JEE Main PYQs 100% Expected Periodic Table MCQs With Solutions pdf download

JEE Main 2022 Question:
Given two statements below:
Statement I: In Cl₂ molecule the covalent radius is double of the atomic radius of chlorine.
Statement II: Radius of anionic species is always greater than their parent atomic radius.
Options:
(A) Both Statement I and Statement II are correct.
(B) Both Statement I and Statement II are incorrect.
(C) Statement I is correct but Statement II is incorrect.
(D) Statement I is incorrect but Statement II is correct.

Answer: Correct Option (D)

Step 1: Concept of covalent radius

• Covalent radius is defined as half the bond length between two identical atoms joined by a covalent bond.
• In Cl₂, the bond length is the distance between the nuclei of the two chlorine atoms.
• Therefore, covalent radius = ½ (bond length of Cl₂).
• It is not double of the atomic radius, so Statement I is incorrect.

Step 2: Concept of ionic radius in anions

• When an atom gains electrons to form an anion, electron-electron repulsion increases.
• The effective nuclear charge per electron decreases, leading to expansion of the electron cloud.
• Hence, radius of an anion > atomic radius of its neutral parent atom.
• So Statement II is correct.

Final Answer
$$\boxed{(D)\ \text{Statement I is incorrect but Statement II is correct.}}$$

Concept Takeaway

• Covalent radius = ½ bond length in diatomic molecule (not related directly to atomic radius values).
• Anions are always larger than their parent atoms due to decreased effective nuclear attraction.
• Important for JEE/Class 11 Chemistry preparation under periodic table and periodicity. For detailed study material, preparation notes, and chapterwise JEE PYQs, refer to Anand Classes.

JEE Main 2022 Question:
The first ionization enthalpies of Be, B, N and O follow the order:|
(A) O < N < B < Be
(B) Be < B < N < O
(C) B < Be < N < O
(D) B < Be < O < N

Answer Correct Option: (D) B < Be < O < N

Step 1: General trend of ionization enthalpy across a period

• Ionization enthalpy increases across a period (left → right) due to increasing effective nuclear charge and decreasing atomic radius.

Step 2: Exception Be vs B

• Be = 1s² 2s² (stable filled 2s orbital).
• B = 1s² 2s² 2p¹ (electron enters 2p orbital).
• 2p electron is easier to remove than 2s electron.
• Therefore, B < Be.

Step 3: Exception N vs O

• N = 1s² 2s² 2p³ (stable half-filled configuration).
• O = 1s² 2s² 2p⁴ (one paired electron in 2p orbital → easier removal).
• Therefore, O < N.

Step 4: Final Order

• Combining trend and exceptions: B < Be < O < N.

Final Answer
$$ \boxed{(D)\ B < Be < O < N} $$

Approximate First Ionization Enthalpies (kJ mol⁻¹):

Concept Takeaway

• Filled (s²) and half-filled (p³) orbitals provide extra stability.
• Hence, B has lower I.E. than Be, and O has lower I.E. than N.
• Important for JEE / NEET questions from Periodic Table and Periodicity.
• For more study material, preparation notes, and JEE PYQs chapterwise, download notes with Anand Classes, Class 11 Chemistry, Periodic Table and Periodicity.

JEE Main 2022 Question:
The IUPAC nomenclature of an element with electronic configuration [Rn] 5f¹⁴ 6d¹ 7s² is:
(A) Unnilbium
(B) Unnilunium
(C) Unnilquadium
(D) Unniltrium

Answer Correct Option: (D) Unniltrium

Step 1: Identify the atomic number

• Radon ([Rn]) has atomic number 86.
• Additional electrons: 5f¹⁴ 6d¹ 7s² = 17.
• Total = 86 + 17 = 103.

Step 2: Locate the element

• Z = 103 corresponds to Lawrencium (Lr).

Step 3: Apply IUPAC temporary nomenclature rules (for Z > 100)

• Digit roots: 1 = un, 0 = nil, 3 = tri.
• Combine + suffix “-ium”: un-nil-tri-ium = Unniltrium (Untr).

Final Answer
$$ \boxed{(D)\ \text{Unniltrium}} $$

Concept Takeaway

• For elements with Z > 100, IUPAC assigned systematic temporary names using Latin/Greek roots.
• Atomic number 103 → Unniltrium (later officially named Lawrencium, Lr).
• Important for Class 11 Chemistry, Periodic Table, JEE PYQs chapterwise preparation notes with Anand Classes.

JEE Main 2022 Question:
The correct order of electron gain enthalpy (−ve value) is:
(A) O > S > Se > Te
(B) O < S < Se < Te
(C) O < S > Se > Te
(D) O < S > Se < Te

Answer Correct Option: (C) O < S > Se > Te

Step 1: General Trend

• In a group, electron gain enthalpy becomes less negative down the group because atomic size increases and attraction for the incoming electron decreases.

Step 2: Anomaly in Oxygen

• Oxygen has a smaller atomic size, so the added electron goes into the compact 2p subshell.
• This causes strong electron–electron repulsion, making its electron gain enthalpy less negative than Sulfur.

Step 3: Compare S, Se, Te

• From S → Se → Te, atomic size increases.
• The added electron feels weaker nuclear attraction, so the enthalpy becomes less negative gradually.

Final Order
$$ O < S > Se > Te $$

Final Answer
$$ \boxed{(C)\ O < S > Se > Te} $$

Concept Takeaway

• Anomaly alert: Oxygen has less negative electron gain enthalpy than Sulfur due to repulsion in its small 2p orbital.
• Trend in Group 16: S > Se > Te.
• Very important for Class 11 Chemistry, Periodic Table and Periodicity, JEE PYQs chapterwise preparation notes with Anand Classes.

JEE Main 2022 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 for oxygen is lower than that of nitrogen.
Reason R: The four electrons in 2p orbitals of oxygen experience more electron-electron repulsion.
Options:
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is NOT the correct explanation of A.
(C) A is correct but R is not correct.
(D) A is not correct but R is correct.

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

Step 1: Recall the concept of ionization enthalpy
Ionization enthalpy generally increases across a period due to rising nuclear charge. However, exceptions occur due to subshell stability and electron repulsion.

Step 2: Compare N and O electronic configurations
N: 1s² 2s² 2p³ → half-filled stable configuration
O: 1s² 2s² 2p⁴ → one 2p orbital contains a paired set of electrons

Step 3: Explain the anomaly

• Nitrogen has extra stability from half-filled 2p³, requiring more energy to remove an electron.
• Oxygen has paired electrons in 2p, leading to greater repulsion → easier removal of an electron.

Final Answer
(A) Both A and R are correct and R is the correct explanation of A

Concept Takeaway

• Half-filled orbitals (N: 2p³) are more stable.
• Paired electrons increase repulsion (O: 2p⁴), lowering ionization enthalpy.
• Anomaly: Ionization energy order → N > O despite the general periodic trend.
• Exam Tip: For JEE/NEET, always check subshell stability when periodic trends show exceptions.

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JEE Main 2022 Question:
Match List – I with List – II

Correct Option: (B) (A) – (IV), (B) – (II), (C) – (I), (D) – (III)

Step 1: Analyze each oxide

• Cl₂O₇ (Dichlorine heptoxide): Non-metal oxide in high oxidation state → acidic.
• Na₂O (Sodium oxide): Alkali metal oxide → basic.
• Al₂O₃ (Aluminium oxide): Can react with both acids and bases → amphoteric.
• N₂O (Nitrous oxide): Neutral oxide (like CO, H₂O) → neutral.

Step 2: Match pairs
(A) Cl₂O₇ → (IV) Acidic
(B) Na₂O → (II) Basic
(C) Al₂O₃ → (I) Amphoteric
(D) N₂O → (III) Neutral

Final Answer
$$ \boxed{(B)} $$

Concept Takeaway

• Non-metal oxides (high oxidation state) → acidic.
• Metal oxides → basic.
• Amphoteric oxides react with both acids and bases (Al₂O₃, ZnO).
• Neutral oxides (N₂O, CO, H₂O) → neither acidic nor basic.

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JEE Main 2022 Question:
Among the following, basic oxide is:
(A) SO₃
(B) SiO₂
(C) CaO
(D) Al₂O₃

Answer: (C) CaO

Step 1: Recall the rule

• Metal oxides → usually basic
• Non-metal oxides → usually acidic
• Some metal oxides (like Al₂O₃, ZnO) → amphoteric

Step 2: Check each option

• SO₃: Non-metal oxide → Acidic
• SiO₂: Non-metal oxide → Acidic
• CaO: Metal oxide → Basic
• Al₂O₃: Metal oxide but shows dual behavior → Amphoteric

Final Answer
$$ \boxed{\text{CaO is the basic oxide.}} $$

Concept Takeaway

• Basic oxides react with acids → form salt + water
• Acidic oxides react with bases → form salt + water
• Amphoteric oxides react with both acids and bases

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JEE Main 2022 Question:
Element “E” belongs to period 4 and group 16 of the periodic table.
The valence shell electron configuration of the element just above “E” in the group is:
(A) 3s² 3p⁴
(B) 3d¹⁰ 4s² 4p⁴
(C) 4d¹⁰ 5s² 5p⁴
(D) 2s² 2p⁴

Answer : The correct option is (A) 3s² 3p⁴

Step 1: Identify element “E”

• Period 4, Group 16 → Selenium (Se).

Step 2: Find the element above “E” in the group

• Just above Se in Group 16 is Sulfur (S) (Period 3, Group 16).

Step 3: Write valence shell configuration of Sulfur

• General configuration for Group 16 elements: ns² np⁴.
• For Sulfur (n = 3): 3s² 3p⁴.

Final Answer
The correct option is (A) 3s² 3p⁴.

Concept Takeaway

• Group 16 elements have outer configuration ns² np⁴.
• Moving down the group, only the principal quantum number (n) increases.

JEE Main 2022 Question:
The correct order of increasing ionic radii is:
(A) Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
(B) N³⁻ < O²⁻ < F⁻ < Na⁺ < Mg²⁺
(C) F⁻ < Na⁺ < O²⁻ < Mg²⁺ < N³⁻
(D) Na⁺ < F⁻ < Mg²⁺ < O²⁻ < N³⁻

Answer : The correct order is: Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻

Step 1: Identify isoelectronic species

• The given ions are Mg²⁺, Na⁺, F⁻, O²⁻, N³⁻.
• All have 10 electrons (same as Neon).

Step 2: Compare nuclear charges (Z)

• N³⁻ → Z = 7
• O²⁻ → Z = 8
• F⁻ → Z = 9
• Na⁺ → Z = 11
• Mg²⁺ → Z = 12

Step 3: Apply the rule

• For isoelectronic species: greater nuclear charge → smaller radius.
• Hence, radius order is the reverse of nuclear charge.

Step 4: Arrange from smallest to largest

• Smallest: Mg²⁺ (Z = 12)
• Then Na⁺ (Z = 11)
• Then F⁻ (Z = 9)
• Then O²⁻ (Z = 8)
• Largest: N³⁻ (Z = 7)

Final Answer
The correct order is: Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
→ Option (A)

Concept Takeaway

• Isoelectronic series radii depend only on nuclear charge (Z).
• Higher Z → stronger attraction → smaller radius.

JEE Main 2022 Question:
Assertion (A): The ionic radii of O²⁻ and Mg²⁺ are same.
Reason (R): Both O²⁻ and Mg²⁺ are isoelectronic species.
Options:
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.

Answer : Correct option is (D) (A) is false but (R) is true.

Step 1: Check Reason (R)

• O²⁻ has 8 protons + 2 extra electrons = 10 electrons.
• Mg²⁺ has 12 protons − 2 electrons = 10 electrons.
• Both are isoelectronic (10 electrons each).
• So, Reason (R) is true.

Step 2: Check Assertion (A)

• Even though they are isoelectronic, ionic radii are not the same.
• O²⁻ has fewer protons (Z = 8) → weaker attraction → larger radius.
• Mg²⁺ has more protons (Z = 12) → stronger attraction → much smaller radius.
• So, Assertion (A) is false.

Final Answer:
Correct option is (D) (A) is false but (R) is true.

Concept Takeaway:
Isoelectronic species have the same number of electrons, but ionic size depends on nuclear charge:

• Higher nuclear charge → smaller radius.

JEE Main 2022 Question:
The correct order of electron gain enthalpies of Cl, F, Te, and Po is:
Options:
(A) F < Cl < Te < Po
(B) Po < Te < F < Cl
(C) Te < Po < Cl < F
(D) Cl < F < Te < Po

Answer: Correct option is (B) Po < Te < F < Cl.

Step 1: Recall the trend

• Across a period: Electron gain enthalpy becomes more negative (due to higher nuclear charge).
• Down a group: Electron gain enthalpy becomes less negative (due to larger atomic size).

Step 2: Halogens (Cl vs. F)

• Exception: Although F is more electronegative, its small size causes strong electron-electron repulsion.
• Result: Chlorine has a more negative electron gain enthalpy than Fluorine.
• Order: F < Cl.

Step 3: Chalcogens (Te vs. Po)

• Down the group, electron gain enthalpy becomes less negative.
• Order: Po < Te.

Step 4: Compare Halogens vs. Chalcogens

• Halogens have much more negative electron gain enthalpies than chalcogens.
• Overall order: Po < Te < F < Cl.

Final Answer:
Correct option is (B) Po < Te < F < Cl.

Concept Takeaway:

• Chlorine has the most negative electron gain enthalpy among these.
• Fluorine is less negative than Cl due to repulsions.
• Heavier chalcogens (Te, Po) have much less negative values.

JEE Main 2022 Question :
Given below are the oxides: Na₂O, As₂O₃, N₂O, NO and Cl₂O₇
Number of amphoteric oxides is:
A) 0
B) 1
C) 2
D) 3

Answer: (B) 1

Given the problem: Identify the number of amphoteric oxides among Na₂O, As₂O₃, N₂O, NO, and Cl₂O₇. Here’s a detailed explanation:

Step 1: Recall the definitions and general trends

• Basic oxides: Typically formed by metals, especially alkali and alkaline earth metals. These oxides react with acids to form salts and water. Example: Na₂O + HCl → NaCl + H₂O.
• Acidic oxides: Typically formed by non-metals, especially in higher oxidation states. These oxides react with bases to form salts and water. Example: Cl₂O₇ + 2 NaOH → 2 NaClO₄ + H₂O.
• Amphoteric oxides: Oxides that can react with both acids and bases to form salts and water. These are usually oxides of metals that are borderline between metallic and non-metallic character (like Al, Zn, Pb, Sn, and some metalloids like As).
• Neutral oxides: Oxides that do not show significant acidic or basic behavior. Common examples: N₂O, NO, CO, CO₂ in some cases.

Step 2: Analyze each oxide

1. Na₂O (Sodium oxide)
   • Sodium is an alkali metal.
   • Alkali metal oxides are strongly basic.
   • Na₂O reacts with acids: Na₂O + 2 HCl → 2 NaCl + H₂O.
   • Conclusion: Basic oxide.
2. As₂O₃ (Arsenic trioxide)
   • Arsenic is a metalloid.
   • Oxides of metalloids often have amphoteric behavior, reacting with both acids and bases:
     • With acid: As₂O₃ + 6 HCl → 2 AsCl₃ + 3 H₂O
     • With base: As₂O₃ + 2 NaOH + H₂O → 2 Na[AsO₂(OH)]
   • Conclusion: Amphoteric oxide.
3. N₂O (Nitrous oxide)
   • Nitrogen in oxidation state +1.
   • This oxide is neutral, not reacting significantly with acids or bases.
   • Conclusion: Neutral oxide.
4. NO (Nitric oxide)
   • Nitrogen in oxidation state +2.
   • Does not react with acids or bases under normal conditions.
   • Conclusion: Neutral oxide.
5. Cl₂O₇ (Dichlorine heptoxide)
   • Chlorine is a non-metal in high oxidation state (+7).
   • Reacts with bases: Cl₂O₇ + 2 NaOH → 2 NaClO₄ + H₂O
   • Does not react with acids significantly.
   • Conclusion: Acidic oxide.

Step 3: Identify amphoteric oxides

• From the above analysis, only As₂O₃ is amphoteric.

Step 4: Count

• Number of amphoteric oxides = 1

Answer: (B) 1

Important Points:

• Amphoteric oxides usually belong to metalloids or metals near the metalloid line (like Al, Zn, Pb, Sn, As).
• Alkali metal oxides are always basic.
• Non-metal oxides in high oxidation states are acidic.
• Neutral oxides are typically low oxidation state non-metals or diatomic molecules like NO, N₂O.

Concept Takeaway:

• Recognizing element type (metal, metalloid, non-metal) and oxidation state helps predict oxide behavior.
• This is a common JEE Main / Advanced trick question testing your grasp of periodic trends.