Previous year questions: Classification of Elements and Periodicity in Properties

Classification of Elements and Periodicity in Properties30 min readLesson 6 of 6

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks) Two statements are given. Statement I: the first ionization enthalpy of the elements Na, Mg, Cl and Ar follows the order Na > Mg > Cl > Ar. Statement II: among Ca, Al, Fe and B, the third ionization enthalpy is very high for Ca. Which is correct? Answer: Statement I is false. Going across a period the first ionization enthalpy generally rises, so the order should be Na below Mg below Cl below Ar; the given order reverses it. Statement II is true. Calcium has the configuration [Ar] 4s2, so removing its first two electrons leaves the stable argon core; the third electron must come from that core and costs far more energy. The correct option is that statement I is false and statement II is true. A statement that gives a descending order across a period is worth checking at once, because the trend almost never runs that way.

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks) A monoatomic anion A- has 45 neutrons and 36 electrons. What are the atomic mass, the group in the periodic table and the physical state at room temperature of the element A? Answer: The anion has 36 electrons and a single negative charge, so the neutral atom has 35 electrons and therefore 35 protons, which makes Z = 35, bromine. The mass number is protons plus neutrons, 35 + 45 = 80. Bromine has the configuration [Ar] 3d10 4s2 4p5, so it lies in period 4, group 17, in the p-block. Bromine is a liquid at room temperature. The correct option is 80, group 17, liquid. Reading the question in the other direction is useful too: the charge of the ion fixes the electron count, the electron count of a neutral atom fixes Z, and the neutron count then completes the mass number.

Q3 (JEE Main 2026, 4 Apr Shift 2, 4 marks) The first ionization enthalpy of magnesium is +737 kJ mol^-1. Which value is the most probable second ionization enthalpy of magnesium: -906, -856, +1450 or +590 kJ mol^-1? Answer: All ionization enthalpies are positive because energy must be supplied to remove an electron from a neutral atom or a positive ion, so the two negative options are impossible. The second value must be larger than the first, which rules out +590. Magnesium loses its first electron from 4s, but after that the ion has the neon core with configuration [Ne] 3s1; the second electron comes from the same 3s subshell, and the ion is already positively charged. The value therefore rises substantially above 737, and +1450 kJ mol^-1 is the only plausible choice. The large second value also warns that the third electron would come from the closed neon core.

Q4 (JEE Main 2026, 5 Apr Shift 1, 4 marks) Statement I: the correct order of electronegativity of fluorine, oxygen and nitrogen is F > O > N. Statement II: the oxidation state of oxygen in OF2 is +2 and in Na2O is -2. Are the statements true? Answer: Statement I is true. Electronegativity rises across a period, so fluorine is above oxygen and oxygen is above nitrogen on the Pauling scale; the values are 4.0, 3.5 and 3.0. Statement II is true. Fluorine is more electronegative than oxygen, so in OF2 oxygen is assigned +2, while in Na2O the more electropositive sodium gives oxygen its usual -2 state. Both statements are correct.

Q5 (JEE Main 2026, 6 Apr Shift 1, 4 marks) The first and second ionization enthalpies of lithium are 520 and 7297 kJ mol^-1. Calculate the energy required to convert 3.5 mg of gaseous lithium into Li2+ ions. (Molar mass of Li = 7 g mol^-1) Answer: Converting one mole of Li to Li2+ requires both ionizations, so the energy per mole is 520 + 7297 = 7817 kJ mol^-1. The amount of lithium is 3.5 mg divided by 7 g mol^-1, which is 3.5 x 10^-3 g divided by 7 g mol^-1, giving 5.0 x 10^-4 mol. The energy is 7817 x 5.0 x 10^-4 = 3.91 kJ, which rounds to 4 kJ. The huge second ionization enthalpy dominates the sum because after the first electron is lost the Li+ ion has the helium core.

E = (520 + 7297) \times \frac{3.5 \times 1 0 ^{- 3}}{7} = 7817 \times 5.0 \times 1 0^{- 4} = 3.91 kJ \approx 4 kJ

Energy for converting 3.5 mg of lithium to Li2+

Q6 (JEE Main 2026, 6 Apr Shift 2, 4 marks) In a period, the first ionization enthalpy of the element at the extreme left and the negative electron gain enthalpy of the extreme right element, excluding the noble gases, are respectively: lowest and lowest, highest and lowest, lowest and highest, or highest and highest? Answer: The extreme left element of a period is an alkali metal with a single loosely held ns electron, so its first ionization enthalpy is the lowest in that period. The extreme right element before the noble gas is a halogen with an ns2 np5 configuration; it needs one electron to complete the octet, so its negative electron gain enthalpy is the highest in magnitude in the period. The correct pair is lowest and highest.

Q7 (NCERT Exemplar, 3 marks) For the isoelectronic species Na+, Mg2+, F- and O2-, the correct order of increasing radii is: (i) F- < O2- < Mg2+ < Na+ (ii) Mg2+ < Na+ < F- < O2- (iii) O2- < F- < Na+ < Mg2+ (iv) O2- < F- < Mg2+ < Na+. Answer: All four species have ten electrons. For isoelectronic species the radius falls as the nuclear charge rises. The atomic numbers are 12 for Mg, 11 for Na, 9 for F and 8 for O, so the radii increase in the order Mg2+, Na+, F-, O2-. The correct option is (ii). The negative ions are larger than the positive ions because the same electron count is held by a smaller nuclear charge. Source: NCERT Exemplar Class 11 Chemistry, Unit 3, MCQ 1.

Q8 (NCERT Exemplar, 3 marks) Which of the following is not an actinoid: curium with Z = 96, californium with Z = 98, uranium with Z = 92, or terbium with Z = 65? Answer: The actinoids are the elements in which the 5f subshell is being filled, and they run from actinium, Z = 89, to lawrencium, Z = 103. Curium, californium and uranium all fall inside that range. Terbium with Z = 65 belongs to the lanthanoid series, where the 4f subshell fills, so it is not an actinoid. The correct option is terbium. Source: NCERT Exemplar Class 11 Chemistry, Unit 3, MCQ 2.

Q9 (NCERT Exemplar, 3 marks) The order of the screening effect of electrons of the s, p, d and f orbitals of a given shell on its outer shell electrons is: (i) s > p > d > f (ii) f > d > p > s (iii) p < d < s > f (iv) f > p > s > d. Answer: Electrons in a more penetrating orbital spend more time close to the nucleus and screen an outer electron more effectively. For the same shell the order of penetration and therefore of screening is s, then p, then d, then f. Statement (i) gives this order correctly. In physical terms, an s electron has the greatest probability of being found close to the nucleus of the four subshells, while an f electron stays farthest out and screens least. This is also why an outer s electron is held more tightly than an outer p electron in the same shell, which explains the beryllium to boron dip in ionization enthalpy. Source: NCERT Exemplar Class 11 Chemistry, Unit 3, MCQ 3.

Q10 (NCERT Exemplar, 3 marks) The first ionization enthalpies of Na, Mg, Al and Si are in the order: (i) Na < Mg > Al < Si (ii) Na > Mg > Al > Si (iii) Na < Mg < Al < Si (iv) Na > Mg > Al < Si. Answer: Across a period the first ionization enthalpy rises in general, so sodium is lowest and silicon is highest. The expected smooth order is interrupted between magnesium and aluminium. Magnesium loses a 3s electron from a filled subshell while aluminium loses a 3p electron from a higher, better shielded orbital, so the value for aluminium is lower than that for magnesium. The order is Na below Mg above Al below Si, which is option (i). The same shape appears in the third period, where aluminium again sits below magnesium while silicon rises above both. Source: NCERT Exemplar Class 11 Chemistry, Unit 3, MCQ 4.