Periodic Table Most Expected JEE Main PYQs MCQs With Solutions pdf download

Anand Classes brings you the Periodic Table Most Expected JEE Main PYQs MCQs With Solutions PDF, specially prepared to help students crack JEE Main with confidence. This study material covers important questions from previous years’ papers along with detailed solutions, making it easier for students to understand patterns and prepare effectively. Ideal for Class 11 and 12 students, this PDF focuses on frequently asked concepts, trends, and numerical problems related to the periodic table. Click the print button to download study material and notes.

JEE Main 2023 Question :
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R)
Assertion (A): The first ionization enthalpy of 3d series elements is more than that of group 2 metals.
Reason (R): In 3d series of elements successive filling of d-orbitals takes place.
Options:
(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) (A) is false but (R) is true

Answer: Correct Option: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)

Step 1: Understanding first ionization enthalpy

• First ionization enthalpy is the energy required to remove the most loosely bound electron from a neutral gaseous atom.
• It depends on effective nuclear charge, atomic radius, shielding effect, and electronic configuration.

Step 2: Comparing Group 2 vs 3d series

• Group 2 metals (alkaline earth metals) have their outer electrons in the ns orbital. These s-electrons are relatively easy to remove due to lower nuclear charge and larger atomic size.
• Transition metals of the 3d series, however, experience a greater effective nuclear charge because the poor shielding of the 3d electrons makes the nucleus pull outer electrons more strongly.
• As a result, their first ionization enthalpy is generally higher than that of Group 2 metals in the same period.

Step 3: Role of successive filling of d-orbitals

• In the 3d series, electrons are added to the 3d subshell.
• These d-electrons shield outer 4s electrons poorly, causing the effective nuclear charge on the 4s electrons to increase.
• This increased nuclear attraction raises the ionization enthalpy.

Final Answer
$$ \boxed{(A)\ \text{Both (A) and (R) are true and (R) is the correct explanation of (A)}} $$

Concept Takeaway

• Across a period, ionization enthalpy increases due to higher nuclear charge and smaller atomic size.
• d-electrons shield poorly, making it harder to remove the outermost s-electrons in transition metals.
• Group 2 elements lose ns electrons more easily, while 3d elements hold their outer electrons more strongly.
• Useful for JEE Chemistry (Periodic Trends, Class 11 NCERT) — remember comparisons between s-block and d-block elements.

JEE Main 2023 Question:
The correct increasing order of the ionic radii is:
(A) Cl⁻ < Ca²⁺ < K⁺ < S²⁻
(B) K⁺ < S²⁻ < Ca²⁺ < Cl⁻
(C) Ca²⁺ < K⁺ < Cl⁻ < S²⁻
(D) S²⁻ < Cl⁻ < Ca²⁺ < K⁺

Answer: Correct option (C) Ca²⁺ < K⁺ < Cl⁻ < S²⁻

Step 1: Concept – Isoelectronic Species
The ions Ca²⁺, K⁺, Cl⁻, and S²⁻ all have 18 electrons (isoelectronic with Argon).
Rule: For isoelectronic species, the greater the nuclear charge (Z), the smaller the ionic radius.

Step 2: Nuclear Charges
Ca²⁺ : Z = 20
K⁺ : Z = 19
Cl⁻ : Z = 17
S²⁻ : Z = 16

Step 3: Comparison

• Higher Z → smaller radius
• Lower Z → larger radius

Step 4: Order of radii
Ca²⁺ (smallest) < K⁺ < Cl⁻ < S²⁻ (largest)

Final Answer:
(C) Ca²⁺ < K⁺ < Cl⁻ < S²⁻

Key Takeaway:

• Isoelectronic species → compare by nuclear charge.
• More protons → smaller radius, fewer protons → larger radius.
• Understanding ionic radii trends is crucial in the periodic table and periodicity chapter. Such questions are common in JEE PYQs and help in building strong class 11 chemistry preparation notes. For detailed study material and practice, refer to Anand Classes resources.

JEE Main 2023 Question:
Bond dissociation energy of “E–H” bond of the H₂E hydrides of group 16 elements (given below), follows order.
(A) O
(B) S
(C) Se
(D) Te
Options
(A) A > B > C > D
(B) D > C > B > A
(C) B > A > C > D
(D) A > B > D > C

Answer: Correct option (A) A > B > C > D

Step 1: Concept – Bond Dissociation Energy Trend
Bond dissociation energy decreases down a group in the periodic table.
Reason: The atomic size of the central atom (E) increases down the group, resulting in longer and weaker E–H bonds, which require less energy to break.

Step 2: Apply to Group 16 Hydrides
Hydrides: H₂O, H₂S, H₂Se, H₂Te
Order of bond strength: O–H > S–H > Se–H > Te–H

Step 3: Final Comparison
Thus, the bond dissociation energy order is:
O > S > Se > Te

Final Answer
$$ \boxed{A > B > C > D} $$

Concept Takeaway

• Bond strength decreases down the group due to increasing atomic radius.
• Strongest bond is in H₂O, weakest in H₂Te.
• Important trend for class 11 chemistry, periodic table, and periodicity in properties.
  For practice, go through JEE PYQs chapterwise and detailed preparation notes from Anand Classes to strengthen this topic.

JEE Main 2023 Question:
Match List I with List II

Options:
(A) A-II, B-IV, C-I, D-III
(B) A-IV, B-III, C-II, D-I
(C) A-IV, B-II, C-I, D-III
(D) A-I, B-III, C-IV, D-II

Answer: Correct option (C) A-IV, B-II, C-I, D-III

Step 1: Recall block classification

• s-block: Elements with last electron in s-orbital (Groups 1 and 2).
• p-block: Last electron in p-orbital (Groups 13–18).
• d-block: Transition metals with last electron in d-orbital (Groups 3–12).
• f-block: Lanthanides and actinides with differentiating electron in f-orbital.

Step 2: Identify elements by atomic number

• (A) 37 → Rubidium (Rb), Group 1 alkali metal → s-block (IV).
• (B) 78 → Platinum (Pt), Group 10 transition metal → d-block (II).
• (C) 52 → Tellurium (Te), Group 16 chalcogen → p-block (I).
• (D) 65 → Terbium (Tb), lanthanide → f-block (III).

Step 3: Final Matching
(A–IV), (B–II), (C–I), (D–III).

Final Answer
$$ \boxed{A \to IV, \; B \to II, \; C \to I, \; D \to III} $$

Concept Takeaway

• Block of element = type of subshell (s, p, d, f) where last electron enters.
• Helps in quickly classifying elements in periodic table and periodicity questions.
  For JEE preparation, practice Anand Classes PYQs chapterwise and revise class 11 chemistry periodic table trends for accuracy in block identification.

JEE Main 2023 Question:
Given below are two statements:
Statement I: The decrease in first ionization enthalpy from B to Al is much larger than that from Al to Ga.
Statement II: The d orbitals in Ga are completely filled.
In the light of the above statements, choose the most appropriate answer from the options given below:
(A) Statement I is correct but Statement II is incorrect
(B) Statement I is incorrect but Statement II is correct
(C) Both the statements I and II are correct
(D) Both the statements I and II are incorrect

Answer: Correct option (C) Both the statements I and II are correct

Step 1: Analyze Statement I

• As we move down Group 13 (B → Al → Ga), ionization enthalpy generally decreases due to increase in atomic size and shielding effect.
• The drop from B to Al is large because the size increases significantly from period 2 to period 3.
• From Al to Ga, the decrease is smaller than expected because the poor shielding of 3d electrons in Ga increases effective nuclear charge.
• ✅ Statement I is correct.

Step 2: Analyze Statement II

• Gallium (Ga, Z = 31) has electronic configuration:
  $$ [Ar] \, 3d^{10} \, 4s^{2} \, 4p^{1} $$
• The 3d orbitals are completely filled (10 electrons).
• ✅ Statement II is correct.

Step 3: Final Evaluation
Both statements are correct.

Final Answer
$$ \boxed{\text{Both the statements I and II are correct}} $$

Concept Takeaway

• Ionization enthalpy decreases down a group, but the effect of d-orbital shielding in Ga reduces the expected drop.
• Completely filled subshells (like 3d¹⁰ in Ga) give extra stability.
  For JEE Chemistry, always check periodic table and periodicity anomalies such as d-block shielding.
  Practice more PYQs chapterwise, study material, and notes from Anand Classes to master periodic trends.

JEE Main 2023 Question:
Inert gases have positive electron gain enthalpy. Its correct order is:
(A) He < Ne < Kr < Xe
(B) He < Xe < Kr < Ne
(C) Xe < Kr < Ne < He
(D) He < Kr < Xe < Ne

Answer: Correct option (B) He < Xe < Kr < Ne

Step 1: Recall the concept

• Electron gain enthalpy is the energy change when an extra electron is added to a gaseous atom.
• For noble gases (He, Ne, Kr, Xe), the outermost shell is completely filled.
• Adding another electron requires energy (due to electronic repulsion), hence they show positive electron gain enthalpy.

Step 2: Trend in noble gases

• Normally, down a group electron gain enthalpy becomes less positive because size increases.
• But in noble gases, the trend is irregular due to differences in atomic orbitals and electron repulsion effects.

Step 3: Experimental data observation

• Order of positive electron gain enthalpy:
  $$ He < Xe < Kr < Ne $$
• Helium: Least positive (small size but strong shielding repulsion).
• Neon: Most positive (adding electron is most difficult).
• Xenon and Krypton: Intermediate values.

Final Answer:
$$ \boxed{\text{(B) He < Xe < Kr < Ne}} $$

A quick comparison table of the approximate electron gain enthalpy values for noble gases:

Key points:

• All values are positive because energy is required to add an electron (repulsion to closed shell).
• Order: He < Xe < Kr ≈ Ar < Ne
• Ne has the highest positive electron gain enthalpy (most difficult to add electron).
• Xe has the lowest among heavier noble gases, explaining why Xe can form compounds like XeF₂.

Concept Takeaway:

• Noble gases always have positive electron gain enthalpies.
• The trend is not regular, but based on experimental values, the order is: He < Xe < Kr < Ne.
• Argon (Ar) and Krypton (Kr): Have nearly equal electron gain enthalpies, which are less positive than neon’s.
• Important for Periodic Table and Periodicity in JEE / NEET Chemistry PYQs.

JEE Main 2023 Question:
The first ionization enthalpy of Na, Mg and Si, respectively, are: 496, 737 and 786 kJ mol⁻¹. The first ionization enthalpy (kJ mol¯¹) of Al is:
(A) 487
(B) 768
(C) 577
(D) 856

Answer Correct Option: (C) 577

Step 1: Analyze the trend of first ionization enthalpy across a period.

First ionization enthalpy generally increases across a period as the effective nuclear charge increases and atomic size decreases, making it harder to remove an electron. However, there are exceptions due to electron configuration stability.

Step 2: Consider the electron configurations and exceptions.

Na (Sodium): [Ne] $3s^{1}$ → Ionization enthalpy is relatively low as removing the single $3s$ electron results in a stable noble gas configuration.
Mg (Magnesium): [Ne] $3s^{2}$ → Ionization enthalpy is higher than Na as it requires removing an electron from a filled $s$-orbital, which is relatively stable.
Al (Aluminum): [Ne] $3s^{2}3p^{1}$ → Ionization enthalpy is expected to be lower than Mg because the electron is removed from a $p$-orbital, which is at a higher energy level and experiences more shielding compared to an $s$-orbital electron. This makes it easier to remove the $3p^{1}$ electron than a $3s^{2}$ electron.
Si (Silicon): [Ne] $3s^{2}3p^{2}$ → Ionization enthalpy is higher than Al as it involves removing an electron from a more stable configuration.

Step 3: Compare the given values and predict Al’s ionization enthalpy.

Given values:
Na: 496 kJ mol⁻¹
Mg: 737 kJ mol⁻¹
Si: 786 kJ mol⁻¹

Based on the exception where Al’s ionization enthalpy is lower than Mg’s due to $p$-orbital removal, the value for Al should be less than 737 kJ mol⁻¹ but higher than Na. Among the options, 577 kJ mol⁻¹ fits correctly.

Final Answer
$$ \boxed{577 \ \text{kJ mol}^{-1}} $$

Concept Takeaway

• Across a period, ionization enthalpy increases but exceptions occur due to $s$ vs $p$ orbital stability.
• Al has lower ionization enthalpy than Mg because the $3p^{1}$ electron is easier to remove than $3s^{2}$.
• This is a very common JEE exam trap question.

For JEE preparation, always revise the periodic table and periodicity trends carefully. You can download notes, study material, and preparation notes for class 11 chemistry, JEE PYQs chapterwise, periodic table and periodicity, Anand Classes for thorough practice.

JEE Main 2022 Question:

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:
(A) (A) and (B) only
(B) (B) and (C) only
(C) (A) and (C) only
(D) (C) and (D) only

Answer Correct Option: (C) (A) and (C) only

Step 1: Recall the periodic trend.
Electron gain enthalpy generally becomes less negative down a group because the added electron enters orbitals farther from the nucleus, experiencing less attraction and more repulsion.

Step 2: Analyze the given pairs.

• (A) Rb and Cs: Both are alkali metals with very low (close to zero or positive) electron gain enthalpies. Their values are nearly the same.
• (B) Na and K: Also alkali metals, but the difference between Na and K is slightly larger compared to Rb and Cs. They are not considered “nearly identical.”
• (C) Ar and Kr: Noble gases have positive electron gain enthalpies (energy must be supplied to add an electron). Since both are inert and stable, their values are relatively similar.
• (D) I and At: Both are halogens, but I has a much more negative value than At. The difference is significant, so they are not nearly the same.

Step 3: Select correct pairs.
The nearly identical pairs are (A) Rb and Cs and (C) Ar and Kr.

Final Answer
$$ \boxed{(C) \ (A) \ \text{and} \ (C) \ \text{only}} $$

Concept Takeaway

• Alkali metals (Rb, Cs) have very low electron gain enthalpies, close to each other.
• Noble gases (Ar, Kr) have highly positive electron gain enthalpies, nearly identical due to their stable octet.
• Halogens show large differences down the group, so I and At are not nearly the same.
• Exam Tip: Watch for exceptions in alkali metals and noble gases when comparing trends in electron gain enthalpy.

For practice and revision, download notes and PYQs from class 11 chemistry, periodic table and periodicity, JEE preparation, Anand Classes.

JEE Main 2022 Question:
The incorrect statement is:
(A) The first ionization enthalpy of K is less than that of Na and Li.
(B) Xe does not have the lowest first ionization enthalpy in its group.
(C) The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38.
(D) The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.

Answer: Correct option (D)

Step 1: Concept – First ionization enthalpy (IE₁) trends
IE₁ generally decreases down a group and increases across a period, but subshell electronic configuration (s, p, d) and shielding effects produce important exceptions.

Step 2: Identify elements and their electronic configurations

• K (Z=19), Na (Z=11), Li (Z=3): all Group 1 → K has the largest size and lowest IE₁ among these three, so (A) is correct.
• Xe is a noble gas (Group 18); IE₁ decreases down the group so a heavier noble gas (like Rn) will have lower IE₁ than Xe → (B) is correct.
• Atomic number 37 (Rb) is a Group 1 element and 38 (Sr) is Group 2 in same period → Rb (Group 1) has lower IE₁ than Sr (Group 2) → (C) is correct.
• Zn (Z=30) electronic configuration: [Ar] 3d¹⁰ 4s². Ga (Z=31): [Ar] 3d¹⁰ 4s² 4p¹. Removing an electron from Ga removes a 4p electron (relatively easier) whereas removing the first electron from Zn involves a 4s electron influenced by a full d¹⁰ core. Empirical IE₁ values show Zn (≈ 906 kJ·mol⁻¹) > Ga (≈ 578 kJ·mol⁻¹), so the statement that Ga IE₁ is higher than Zn is false.

Step 3: Why (D) is incorrect (focused explanation)

• Ga has a 4p electron as the outermost electron — this 4p electron is easier to remove than the 4s/3d-influenced electron in Zn.
• Zn’s filled 3d¹⁰ subshell and higher effective nuclear attraction make its first ionization enthalpy significantly larger than Ga’s.
• Therefore (D) contradicts both electronic-structure reasoning and experimental IE₁ values.

Final Answer : The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30 (is incorrect).

Concept Takeaway

• Remember: compare elements by electronic configuration (which subshell the electron is removed from) not only by position.
• Filled d-subshells (like 3d¹⁰) often lead to higher IE₁ values for the following elements compared to a neighbor with a loosely held p-electron.
• Useful for JEE/Class 11 Chemistry revision on periodic trends—practice with IE numerical values and configs for quick checks. For concise notes and chapterwise PYQs on periodic table and periodicity, refer to Anand Classes study material.

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