Prepare effectively for JEE Main with the most expected PYQs and MCQs from Classification of Elements and Periodicity in Properties. This collection includes carefully selected previous year questions and important multiple-choice questions based on periodic trends, electronic configuration, ionization enthalpy, electron gain enthalpy, atomic and ionic radii, electronegativity, and the modern periodic table. Detailed step-by-step solutions help students understand concepts clearly, improve problem-solving skills, and strengthen their preparation for JEE Main, CBSE Class 11 Chemistry, and other engineering entrance examinations.
JEE Main PYQs Most Important 100% Ex[ected MCQs With Answers Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties
Practice these JEE Main Most Expected PYQs and MCQs from Classification of Elements and Periodicity in Properties to strengthen your conceptual understanding and improve exam performance. Regular practice of important questions with detailed solutions helps students identify recurring patterns, revise key concepts efficiently, and build confidence for JEE Main and other competitive examinations.
For complete preparation, also study JEE Main PYQs Previous Year Questions MCQs : Classification of Elements and Periodicity in Properties
How can students improve their performance in JEE Main Chemistry?
Students can improve their performance by building strong conceptual understanding, revising topics regularly, solving a variety of MCQs, and practicing previous year questions. Consistent study and proper time management are also important for achieving better results.
Frequently linked concepts include IMU CET Merchant Navy Sponsorship Exam MCQs (PYQs) on Periodic Table and Periodicity
What is the importance of solving previous year questions for competitive exams?
Previous year questions help students become familiar with the examination pattern, understand frequently tested concepts, and develop effective problem-solving skills. They also help in identifying strengths and areas that need additional practice.
For complete preparation, also study NEET PYQs Previous Year Questions MCQs Classification of Elements and Periodicity in Properties
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)
Answer: Option (D) is correct — Statement I is incorrect but Statement II is correct.
To determine the correct option, let us examine each statement separately.
Statement I: “In Cl₂ molecule the covalent radius is double of the atomic radius of chlorine.”
This statement is incorrect. The covalent radius of an atom is defined as half of the distance between the nuclei of two identical atoms joined by a covalent bond. In a chlorine molecule (Cl₂), the bond length represents the distance between the two chlorine nuclei. Therefore, the covalent radius of chlorine is equal to half of the Cl–Cl bond length.
Covalent radius = ½ (bond length of Cl₂)
The statement incorrectly suggests that the covalent radius is double the atomic radius. In reality, the bond length is approximately twice the covalent radius, not the other way around. Hence, Statement I is false.
Statement II: “Radius of anionic species is always greater than their parent atomic radius.”
This statement is correct. When an atom gains one or more electrons to form an anion, the number of electrons increases while the nuclear charge remains unchanged. As a result, electron-electron repulsion within the electron cloud increases, causing the outer electrons to spread farther from the nucleus.
Because of this increased repulsion and reduced effective attraction per electron, the size of the anion becomes larger than that of the corresponding neutral atom. Therefore, Statement II is true.
Since Statement I is incorrect and Statement II is correct, the correct answer is :
Final Answer: (D) Statement I is incorrect but Statement II is correct.
Important exam-related topics include JEE Main Solved Question Papers PYQs Important MCQs Periodic Table and Periodicity
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 : Option (D) is correct – B < Be < O < N
Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. In general, ionization enthalpy increases across a period from left to right because the effective nuclear charge increases and atomic size decreases. As a result, electrons are held more strongly by the nucleus.
However, there are two important exceptions in the second period involving the pairs Be–B and N–O.
Beryllium has the electronic configuration 1s² 2s². The 2s subshell is completely filled, which provides extra stability to the atom. Boron has the electronic configuration 1s² 2s² 2p¹, where the outermost electron is present in a 2p orbital. Since a 2p electron is higher in energy and less tightly held than a 2s electron, it can be removed more easily. Therefore, boron has a lower ionization enthalpy than beryllium.
Another exception occurs between nitrogen and oxygen. Nitrogen has the electronic configuration 1s² 2s² 2p³, which represents a stable half-filled p-subshell. Oxygen has the electronic configuration 1s² 2s² 2p⁴, where one of the p orbitals contains a pair of electrons. The repulsion between paired electrons makes it easier to remove an electron from oxygen than from nitrogen. Consequently, oxygen has a lower ionization enthalpy than nitrogen.
Taking into account the general periodic trend and these exceptions, the increasing order of first ionization enthalpies is:
B < Be < O < N
Final Answer: (D) B < Be < O < N
Approximate First Ionization Enthalpies (kJ mol⁻¹):
| Element | Electronic Configuration | I.E. (kJ mol⁻¹) |
|---|---|---|
| B (Z=5) | 1s² 2s² 2p¹ | 800 |
| Be (Z=4) | 1s² 2s² | 900 |
| O (Z=8) | 1s² 2s² 2p⁴ | 1310 |
| N (Z=7) | 1s² 2s² 2p³ | 1400 |
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: Option (D) is correct – Unniltrium
To determine the IUPAC name of the element, we must first calculate its atomic number from the given electronic configuration.
The noble gas core [Rn] represents radon, which has an atomic number of 86. Beyond radon, the given configuration contains 14 electrons in the 5f subshell, 1 electron in the 6d subshell, and 2 electrons in the 7s subshell. Thus, the total number of additional electrons is 17.
Adding these electrons to the atomic number of radon gives : 86 + 17 = 103
Therefore, the element has atomic number 103, which corresponds to the element Lawrencium (Lr).
Before an official name is assigned, IUPAC uses a systematic temporary naming scheme based on the digits of the atomic number. For atomic number 103, the digit roots are :
1 is for un
0 is for nil
3 is for tri
Combining these roots and adding the suffix “ium” gives the temporary IUPAC name:
Unniltrium
Hence, the correct answer is Option (D) Unniltrium.
Note: The currently accepted name for the element with atomic number 103 is Lawrencium (Lr), but according to the systematic IUPAC temporary nomenclature, it is called Unniltrium.
Explore more concepts related to What is Chemical Bond ? Why do Atoms Combine ? How do Atoms Combine ?
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
Electron gain enthalpy refers to the amount of energy released when a neutral gaseous atom accepts an electron to form a negative ion. As a general rule, when moving down a group in the periodic table, atomic size increases because new electron shells are added. Because the valence shell moves further away from the nucleus, the incoming electron experiences a weaker electrostatic attraction to the nuclear charge, which typically causes the electron gain enthalpy to become less negative down the group.
However, oxygen presents a significant exception to this rule due to its exceptionally small atomic radius. In an oxygen atom, the incoming electron must enter the highly compact second electron shell (2p subshell), where the existing electrons are packed tightly together. This creates intense electron-electron repulsion that actively resists the addition of another electron, meaning significantly less energy is released than expected.
In contrast, sulfur has a larger atomic volume with a more spacious 3p subshell, which easily accommodates an incoming electron with minimal inner-electron repulsion. Consequently, sulfur releases more energy and has a much more negative electron gain enthalpy than oxygen, giving us the initial relationship of O < S.
After sulfur down the rest of the group from sulfur to selenium and then to tellurium (S → Se→ Te), the traditional periodic trend resumes normally. The atomic size steadily expands with each successive element, causing the nuclear pull on any newly added electron to weaken. As a result, the electron gain enthalpy becomes progressively less negative from sulfur down to tellurium (S > Se > Te).
Therefore combine both the above orders, we get
O < S > Se > Te
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
The first ionization enthalpy increases as move from left to right across a period in the periodic table. This general trend occurs because the nuclear charge increases while the electrons are added to the same principal energy level, pulling the outer electrons closer and holding them more tightly. However, the transition from nitrogen to oxygen presents a well-known exception to this rule, validating Assertion A that the first ionization enthalpy of oxygen is actually lower than that of nitrogen.
The explanation for this anomaly lies directly within the specific arrangement of electrons in their outermost subshells. Nitrogen has an atomic number of 7, giving it a electronic configuration of 1s² 2s² 2p³. According to Hund’s rule, the three electrons in the 2p subshell occupy three separate orbitals ($2p_x^1, 2p_y^1, 2p_z^1$) with parallel spins. This symmetrical, half-filled configuration imparts extra stability to the nitrogen atom, making it highly resistant to losing an electron.
On the other hand, oxygen has an atomic number of 8 with a electronic configuration of 1s² 2s² 2p⁴. In oxygen’s 2p subshell, the fourth electron must pair up with an existing electron in one of the orbitals (resulting in $2p_x^2, 2p_y^1, 2p_z^1$). Forcing two negatively charged electrons into the same small region of space creates significant inter-electronic repulsion. This mutual repulsion makes it energetically easier to remove one of these paired electrons from oxygen compared to removing an electron from nitrogen’s stable, half-filled subshell. Because this electron-electron repulsion makes the easier removal of oxygen’s electron, Reason R is not only factually correct but serves as the precise and correct explanation for Assertion A. Thus, option (A) is the correct choice.
JEE Main 2022 Question:
Match List – I with List – II
| Oxide | Nature |
|---|---|
| Cl₂O₇ | Acidic |
| Na₂O | Basic |
| Al₂O₃ | Amphoteric |
| N₂O | Neutral |
Correct Option: (B) (A) – (IV), (B) – (II), (C) – (I), (D) – (III)
First Analyze each oxide
Cl₂O₇ (Dichlorine heptoxide) : Non-metal oxide in high oxidation state, which is acidic.
Na₂O (Sodium oxide): Alkali metal oxide, which is basic.
Al₂O₃ (Aluminium oxide): Can react with both acids and bases, which is amphoteric.
N₂O (Nitrous oxide): Neutral oxide (like CO, H₂O), which is neutral.
Now Match pairs
(A) Cl₂O₇ (IV) Acidic
(B) Na₂O (II) Basic
(C) Al₂O₃ (I) Amphoteric
(D) N₂O (III) Neutral
JEE Main 2022 Question:
Among the following, basic oxide is :
(A) SO₃
(B) SiO₂
(C) CaO
(D) Al₂O₃
Answer: (C) CaO
According to basic rule of oxides : Metal oxides are usually basic, Non-metal oxides are usually acidic and some metal oxides (like Al₂O₃, ZnO) are amphoteric.
SO₃ a Non-metal oxide is Acidic, SiO₂ a Non-metal oxide is Acidic, CaO is Metal oxide is Basic and Al₂O₃ is Metal oxide but shows dual behavior and is Amphoteric in nature.
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
The element “E” belongs to Period 4, Group 16 is Selenium (Se). Just above Se in Group 16 is Sulfur (S) (Period 3, Group 16). For Sulfur (n = 3), the valence shell electronic configuration is 3s² 3p⁴.
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⁻
To determine the correct order, we first observe that all the given ions contain the same number of electrons. Each ion has 10 electrons, making them isoelectronic species. Since they have the same electronic configuration as neon, the size difference among them depends mainly on the nuclear charge of the atom.
In an isoelectronic series, the ion with the greater nuclear charge attracts its electrons more strongly toward the nucleus. As a result, the electron cloud is pulled closer, leading to a smaller ionic radius. Conversely, a lower nuclear charge results in a weaker attraction and a larger ionic radius.
Among the given ions, magnesium ion (Mg²⁺, Z = 12) has the highest nuclear charge, followed by sodium ion (Na⁺, Z = 11), fluoride ion (F⁻, Z = 9), oxide ion (O²⁻, Z = 8), and nitride ion (N³⁻, Z = 7). Therefore, Mg²⁺ has the smallest radius, while N³⁻ has the largest radius. Arranging the ions in order of increasing ionic radii gives :
Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
Hence, Option (A) is the correct answer.
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: Option (D) is correct – Assertion (A) is false but Reason (R) is true.
To evaluate the given assertion and reason, let us first examine the reason. The oxide ion (O²⁻) contains 8 protons and gains 2 electrons, giving it a total of 10 electrons. Similarly, the magnesium ion (Mg²⁺) contains 12 protons and loses 2 electrons, also resulting in 10 electrons. Since both ions have the same number of electrons, they are classified as isoelectronic species. Therefore, the reason is correct.
Now consider the assertion. Although O²⁻ and Mg²⁺ are isoelectronic, their ionic radii are not the same. In an isoelectronic series, the size of an ion depends on the nuclear charge. A higher nuclear charge exerts a stronger attraction on the electrons, pulling them closer to the nucleus and reducing the ionic radius.
Magnesium ion (Mg²⁺) has 12 protons, whereas oxide ion (O²⁻) has only 8 protons. Because Mg²⁺ has a much greater nuclear charge, it attracts its electrons more strongly and is therefore significantly smaller than O²⁻. As a result, the ionic radii of O²⁻ and Mg²⁺ are not equal.
Hence, the assertion is false, while the reason is true.
Final Answer:
Option (D): Assertion (A) is false but Reason (R) is true.
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: Option (B) is correct – Po < Te < F < Cl.
Electron gain enthalpy is the energy change that occurs when an electron is added to an isolated gaseous atom. A more negative value of electron gain enthalpy indicates a greater tendency of an atom to accept an electron.
As we move across a period from left to right, electron gain enthalpy generally becomes more negative because the effective nuclear charge increases. On the other hand, as we move down a group, atomic size increases and the attraction between the nucleus and the incoming electron decreases. Therefore, electron gain enthalpy generally becomes less negative down a group.
When comparing fluorine and chlorine, an important exception is observed. Although fluorine is more electronegative, its very small atomic size causes significant repulsion between the incoming electron and the electrons already present in the compact 2p orbitals. As a result, chlorine releases more energy on gaining an electron and has a more negative electron gain enthalpy than fluorine.
For tellurium (Te) and polonium (Po), both belong to the chalcogen family. Moving down the group from tellurium to polonium, atomic size increases and electron gain enthalpy becomes less negative. Therefore, polonium has a less negative electron gain enthalpy than tellurium.
Since halogens generally have much more negative electron gain enthalpies than chalcogens, chlorine and fluorine both have more negative values than tellurium and polonium. Combining all these comparisons, the increasing order of electron gain enthalpy is :
Po < Te < F < Cl
Hence, Option (B) is the correct answer.
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
Answer: Option (B) is correct.
The given oxides are Na₂O, As₂O₃, N₂O, NO, and Cl₂O₇. To determine the number of amphoteric oxides, we must first understand the nature of each oxide.
Sodium oxide (Na₂O) is an oxide of an alkali metal. Alkali metal oxides are strongly basic in nature and readily react with acids to form salts and water. Therefore, Na₂O is a basic oxide.
Arsenic trioxide (As₂O₃) is an oxide of arsenic, which is a metalloid. It exhibits both acidic and basic properties. It can react with acids as well as bases, making it an amphoteric oxide. This dual behavior is characteristic of many metalloid oxides.
Nitrous oxide (N₂O) and nitric oxide (NO) are neutral oxides of nitrogen. These oxides do not show significant acidic or basic behavior and generally do not react with acids or bases under ordinary conditions. Therefore, neither of them is amphoteric.
Dichlorine heptoxide (Cl₂O₇) is an oxide of chlorine in its highest oxidation state. It is a strongly acidic oxide and acts as the anhydride of perchloric acid. It reacts with bases but does not exhibit amphoteric behavior.
From the above analysis, only As₂O₃ is amphoteric. Therefore, the number of amphoteric oxides among the given compounds is one.
Final Answer: (B) 1
| Exam Tip |
|---|
| Amphoteric oxides commonly include Al₂O₃, ZnO, BeO, SnO₂, PbO, and As₂O₃. Oxides of alkali metals are generally basic, while non-metal oxides in high oxidation states are usually acidic. |
JEE Main 2021 Question:
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Barium carbonate is insoluble in water and is highly stable.
Reason (R): The thermal stability of the carbonates increases with increasing cationic size.
(A) Both (A) and (R) are true and (R) is the true explanation of (A)
(B) (A) is true but (R) is false
(C) Both (A) and (R) are true but (R) is not the true 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).
The thermal stability of Group 2 carbonates increases as we move down the group from magnesium to barium. This trend is related to the increasing size of the metal cations. Larger cations possess lower polarizing power and therefore cause less distortion of the carbonate ion. As a result, the carbonate ion remains more stable and decomposition becomes more difficult. Hence, the thermal stability of carbonates increases with increasing cationic size. Therefore, the reason is correct.
Now consider the assertion. Barium carbonate is a carbonate of a large alkaline earth metal cation, Ba²⁺. Due to its large ionic size and low polarizing power, the carbonate ion remains relatively undistorted, making barium carbonate highly thermally stable. In addition, barium carbonate is only sparingly soluble in water and is generally regarded as insoluble in qualitative chemistry.
The reason directly explains why barium carbonate is highly stable. The large size of the Ba²⁺ ion reduces polarization of the carbonate ion, resulting in greater ionic character and enhanced thermal stability.
Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.
Final Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
JEE Main 2021 Question:
The ionic radii of K⁺, Na⁺, Al³⁺ and Mg²⁺ are in the order:
(A) Na⁺ < K⁺ < Mg²⁺ < Al³⁺
(B) Al³⁺ < Mg²⁺ < K⁺ < Na⁺
(C) Al³⁺ < Mg²⁺ < Na⁺ < K⁺
(D) K⁺ < Al³⁺ < Mg²⁺ < Na⁺
Answer : Option (C) is correct – Al³⁺ < Mg²⁺ < Na⁺ < K⁺
The size of an ion depends on factors such as nuclear charge, number of electrons, and electronic configuration. For ions having the same number of electrons, known as isoelectronic ions, the ion with the greater nuclear charge attracts its electrons more strongly and therefore has a smaller ionic radius.
Let us first examine the electronic configurations of the given ions. Sodium ion (Na⁺), magnesium ion (Mg²⁺), and aluminium ion (Al³⁺) each contain 10 electrons and are isoelectronic with neon. Among these ions, Al³⁺ has the highest nuclear charge, followed by Mg²⁺ and then Na⁺. Since a higher nuclear charge results in a stronger attraction between the nucleus and electrons, the ionic radii increase in the order:
Al³⁺ < Mg²⁺ < Na⁺
Potassium ion (K⁺), however, contains 18 electrons and has the electronic configuration of argon. Because it possesses an additional electron shell compared to the other three ions, its ionic radius is significantly larger.
Therefore, when all four ions are arranged in increasing order of ionic radii, the correct sequence is:
Al³⁺ < Mg²⁺ < Na⁺ < K⁺
Hence, Option (C) is the correct answer.
| Exam Tip |
|---|
| In an isoelectronic series, ionic radius decreases as nuclear charge increases. Also, ions with more electron shells are generally larger than ions with fewer shells. |
JEE Main 2021 Question:
The incorrect statement about D.I. Mendeleev is:
(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
Answer : Correct Option: (B) At the time, he proposed Periodic Table of elements, structure of atom was known
Dmitri Mendeleev published the first version of the periodic table in 1869. At that time, the atomic structure was not yet understood; concepts like electrons, protons, neutrons, and their arrangement were unknown. Mendeleev organized elements based on atomic weights and recurring chemical properties, not atomic numbers. The relationship between atomic number and number of protons was established later by Henry Moseley in the early 20th century.
(A) Correct – Mendeleev authored Principles of Chemistry, a widely influential textbook.
(C) Correct – Element 101, Mendelevium (Md), is named in his honor.
(D) Minor clarification – Mendeleev did not invent the barometer; that was Evangelista Torricelli in the 17th century.
Final Answer: Structure of atom was not known when Mendeleev proposed the periodic table.
JEE Main 2021 Question:
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) neon, sodium and chlorine
(B) argon, chlorine and sodium
(C) chlorine, lithium and sodium
(D) argon, lithium and sodium
Answer: Option (B) is correct – Argon, Chlorine and Sodium
Ionization energy generally increases across a period from left to right because the effective nuclear charge increases and atomic size decreases. On the other hand, ionization energy decreases down a group due to increasing atomic size and shielding effect. Noble gases possess exceptionally high ionization energies because of their stable electronic configurations, whereas alkali metals have very low ionization energies since they can easily lose their single valence electron.
Magnesium is a Group 2 element with atomic number 12. Its first ionization energy is higher than that of sodium, an alkali metal in the same period, but lower than that of chlorine and argon, which are located further to the right in Period 3. Argon, being a noble gas, has a particularly high ionization energy, while chlorine also has a significantly higher ionization energy than magnesium.
Therefore :
Argon has a higher ionization energy than magnesium.
Chlorine has a higher ionization energy than magnesium.
Sodium has a lower ionization energy than magnesium.
This perfectly satisfies the given condition.
Final Answer: (B) Argon, Chlorine and Sodium
In Period 3, the general order of first ionization energies is:
Na < Mg < Cl < Ar
This makes it easy to identify elements having ionization energies greater or smaller than magnesium.
JEE Main 2021 Question :
The ionic radius of Na⁺ ion is 1.02 Å. The ionic radii (in Å) of Mg²⁺ and Al³⁺, 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
Answer: Option (B) is correct – 0.72 Å and 0.54 Å
To determine the ionic radii of Mg²⁺ and Al³⁺, we use the concept of isoelectronic species. Isoelectronic ions have the same number of electrons but different nuclear charges. In such a series, the ion with the greater nuclear charge has the smaller ionic radius because its electrons are attracted more strongly toward the nucleus.
The ions Na⁺, Mg²⁺, and Al³⁺ each contain 10 electrons and therefore are isoelectronic with neon. Since they all have the same number of electrons, their sizes depend mainly on the number of protons present in the nucleus.
Among these ions, Na⁺ has 11 protons, Mg²⁺ has 12 protons, and Al³⁺ has 13 protons. As the nuclear charge increases from sodium to aluminium, the attraction between the nucleus and the electron cloud becomes stronger, causing the ionic radius to decrease.
Therefore, the order of ionic radii is :
Al³⁺ < Mg²⁺ < Na⁺
Since the ionic radius of Na⁺ is given as 1.02 Å, both Mg²⁺ and Al³⁺ must have smaller radii, with Mg²⁺ larger than Al³⁺.
Among the given options, only 0.72 Å for Mg²⁺ and 0.54 Å for Al³⁺ satisfy this trend.
Final Answer: (B) 0.72 Å and 0.54 Å
JEE main 2021 Question:
The characteristics of elements X, Y and Z with atomic numbers 33, 53, and 83, respectively, are :
(A) X and Y are metalloids and Z is a metal
(B) X and Z are non-metals and Y is a metalloid
(C) X is a metalloid, Y is a non-metal and Z is a metal
(D) X, Y and Z are metals
Answer: Option (C) is correct – X is a metalloid, Y is a non-metal and Z is a metal
Element X has atomic number 33, which corresponds to Arsenic (As). Arsenic is located along the zigzag or staircase line that separates metals from non-metals in the periodic table. Elements in this region generally exhibit properties intermediate between metals and non-metals and are classified as metalloids.
Element Y has atomic number 53, which corresponds to Iodine (I). Iodine belongs to Group 17, the halogen family. Halogens are typical non-metals and possess high electronegativity and electron affinity. Therefore, iodine is classified as a non-metal.
Element Z has atomic number 83, which corresponds to Bismuth (Bi). Bismuth is a member of Group 15 and exhibits metallic properties such as electrical conductivity and metallic lustre. It is classified as a metal.
Thus, the classifications are :
X (Arsenic) is a Metalloid
Y (Iodine) is a Non-metal
Z (Bismuth) is a Metal
Therefore, the correct option is : (C) X is a metalloid, Y is a non-metal and Z is a metal
JEE Main 2021 Question:
The correct order of bond dissociation enthalpy for halogens is :
(A) F₂ > Cl₂ > Br₂ > I₂
(B) Cl₂ > Br₂ > F₂ > I₂
(C) Cl₂ > Br₂ > F₂ > I₂
(D) Cl₂ > F₂ > Br₂ > I₂
Answer: Option (C) is correct – Cl₂ > Br₂ > F₂ > I₂
Bond dissociation enthalpy (BDE) is the amount of energy required to break one mole of a covalent bond in the gaseous state. A higher bond dissociation enthalpy indicates a stronger bond.
As we move down the halogen group, atomic size increases and the bond length between two halogen atoms becomes larger. Longer bonds are generally weaker and require less energy to break. Based on this trend alone, one might expect the bond strength to decrease from fluorine to iodine.
However, fluorine shows an important exception. The fluorine atom is extremely small, and each fluorine atom contains three lone pairs of electrons. In the F₂ molecule, these lone pairs are very close to each other, resulting in strong electron-electron repulsions. These repulsive forces weaken the F–F bond significantly.
Because of this unusual weakening, the F–F bond is less strong than the Cl–Cl and Br–Br bonds, despite fluorine being the smallest halogen. Chlorine has the highest bond dissociation enthalpy among the halogens because it has an optimal balance of bond length and minimal lone-pair repulsion.
Therefore, the correct order of bond dissociation enthalpy is:
Cl₂ > Br₂ > F₂ > I₂
Hence, Option (C) is the correct answer.
| Exam Tip |
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| Fluorine is an exception to the normal trend because strong lone pair-lone pair repulsions weaken the F–F bond, making its bond dissociation enthalpy lower than that of Cl₂ and Br₂. |
JEE Main 2021 Question:
Consider the elements Mg, Al, Si, S, and P. The correct increasing order of first ionization enthalpy is:
(A) Al < Mg < Si < S < P
(B) Al < Mg < S < Si < P
(C) Mg < Al < Si < S < P
(D) Mg < Al < Si < P < S
Answer: Option (A) is correct – Al < Mg < Si < S < P
First ionization enthalpy is the energy required to remove the outermost electron from an isolated gaseous atom. In general, ionization enthalpy increases across a period because the effective nuclear charge increases and atomic size decreases, causing electrons to be held more strongly by the nucleus.
The given elements, Mg, Al, Si, P, and S, all belong to the third period. While the general trend is an increase in ionization enthalpy from left to right, there are important exceptions due to electronic configuration.
Magnesium has the electronic configuration 3s², which represents a completely filled s-subshell and provides extra stability. Aluminium has the configuration 3s² 3p¹, where the outermost electron is present in a higher-energy p-orbital. This p-electron is easier to remove than an s-electron, so aluminium has a lower ionization enthalpy than magnesium.
Moving further across the period, silicon has a higher ionization enthalpy than both aluminium and magnesium because of its increased nuclear charge. Phosphorus possesses a half-filled 3p³ configuration, which is particularly stable and results in a relatively high ionization enthalpy.
Sulfur has the configuration 3p⁴, where one of the p-orbitals contains a pair of electrons. The repulsion between these paired electrons makes it slightly easier to remove an electron from sulfur than from phosphorus. Therefore, sulfur has a lower ionization enthalpy than phosphorus.
Combining the general trend and these exceptions, the increasing order of first ionization enthalpy is :
Al < Mg < Si < S < P
Hence, Option (A) is the correct answer.
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. This section also includes detailed study of What is Ionization Enthalpy? Definition, Units, Factors and Successive IE and Ionization Enthalpy Trends Along a Period and Down a Group for better understanding of periodic properties and reactivity of elements.