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Classification of Elements and Periodicity — JEE Main Chemistry MCQs with Solutions
Free JEE Main Chemistry Classification of Elements and Periodicity MCQs with step-by-step solutions covering History, Laws & Modern Periodic Table, Electronic Configuration, IUPAC & Blocks, Atomic & Ionic Radii, Screening & Isoelectronic, Ionization Enthalpy & Electron Gain Enthalpy, Electronegativity, Reactivity & Anomalous Properties. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — History, Laws & Modern Periodic Table · easy · theory
The modern periodic law states that the properties of elements are a periodic function of their:
A. Density
B. Number of neutrons
C. Atomic mass
D. Atomic number ✓ Correct
Solution: Moseley established that atomic number (Z), not atomic mass, is the fundamental property; properties repeat periodically with Z.
Q2 — History, Laws & Modern Periodic Table · easy · theory
Mendeleev's periodic law was based on the:
A. Atomic mass ✓ Correct
B. Atomic number
C. Number of valence electrons
D. Density
Solution: Mendeleev arranged elements in order of increasing atomic mass.
Q3 — History, Laws & Modern Periodic Table · easy · theory
Döbereiner's triads are groups of three elements in which the atomic mass of the middle element is approximately:
A. The arithmetic mean of the other two ✓ Correct
B. The sum of the other two
C. Double the lightest
D. The product of the other two
Solution: In a triad, m(middle) ≈ [m(1) + m(3)]/2.
Q4 — History, Laws & Modern Periodic Table · easy · theory
Newlands' law of octaves compared the repetition of properties (every 8th element) to:
A. Days of the week
B. Colours of light
C. Musical notes ✓ Correct
D. Planetary orbits
Solution: Every eighth element had similar properties, like the octaves in music.
Q5 — History, Laws & Modern Periodic Table · easy · theory
A major achievement of Mendeleev's table was that he:
A. Explained isotopes
B. Placed hydrogen correctly
C. Left gaps and correctly predicted properties of undiscovered elements (eka-aluminium, eka-silicon) ✓ Correct
D. Used atomic number
Solution: He predicted eka-aluminium (Ga) and eka-silicon (Ge) with remarkable accuracy.
Q6 — History, Laws & Modern Periodic Table · easy · theory
The long form (modern) periodic table has:
A. 7 periods and 8 groups
B. 6 periods and 16 groups
C. 8 periods and 18 groups
D. 7 periods and 18 groups ✓ Correct
Solution: The long form has 7 horizontal periods and 18 vertical groups, arranged by atomic number.
Q7 — History, Laws & Modern Periodic Table · easy · theory
The period number of an element corresponds to the:
A. Number of valence electrons
B. Number of protons
C. Group number
D. Principal quantum number (n) of its outermost shell ✓ Correct
Solution: The highest principal quantum number of the valence shell equals the period number.
Q8 — History, Laws & Modern Periodic Table · easy · theory
Elements belonging to the same group have the same:
A. Number of neutrons
B. Number of valence electrons (similar valence-shell configuration) ✓ Correct
C. Atomic mass
D. Number of shells
Solution: Same group ⇒ same outer-shell configuration ⇒ similar chemical properties.
Q9 — History, Laws & Modern Periodic Table · easy · numerical
In the triad Li (7), Na (23), K (39), the predicted atomic mass of Na from the triad rule is:
A. 23 ✓ Correct
B. 46
C. 32
D. 16
Solution: Mean of Li and K = (7 + 39)/2 = 23, matching Na.
Q10 — History, Laws & Modern Periodic Table · easy · numerical
For the triad Cl (35.5), Br (?), I (127), the atomic mass of Br predicted by the triad rule is about:
A. 91.5
B. 46
C. 162
D. 81 ✓ Correct
Solution: Mean = (35.5 + 127)/2 = 81.25 ≈ 80 (actual Br).
Q11 — History, Laws & Modern Periodic Table · easy · numerical
For the triad Ca (40), Sr (?), Ba (137), the predicted mass of Sr is:
A. 88.5 ✓ Correct
B. 48.5
C. 177
D. 97
Solution: Mean = (40 + 137)/2 = 88.5 ≈ 88 (actual Sr).
Q12 — History, Laws & Modern Periodic Table · easy · numerical
The element with atomic number 11 lies in:
A. Period 3, Group 11
B. Period 2, Group 1
C. Period 1, Group 3
D. Period 3, Group 1 ✓ Correct
Solution: Na: 1s²2s²2p⁶3s¹ ⇒ period 3, group 1.
Q13 — History, Laws & Modern Periodic Table · easy · numerical
For the triad S (32), Se (?), Te (127.6), the predicted atomic mass of Se is about:
A. 80 ✓ Correct
B. 160
C. 48
D. 95.8
Solution: Mean = (32 + 127.6)/2 ≈ 79.8 ≈ 79 (actual Se).
Q14 — History, Laws & Modern Periodic Table · easy · numerical
The element with atomic number 12 is in:
A. Period 3, Group 12
B. Period 4, Group 2
C. Period 2, Group 2
D. Period 3, Group 2 ✓ Correct
Solution: Mg: [Ne]3s² ⇒ period 3, group 2.
Q15 — History, Laws & Modern Periodic Table · easy · numerical
The total number of groups in the long form of the periodic table is:
A. 9
B. 8
C. 16
D. 18 ✓ Correct
Solution: The modern long form has 18 vertical groups.
Q16 — History, Laws & Modern Periodic Table · easy · numerical
The element with atomic number 19 (K) is found in:
A. Period 4, Group 19
B. Period 1, Group 4
C. Period 4, Group 1 ✓ Correct
D. Period 3, Group 1
Solution: K: [Ar]4s¹ ⇒ period 4, group 1.
Q17 — History, Laws & Modern Periodic Table · easy · numerical
The number of elements in the 3rd period is:
A. 2
B. 32
C. 18
D. 8 ✓ Correct
Solution: The 3rd period fills 3s and 3p ⇒ 2 + 6 = 8 elements.
Q18 — Electronic Configuration, IUPAC & Blocks · easy · theory
The general valence-shell configuration of the s-block elements is:
A. ns² np¹⁻⁶
B. (n−2)f¹⁻¹⁴
C. ns¹⁻² ✓ Correct
D. (n−1)d¹⁻¹⁰ ns⁰⁻²
Solution: s-block (groups 1–2) have ns¹ or ns² valence configurations.
Q19 — Electronic Configuration, IUPAC & Blocks · easy · theory
The general valence-shell configuration of the p-block elements is:
A. (n−2)f¹⁻¹⁴
B. (n−1)d¹⁻¹⁰ ns¹⁻²
C. ns¹⁻²
D. ns² np¹⁻⁶ ✓ Correct
Solution: p-block (groups 13–18): ns² np¹ to ns² np⁶.
Q20 — Electronic Configuration, IUPAC & Blocks · easy · theory
The general configuration of the d-block (transition) elements is:
A. (n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹
B. ns¹⁻²
C. (n−1)d¹⁻¹⁰ ns⁰⁻² ✓ Correct
D. ns² np¹⁻⁶
Solution: Transition elements fill the (n−1)d subshell: (n−1)d¹⁻¹⁰ ns⁰⁻².
Q21 — Electronic Configuration, IUPAC & Blocks · easy · theory
The f-block (inner-transition) elements have the general configuration:
A. (n−1)d¹⁻¹⁰ ns²
B. ns² np¹⁻⁶
C. (n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns² ✓ Correct
D. ns¹⁻²
Solution: Inner-transition elements fill the (n−2)f subshell.
Q22 — Electronic Configuration, IUPAC & Blocks · easy · theory
In IUPAC nomenclature for Z > 100, the name always ends with the suffix:
A. -ous
B. -ium ✓ Correct
C. -ate
D. -ide
Solution: The digit-root name is completed with the suffix "-ium".
Q23 — Electronic Configuration, IUPAC & Blocks · easy · theory
The inert (noble) gases have the general valence configuration:
A. (n−1)d¹⁰ ns²
B. ns² np⁶ (except He: 1s²) ✓ Correct
C. ns¹
D. ns² np⁵
Solution: Group 18 elements have completely filled ns²np⁶ shells (He is 1s²).
Q24 — Electronic Configuration, IUPAC & Blocks · easy · theory
Transition elements are also called d-block elements because their differentiating electron enters the:
A. np subshell
B. ns subshell
C. (n−2)f subshell
D. (n−1)d subshell ✓ Correct
Solution: The last electron goes into the (n−1)d orbitals.
Q25 — Electronic Configuration, IUPAC & Blocks · easy · theory
Elements with configuration ending in ns² np⁶ (group 18) are chemically:
A. Highly reactive metals
B. Strong reducing agents
C. Strong oxidising acids
D. Inert / very unreactive (stable octet) ✓ Correct
Solution: A complete octet gives noble gases their characteristic inertness.
Q26 — Electronic Configuration, IUPAC & Blocks · easy · theory
The lanthanides and actinides are placed in the ___ block:
A. s
B. f ✓ Correct
C. p
D. d
Solution: Inner-transition elements (4f and 5f filling) constitute the f-block.
Q27 — Electronic Configuration, IUPAC & Blocks · easy · numerical
The IUPAC name of the element with Z = 104 is:
A. Ununquadium (Uuq)
B. Unniltrium (Unt)
C. Unnilpentium (Unp)
D. Unnilquadium (Unq) ✓ Correct
Solution: 1-0-4 → un-nil-quad-ium = Unnilquadium, symbol Unq.
Q28 — Electronic Configuration, IUPAC & Blocks · easy · numerical
The IUPAC name of the element with Z = 105 is:
A. Ununpentium (Uup)
B. Nilunpentium (Nup)
C. Unnilquadium (Unq)
D. Unnilpentium (Unp) ✓ Correct
Solution: 1-0-5 → un-nil-pent-ium = Unnilpentium, symbol Unp.
Q29 — Electronic Configuration, IUPAC & Blocks · easy · numerical
The number of valence electrons in an element of Group 16 is:
A. 4
B. 6 ✓ Correct
C. 8
D. 16
Solution: Group 16 ⇒ ns²np⁴ ⇒ 6 valence electrons.
Q30 — Electronic Configuration, IUPAC & Blocks · easy · numerical
The period and group of the element with Z = 15 are:
A. Period 3, Group 5
B. Period 2, Group 15
C. Period 3, Group 15 ✓ Correct
D. Period 15, Group 3
Solution: P: [Ne]3s²3p³ ⇒ period 3, group 15.