p Block Elements (Group 15, 16, 17 & 18) JEE Main previous year questions with solutions

5 solved JEE Main questions on p Block Elements (Group 15, 16, 17 & 18), free to read — no sign-in needed. The full chapter has 192 questions; sign in to attempt the remaining 187 in the exam simulator.

1. Q1JEE Main 2026 (02 Apr, Shift 1)Halogens and Its Compounds
   Given below are two statements : Statement (I) : Oxidising power of halogens decreases in the order $F_2 \gt Cl_2 \gt Br_2 \gt I_2$, 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. A.Both Statement I and Statement II are true
   2. B.Both Statement I and Statement II are false
   3. C.Statement I is true but Statement II is false
   4. D.Statement I is false but Statement II is true
   Show answer & solution

   Answer: (A)

   Statement I is true. The oxidising power of halogens decreases down the group in the order $F_2 \gt Cl_2 \gt Br_2 \gt I_2$. A stronger oxidising halogen can displace a weaker oxidising halogen from its aqueous halide solution. This principle forms the basis of the layer test. Statement II is true. In the layer test, chlorine water ($Cl_2$) is added to an aqueous solution containing bromide ($Br^-$) or iodide ($I^-$) ions. Since $Cl_2$ is a stronger oxidising agent, it oxidises $Br^-$ to $Br_2$ and $I^-$ to $I_2$. The reaction is given by: $Cl_2 + 2X^- \rightarrow 2Cl^- + X_2$ (where $X = Br, I$) This is a non-metal displacement redox reaction. The liberated $Br_2$ or $I_2$ is then extracted into an organic solvent layer (like $CHCl_3$ or $CCl_4$), where it imparts a characteristic orange-red colour for $Br_2$ or a violet colour for $I_2$, thereby identifying them. Therefore, both Statement I and Statement II are true. Answer: Both Statement I and Statement II are true
2. Q2JEE Main 2025 (03 Apr, Shift 1)Nitrogen and Its Compounds
   Given below are two statements Statement I : The N-N single bond is weaker and longer than that of $\mathrm{P}-\mathrm{P}$ single bond Statement II : Compounds of group 15 elements in +3 oxidation states readily undergo disproportionation reactions. In the light of above statements, choose the correct answer from the options given below
   1. A.Statement I is true but statement II is false
   2. B.Both statement I and statement II are false
   3. C.Statement I is false but statement II is true
   4. D.Both statement I and statement II are true
   Show answer & solution

   Answer: (B)

   $\ddot{\mathrm{N}}-\ddot{\mathrm{N}}$ single bond weaker than $\ddot{\mathrm{P}}-\ddot{\mathrm{P}}$ due to more $\ell \mathrm{p}-\ell \mathrm{p}$ repulsion. Bond length $\Rightarrow d_{p-p} \gt d_{N-N}(\operatorname{size} \uparrow$, B.L. $\uparrow)$ In group 15 elements only N \& P show disproportionation in +3 oxidation state, As, Sb \& Bi have almost inert for disproportionation in +3 oxidation state. So both statements are false.
3. Q3JEE Main 2024 (31 Jan, Shift 1)Noble Gases
   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:
   1. A.Statement I is false but Statement II is true
   2. B.Both Statement I and Statement II are true
   3. C.Statement I is true but Statement II is false
   4. D.Both Statement I and Statement II are false
   Show answer & solution

   Answer: (D)

   As noble gases do not form compounds, hence they cannot occur in the form of diatomic molecules. Noble gases, being monoatomic, have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points. Noble gases are held together by weak dispersion forces. Statement I and II are False
4. Q4JEE Main 2023 (11 Apr, Shift 2)Phosphorus and Its Compounds
   One mole of ${P}_{4}$ reacts with $8$ moles of ${SOCl}_{2}$ to give $4$ moles of $A$, $x$ mole of ${SO}_{2}$ and $2$ moles of $B$. $A,B$ and $x$ respectively are
   1. A.${POCl}_{3},{S}_{2}{Cl}_{2}$ and $2$
   2. B.${PCl}_{3},{S}_{2}{Cl}_{2}$ and $4$
   3. C.${PCl}_{3},{S}_{2}{Cl}_{2}$ and $2$
   4. D.${POCl}_{3},{S}_{2}{Cl}_{2}$ and $4$
   Show answer & solution

   Answer: (B)

   Thionyl chloride reacts with white phosphorus to produce phosphorus trichloride. The other side products in this reaction are sulphur dioxide and Sulphur monochloride. ${P}_{4}+8{SOCl}_{2}\rightarrow 4{PCl}_{3}+4{SO}_{2}+2{S}_{2}{Cl}_{2}$ Hence $A={PCl}_{3},x=4,B={S}_{2}{Cl}_{2}$
5. Q5JEE Main 2022 (26 Jun, Shift 1)Sulphur and Its Compounds
   The correct order of melting points of hydrides of group $16$ elements is
   1. A.${H}_{2}S\lt {H}_{2}Se\lt {H}_{2}Te\lt {H}_{2}O$
   2. B.${H}_{2}O\lt {H}_{2}S\lt {H}_{2}Se\lt {H}_{2}Te$
   3. C.${H}_{2}S\lt {H}_{2}Te\lt {H}_{2}Se\lt {H}_{2}O$
   4. D.${H}_{2}Se\lt {H}_{2}S\lt {H}_{2}Te\lt {H}_{2}O$
   Show answer & solution

   Answer: (A)

   In general molar mass $↑,MP↑$ but in ${H}_{2}O$ due to $H$-bonding, it is maximum among the following: Property ${H}_{2}O$ ${H}_{2}S$ ${H}_{2}Se$ ${H}_{2}Te$ $m.p./K$ $273$ $188$ $208$ $222$