JEE Main PYQs | Important MCQs With Solutions Periodic Table pdf Download

Anand Classes brings you the most important and frequently asked MCQs with solutions from previous years’ JEE Main papers based on the Periodic Table. This study material is carefully curated to help students strengthen their concepts, revise efficiently, and score better in exams. Each question comes with a detailed solution to enhance understanding and problem-solving skills. Click the print button to download study material and notes.

JEE Main 2025 – Prediction of Group From Successive ionisation Enthalpies

JEE Main 2025 Question :
The successive 5 ionisation energies of an element are 800, 2427, 3658, 25024 and 32824 kJ mol⁻¹, respectively. By using the above values predict the group in which the above element is present:
(A) Group 4
(B) Group 2
(C) Group 13
(D) Group 14

Answer : Correct Option: (C) Group 13

Step 1: Identify the large jump in ionisation energies
The ionisation energies are: 800, 2427, 3658, 25024, 32824 kJ mol⁻¹.
A sharp increase occurs between the 3rd and 4th values (3658 → 25024). This means the element has 3 valence electrons. After losing 3, the next electron comes from the stable inner shell, requiring much more energy.

Step 2: Connect valence electrons to group number
An element with 3 valence electrons (configuration $ns^2np^1$) belongs to Group 13 in the periodic table. Examples: B, Al, Ga, In, Tl.
Thus, the given element lies in Group 13.

Step 3: Link with study material and exam use
This concept belongs to periodicity, ionisation enthalpy, and periodic table chapter of Class 11 Chemistry. In Anand Classes notes, study material, and pdf download, this topic is explained with solved MCQs, PYQs, and practice questions for JEE.
Recognising the jump in successive ionisation enthalpies is a standard trick for solving such MCQs quickly.

Final Answer
$$\boxed{\text{Group 13}}$$

Concept Takeaway

• Periodic trend: A large jump indicates the number of valence electrons.
• 3 valence electrons → Group 13 → $ns^2np^1$ electronic configuration.
• Very useful for MCQs, PYQs, and exam revision notes.

Exam Tip (Anand Classes): In JEE/NEET MCQs, whenever you see successive ionisation energies, mark the place of the sharp rise. That number tells you the count of valence electrons and hence the group number directly.

JEE Main 2025 – Ionization Energy Trend in Carbon family

JEE Main 2025 Question :
Given below are two statements:
Statement (I): The first ionization energy of Pb is greater than that of Sn.
Statement (II): The first ionization energy of Ge is greater than that of Si.
In the light of the above statements, choose the correct answer from the options given below:
(A) Statement I is false but Statement II is true
(B) Both Statement I and Statement II are false
(C) Both Statement I and Statement II are true
(D) Statement I is true but Statement II is false

Answer : Correct Option: (D) Statement I is true but Statement II is false

Explanation:

• Statement I: The first ionization energy of Pb is greater than that of Sn.
  Normally, ionization energy decreases down a group. However, in Group 14, Lead (Pb) shows slightly higher ionization energy than Sn. This anomaly arises due to the poor shielding effect of 4f and 5d electrons in Pb, which increases effective nuclear charge, pulling valence electrons more strongly.
  ⇒ Statement I is true.
• Statement II: The first ionization energy of Ge is greater than that of Si.
  Moving down the group from Si to Ge, atomic size increases and shielding effect is stronger. Thus, it becomes easier to remove an electron, meaning ionization energy decreases. Therefore, Ge has a lower ionization energy than Si.
  ⇒ Statement II is false.

Final Answer
$$\boxed{\text{(D) Statement I is true but Statement II is false}}$$

Concept Notes (Anand Classes Study Material, MCQs, PYQs):

• General periodic trend: Ionization energy increases across a period and decreases down a group.
• Exception: Pb > Sn due to ineffective shielding (inert pair effect, f-block contraction).
• Useful for NEET/JEE exam MCQs and PYQs.

Exam Tip (Anand Classes): Always check for anomalies caused by f- and d-electron shielding. These exceptions are frequently asked in MCQs and previous year questions (PYQs).

JEE Main 2025 – Information About Periodic Table

JEE Main 2025 Question :
Which of the following statements are NOT true about the periodic table?
Options:
(A) The properties of elements are function of atomic weights.
(B) The properties of elements are function of atomic numbers.
(C) Elements having similar outer electronic configurations are arranged in same period.
(D) An element’s location reflects the quantum numbers of the last filled orbital.
(E) The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled.
Choose the correct answer from the options given below:
(A) A, C and E Only
(B) A and E Only
(C) B, C and E Only
(D) D and E Only

Answer : (A) A, C and E Only

Statement A:
The properties of elements are function of atomic weights.
❌ Not true. Mendeleev arranged elements by atomic weight, but the modern periodic law states that the properties of elements are a function of atomic numbers.

Statement B:
The properties of elements are function of atomic numbers.
✅ True. This is the basis of the modern periodic table.

Statement C:
Elements having similar outer electronic configurations are arranged in the same period.
❌ Not true. They are arranged in the same group, not the same period.

Statement D:
An element’s location reflects the quantum numbers of the last filled orbital.
✅ True. The position (period & block) reflects the principal quantum number n and the subshell type (s, p, d, f).

Statement E:
The number of elements in a period is same as the number of atomic orbitals available in that energy level.
❌ Not true. The number of elements in a period equals the number of electrons that can occupy orbitals being filled, not just the number of orbitals. Example: Period 2 has 8 elements (2s + 2p), not 4 orbitals.

✅ Final Answer

The statements that are NOT true are A, C, and E only.
Hence, the correct option is:
(A) A, C and E Only

📘 Study Tip (Anand Classes Notes)

This type of question is frequently asked in MCQs, PYQs, and NEET/JEE mains exams. Remember:

• Periodic law → atomic number, not atomic weight.
• Similar configurations → same group, not period.
• Period length → based on electrons filled, not orbitals alone.

For more structured study material, notes, and PDF downloads, follow Anand Classes to revise periodic table concepts effectively.

JEE Main 2025 – Group 14 Elements Melting Point Trend

JEE Main 2025 Question :
Given below are the atomic numbers of some group 14 elements. The atomic number of the element with lowest melting point is:
(A) 14
(B) 50
(C) 6
(D) 82

Answer: (B) 50

Step 1: Concept Explanation / Rule / Definition
Melting point is determined by the type and strength of bonding and by the solid-state structure (cohesive energy).
In Group 14 the nature of bonding changes as you move down the group: from strong covalent network solids (carbon, silicon, germanium) to metallic bonding (tin, lead). Allotropy (different structural forms of the same element) also matters — different allotropes can have very different thermal stabilities.

Step 2: Identify Elements or Given Data
Group 14 elements and the options here are:

• (A) $Z=14 \Rightarrow \text{Si (Silicon)}$
• (B) $Z=50 \Rightarrow \text{Sn (Tin)}$
• (C) $Z=6 \Rightarrow \text{C (Carbon)}$
• (D) $Z=82 \Rightarrow \text{Pb (Lead)}$

Representative melting / sublimation temperatures (approx., standard pressure):

• Carbon (graphite): $T_{\text{sub}}\approx 3642^\circ\text{C}$ (graphite sublimes rather than melting at 1 atm).
• Silicon: $T_m \approx 1414^\circ\text{C}$.
• Tin: $T_m \approx 231.9^\circ\text{C}$.
• Lead: $T_m \approx 327.5^\circ\text{C}$.

Step 3: Comparison / Solving Process (detailed)

1. Compare numerical values:

• Carbon (graphite) $ \sim 3642^\circ\text{C}$ (very high — graphite/diamond are covalent-network solids).
• Silicon $1414^\circ\text{C}$ (covalent diamond-like network).
• Lead $327.5^\circ\text{C}$ (metallic bonding).
• Tin $231.9^\circ\text{C}$ (metallic bonding) — lowest of the four. So numerically, tin ($Z=50$) has the smallest melting temperature among the options.

2. Why are the numbers arranged this way? — bonding & structure explanation:

• Carbon (C, $Z=6$): In its common forms carbon forms very strong covalent bonds. Diamond is a three-dimensional covalent network with extremely high cohesive energy; graphite is layered with strong in-plane covalent bonds and weaker interlayer forces, but even graphite requires very high temperature to break down (it sublimes at very high temperature). Hence carbon’s thermal stability is the highest in Group 14.
• Silicon (Si, $Z=14$) and Germanium (Ge, $Z=32$): Both adopt diamond-type (covalent) lattices with substantial bond strengths, so their melting points are high (Si > Ge but both >> Sn, Pb).
• Tin (Sn, $Z=50$): At room temperature the stable form is metallic β-Sn (white tin) which has metallic bonding that is relatively weak compared with the strong covalent networks of C/Si/Ge. This gives tin a low melting point (~$232^\circ\text{C}$). (Note: tin also has a low-temperature allotrope, α-Sn or “gray tin”, with a diamond-like structure, but at ordinary temperatures β-Sn is the relevant metallic form.)
• Lead (Pb, $Z=82$): Lead is also metallic; its metallic bonding and crystal packing give it a higher melting point than tin but still much lower than silicon or carbon.

3. Trend summary and exceptions:

• A simple “melting point decreases down the group” rule does not hold strictly for Group 14 because the type of bonding and allotropy changes (covalent network → metallic). Carbon is an extreme case (very high mp). Tin shows the lowest mp in this list because of its metallic β-Sn structure with relatively low cohesive energy. Lead’s mp is higher than tin’s because of stronger metallic bonding/packing in lead’s structure.

Final Answer
$$\boxed{50\ (\text{Sn})}$$

Correct Option: (B) 50

Concept Takeaway

• Exam Tip: When comparing properties like melting point, always identify the type of bonding and most stable allotrope at room conditions — periodic trends may be overridden by changes in bonding.
• Important Point: Covalent network solids (diamond, silicon) have very high melting/sublimation temperatures; metallic forms (Sn, Pb) have much lower melting points — tin is the lowest among the given options.
• Questions like this commonly appear in class 11 chemistry and in JEE PYQs chapterwise tests on the periodic table and periodicity. Students often find it helpful to download notes and study material and use preparation notes from sources such as Anand Classes to practise exceptions and allotropy-based reasoning.

JEE Main 2025 – Rule for elements in same period

JEE Main 2025 Question :
The element that does not belong to the same period of the remaining elements (modern periodic table) is:
(A) Platinum
(B) Osmium
(C) Iridium
(D) Palladium

Answer: (D) Palladium

Step 1: Concept Explanation / Rule / Definition
A period in the periodic table is a horizontal row. Elements in the same period are arranged in the same row of the table. For the transition (d-block) elements you should remember the nth row often corresponds to filling the $(n-1)d$ subshell (for example, the 4d series occupies Period 5, the 5d series occupies Period 6).

Step 2: Identify Elements or Given Data
List the elements with their atomic numbers and the d-series they belong to:

• Osmium (Os), Z = 76 — 5d series → Period 6
• Iridium (Ir), Z = 77 — 5d series → Period 6
• Platinum (Pt), Z = 78 — 5d series → Period 6
• Palladium (Pd), Z = 46 — 4d series → Period 5

You can also check typical ground-state electron configurations (showing which d-series is being filled):

• $\mathrm{Os}:\ [\mathrm{Xe}]\,4f^{14}\,5d^{6}\,6s^{2}$ (5d electrons — Period 6)
• $\mathrm{Ir}:\ [\mathrm{Xe}]\,4f^{14}\,5d^{7}\,6s^{2}$ (5d electrons — Period 6)
• $\mathrm{Pt}:\ [\mathrm{Xe}]\,4f^{14}\,5d^{9}\,6s^{1}$ (5d electrons — Period 6)
• $\mathrm{Pd}:\ [\mathrm{Kr}]\,4d^{10}$ (4d electrons — 4d series → Period 5)

Step 3: Comparison / Solving Process (detailed)

• How to decide quickly: identify which d-series the element belongs to (4d or 5d). The 4d series elements occupy Period 5 (row 5); the 5d series occupy Period 6 (row 6).
• From the atomic numbers / configurations above: Os, Ir and Pt are members of the 5d series (Period 6), while Pd is a member of the 4d series (Period 5).
• Therefore Palladium (Pd) is in a different period (Period 5) from Platinum, Osmium and Iridium (all in Period 6).

Final Answer
$$\boxed{\text{Palladium (Pd)}}$$

Correct Option: (D) Palladium

Concept Takeaway

• Exam Tip: For d-block “odd one out” period questions, check the d-series (4d vs 5d): the 4d series = Period 5, the 5d series = Period 6. If you remember which series each element belongs to, you can answer these questions instantly.
• Important Point: Electron configurations help explain the placement (e.g. Pd is $[\mathrm{Kr}]\,4d^{10}$ so it sits in the 4d series / 5th row), while Os/Ir/Pt have 5d electrons and sit in the 6th row.
• This type of classification is commonly asked in class 11 chemistry and in JEE PYQs chapterwise practice on the periodic table and periodicity. Students preparing with Anand Classes study material and preparation notes often practise 4d vs 5d series identification to avoid mistakes when spotting the odd element out.

JEE Main 2025 – Acidic Oxides in Periodic Table

JEE Main 2025 Question :
Statement (I): An element in the extreme left of the periodic table forms acidic oxides.
Statement (II): Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the periodic table.
Options:
(A) Both Statement I and Statement II are false
(B) Both Statement I and Statement II are true
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is true

Answer: (D) Statement I is false but Statement II is true

Step 1: Concept Explanation / Rule / Definition

• The nature of oxides depends on the position of the element in the periodic table.
• Metals (on the left) form basic oxides because they donate electrons and form ionic compounds with oxygen. These oxides, when dissolved in water, form bases (alkalis).
• Nonmetals (on the right) form acidic oxides because they accept electrons and form covalent oxides. These oxides, when dissolved in water, form acids.
• Some elements in the center of the periodic table (like Al, Zn, Sn, Pb) form amphoteric oxides that can react with both acids and bases.

Step 2: Identify Elements or Given Data

• Extreme left elements: Alkali metals (Na, K) and alkaline earth metals (Mg, Ca).
• Examples:
  • $Na_2O + H_2O \rightarrow 2NaOH$
  • $CaO + H_2O \rightarrow Ca(OH)_2$
    These produce strongly basic hydroxides.
• Extreme right elements: Nonmetals such as sulfur, phosphorus, chlorine.
• Examples:
  • $SO_3 + H_2O \rightarrow H_2SO_4$
  • $P_2O_5 + 3H_2O \rightarrow 2H_3PO_4$
  • $Cl_2O_7 + H_2O \rightarrow 2HClO_4$
    These produce acids when their oxides react with water.

Step 3: Comparison / Solving Process (Detailed)

• Statement (I): “An element in the extreme left forms acidic oxides.”
• This is false because alkali and alkaline earth metals form basic oxides. For example, $Na_2O$ forms NaOH (a base) in water.
• Statement (II): “Acid is formed during the reaction between water and oxide of a reactive element in the extreme right.”
• This is true because nonmetal oxides (like $SO_3, P_2O_5, Cl_2O_7$) dissolve in water to form acids like $H_2SO_4, H_3PO_4, HClO_4$.

Thus:

• Statement I → False
• Statement II → True

Final Answer
$$\boxed{\text{Statement I is false, Statement II is true}}$$

Correct Option: (D) Statement I is false but Statement II is true

Concept Takeaway

• Exam Tip: When solving oxide nature questions, always remember the basic → amphoteric → acidic trend across a period.
• Important Point:
• Metals (left side) → basic oxides
• Metalloids/transition elements (center) → amphoteric oxides
• Nonmetals (right side) → acidic oxides
• This trend explains why sodium oxide forms NaOH (basic), aluminium oxide behaves amphoteric, and sulfur trioxide forms H₂SO₄ (acidic).
• Such questions appear frequently in class 11 chemistry and JEE PYQs chapterwise tests on Periodic Table and Periodicity. With Anand Classes study material and preparation notes, students can easily master exceptions and tricky conceptual problems.

JEE Main 2025 – Periodic Table Properties Trend

JEE Main 2025 Question :
Match List-I with List-II.

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

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

Step 1: Analyze (A) — Al³⁺ < Mg²⁺ < Na⁺ < F⁻

• This sequence shows the order of ionic radii.
• More positive charge → smaller radius; anions (F⁻) are the largest.
• Hence, (A) → (IV).

Step 2: Analyze (B) — B < C < O < N

• This is the order of first ionisation enthalpy.
• General trend: ionisation enthalpy increases across a period.
• Exception: N > O due to half-filled stability of N (2p³).
• Hence, (B) → (I).

Step 3: Analyze (C) — B < Al < Mg < K

• This shows increasing metallic character.
• Metallic nature increases down a group and from right to left across a period.
• Hence, (C) → (II).

Step 4: Analyze (D) — Si < P < S < Cl

• This is the order of electronegativity across a period.
• Electronegativity increases left to right.
• Hence, (D) → (III).

Final Answer
$$\boxed{\text{Option (A): (A)-(IV), (B)-(I), (C)-(II), (D)-(III)}}$$

Concept Takeaway

• Exam Tip: Always connect sequences with periodic trends:
• Ionic radii: higher charge → smaller radius.
• Ionisation enthalpy: increases across, decreases down (watch exceptions).
• Metallic character: increases down a group, decreases across a period.
• Electronegativity: increases across, decreases down.
• These periodic properties are frequently tested in class 11 chemistry and JEE PYQs from Periodic Table and Periodicity.

JEE Main 2025 – Electronegativity Trend

JEE Main 2025 Question :
Which of the following electronegativity orders is incorrect?
A) S < Cl < O < F
B) Al < Si < C < N
C) Al < Mg < B < N
D) Mg < Be < B < N

Answer: (C) Al < Mg < B < N

Step 1: Recall Electronegativity Trends
Electronegativity is the ability of an atom to attract a shared pair of electrons in a chemical bond.

• Across a period (left → right): increases because of higher nuclear charge and smaller atomic radius.
• Down a group (top → bottom): decreases because atoms get larger and the outer electrons are farther from the nucleus.
• Fluorine (F): most electronegative element with a value of 3.98 on the Pauling scale.

Step 2: Check with Actual Pauling Scale Values

Step 3: Evaluate Each Option in Detail

• Option A: S < Cl < O < F
  Values: S (2.58), Cl (3.16), O (3.44), F (3.98).
  This order is correct because Cl is less electronegative than O, and F is the highest.
• Option B: Al < Si < C < N
  Values: Al (1.61), Si (1.90), C (2.55), N (3.04).
  Across Period 2/3 → electronegativity increases left to right.
  This order is correct.
• Option C: Al < Mg < B < N
  Values: Mg (1.31), Al (1.61), B (2.04), N (3.04).
  The given order suggests Al < Mg, but actually Mg (1.31) < Al (1.61).
  This makes the order incorrect.
• Option D: Mg < Be < B < N
  Values: Mg (1.31), Be (1.57), B (2.04), N (3.04).
  Correct because Be > Mg, and the order increases further through B to N.

Step 4: Conclusion
The incorrect electronegativity order is option (C).

Final Answer
$$\boxed{\text{(C) Al < Mg < B < N}}$$

Concept Takeaway

• Exam Tip: Always verify doubtful orders using Pauling electronegativity values.
• Important Point:
• Across a period: electronegativity increases.
• Down a group: electronegativity decreases.
• Alkali metals have the lowest electronegativity in their periods, halogens among the highest.
• This type of question is common in class 11 chemistry and JEE PYQs under Periodic Table and Periodicity.
• With Anand Classes preparation notes and study material, you can remember such orders easily and avoid confusion in competitive exams.

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