Periodic Table Previous year Questions with Solutions
Frequently Asked Questions
Yes, solving previous year's JEE questions on the periodic table helps you understand the question pattern, important topics, and the conceptual depth expected by examiners.
Commonly repeated topics include periodic trends (ionization energy, electronegativity, atomic size), classification of elements, anomalies, and group properties (especially s, p, d, f block elements).
Focus on group and period trends, element classification, and key exceptions. Use mnemonics, practice past questions, and revise NCERT textbooks thoroughly.
You can access them through standard JEE preparation resources such as NCERT Exemplar, the official NTA website, or trusted platforms like ALLEN.
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Periodic Table Previous Year Questions with Solutions
1.0Introduction
Mastering the Periodic Table is crucial for excelling in competitive exams like JEE and NEET. Practicing JEE Periodic Table previous year questions with solutions not only strengthens your conceptual understanding but also familiarizesyou with the exam pattern and frequently asked question types. These solved questions help students identify important topics, improve accuracy, and build exam confidence—making them an essential part of smart preparation.
In the JEE Main exam, topics related to the Periodic Table—such as periodic properties, element classification, and periodic trends—typically account for 1 to 2 questions in the Chemistry section.
JEE Main Previous Year Solved Questions on Periodic Table
JEE Adv Previous Year Solved Questions on Periodic Table
2.0Classification of Elements and Periodicity in Properties
The Periodic Table is an organized arrangement of elements based on their atomic number, electron configuration, and recurring chemical properties. Initially proposed by Dmitri Mendeleev, it has been refined over time to align with modern atomic theory.
The periodic table is a systematic arrangement of elements, grouping those with similar properties together.
3.0Key Concepts to Remember
Groups in the Periodic Table
The periodic table consists of 18 vertical columns known as groups. Elements in the same group form a family and are often named after the first element in the group. Some groups also have specific names, including:
Group 1 – Alkali Metals
Group 2 – Alkaline Earth Metals
Group 15 – Pnicogens
Group 16 – Chalcogens
Group 17 – Halogens
Group 18 – Noble Gases (or Aerogens)
All other groups are generally named after their first element.
Periods in the Periodic Table
The horizontal rows in the periodic table are called periods. There are seven periods in the long-form periodic table:
1st period (H → He) contains only 2 elements, making it the shortest period.
2nd period (Li → Ne) and 3rd period (Na → Ar) contain 8 elements each and are referred to as short periods.
4th period (K → Kr) and 5th period (Rb → Xe) contain 18 elements each and are considered long periods.
6th period (Cs → Ra) is the longest period, with 32 elements.
7th period starts with Fr (Francium) and remains incomplete, containing 19 elements so far.
Shielding Effect (Screening Effect)
In multi-electron atoms, the outermost (valence) electrons experience both:
Attraction from the positively charged nucleus.
Repulsion from inner-shell electrons.
This reduces the effective nuclear attraction on valence electrons, a phenomenon called the shielding or screening effect.
Effective Nuclear Charge (Zeff)
The effective nuclear charge (Zeff) is the net positive charge experienced by an electron. It is calculated using the formula:
Zeff = Z − 𝛔
Where:
Z = Actual nuclear charge
𝛔 = Screening constant
Across a period (left to right), the effective nuclear charge increases, due to greater nuclear attraction.
Related Video:
4.0Periodic Trends in Properties
Ionization Energy
Definition: The energy required to remove an electron from an atom.
Trend in a Group:Decreases down a group as atomic size increases.
Trend in a Period:Increases across a period due to stronger nuclear attraction.
Electron Affinity
The energy released when an electron is added to a neutral gaseous atom.
Electronegativity
Definition: The tendency of an atom to attract a bonded pair of electrons.
Trend in a Period:Increases from left to right due to decreasing atomic size and increasing nuclear charge. Alkali metals have the lowest electronegativity. Halogens have the highest electronegativity. Noble gases have zero electronegativity.
Trend in a Group: Decreases from top to bottom due to increasing atomic size.
Electronegativity Scales
Mulliken’s Scale: Defines electronegativity as the average of ionization potential and electron affinity.
Electronegativity= Ionization Potential+Electron Affinity2 Pauling’s Scale: The most commonly used scale, based on bond energy differences. ΔE=(Actual Bond Energy)−(EA−A * EB−B)
The relationship between Pauling and Mulliken electronegativity values is: 𝜒(Pauling) = 0.34×𝜒(Mulliken) − 0.2
5.0Other Periodic Properties
Atomic Volume
The volume occupied by one gram-atom of an element.
Calculated as: Atomic Volume=Gram Atomic Weight/Density (in solid state)
Measured in c.c./mole.
Density
Varies periodically with atomic number.
Generally, increases within a period, reaching a maximum at central elements, then decreases.
Melting and Boiling Points
Exhibit periodic trends across the periodic table.
Generally, increase across a period and decrease down a group.
Oxidation State (Oxidation Number, O.N.)
Indicates the total number of electrons gained or lost by an atom when forming a compound.
Magnetic Properties
Depends on the electronic structure of individual atoms.
6.0JEE Main Past Year Questions with Solutions on Periodic Table
Q.1 Given below are two statements :
Statement (I) : The 4f and 5f - series of elements are placed separately in the Periodic table to preserve the principle of classification.
Statement (II) :S-block elements can be found in pure form in nature. In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are false
Ans. (3)
Sol.s-block elements are highly reactive and found in combined state.
Q.2 The correct order of first ionization enthalpy values of the following elements is :
(A) O
(B) N
(C) Be
(D) F
(E) B
Choose the correct answer from the options given below :
(1) B < D < C < E < A (2) E < C < A < B < D
(3) C < E < A < B < D (4) A < B < D < C < E
Ans. (2)
Sol.Correct order of Ist IE
Li < B < Be < C < O < N < F < Ne
↓ ↓ ↓ ↓ ↓
E < C < A < B < D
Q.3 The transition metal having highest 3rd ionisation enthalpy is :
So Mn has the highest 3rd IE among all the given elements due to d5 configuration.
Q.4 Given below are two statements :
Statement (I) : Both metal and non-metal exist in p and d-block elements.
Statement (II) : Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.
In the light of the above statements, choose the most appropriate answer from the option given below:
(1) Both Statement I and Statement II are false
(2) Statement I is false but Statement II is true
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true
[2024]
Ans. (2)
Sol. I. In p-Block both metals and non metals are present but in d-Block only metals are present.
II. EN and IE of non metals are greater than that of metals
I - False, II-True
Q.5 It is observed that characteristic X-ray spectra of elements show regularity. When frequency to the power 'n' i.e. vn of X-rays emitted is plotted against atomic number 'Z', the following graph is obtained.
The value of 'n' is
(1) 1
(2) 2
(3) 12
(4) 3
Ans. (3)
Sol. According to Henry Moseley √v∝ z – b
So n = 12
Q.6 For electron gain enthalpies of the elements denoted as DegH, the incorrect option is :
(1) ΔegH (Cl) < ΔegH (F)
(2) ΔegH (Se) < ΔegH (S)
(3) ΔegH (I) < ΔegH (At)
(4) ΔegH (Te) < ΔegH (Po)
Ans. (2)
Sol.
(1) ΔegH (Cl) < Δeg H (F)
(–345) (– 328) Correct
(2) ΔegH (Se) < ΔegH (S)
(–195) (– 200) Incorrect
(3) ΔeggH (I) < ΔegH (At)
(–295) (– 270) Correct
(4) ΔegH (Te) < ΔegH (Po)
(–190) (– 183) Correct
Q.7 Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: 5f electrons can participate in bonding to a far greater extent than 4f electrons
Reason R: 5f orbitals are not as buried as 4f orbitals
In the light of the above statements, choose the correct answer from the options given below
(1) Both A and R are true but R is NOT the correct explanation of A
(2) Both A and R are true and R is the correct explanation of A
(3) A is false but R is true
(4) A is true but R is false
[2023]
Ans. (2)
Sol. 5f orbital not buried as 4f orbitals so e– present in 5f orbital experience less nuclear attraction than e– present in 4f orbital. Hence electrons of 5f orbital can take part in bonding to a far greater extent.
Q.8 Outermost electronic configurations of four elements A, B, C, D are given below:
3s2 (B) 3s23p1 (C) 3s23p3 (D) 3s23p4
The correct order of first ionization enthalpy for them is:
(1) (A) < (B) < (C) < (D)
(2) (B) < (A) < (D) < (C)
(3) (B) < (D) < (A) < (C)
(4) (B) < (A) < (C) < (D)
Ans. (3)
(A) 3s2→ Mg
(B) 3s23p1→ Al
(C) 3s23p3 → P
(D) 3s23p4→ S
Ans. (3) C > D > A > B.
P > S > Mg > Al
Q.9 The first ionization enthalpy of Na, Mg and Si, respectively, are: 496, 737 and 786 mo1-1. The first ionization enthalpy (kJ mol-1) of Al is:
(1) 487 (2) 768 (3) 577 (4) 856
Ans. (3)
Sol. I.E. : Na < Al < Mg < Si
∴ 496 < IE (Al) < 737
i.e. IE(Al) = 577 kJ mol-1
Q.10 In which of the following pairs, electron gain enthalpies of constituent elements are nearly the same or identical ?
(A) Rb and Cs (B) Na and K (C) Ar and Kr (D) I and At
Choose the correct answer from the options given below :
(1) (A) and (B) only (2) (B) and (C) only
(3) (A) and (C) only (4) (C) and (D) only
Ans. (3)
Sol. Rb & Cs have nearly the same electron gain enthalpy electron gain enthalpy = – 46 kj/ml.
Ar & Kr have the same . Value is ΔHeq + 96 kj/ml
Q.11. Match List -I with List - II
List - I
List - II
Electronic configuration of elements
Δi in kJ mol–1
(a) 1s22s2
(i) 801
(b) 1s22s22p4
(ii) 899
(c) 1s22s22p3
(iii)1314
(d) 1s22s22p1
(iv) 1402
Choose the most appropriate answer from the options given below -
Be has more IE compared to B due to extra stability & N has more IE compared to oxygen due to extra stability
Hence, N → 1402 kJ/mol
O → 1314 kJ/mol
B → 801 k
Be → 899 kJ/mol
Q.12. The ionic radii of F– and O2– respectively are 1.33 Å and 1.4 Å, while the covalent radius of N is 0.74 Å. The correct statement for the ionic radius of N3– from the following is :
(1) It is smaller than F– and N
(2) It is bigger than O2– and
(3) It is bigger than F– and N, but smaller than of O2–
(4) It is smaller than O2– and F–, but bigger than of N
Ans. (2)
Sol. F–, O2– and N3– all are isoelectronic species in which N3– have the least number of protons due to which its size increases as least nuclear attraction is experienced by the outer shell electrons.
Size order N3– > O2– > F–
Q.13 The correct order of conductivity of ions in water is :
(1) Na+ > K+ > Rb+ > Cs+
(2) Cs+ > Rb+ > K+ > Na+
(3) K+ > Na+ > Cs+ > Rb+
(4) Rb+ > Na+ > K+ > Li+
Ans.(2)
Sol.
∴ Correct option is Na+ > K+ > Rb+ > Cs+.
OR
Sol.
As the size of gaseous ions decreases, it gets more hydrated in water and hence, the size of aqueous ions increases. When this bulky ion moves in solution, it experiences greater resistance and hence lower conductivity.
