Prepare for JEE Main with a comprehensive collection of solved question papers, previous year questions (PYQs), and important MCQs from Classification of Elements and Periodicity in Properties (Chapter 3, Class 11 Chemistry). These carefully selected questions cover all major concepts of the modern periodic table, periodic trends, ionization enthalpy, electron gain enthalpy, atomic and ionic radii, electronegativity, and chemical reactivity. Detailed solutions help students strengthen conceptual understanding, improve problem-solving skills, and gain confidence for JEE Main and other engineering entrance examinations.
JEE Main Solved Question Papers, Topicwise PYQs & Most Important 100% Expected MCQs – Classification of Elements and Periodicity in Properties
Strengthen your JEE Main Chemistry preparation with solved question papers, previous year questions (PYQs), and important MCQs from Classification of Elements and Periodicity in Properties. These questions are designed to help students revise key concepts, understand the exam pattern, and improve accuracy through regular practice and detailed solutions.
Explore more concepts related to What is Chemical Bond ? Why do Atoms Combine ? How do Atoms Combine ?
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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: Option (A) is correct – Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
First ionization enthalpy is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom. Its magnitude depends on factors such as effective nuclear charge, atomic size, shielding effect, and electronic configuration.
The assertion states that the first ionization enthalpy of 3d series elements is greater than that of Group 2 metals. This statement is generally true. Compared to alkaline earth metals, transition elements of the 3d series experience a higher effective nuclear charge. As a result, their outer electrons are held more strongly by the nucleus and require more energy for removal.
Now consider the reason. In the 3d transition series, electrons are progressively added to the 3d subshell. These 3d electrons do not shield the nuclear charge very effectively. Consequently, as the atomic number increases across the series, the effective nuclear charge experienced by the outer electrons also increases.
Because of this poor shielding by the 3d electrons, the attraction between the nucleus and the outer electrons becomes stronger. This increased attraction raises the first ionization enthalpy of the 3d series elements compared to Group 2 metals.
Therefore, the reason correctly explains why the first ionization enthalpy of 3d transition elements is generally higher.
Final Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
| Exam Tip |
|---|
| The poor shielding effect of d-electrons leads to a higher effective nuclear charge in transition elements, resulting in higher ionization enthalpies than those of Group 2 metals. |
Continue learning with JEE Main Most Expected PYQs MCQs With Solutions Classification of Elements and Periodicity in Properties
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: Option (C) is correct – Ca²⁺ < K⁺ < Cl⁻ < S²⁻
To determine the correct order of ionic radii, we first note that all four ions Ca²⁺, K⁺, Cl⁻, and S²⁻contain 18 electrons. Therefore, they are isoelectronic species, having the same electronic configuration as argon.
For isoelectronic ions, the size depends primarily on the nuclear charge. When the number of electrons remains the same, an ion with a greater number of protons exerts a stronger attraction on its electron cloud. This stronger attraction pulls the electrons closer to the nucleus, resulting in a smaller ionic radius.
The nuclear charges of the given ions are :
Ca²⁺ : 20 protons
K⁺ : 19 protons
Cl⁻ : 17 protons
S²⁻ : 16 protons
Since Ca²⁺ has the highest nuclear charge, it attracts its electrons most strongly and therefore has the smallest radius. As the nuclear charge decreases from Ca²⁺ to S²⁻, the ionic radius increases. Thus, the increasing order of ionic radii is:
Ca²⁺ < K⁺ < Cl⁻ < S²⁻
Hence, Option (C) is the correct answer.
To strengthen your concepts, learn about NCERT Exemplar Solutions Classification of Elements and Periodicity in Properties Class 11 Chemistry
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 : Option (A) is correct – O > S > Se > Te
Bond dissociation energy is the amount of energy required to break a chemical bond between two atoms in the gaseous state. A stronger bond has a higher bond dissociation energy, while a weaker bond requires less energy to break.
In Group 16, the hydrides are H₂O, H₂S, H₂Se, and H₂Te. As we move down the group from oxygen to tellurium, the size of the central atom increases. The increase in atomic size leads to a longer E–H bond because the bonding electrons are located farther from the nucleus.
Longer bonds are generally weaker than shorter bonds because the overlap between atomic orbitals decreases. As a result, the bond strength decreases from O–H to Te–H. Since bond dissociation energy is directly related to bond strength, the energy required to break the bond also decreases down the group.
Therefore, the order of bond strengths is :
O–H > S–H > Se–H > Te–H
Consequently, the order of bond dissociation energies is :
H₂O > H₂S > H₂Se > H₂Te
or simply,
O > S > Se > Te
Hence, Option (A) is the correct answer.
| Exam Tip |
|---|
| Down a group, atomic size increases, bond length increases, bond strength decreases, and therefore bond dissociation energy decreases. |
Read more about NCERT Solutions: Classification of Elements and Periodicity in Properties Class 11 Chemistry
JEE Main 2023 Question:
Match List I with List II
| List I (Atomic number) | List II (Block of periodic table) |
|---|---|
| (A) 37 (B) 78 (C) 52 (D) 65 | (1) p-block (II) d-block (III) f-block (IV) s-block |
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: Option (C) is correct – A-IV, B-II, C-I, D-III
To match the atomic numbers with their respective blocks in the periodic table, we first identify the elements corresponding to the given atomic numbers and then determine the subshell in which the differentiating electron enters.
Atomic number 37 corresponds to Rubidium (Rb). Rubidium belongs to Group 1 and has its outermost electron in an s-orbital. Therefore, it is an s-block element.
Atomic number 78 corresponds to Platinum (Pt). Platinum is a transition metal belonging to the d-block because its differentiating electron enters a d-orbital. Hence, it is a d-block element.
Atomic number 52 corresponds to Tellurium (Te). Tellurium belongs to Group 16 and its valence electrons occupy p-orbitals. Therefore, it is a p-block element.
Atomic number 65 corresponds to Terbium (Tb). Terbium is a lanthanide, and lanthanides are classified as f-block elements because their differentiating electron enters an f-orbital.
Thus, the correct matching is :
A (37) belongs to s-block (IV)
B (78) belongs to d-block (II)
C (52) belongs to p-block (I)
D (65) belongs to f-block (III)
Therefore, the correct option is : (C) A-IV, B-II, C-I, D-III
| Exam Tip |
|---|
| The block of an element is determined by the subshell into which the last or differentiating electron enter. s-block belongs to Groups 1 and 2 p-block belongs to Groups 13 to 18 d-block belongs to Transition elements f-block belongs to Lanthanides and Actinides |
Build strong concepts by studying NCERT Intext Solutions: Classification of Elements and Periodicity in Properties
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: Option (C) is correct – Both Statement I and Statement II are correct.
Statement I : The decrease in first ionization enthalpy from B to Al is much larger than that from Al to Ga.
This statement is correct. As we move down Group 13, ionization enthalpy generally decreases because atomic size increases and the outermost electrons experience greater shielding. Therefore, a noticeable decrease is observed from boron to aluminium.
However, the trend does not continue smoothly from aluminium to gallium. Gallium contains a filled 3d subshell, and the 3d electrons are poor at shielding the nuclear charge. As a result, the effective nuclear charge experienced by the outer electrons in gallium is higher than expected. This counteracts the effect of increased atomic size and causes only a small decrease in ionization enthalpy from aluminium to gallium.
Hence, Statement I is true.
Statement II : The d orbitals in Ga are completely filled.
This statement is also true. Gallium has atomic number 31 and its electronic configuration is [Ar] 3d¹⁰ 4s² 4p¹
The 3d subshell contains ten electrons and is therefore completely filled. These filled d-orbitals play an important role in the anomalous ionization enthalpy trend observed in Group 13.
Thus, Statement II is also correct.
Since both statements are true, the correct answer is:
Final Answer: (C) Both Statement I and Statement II are correct.
| Exam Tip |
|---|
| The poor shielding effect of the filled 3d electrons in gallium increases the effective nuclear charge, making the decrease in ionization enthalpy from Al to Ga much smaller than expected. |
Practice more questions from NEET PYQs Previous Year Questions MCQs Classification of Elements and Periodicity in Properties
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: Option (B) is correct – He < Xe < Kr < Ne
Electron gain enthalpy is the energy change associated with the addition of an electron to an isolated gaseous atom. For most elements, energy is released when an electron is added, resulting in a negative electron gain enthalpy. However, noble gases are an exception.
Noble gases possess completely filled valence shells, which are highly stable. Adding an extra electron would require the electron to enter a higher-energy shell or subshell. This process is energetically unfavorable and therefore requires an input of energy. As a result, noble gases have positive electron gain enthalpy values.
The magnitude of this positive electron gain enthalpy depends on factors such as atomic size, electronic configuration, and the energy level into which the additional electron must enter. Among the noble gases given, neon offers the greatest resistance to accepting an extra electron and therefore has the highest positive electron gain enthalpy.
Experimental measurements show that the correct increasing order of positive electron gain enthalpy for these noble gases is :
He < Xe < Kr < Ne
Thus, helium has the smallest positive electron gain enthalpy, while neon has the largest.
Final Answer: (B) He < Xe < Kr < Ne
| Exam Tip |
|---|
| Noble gases have positive electron gain enthalpy because their completely filled electronic configurations make the addition of an extra electron energetically unfavorable. |
A quick comparison table of the approximate electron gain enthalpy values for noble gases is as follows :
| Noble Gas | Atomic Number | Approx. Electron Gain Enthalpy (kJ/mol) | Remark |
|---|---|---|---|
| He | 2 | +48 | Least positive (difficult but not maximum) |
| Ne | 10 | +116 | Most positive (very stable configuration) |
| Ar | 18 | +96 | High positive value, but less than Ne |
| Kr | 36 | +96 | Similar to Ar, slightly lower |
| Xe | 54 | +77 | Lower positive value, more accessible to add electron |
Important concepts connected to this topic are JEE Main PYQs Previous Year Questions MCQs : Classification of Elements and Periodicity in Properties
JEE Main 2023 Question:
The first ionization enthalpy of Na, Mg and Si, respectively, are: 496, 737 and 786 kJ/mol. The first ionization enthalpy (kJ/mol) of Al is :
(A) 487
(B) 768
(C) 577
(D) 856
Answer: Option (C) is correct – 577 kJ mol⁻¹
First ionization enthalpy is the energy required to remove the outermost electron from an isolated gaseous atom. Across a period, ionization enthalpy generally increases because the effective nuclear charge increases and atomic size decreases. However, certain exceptions arise due to the stability of electronic configurations.
The electronic configuration of sodium is [Ne] 3s¹. Since the single 3s electron can be removed relatively easily, sodium has a low first ionization enthalpy of 496 kJ mol⁻¹.
Magnesium has the configuration [Ne] 3s². The filled 3s subshell provides extra stability, so more energy is required to remove an electron. Therefore, magnesium has a significantly higher ionization enthalpy of 737 kJ mol⁻¹.
Aluminium has the configuration [Ne] 3s² 3p¹. The outermost electron is present in a 3p orbital, which is higher in energy and less tightly held than a 3s electron. As a result, removing the 3p electron from aluminium is easier than removing a 3s electron from magnesium. This causes aluminium to have a lower ionization enthalpy than magnesium, despite being placed to the right of magnesium in the periodic table.
Silicon has the configuration [Ne] 3s² 3p² and exhibits a higher ionization enthalpy than aluminium due to the increased effective nuclear charge.
Since the ionization enthalpy of aluminium must be lower than that of magnesium (737 kJ mol⁻¹) but higher than that of sodium (496 kJ mol⁻¹), the value that best fits is 577 kJ mol⁻¹
Therefore, the correct answer is (C) 577 kJ mol⁻¹
Important exam-related topics include IMU CET Merchant Navy Sponsorship Exam MCQs (PYQs) on Periodic Table and Periodicity
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: Option (C) is correct – (A) and (C) only
Electron gain enthalpy is the energy change that occurs when an electron is added to an isolated gaseous atom. In general, electron gain enthalpy becomes less negative down a group because atomic size increases and the added electron experiences less attraction from the nucleus.
Let us examine each pair individually.
Rb and Cs : Both are heavy alkali metals with large atomic sizes. Their electron gain enthalpy values are very small and quite close to each other. Therefore, they are considered nearly identical.
Na and K : Although both belong to Group 1, the difference in their electron gain enthalpy values is more noticeable than that between Rb and Cs. Hence, they are not regarded as nearly identical.
Ar and Kr : Both are noble gases with completely filled valence shells. Adding an extra electron is energetically unfavorable, so they have positive electron gain enthalpy values. Their values are very similar and are considered nearly the same.
I and At : Both are halogens, but the electron gain enthalpy of astatine is significantly less negative than that of iodine because of its larger size and weaker attraction for the incoming electron. Therefore, their values are not nearly identical.
Thus, the pairs having nearly the same electron gain enthalpy are : Rb and Cs, Ar and Kr.
Therefore, the correct answer is (C) : (A) and (C) only.
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: Option (D) is correct – it is the incorrect statement.
First ionization enthalpy is the energy required to remove the outermost electron from an isolated gaseous atom. Although ionization enthalpy generally decreases down a group and increases across a period, several exceptions arise due to electronic configuration and shielding effects.
Let us examine each statement.
Statement (A) : The first ionization enthalpy of potassium is less than that of sodium and lithium. This statement is correct. Potassium lies below sodium and lithium in Group 1. As atomic size increases down the group, the outermost electron is farther from the nucleus and can be removed more easily. Therefore, potassium has the lowest first ionization enthalpy among the three.
Statement (B) : Xenon does not have the lowest first ionization enthalpy in its group. This statement is also correct. In Group 18, ionization enthalpy decreases down the group. Since radon lies below xenon, radon has a lower first ionization enthalpy than xenon. Hence, xenon is not the element with the lowest ionization enthalpy in the noble gas group.
Statement (C) : The first ionization enthalpy of the element with atomic number 37 is lower than that of the element with atomic number 38. Atomic number 37 corresponds to rubidium (Rb) and atomic number 38 corresponds to strontium (Sr). Rubidium is a Group 1 element, whereas strontium is a Group 2 element. Group 1 elements have lower ionization enthalpies than Group 2 elements in the same period. Therefore, this statement is correct.
Statement (D) : The first ionization enthalpy of gallium is higher than that of the d-block element with atomic number 30. Atomic number 30 corresponds to zinc (Zn). Gallium (Ga, Z = 31) has the electronic configuration [Ar] 3d¹⁰ 4s² 4p¹, and the electron removed during ionization is a 4p electron, which is relatively easy to remove. Zinc has the stable configuration [Ar] 3d¹⁰ 4s², and its first ionization enthalpy is significantly higher than that of gallium. Therefore, the statement that gallium has a higher first ionization enthalpy than zinc is false.
Final Answer: (D) The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30 is the incorrect statement.
Important Chapter Interlinks
This section provides a complete and interconnected study of Classification of Elements and Periodicity in Properties, starting with detailed theory and notes for Class 11 Chemistry to build a strong conceptual foundation. You can explore atomic radius and its types including covalent, van der Waals, metallic, and ionic radii to understand periodic trends in atomic size. It also includes Screening Effect (Shielding Effect) : Calculation of Effective or Reduced Nuclear Charge (Slater’s Rules), which explains how inner electrons reduce the nuclear attraction on outer electrons and influence periodic trends. In addition, topics like Radius of Cation is Less and Anion is More Than Its Parent Atom, Size Variation in Isoelectronic Series help explain how ionic size changes due to gain or loss of electrons and how nuclear charge affects size in species with the same number of electrons. The causes of periodicity explain why elements show repeating properties based on electronic configuration, which is further supported by the modern periodic law and structure of the modern periodic table including groups, periods, and blocks for elements even beyond atomic number 100. The historical development is covered through Mendeleev’s periodic law and table, leading to the modern classification of elements into s, p, d, and f blocks with prediction of period, group, and block. To strengthen exam preparation, you can practice JEE Main PYQs, IMU CET PYQs and Merchant Navy sponsorship exam MCQs, and other previous year questions with solutions, along with solved examples, conceptual questions, and practice problems on the modern periodic table. Learn more in this section also to radius of cation is less and anion is more than its parent atom and size variation in Isoelectronic Series. Additionally, complete study material, mock tests, and guidance are provided under Anand Classes Chemistry notes, along with expert support from Er Neeraj Anand, making this section a comprehensive resource for competitive exam preparation.