Periodic Classification: JEE Main 2026 Previous Year Questions (Solved)
Solved JEE Main 2026 previous year questions on periodic classification with step-by-step solutions covering ionisation enthalpy, electron gain enthalpy, electronegativity and periodic trends.
Practice the most recent JEE Main 2026 questions from Periodic Classification, each with a concise, worked-out solution.
Solutions are AI-generated and pending review.
Solution
The anion $A^-$ carries one extra electron, so the neutral atom has
$$Z = \text{electrons in } A^- - 1 = 36 - 1 = 35$$An atomic number $Z = 35$ corresponds to bromine (Br).
$$\text{Atomic mass} = Z + \text{neutrons} = 35 + 45 = 80$$Bromine’s configuration is $[\text{Ar}]\,3d^{10}\,4s^2\,4p^5$, placing it in Group 17 (halogens). Elemental bromine $\left(\text{Br}_2\right)$ is a reddish-brown liquid at room temperature.
Answer: A (80, 17, liquid)
Solution
Statement I: First ionisation enthalpy increases across a period. Approximate values (kJ/mol):
$$\text{Na} = 496,\quad \text{Mg} = 737,\quad \text{Cl} = 1251,\quad \text{Ar} = 1521$$So the true order is $\text{Ar} > \text{Cl} > \text{Mg} > \text{Na}$, the reverse of what is stated. Statement I is false.
Statement II: Calcium is $[\text{Ar}]\,4s^2$. Removing two electrons gives $\text{Ca}^{2+} = [\text{Ar}]$, a stable noble-gas core. The third electron must come from this core, so
$$IE_3(\text{Ca}) \gg IE_3(\text{Al, Fe, B})$$Statement II is true.
Answer: D (Statement I is false but Statement II is true)
Solution
Ionisation enthalpies are always endothermic (positive), so the answer cannot be negative — this rules out $-906$ and $-856$.
The second ionisation removes an electron from a positive ion $\left(\text{Mg}^+\right)$, where the remaining electrons are held more tightly by the same nuclear charge. Hence
$$IE_2 > IE_1 = 737 \text{ kJ/mol}$$This rules out $+590$. The only value that is both positive and greater than $737$ is $+1450$ kJ/mol. (The experimental value of $IE_2$ for Mg is $\approx 1451$ kJ/mol.)
Answer: C ($+1450$ kJ/mol)
Solution
Statement I: After losing one electron, the second electron is removed from Group-13 cations. Approximate $IE_2$ values (kJ/mol):
$$\text{B} = 2427,\quad \text{Al} = 1817,\quad \text{Ga} = 1979$$The true order is $\text{B} > \text{Ga} > \text{Al}$ (Ga’s higher value comes from poor shielding by the intervening $3d^{10}$ electrons). This is not $B > Al > Ga$, so Statement I is false.
Statement II: First ionisation enthalpies of Group-14 elements (kJ/mol):
$$\text{Si} = 786,\quad \text{Ge} = 762,\quad \text{Sn} = 709,\quad \text{Pb} = 716$$The true order is $\text{Sn} < \text{Pb} < \text{Ge} < \text{Si}$, not $Si < Ge < Pb < Sn$. Statement II is false.
Answer: B (Both Statement I and Statement II are false)
Solution
Extreme left element of a period is an alkali metal (Group 1). It has a single loosely held $ns^1$ electron, so its first ionisation enthalpy is the lowest in the period.
Extreme right element (excluding noble gases) is a halogen (Group 17). It needs just one electron to complete its octet, releasing a large amount of energy, so its negative electron gain enthalpy is the highest (most negative) in the period.
Answer: C (lowest and highest)
Solution
Statement I: Electronegativity increases across Period 2. On the Pauling scale:
$$\text{F} = 4.0 > \text{O} = 3.5 > \text{N} = 3.0$$So $F > O > N$ is correct. Statement I is true.
Statement II: In $OF_2$, fluorine is more electronegative than oxygen, so F takes $-1$ each:
$$x + 2(-1) = 0 \implies x = +2$$In $Na_2O$, sodium is $+1$ each:
$$2(+1) + y = 0 \implies y = -2$$Both assignments are correct. Statement II is true.
Answer: A (Both Statement I and Statement II are true)
Solution
For $n = 2$, identify each element and its known first ionisation energy (kJ/mol):
$$\text{A: } 2s^2 = \text{Be} \;(899) \rightarrow \text{II}$$$$\text{B: } 2s^2 2p^1 = \text{B} \;(800) \rightarrow \text{III}$$$$\text{C: } 2s^2 2p^3 = \text{N} \;(1402) \rightarrow \text{IV}$$$$\text{D: } 2s^2 2p^6 = \text{Ne} \;(2080) \rightarrow \text{I}$$Note nitrogen’s half-filled $2p^3$ gives extra stability, raising its $IE_1$ above that of oxygen; neon (noble gas) has the highest value.
Answer: A (A-II, B-III, C-IV, D-I)