JEE Periodic Table & Periodicity PYQs With Video Solutions PDF

JEE Periodic Table & Periodicity PYQ

Periodic Table & Periodicity is an important chapter in JEE Chemistry. It helps you understand how the properties of elements change across periods and down groups in the periodic table. These trends are based on electronic configuration, effective nuclear charge, shielding effect, and atomic size. In this chapter, you study atomic radius, ionic radius, ionisation enthalpy, electron gain enthalpy, electronegativity, valency, metallic character, and periodic trends. These ideas are also useful in chemical bonding, coordination chemistry, p-block chemistry, and inorganic chemistry. That is why an understanding of this chapter is important for later topics as well. Most questions from Periodic Table & Periodicity are based on comparing properties, arranging elements in the correct order, identifying exceptions, and using electronic configuration. The calculations are usually limited, but a small mistake in group position, shell number, or trend can change the final answer. Solving previous-year questions helps you understand question patterns. It also improves your speed and makes trend-based questions easier to solve.

On this page, you can revise the key concepts, download the PYQ PDF, learn important formulas and trends, avoid common mistakes, and practise chapter-wise questions.

With regular practice, Periodic Table & Periodicity can become an easy and scoring chapter in JEE Chemistry.

JEE Periodic Table & Periodicity Previous-Year Questions

Previous-year questions help you understand which concepts are asked most often. In this chapter, questions commonly come from atomic size, ionisation enthalpy, electron gain enthalpy, electronegativity, effective nuclear charge, shielding, and electronic configuration.

While solving JEE Advanced PYQ, first identify the period, group, and electronic configuration of the elements being compared.

Across a period, effective nuclear charge generally increases.

Down a group, the number of electron shells increases.

These two ideas explain many important periodic trends.

Atomic radius generally decreases from left to right across a period and increases down a group.

Ionisation enthalpy generally increases across a period and decreases down a group.

For trend-based questions, do not apply a rule blindly.

There are important exceptions caused by extra stability of completely filled and half-filled subshells, differences in atomic size, and electron-electron repulsion.

For example, the first ionisation enthalpy of nitrogen is higher than oxygen because nitrogen has a stable half-filled 2p³ configuration. Similarly, beryllium has a higher first ionisation enthalpy than boron.

Electron gain enthalpy also shows exceptions. Chlorine has a more negative electron gain enthalpy than fluorine because the small size of fluorine increases repulsion in its compact 2p subshell.

Try to solve each question within a fixed time. After completing the set, check every wrong answer and find the exact reason behind the mistake.

Common reasons may include:

• You followed a general trend without checking an exception.
• You confused atomic radius with ionic radius.
• You ignored electronic configuration.
• You compared elements from different periods incorrectly.
• You forgot the effect of effective nuclear charge.

Keeping a short list of trends and major exceptions can make these questions much easier.

JEE Periodic Table & Periodicity PYQs PDF

You can download the JEE Periodic Table & Periodicity PYQs PDF from the section below.

First, solve the PDF without looking at the answers. Try to complete it in one sitting so that you can understand your current speed and accuracy.

After solving the questions, divide them into three groups:

• Questions you solved easily
• Questions that took more time
• Questions you could not solve

Revise the periodic trend or exception used in the difficult questions. Try those questions again after a few days.

While practising these JEE Periodic Table & Periodicity Questions, do not only compare the final answer with the solution.

Check the group, period, valence-shell configuration, number of shells, effective nuclear charge, and relevant exception.

When comparing isoelectronic species, remember that the species with the greater nuclear charge has the smaller radius.

For example, among O²⁻, F⁻, Na⁺, and Mg²⁺, all species have ten electrons. Their size decreases as nuclear charge increases.

For ionisation enthalpy questions, first check the electronic configuration.

Removing an electron from a stable filled or half-filled subshell may require more energy than a simple periodic trend suggests.

For electronegativity, remember that it generally increases across a period and decreases down a group.

Noble gases are usually not included in simple electronegativity comparisons.

When you solve the PDF again, focus on the questions that were incorrect or took too much time.

Important Formulas and Trends for JEE Periodic Table PYQs

Periodic Table and Periodicity questions are mainly based on periodic trends, electronic configurations and important exceptions. The following table summarises the most useful formulas and rules for quick revision.

These periodic trends are general and may have exceptions due to half-filled or completely filled subshells. For example, the first ionisation enthalpy of magnesium is higher than that of aluminium because magnesium has a stable filled 3s subshell. Similarly, nitrogen has a higher first ionisation enthalpy than oxygen because nitrogen has a stable half-filled 2p subshell.

Before applying a trend, identify the element’s group, period, valence-shell configuration and any important electronic-configuration exceptions.

Common Mistakes to Avoid in JEE Periodic Table & Periodicity PYQs

One common mistake is assuming that every periodic trend is perfectly regular.

General trends are useful, but JEE questions often test important exceptions.

Students also confuse atomic radius, covalent radius, van der Waals radius, and ionic radius.

Remember that a cation is usually smaller than its parent atom, while an anion is usually larger than its parent atom.

While solving JEE Periodic Table & Periodicity Questions, do not compare elements only by their position in the table.

Write the valence-shell electronic configuration when unclear.

Another common mistake is assuming that fluorine has the most negative electron gain enthalpy.

Chlorine has a more negative electron gain enthalpy than fluorine because the added electron experiences greater repulsion in fluorine's compact 2p orbital.

Students may also assume that oxygen has a higher first ionisation enthalpy than nitrogen because oxygen is placed to the right.

Nitrogen actually has a higher value due to the extra stability of its half-filled 2p³ configuration.

While comparing isoelectronic ions, students sometimes focus on charge signs instead of nuclear charge.

For isoelectronic species, a higher number of protons means a smaller radius.

Another mistake is treating electronegativity and electron gain enthalpy as the same property.

Electronegativity describes attraction for shared electrons in a bond, while electron gain enthalpy refers to the enthalpy change when an isolated gaseous atom gains an electron.

Before finalising your answer, check the electronic configuration, periodic table position, relevant trend, and possible exception once more.

List of JEE Periodic Table & Periodicity PYQs

Use the questions given below as a short chapter test.

Try to solve them like a JEE Advanced Mock Test. Set a time limit, avoid using notes, and check the answers only after completing the full set.

Question 1

Three elements X, Y and Z are in the 3$$^{rd}$$ period of the periodic table. The oxides of X, Y and Z, respectively, are basic, amphoteric and acidic. The correct order of the atomic numbers of X, Y and Z is:

A

$$Z < Y < X$$

B

$$X < Y < Z$$

C

$$X < Z < Y$$

D

$$Y < X < Z$$

Show Answer Explanation Video Solution

correct answer:- 2

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Question 2

The atomic number of the element unnilennium is:

A

109

B

102

C

108

D

119

Show Answer Explanation Video Solution

correct answer:- 1

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Question 3

The correct order of electron gain enthalpy is:

A

S > O > Se > Te

B

O > S > Se > Te

C

S > Se > Te > O

D

Te > Se > S > O

Show Answer Explanation

correct answer:- 3

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Question 4

The ionic radius of Na$$^+$$ ions is 1.02 angstrom. The ionic radii (in angstrom) of Mg$$^{2+}$$ and Al$$^{3+}$$, respectively, are:

A

1.05 and 0.99

B

0.72 and 0.54

C

0.85 and 0.99

D

0.68 and 0.72

Show Answer Explanation

Solution

Na$$^+$$, Mg$$^{2+}$$, and Al$$^{3+}$$ are isoelectronic species, each having 10 electrons. However, they differ in their nuclear charges: Na has 11 protons, Mg has 12, and Al has 13. As nuclear charge increases across this isoelectronic series, the electrons are pulled more tightly toward the nucleus, causing the ionic radius to decrease.

The ionic radius of Na$$^+$$ is given as 1.02 angstrom. Since Mg$$^{2+}$$ has a higher nuclear charge, its ionic radius is smaller, approximately 0.72 angstrom. Al$$^{3+}$$, with an even greater nuclear charge, has the smallest radius of about 0.54 angstrom.

Therefore, the ionic radii of Mg$$^{2+}$$ and Al$$^{3+}$$ are $$\textbf{0.72}$$ and $$\textbf{0.54}$$ angstrom respectively, which corresponds to option (2).

correct answer:- 2

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Question 5

The ionic radius of Na$$^+$$ ions is 1.02 angstrom. The ionic radii (in $$\mathring{A}$$) of Mg$$^{2+}$$ and Al$$^{3+}$$, respectively, are:

A

1.05 and 0.99

B

0.72 and 0.54

C

0.85 and 0.99

D

0.68 and 0.72

Show Answer Explanation

correct answer:- 2

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Question 6

Which one of the following statements for D.I. Mendeleeff, is incorrect?

A

He authored the textbook - Principles of Chemistry.

B

At the time, he proposed Periodic Table of elements structure of atom was known.

C

Element with atomic number 101 is named after him.

D

He invented accurate barometer.

Show Answer Explanation

correct answer:- 2

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Question 7

The set of elements that differ in mutual relationship from those of the other sets is:

A

Li - Mg

B

B - Si

C

Be - Al

D

Li - Na

Show Answer Explanation

correct answer:- 4

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Question 8

What are the products formed in sequence when excess of CO$$_2$$ is passed in slaked lime?

A

CaO, CaCO$$_3$$

B

Ca(HCO$$_3$$)$$_2$$, CaCO$$_3$$

C

CaCO$$_3$$, Ca(HCO$$_3$$)$$_2$$

D

CaO, Ca(HCO$$_3$$)$$_2$$

Show Answer Explanation

correct answer:- 3

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Question 9

The absolute value of the electron gain enthalpy of halogens satisfies:

A

$$I > Br > Cl > F$$

B

$$Cl > Br > F > I$$

C

$$Cl > F > Br > I$$

D

$$F > Cl > Br > I$$

Show Answer Explanation

correct answer:- 3

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Question 10

A s-block element (M) reacts with oxygen to form an oxide of the formula MO$$_2$$. The oxide is pale yellow in colour and paramagnetic. The element (M) is:

A

Mg

B

Na

C

Ca

D

K

Show Answer Explanation

correct answer:- 4

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Question 11

Match List - I with List - II.

List-I (Salt)	List-II (Flame colour wavelength)
(a) LiCl	(i) 455.5 nm
(b) NaCl	(ii) 670.8 nm
(c) RbCl	(iii) 780.0 nm
(d) CsCl	(iv) 589.2 nm

Choose the correct answer from the options given below:

A

(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)

B

(a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)

C

(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

D

(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)

Show Answer Explanation

correct answer:- 4

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Question 12

The correct decreasing order for metallic character is

A

$$Na > Mg > Be > Si > P$$

B

$$P > Si > Be > Mg > Na$$

C

$$Si > P > Be > Na > Mg$$

D

$$Be > Na > Mg > Si > P$$

Show Answer Explanation

correct answer:- 1

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Question 13

The first ionization energy of magnesium is smaller, as compared to that of elements X and Y, but higher than that of Z. The elements X, Y and Z, respectively, are :

A

chlorine, lithium and sodium

B

argon, lithium and sodium

C

argon, chlorine and sodium

D

neon, sodium and chlorine

Show Answer Explanation

correct answer:- 3

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Question 14

Two elements A and B have similar chemical properties. They don't form solid hydrogencarbonates, but react with nitrogen to form nitrides. A and B, respectively, are:

A

Na and Rb

B

Na and Ca

C

Cs and Ba

D

Li and Mg

Show Answer Explanation Video Solution

correct answer:- 4

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Question 15

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R .
Assertion A : Sodium is about 30 times as abundant as potassium in the oceans.
Reason R : Potassium is bigger in size than sodium.
In the light of the above statements, choose the correct answer from the options given below

A

Both A and R are true and R is the correct explanation of A

B

A is true but R is false

C

Both A and R are true but R is NOT the correct explanation of A

D

Both A and R are false

Show Answer Explanation Video Solution

correct answer:- 1

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Question 16

Inert gases have positive electron gain enthalpy. Its correct order is

A

Xe < Kr < Ne < He

B

He < Ne < Kr < Xe

C

He < Xe < Kr < Ne

D

He < Kr < Xe < Ne

Show Answer Explanation

correct answer:- 2

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Question 17

The total number of isoelectronic species from the given set is _____.
O$$^{2-}$$, F$$^-$$, Al, Mg$$^{2+}$$, Na$$^+$$, O$$^+$$, Mg, Al$$^{3+}$$, F

Show Answer Explanation Video Solution

correct answer:- 5

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Question 18

Number of water molecules in washing soda and soda ash respectively are:

A

1 and 10

B

10 and 1

C

10 and 0

D

1 and 0

Show Answer Explanation

correct answer:- 3

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Question 19

Match List I with List II.
List I With Ore List II Composition
(A) Siderite             (I) FeCO$$_3$$
(B) Malachite         (II) CuCO$$_3$$ · Cu(OH)$$_2$$
(C) Sphalerite         (III) ZnS
(D) Calamine          (IV) ZnCO$$_3$$
Choose the correct answer from the options given below:

A

A-I, B-II, C-III, D-IV

B

A-III, B-IV, C-E, D-I

C

A-IV, B-III, C-I, D-II

D

A-I, B-II, C-IV, D-III

Show Answer Explanation Video Solution

correct answer:- 1

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Question 20

Given below are two statements:
Statement - I: Along the period, the chemical reactivity of the element gradually increases from group 1 to group 18.
Statement - II: The nature of oxides formed by group 1 element is basic while that of group 17 elements is acidic.
In the light of the above statements, choose the most appropriate from the options given below:

A

Both Statement I and Statement II are true.

B

Statement I is true but Statement II is false.

C

Statement I is false but Statement II is true.

D

Both Statement I and Statement II are false.

Show Answer Explanation

correct answer:- 3

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Question 21

Ion having highest hydration enthalpy among the given alkaline earth metal ions is:

A

Be$$^{2+}$$

B

Sr$$^{2+}$$

C

Ba$$^{2+}$$

D

Ca$$^{2+}$$

Show Answer Explanation

correct answer:- 1

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Question 22

The correct statement(s) regarding the periodic properties of elements is(are)

A

Second ionization enthalpy of carbon atom is less than that of boron atom.

B

Increasing order of ionic radii: $$\mathrm{Al^{3+}} < \mathrm{Mg^{2+}} < \mathrm{Na^{+}}$$

C

Under identical conditions, in solid state, the density of potassium metal is more than density of sodium metal.

D

The H$$-$$H bond is weaker than F$$-$$F bond.

Show Answer

correct answer:- 1,2

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Question 23

The first ionization enthalpies of $$Be$$, $$B$$, $$N$$ and $$O$$ follow the order

A

$$B < Be < O < N$$

B

$$O < N < B < Be$$

C

$$Be < B < N < O$$

D

$$B < Be < N < O$$

Show Answer Explanation

correct answer:- 1

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Question 24

The correct trend in the first ionization enthalpies of the elements in the 2nd period of the periodic table is:

A

$$Li < B < Be < C < N < O < F < Ne$$

B

$$Li < Be < B < C < O < N < F < Ne$$

C

$$Li < B < Be < C < O < N < F < Ne$$

D

$$Li < Be < B < C < N < O < F < Ne$$

Show Answer Explanation

Solution

Ionization enthalpy generally increases across a period due to decreasing atomic size and increasing effective nuclear charge. However, there are two important exceptions in the second period.

Beryllium ($$1s^2 2s^2$$) possesses a completely filled $$2s$$ subshell, which is particularly stable. Boron ($$1s^2 2s^2 2p^1$$) has its outermost electron in the higher-energy $2p$ orbital, making it easier to remove. Hence, $IE(Be) > IE(B)$.

Nitrogen ($$1s^2 2s^2 2p^3$$) has a half-filled $2p$ subshell, which provides extra stability. Oxygen ($$1s^2 2s^2 2p^4$$) contains one paired electron in the $2p$ subshell, resulting in increased electron-electron repulsion and making electron removal easier. Therefore, $$IE(N) > IE(O)$$.

Taking these exceptions into account, the correct order is:

$$Li < B < Be < C < O < N < F < Ne$$

Hence, option C is the correct option.

correct answer:- 3

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Instruction for set :

Question 25

Match the chemical elements listed in List-I with their corresponding characteristic atomic or physical properties described in List-II according to the trends in the periodic table:

Select the correct matching option from the choices given below:

A

P-1, Q-3, R-2, S-4

B

P-1, Q-3, R-4, S-2

C

P-3, Q-1, R-2, S-4

D

P-3, Q-1, R-4, S-2

Show Answer Explanation

Solution

Let us analyze the periodic properties of each element systematically to establish the correct match:

1. Fluorine (P):
Fluorine is the first element of Group 17 (halogens). Due to its extremely small atomic radius and high effective nuclear charge ($$Z_{\text{eff}}$$), it has the maximum power to attract shared pairs of bonding electrons towards itself. Hence, Fluorine possesses the highest electronegativity value ($$4.0$$) on the Pauling scale .
$$\implies \text{P matches with 3}$$

2. Chlorine (Q):
Although Fluorine is more electronegative, its extremely small atomic size leads to strong inter-electronic repulsions in its relatively compact $$2p$$ subshell when a new electron is added. Chlorine ($$3p$$ subshell) has a larger volume, which easily accommodates the incoming electron with minimal repulsion. Thus, Chlorine releases more energy upon electron capture and has the highest negative electron gain enthalpy in the entire periodic table.
$$\implies \text{Q matches with 1}$$

3. Helium (R):
Helium is located at the top-right corner of the periodic table (Group 18). It features a stable, completely filled closed-shell electronic configuration ($$1s^2$$) combined with the smallest atomic size among all elements. Because of this ultimate nuclear hold, removing an electron requires an extraordinary amount of energy, giving Helium the highest first ionization enthalpy .
$$\implies \text{R matches with 4}$$

4. Cesium (S):
Cesium belongs to Group 1 (alkali metals) near the bottom of the table. Its valence electron sits far away from the nucleus in the $$6s$$ orbital, shielded heavily by inner shells. It loses this electron with incredible ease, making it the most electropositive stable non-radioactive element .
$$\implies \text{S matches with 2}$$

Conclusion:
Combining our individual matches yields the sequence: P-3, Q-1, R-4, S-2 .

correct answer:- 4

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Instruction for set :

Question 26

Show Answer Explanation

Solution

Let us analyze the acid-base behavior of each of the given alkaline earth metal oxides and hydroxides based on periodic properties:

1. Analysis of Beryllium Compounds:
Beryllium ($$\text{Be}$$) exhibits a diagonal relationship with Aluminum ($$\text{Al}$$). Due to its exceptionally small ionic size and high charge density, the beryllium ion ($$\text{Be}^{2+}$$) has high polarizing power, which imparts significant covalent character to its bonds.

• $$\text{BeO}$$ (Beryllium Oxide): It is insoluble in water but dissolves in both acids (forming salts) and strong bases (forming beryllates), making it amphoteric .
  $$\text{BeO} + 2\text{HCl} \rightarrow \text{BeCl}_2 + \text{H}_2\text{O}$$
  $$\text{BeO} + 2\text{NaOH} + \text{H}_2\text{O} \rightarrow \text{Na}_2[\text{Be(OH)}_4]$$
• $$\text{Be(OH)}_2$$ (Beryllium Hydroxide): Similarly, it reacts with both mineral acids and alkalis, proving its amphoteric nature.
  $$\text{Be(OH)}_2 + 2\text{HCl} \rightarrow \text{BeCl}_2 + 2\text{H}_2\text{O}$$
  $$\text{Be(OH)}_2 + 2\text{NaOH} \rightarrow \text{Na}_2[\text{Be(OH)}_4]$$

2. Analysis of Barium and Strontium Compounds:
As we move down Group 2 from Beryllium to Barium, the atomic radius increases, ionization enthalpy drops significantly, and the electropositive (metallic) character increases. Consequently, the oxides and hydroxides become increasingly ionic and highly basic.

• $$\text{BaO}$$ (Barium Oxide): It is a strongly basic metal oxide that reacts violently with water to form a highly alkaline barium hydroxide solution.
• $$\text{Sr(OH)}_2$$ (Strontium Hydroxide): It is a strong, basic diacidic base that dissociates completely in aqueous solutions.

Summary of Classification:
* $$\text{BeO}$$: Amphoteric * $$\text{BaO}$$: Basic * $$\text{Be(OH)}_2$$: Amphoteric * $$\text{Sr(OH)}_2$$: Basic

Therefore, there are exactly 2 amphoteric compounds ($$\text{BeO}$$ and $$\text{Be(OH)}_2$$) in the given list.

Answer:
The correct integer value is 2 .

correct answer:- 2

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Question 27

A monoatomic anion $$(A^-)$$ has 45 neutrons and 36 electrons. Atomic mass, group in the periodic table and physical state at room temperature of the element $$(A)$$ respectively are :

A

$$80, 17, {liquid}$$

B

$$81, 16, {solid}$$

C

$$80, 16, {gas}$$

D

$$81, 15, {gas}$$

Show Answer Explanation Video Solution

correct answer:- 1

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Question 28

The IUPAC symbol for the element with atomic number 119 would be:

A

Unh

B

Uun

C

Une

D

Uue

Show Answer Explanation Video Solution

correct answer:- 4

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Question 29

The electron gain enthalpy (in kJ/mol) of fluorine, chlorine, bromine and iodine, respectively, are

A

$$-296, -325, -333$$ and $$-349$$

B

$$-349, -333, -325$$ and $$-296$$

C

$$-333, -349, -325$$ and $$-296$$

D

$$-333, -325, -349$$ and $$-296$$

Show Answer Explanation Video Solution

correct answer:- 3