 1: Option Elimination

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Stoichiometry

Step 2: Final Concentrations

Pattern Recognition

Chapter Mix

Solution

Core Logic

Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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