A 'p'-block element (E) and hydrogen form a binary cation (EHx)+, while EH3 on treatment with K2HgI4 in alkaline medium gives a precipitate of basic mercury(II)amido-iodine. Given below are first ioni

The reagent K2HgI4 in alkaline medium is Nessler's reagent. It gives a brown precipitate (iodide of Millon's base) with ammonia (NH3). Therefore, the compound EH3 is NH3, and the element (E) is Nitrogen (N). The binary cation is the ammonium ion (NH4^+). 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 2p3 configuration, Nitrogen has a remarkably high first ionization energy, higher than

A 'p'-block element (E) and hydrogen form a binary cation (EHx)+, while EH3 on treatment with K2HgI4 in alkaline medium gives a precipitate of basic mercury(II)amido-iodine. Given below are first ioni

The reagent K2HgI4 in alkaline medium is Nessler's reagent. It gives a brown precipitate (iodide of Millon's base) with ammonia (NH3). Therefore, the compound EH3 is NH3, and the element (E) is Nitrogen (N). The binary cation is the ammonium ion (NH4^+). 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 2p3 configuration, Nitrogen has a remarkably high first ionization energy, higher than

A 'p'-block element (E) and hydrogen form a binary cation (EHx)+, while EH3 on treatment with K2HgI4 in alkaline medium…

Year	2026	2025	2024	Total
Questions	16	14	13	43

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

NH₃ + K₂HgI₄ + KOH arrow HgO\cdot Hg(NH₂)I + KI + H₂O

Core Logic

The reagent $K₂HgI₄$ $K_2HgI_4$ in alkaline medium is Nessler's reagent. It gives a brown precipitate (iodide of Millon's base) with ammonia ( $NH₃$ $NH_3$). Therefore, the compound $EH₃$ $EH_3$ is $NH₃$ $NH_3$, and the element (E) is Nitrogen (N). The binary cation is the ammonium ion $(NH₄^+)$ $(NH_4^+)$.

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 $2p³$ $2p^3$ 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 $kJ mol⁻¹$ $kJ\text{ mol}^{-1}$ is the highest and corresponds to Nitrogen.

Step 1: Final Conclusion

Element E is Nitrogen, and its correct first ionization enthalpy is 1402 $kJ mol⁻¹$ $kJ\text{ mol}^{-1}$.

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.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: p-Block Elements

More p-Block Elements Previous-Year Questions — Page 9

Q64 jee_main_2024_31_jan_evening Properties of Sulphur and Halogens

Given below are two statements: Statement I:

S₈

solid undergoes disproportionation reaction under alkaline conditions to form

S²⁻

and

S₂O₃²⁻

Statement II:

ClO₄^-

can undergo disproportionation reaction under acidic condition. In the light of the above statements, choose the most appropriate answer from the options given below:

A. $Statement I is correct but statement II is incorrect.$
B. $Statement I is incorrect but statement II is correct$
C. $Both statement I and statement II are incorrect.$
D. $Both statement I and statement II are correct.$

Solution

Core Logic

Statement I: $S₈$ $S_8$ disproportionates in alkaline medium to give sulphide ( $S²⁻$ $S^{2-}$) and thiosulphate ( $S₂O₃²⁻$ $S_2O_3^{2-}$). The reaction is:

S₈ + 12OH^- arrow 4S²⁻ + 2S₂O₃²⁻ + 6H₂O

This is a correct statement.

Statement II: $ClO₄^-$ $ClO_4^-$ contains chlorine in its highest possible oxidation state of $+7$ $+7$. A species can undergo disproportionation only if the element can exist in both higher and lower oxidation states than the present one. Since chlorine cannot exceed $+7$ $+7$, $ClO₄^-$ $ClO_4^-$ cannot undergo disproportionation. This is an incorrect statement.

Step 1: Final Conclusion

Statement I is correct, and Statement II is incorrect. Therefore, option (1) is the correct answer.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The p-Block Elements

Q61 jee_main_2024_31_jan_morning Group 18 Elements

Give below are two statements: Statement-I: Noble gases have very high boiling points. Statement-II: Noble gases are monoatomic gases. They are held together by strong dispersion forces. Because of this they are liquefied at very low temperature. Hence, they have very high boiling points. In the light of the above statements, choose the correct answer from the options given below:

A. $Statement I is false but Statement II is true.$
B. $Both Statement I and Statement II are true.$
C. $Statement I is true but Statement II is false.$
D. $Both Statement I and Statement II are false.$

Solution

Core Logic

Statement I and II are False.

Noble gases have low boiling points. Noble gases are held together by weak dispersion forces.

Pattern Recognition

Noble gases are characterized by extremely weak intermolecular forces (London dispersion forces) because they are monoatomic and non-polar, which directly results in very low boiling and melting points.

Chapter Mix

Class 12 Chemistry: The p-Block Elements

Q79 jee_main_2024_31_jan_morning Group 14 Elements

Consider the oxides of group 14 elements

SiO₂

GeO₂

SnO₂

PbO₂

, CO and GeO. The amphoteric oxides are

A. $GeO, GeO₂$ $GeO, GeO_2$
B. $SiO₂, GeO₂$ $SiO_2, GeO_2$
C. $SnO₂, PbO₂$ $SnO_2, PbO_2$
D. $SnO₂, CO$ $SnO_2, CO$

Solution

Core Logic

In Group 14 elements, the nature of oxides varies down the group: $CO$ $CO$ is neutral. $CO₂$ $CO_2$ and $SiO₂$ $SiO_2$ are acidic. $GeO₂$ $GeO_2$ is distinctly acidic. $SnO₂$ $SnO_2$ and $PbO₂$ $PbO_2$ are amphoteric.

Pattern Recognition

Tin ( $Sn$ $Sn$) and Lead ( $Pb$ $Pb$) mostly form amphoteric oxides (e.g., $SnO, SnO₂, PbO, PbO₂$ $SnO, SnO_2, PbO, PbO_2$), distinguishing them from the acidic oxides of Carbon and Silicon.

Chapter Mix

Class 11 Chemistry: The p-Block Elements

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More p-Block Elements Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Final Conclusion

Chapter Mix

Solution

Core Logic

Pattern Recognition

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Core Logic

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