Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Given below are two statements:
Statement (I): Oxidising power of halogens decreases in the order $F_2 > Cl_2 > Br_2 > I_2$, which is the basis of "Layer test".
Statement (II): "Layer test" to identify $Br_2$ and $I_2$ in aqueous solution involves the oxidation of bromide or iodide into $Br_2$ or $I_2$ respectively with $Cl_2$, which is a type of displacement redox reaction.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer: Option (1)
Key idea: Oxidizing power of halogens and Layer test for halide identification
Analyze Statement I: Oxidizing power of halogens decreases as $F_2 > Cl_2 > Br_2 > I_2$ based on reduction potentials (+2.87V, +1.36V, +1.07V, +0.54V respectively)
This order is indeed the basis of layer test since $Cl_2$ can oxidize $Br^-$ and $I^-$ due to higher oxidizing power. Statement I is TRUE
Analyze Statement II: Layer test involves reactions like $Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2$ and $Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$
These are displacement redox reactions where $Cl_2$ displaces $Br_2$ or $I_2$ from their salts. Statement II is TRUE
The liberated $Br_2$ (orange-brown) or $I_2$ (violet) is extracted in organic layer for identification
Since both statements are correct, the answer is option (1)
Given below are two statements:
**Statement I:** The covalency of oxygen is generally two but it can exceed upto four. The oxidation state of oxygen in $SO_2$ is $-2$ and in $OF_2$ it is $+2$.
**Statement II:** The anomalous behaviour of oxygen when compared to the other elements of group 16 is due to its small size and high electronegativity.
In the light of the above statements, choose the *correct* answer from the options given below
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Anomalous behavior of oxygen, covalency limitations, and oxidation states in p-block elements
Correct statements from the following are:
A. Nitrogen in oxidation states from $+1$ to $+4$ disproportionates in acid medium.
B. Nitrogen has the ability to form $d\pi - p\pi$ multiple bonds with itself and other elements with small size and high electronegativity.
C. N-N single bond is stronger than P-P single bond.
D. Nitrogen has highest density in its group due to small size.
E. The maximum covalency of nitrogen is four since it has only four valence orbitals for bonding.
Choose the correct answer from the options given below:
(1) B, C and D Only
(2) C, D and E Only
(3) A, C and E Only
(4) A and E Only
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
What volume of hydrogen gas at STP would be liberated by action of 50 mL of $H_2SO_4$ of 50% purity (density = 1.3 g mL$^{-1}$) on 20 g of zinc?
Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 g mol$^{-1}$ respectively.
(1) 5.824 L
(2) 7.428 L
(3) 6.892 L
(4) 8.375 L
Show answer and solution
Answer: Option (3)
Key idea: Stoichiometry, Limiting Reagent, Gas Volume at STP
Calculate mass of $H_2SO_4$ solution: $50 \text{ mL} \times 1.3 \text{ g/mL} = 65$ g
Find mass of pure $H_2SO_4$: $65 \times 0.5 = 32.5$ g (50% purity)
Given below are two statements:
**Statement I:** The number of pairs among [$Al_2O_3$, $Cr_2O_3$], [$Cl_2O_7$, $Mn_2O_7$], [$Na_2O$, $V_2O_3$] and [$CO$, $N_2O$] that contain oxides of same nature (acidic, basic, neutral or amphoteric) is 4.
**Statement II:** Among $Na_2O$, $Al_2O_3$, $CO$ and $Cl_2O_7$, the most basic and acidic oxides are $Na_2O$ and $Cl_2O_7$, respectively.
In the light of the above statements, choose the **correct** answer from the options given below:
(1) Both **Statement I** and **Statement II** are true
(2) Both **Statement I** and **Statement II** are false
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Nature of oxides (acidic, basic, amphoteric, neutral) and their classification based on metal oxidation states
RMgI when treated with ice cold water liberated a gas which occupied $1.4 \text{ dm}^3/\text{g}$ at STP. The gas produced is further reacted with iodine in presence of $HIO_3$ to give compound (X). Compound (X) in presence of Na and dry ether produced compound (Y). Molar mass of compound (Y) is ______ $g \text{ mol}^{-1}$. (Nearest integer)
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Given below are two statements:
**Statement I:** Aluminium is more electropositive than thallium as the standard electrode potential value of $E^\circ_{Al^{3+}/Al}$ is negative and $E^\circ_{Tl^{3+}/Tl}$ is positive.
**Statement II:** The sum of first three ionization enthalpies of boron is very high when compared to that of aluminium. Due to this reason boron forms covalent compounds only and aluminium forms $Al^{3+}$ ion.
In the light of the above statements, choose the *correct* answer from the options given below
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer: Option (1)
Key idea: Electrode potentials, ionization enthalpies, and bonding trends in p-block elements
**Statement I:** Compare electrode potentials: $E°_{Al^{3+}/Al} = -1.66$ V (negative) and $E°_{Tl^{3+}/Tl} = +1.26$ V (positive)
More negative $E°$ means greater tendency to lose electrons, hence more electropositive. Al is more electropositive than Tl. **Statement I is TRUE**
**Statement II:** Calculate sum of first three ionization enthalpies: Boron ≈ 6888 kJ/mol, Aluminium ≈ 5140 kJ/mol
Boron has higher ionization enthalpies due to smaller size and higher effective nuclear charge
High ionization enthalpies prevent B³⁺ formation, so boron forms covalent compounds. Al can form Al³⁺ ions. **Statement II is TRUE**
Since both statements are true, the correct answer is option (1)
6 April 2026 Shift 2, Q62 · Q62 · MCQ · General Principles and Processes of Isolation of Elements · Medium
Given below are two statements:
**Statement I:** A mixture of $C_{12}H_{22}O_{11}$ (sugar) and NaCl can be separated by dissolving sugar in alcohol, due to differential solubility.
**Statement II:** Rose essence from rose petals is separated by steam distillation due to its high volatility and insolubility in $H_2O$.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Separation techniques based on differential solubility and steam distillation
Find the correct statements related to group 15 hydrides.
A. Reducing nature increases from $NH_3$ to $BiH_3$
B. Tendency to donate lone pair of electrons decreases from $NH_3$ to $BiH_3$
C. The stability of hydrides decreases from $NH_3$ to $BiH_3$
D. HEH bond angle decreases from $NH_3$ to $SbH_3$ (E = Elements of group 15)
Choose the correct answer from the options given below:
(1) A and B only
(2) B and C only
(3) A, B, C and D
(4) A, C and D Only
Show answer and solution
Answer: Option (3)
Key idea: Trends in Group 15 Hydrides - Stability, Basicity, Reducing Nature, and Bond Angles
**Statement A**: As we go from $NH_3$ to $BiH_3$, E-H bond strength decreases, making it easier to release hydrogen. Reducing nature increases down the group. ✓
**Statement B**: Basicity decreases down the group as atomic size increases and electron density decreases. Tendency to donate lone pair: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$. ✓
**Statement C**: Stability depends on E-H bond strength. As atomic size increases down the group, orbital overlap decreases, weakening bonds. Stability: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$. ✓
**Statement D**: Bond angles are $NH_3$ (107°) > $PH_3$ (93.5°) > $AsH_3$ (91.8°) > $SbH_3$ (91.3°). Decreasing hybridization and increasing p-character causes this trend. ✓
All four statements A, B, C, and D are correct.
FAQs
Frequently asked questions
How many questions come from p-Block Elements in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 11 questions on it, about 1.2 per shift.
Where can I practise p-Block Elements PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
