p-Block Elements Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on the p-block elements with step-by-step solutions covering oxidation states, oxide classification, bond-angle trends, inert-pair effect, group-15 hydrides and the brown-ring test.
A curated set of JEE Main 2026 previous-year questions on the p-block elements, each solved step by step so you can verify both the final answer and the reasoning behind it.
Solutions are AI-generated and pending review.
Solution
Statement I: Oxygen usually shows a covalency of two, but it can extend up to four (e.g. in $\text{H}_3\text{O}^+$ and coordinate-bonded species). Checking the oxidation states:
- In $\text{SO}_2$: oxygen is more electronegative than sulphur, so each O is $-2$ (and S is $+4$). Correct.
- In $\text{OF}_2$: fluorine is more electronegative than oxygen, so F is $-1$ each and O is $+2$. Correct.
So Statement I is true.
Statement II: The anomalous behaviour of oxygen (relative to S, Se, Te) arises from its small atomic size, high electronegativity and the absence of $d$-orbitals in its valence shell. True.
Both statements are true.
Answer: A
Solution
Statement I: A more negative reduction potential means a more electropositive (more easily oxidised) metal. Here $E^\circ_{\text{Al}^{3+}/\text{Al}} = -1.66\ \text{V}$ (negative) while $E^\circ_{\text{Tl}^{3+}/\text{Tl}}$ is positive. Hence aluminium is more electropositive than thallium. True.
Statement II: The sum of the first three ionization enthalpies of boron ($\Delta_i H_1 + \Delta_i H_2 + \Delta_i H_3$) is much larger than that of aluminium because of boron’s very small size. This large energy cannot be compensated by lattice/hydration energy, so boron cannot form $\text{B}^{3+}$ and bonds covalently, whereas aluminium readily forms $\text{Al}^{3+}$. This is the standard NCERT reasoning. True.
Both statements are true.
Answer: A
Solution
Statement I: The oxidising power (and standard reduction potential) of the halogens decreases down the group: $F_2 > Cl_2 > Br_2 > I_2$. Because $Cl_2$ is a stronger oxidant than $Br_2$ and $I_2$, it can displace them from their salts — the principle used in the layer test. True.
Statement II: In the layer test, chlorine water is added to the solution containing bromide/iodide:
$$Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2 \qquad Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$$The liberated $Br_2$ (orange) or $I_2$ (violet) dissolves in the added organic layer (e.g. $CS_2/CCl_4$), giving a characteristic coloured layer. This is a displacement redox reaction (Cl is reduced, $Br^-/I^-$ is oxidised). True.
Both statements are true.
Answer: A
Solution
Germanium has a Pauling electronegativity of $2.0$ (to one decimal). Among the group 13 elements, thallium matches this ($\text{EN}_{\text{Tl}} \approx 2.0$), so $E = \text{Tl}$.
For thallium the inert-pair effect is strongest in the group: the $+1$ state is more stable than the $+3$ state, i.e. $\text{Tl}^+ $ is more stable than $\text{Tl}^{3+}$. Consequences for $E^{3+} = \text{Tl}^{3+}$:
- $\text{Tl}^{3+}$ tends to gain electrons and revert to the more stable $\text{Tl}^+$, so it is an oxidizing agent — statement B is correct; it is not a reducing agent, so A is wrong.
- Since $E^+$ is the more stable state, $E^{3+}$ is not more stable than $E^+$ — C is wrong.
- The strong oxidising tendency of $\text{Tl}^{3+}$ gives a positive $E^\circ_{E^{3+}/E}$ — D is correct.
Correct statements: B and D.
Answer: C
Solution
Classify the oxides in each option; the correct set must have one of each type — basic, acidic, amphoteric and neutral.
- Option 1: $Na_2O$ (basic), $K_2O$ (basic), $Al_2O_3$ (amphoteric), $As_2O_3$ (amphoteric) — no acidic, no neutral. Wrong.
- Option 2: $Al_2O_3$ (amphoteric), $As_2O_3$ (amphoteric), $CO$ (neutral), $NO$ (neutral) — no basic, no acidic. Wrong.
- Option 3: $K_2O$ (basic), $Cl_2O_7$ (acidic), $As_2O_3$ (amphoteric), $NO$ (neutral) — all four types present. Correct.
- Option 4: $Na_2O$ (basic), $N_2O$ (neutral), $Al_2O_3$ (amphoteric), $CO$ (neutral) — no acidic. Wrong.
Only option 3 contains all four kinds of oxides.
Answer: C
Solution
Statement I: Comparing the central-oxygen bond angles:
$$F_2O\ (\approx 103^\circ) < H_2O\ (104.5^\circ) < Cl_2O\ (\approx 111^\circ)$$Fluorine is highly electronegative and pulls the bonding pairs away from O, reducing repulsion between bond pairs and compressing the angle; chlorine is larger and less electronegative, so bond pairs stay near O and, with added bp–bp / steric repulsion of the large Cl atoms, the angle opens up. The stated trend is true.
Statement II: $SiF_4$ is a covalent, molecular (gaseous) compound, not ionic. While the heavier $SnF_4$ and $PbF_4$ have considerable ionic character, grouping all three as ionic is incorrect. False.
Statement I is true, Statement II is false.
Answer: C
Solution
This is the classic brown-ring test for nitrate/nitrite. The nitrate is first reduced by $\text{Fe}^{2+}$ in the presence of concentrated $\text{H}_2\text{SO}_4$ to nitric oxide:
$$3\text{Fe}^{2+} + 4\text{H}^+ + \text{NO}_3^- \rightarrow 3\text{Fe}^{3+} + \text{NO} + 2\text{H}_2\text{O}$$The liberated gas $X = \text{NO}$ then combines with fresh $\text{Fe}^{2+}$ (as $\text{FeSO}_4$) to form the brown nitrosyl-iron complex at the interface:
$$[\text{Fe(H}_2\text{O)}_6]^{2+} + \text{NO} \rightarrow [\text{Fe(H}_2\text{O)}_5(\text{NO})]^{2+} + \text{H}_2\text{O}$$Written as the sulphate, $Y = [\text{Fe(NO)}]\text{SO}_4$, the brown ring.
Answer: A
Solution
Evaluate each statement:
- A. Nitrogen in the intermediate oxidation states $+1$ to $+4$ does disproportionate in acidic medium (NCERT fact). Correct.
- B. Nitrogen has no $d$-orbitals, so it forms $p\pi$–$p\pi$ multiple bonds, not $d\pi$–$p\pi$. Wrong.
- C. The N–N single bond ($\approx 159\ \text{kJ mol}^{-1}$) is weaker than the P–P single bond ($\approx 209\ \text{kJ mol}^{-1}$) because of high lone-pair repulsion in the small N atom. Wrong.
- D. Nitrogen is a diatomic gas and has the lowest density in group 15 (Bi has the highest). Wrong.
- E. Nitrogen’s valence shell has only four orbitals ($2s, 2p_x, 2p_y, 2p_z$), so its maximum covalency is four. Correct.
Correct statements: A and E only.
Answer: D
Solution
Statement I: Classify each pair:
- $[Al_2O_3,\ Cr_2O_3]$: both amphoteric — same.
- $[Cl_2O_7,\ Mn_2O_7]$: both acidic (highest-oxidation-state oxides) — same.
- $[Na_2O,\ V_2O_3]$: $Na_2O$ basic; $V_2O_3$ (low oxidation state $+3$) is basic — same.
- $[CO,\ N_2O]$: both neutral — same.
All 4 pairs contain oxides of the same nature, so the count is 4. True.
Statement II: $Na_2O$ (a strongly electropositive metal oxide) is the most basic, and $Cl_2O_7$ (a high-oxidation-state non-metal oxide) is the most acidic among the four. True.
Both statements are true.
Answer: A
Solution
Statement I: Aluminium dissolves in aqueous NaOH to give the tetrahydroxoaluminate(III) ion, not a hexahydroxo species:
$$2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2\uparrow$$The product is $[Al(OH)_4]^-$ (tetrahedral), so the statement claiming $[Al(OH)_6]^{3-}$ is false.
Statement II: Apply VSEPR:
- $ICl_4^-$: I has 4 bond pairs + 2 lone pairs ($sp^3d^2$) $\Rightarrow$ square planar. Correct.
- $ClO_3^-$: Cl has 3 bond pairs + 1 lone pair ($sp^3$) $\Rightarrow$ pyramidal. Correct.
- $IBr_2^-$: I has 2 bond pairs + 3 lone pairs ($sp^3d$) $\Rightarrow$ linear. Correct.
Statement II is true.
Statement I is false, Statement II is true.
Answer: D
Solution
Down group 15 the E–H bond becomes weaker (E gets larger, bond enthalpy falls):
- A. As E–H bonds weaken, the hydrides more readily give up hydrogen, so reducing nature increases $NH_3 \rightarrow BiH_3$. Correct.
- B. The lone pair becomes more diffuse and less available down the group, so the electron-donating (basic) tendency decreases $NH_3 \rightarrow BiH_3$ ($NH_3$ is the strongest donor). Correct.
- C. With decreasing bond dissociation enthalpy, thermal stability decreases $NH_3 \rightarrow BiH_3$. Correct.
- D. Bond angles fall as EN of the central atom drops (less bp–bp repulsion): $NH_3\,(107^\circ) > PH_3\,(93.6^\circ) > AsH_3\,(91.8^\circ) > SbH_3\,(91.3^\circ)$, so the angle decreases $NH_3 \rightarrow SbH_3$. Correct.
All four statements A, B, C and D are correct.
Answer: C