NCERT Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties provide detailed and easy-to-understand explanations of all textbook questions according to the latest CBSE syllabus. This chapter explains the modern periodic table, periodic trends, electronic configuration, atomic radius, ionization enthalpy, electron gain enthalpy, electronegativity, metallic character, and non-metallic character of elements. These NCERT solutions help students develop strong conceptual understanding and improve problem-solving skills for school examinations as well as competitive exams like JEE and NEET. Step-by-step answers make learning chemistry simple, clear, and effective for students.
NCERT Textbook Practice Questions and Answers for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties
NCERT Textbook Questions and Solutions for Class 11 Chemistry Chapter Classification of Elements and Periodicity in Properties help students understand the modern periodic table and important periodic trends in a simple and conceptual manner. Students can use these solutions for board exam preparation, revision, and strengthening their chemistry concepts.
Related topics include Conceptual Questions Based On Atomic Radii, Ionic Radii and Isoelectronic Ions Trends of Periodic Table
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Yes, NCERT questions are very important for Class 11 Chemistry exams because many conceptual and board examination questions are directly based on NCERT textbook exercises and examples.
Students should also study IMU CET Merchant Navy Sponsorship Exam MCQs (PYQs) on Periodic Table and Periodicity
Can NCERT exercise solutions help in competitive exam preparation?
Yes, NCERT exercise solutions help students build strong basic concepts which are useful for competitive exams like JEE Main, NEET, and other entrance examinations.
Students must solve NEET PYQs Previous Year Questions MCQs Classification of Elements and Periodicity in Properties
NCERT 3.1 : What is the basic theme of organisation in the periodic table?
Answer :
The basic theme of organisation of elements in the periodic table is to simplify and systematize the study of the properties of all the elements and millions of their compounds.
On the basis of similarities in chemical properties, the various elements have now been divided into different groups. This has made the study simple because the properties of elements are now studied in the form of groups rather than individually.
For better understanding, also read What is Diagonal Relationship in Periodic Table ?
NCERT 3.2 : Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Answer :
Mendeleev used the atomic weight of elements as the fundamental property for classification in his periodic table. He arranged all the known elements in order of increasing atomic weights, and while doing so, he placed elements with similar chemical properties in the same vertical columns, which he called groups. This systematic arrangement allowed him to show a clear periodicity in the properties of elements.
However, Mendeleev did not always stick strictly to the order of increasing atomic weights. In some cases, he placed an element with a slightly higher atomic weight before another element with a lower atomic weight if the chemical properties demanded it. For example, he placed iodine (atomic weight 127) after tellurium (atomic weight 128), even though iodine has a lower atomic weight, because iodine’s properties matched better with fluorine, chlorine, and bromine in the halogen group. This shows that Mendeleev gave more importance to the chemical properties of elements rather than blindly following atomic weight order.
This flexibility was one of the greatest strengths of Mendeleev’s periodic table. It not only explained the properties of known elements more accurately but also helped in predicting the existence and properties of elements that had not yet been discovered.
Read more about What are s-block Elements of the Modern Periodic Table ?
NCERT 3.3 : What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law?
Answer :
Mendeleev’s Periodic Law states that the physical and chemical properties of the elements are the periodic functions of their atomic weights. In simple terms, he arranged the elements in order of increasing atomic weights and observed that similar properties repeated at regular intervals. This principle helped him to classify the elements into groups and periods.
The Modern Periodic Law, however, states that the physical and chemical properties of the elements are the periodic functions of their atomic numbers. This means that elements are arranged on the basis of their atomic numbers (the number of protons in the nucleus) instead of their atomic weights. Since atomic number is the true fundamental property of an element, this law gave a more accurate arrangement of elements.
The difference becomes clear with examples. In Mendeleev’s table, argon (atomic weight 40) was placed before potassium (atomic weight 39), which seemed wrong because potassium has a lower atomic weight. But when arranged according to atomic number, argon (atomic number 18) comes before potassium (atomic number 19), which is correct. Similarly, cobalt (atomic weight 59) and nickel (atomic weight 59.9) could not be explained properly in Mendeleev’s arrangement, but in the modern table cobalt (atomic number 27) is placed before nickel (atomic number 28) without any problem.
Thus, the basic difference in approach is that Mendeleev used atomic weight as the basis of classification, while the modern periodic law uses atomic number. This shift solved the anomalies and made the arrangement of elements more scientific and accurate.
The fundamental difference between Mendeleev’s Periodic Law and the Modern Periodic Law lies in the foundational property used to classify and arrange the elements as follows :
| Feature | Mendeleev’s Periodic Law | Modern Periodic Law |
| Basic Statement | The properties of elements are a periodic function of their atomic weights. | The properties of elements are a periodic function of their atomic numbers. |
| Basis of Classification | It is based on the atomic mass (the total number of protons and neutrons in the nucleus). | It is based on the atomic number (the total number of protons, which equals the number of electrons in a neutral atom). |
| Scientific Origin | Formulated by Dmitri Mendeleev in 1869 based on macroscopic chemical properties and formulas of oxides/hydrides. | Formulated by Henry Moseley in 1913 based on X-ray frequency experiments that linked atomic structures to properties. |
| Position of Isotopes | Could not explain or accommodate isotopes, because isotopes of the same element have different atomic weights but identical chemical properties. | Easily accommodates isotopes because they share the exact same atomic number, putting them in the same spot. |
| Anomalous Pairs | Led to subshell placement errors where heavier elements sometimes had to be placed before lighter ones (e.g., Argon before Potassium) to match properties. | Resolved all anomalous pairs naturally because atomic number increases sequentially without inversion. |
| Electronic Configuration | Does not relate the arrangement of elements to their underlying electronic configurations. | Directly correlates the positions of elements to their valence shell electronic configurations. |
NCERT 3.4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer.
In the modern periodic table, each period corresponds to the filling of a new principal energy level. The sixth period begins with the filling of the principal quantum number n = 6. For n = 6, the possible values of the azimuthal quantum number l are : l = 0 (6s), l = 1 (6p), l = 2 (6d), l = 3 (6f). However, according to the Aufbau principle, the order of filling is not simple. The relative energies of these subshells are :
6s < 4f < 5d < 6p
Thus, in the sixth period, the electrons are filled in the 6s, 4f, 5d, and 6p subshells. After that seventh period starts becuase next electron is enter into 7s subshell and 6d, 6f subshells are belongs to higher periods. Now let us count the orbitals of 6s, 4f, 5d, 6p subshells :
- s-subshell has 1 orbital
- p-subshell has 3 orbitals
- d-subshell has 5 orbitals
- f-subshell has 7 orbitals
Therefore, the total number of orbitals is : 1 + 3 + 5 + 7 = 16 orbitals. Since each orbital can accommodate a maximum of 2 electrons (Pauli’s Exclusion Principle), the total number of electrons that can be filled is : 16 x 2 = 32. Hence, the sixth period of the periodic table should have 32 elements.
Build strong concepts by studying What are p-block Elements of the Modern Periodic Table ?
NCERT 3.5 In terms of period and group, where would you locate the element with Z = 114 ?
Ans.
The filling of the 6th period ends at Rn (Z = 86). Thereafter, the filling of the 7th period starts. Just like in the 6th period, in the 7th period also, the filling of four subshells occurs : 7s, 5f, 6d, and 7p. According to the Aufbau principle, their relative energies increase in the order :
7s < 5f < 6d < 7p
After Rn (Z = 86), the next two elements with Z = 87 and Z = 88 are s-block elements.
The next fourteen elements, i.e. Z = 90 to Z = 103, are f-block elements.
The following ten elements, i.e. Z = 104 to Z = 112, are d-block elements.
The last six elements, i.e. Z = 113 to Z = 118, are p-block elements.
Therefore, the element with Z = 114 is the second p-block element (i.e., in Group 14) of the 7th period. Thus, the location of the element with Z = 114 in the periodic table is :
Period = 7th, Block = p, Group = 14
Enhance your preparation with Electron Gain Enthalpy : Definition, Units, Factors, Trends, Successive Electron Gain Enthalpy
NCERT 3.6: Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer :
The third period corresponds to principal quantum number n = 3, so only the 3s and 3p subshells are filled in this period. That means the third period contains 2 s-block elements and 6 p-block elements, a total of 8 elements. The third period therefore starts at Z = 11 (sodium, Na) and ends at Z = 18 (argon, Ar). After the two s-block elements (Z = 11, 12), the p-block elements occupy Z = 13 to Z = 18, which correspond respectively to groups 13 – 18. Therefore group 17 in the third period corresponds to Z = 17.
Z = 11 : Na (3s¹) belongs to group 1
Z = 12 : Mg (3s²) belongs to group 2
Z = 13 : Al (3p¹) belongs to group 13
Z = 14 : Si (3p²) belongs to group 14
Z = 15 : P (3p³) belongs to group 15
Z = 16 : S (3p⁴) belongs to group 16
Z = 17 : Cl (3p⁵) belongs to group 17
Z = 18 : Ar (3p⁶) belongs to group 18
The element with atomic number is 17 is chlorine, Cl). The electron configuration of the element with Z = 17 is [Ne] 3s2 3p5.
Explore detailed notes on Electronegativity : Pauling and Mulliken Scale, Periodic Trends, Factors Affecting Electronegativity
Question 3.7: Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory (ii) Seaborg’s group?
Answer :
(i) The Lawrence Berkeley Laboratory in California has contributed to the discovery of many transuranium elements. Two elements named in its honor are :
Lawrencium (Lr, Z = 103), named after Ernest O. Lawrence, the inventor of the cyclotron and founder of the Lawrence Berkeley Laboratory.
Berkelium (Bk, Z = 97), named after the city of Berkeley, where the laboratory is located.
(ii) Seaborg’s group at the University of California was instrumental in the discovery of several heavy elements. One element was directly named to honor Glenn T. Seaborg, the famous American chemist who contributed greatly to actinide chemistry. That element is Seaborgium (Sg, Z = 106).
Practice more questions from Why does atomic radius increase down a group?
NCERT 3.8: Why do elements in the same group have similar physical and chemical properties?
Answer :
Elements in the same group of the periodic table have similar physical and chemical properties because they have the same number of electrons in their outermost shell (valence shell). Since chemical behavior mainly depends on the valence electrons, similar configurations result in similar properties.
Example: Group 1 Elements (Alkali Metals)
All alkali metals have the general configuration ns¹. Each of them has one electron in the outermost shell, which they readily lose to form M⁺ ions. This explains why all alkali metals are soft, highly reactive, good reducing agents, and form similar compounds like oxides, hydroxides, and halides.
Example: Group 17 Elements (Halogens)
All halogens have the outer electronic configuration ns²np⁵. With seven valence electrons, they need only one more electron to complete their octet. This explains why halogens are highly electronegative, reactive, and form salts such as NaCl, KBr, etc. All halogens also exist naturally as diatomic molecules (Cl₂, Br₂, I₂).
Physical Properties Similarity
Because of similar electronic configurations, elements in the same group also exhibit similarities in physical properties such as atomic radius trends, ionization enthalpy patterns, and electronegativity.
Thus, the repetition of the same type of valence shell electronic configuration down a group is the fundamental reason why elements in the same group have similar physical and chemical properties.
For complete preparation, also study complete chapter notes of Classification of Elements and Periodicity in Properties
NCERT 3.9 : What does atomic radius and ionic radius really mean to you?
Answer.
Atomic radius literally means the size of the atom. It can be measured either by X–ray diffraction or by spectroscopic methods.
In case of non–metals, atomic radius is called covalent radius. It is defined as one–half the distance between the nuclei of two covalently bonded atoms of the same element in a molecule.
For example, the internuclear distance between two chlorine atoms in a chlorine molecule is 198 pm. Therefore, the covalent radius of a chlorine atom is :
$$
r_{\text{cov}} = \frac{198}{2} = 99 \ \text{pm} \ (0.99 \ \text{Å})
$$
In case of metals, atomic radius is called metallic radius. It is defined as one–half the distance between two adjacent atoms in the crystal lattice.
For example, the distance between two adjacent copper atoms in solid copper is $256 \ \text{pm}$. Therefore, the metallic radius of copper is:
$$
r_{\text{met}} = \frac{256}{2} = 128 \ \text{pm} \ (1.28 \ \text{Å})
$$
In case of Ionic Radius, Ionic radius means the size of the ion. An ion can be either a cation or an anion.
The size of a cation is always smaller than that of the parent atom because loss of one or more electrons increases the effective nuclear charge ($Z_{\text{eff}}$). As a result, the force of attraction increases and the ionic size decreases.
The size of an anion is always larger than that of the parent atom because the addition of one or more electrons decreases the effective nuclear charge ($Z_{\text{eff}}$). As a result, the force of attraction decreases and the ionic size increases.
Practice more questions from NCERT Exemplar Solutions of Chapter 3 Classification of Elements and Periodicity in Properties for Class 11 Chemistry
NCERT 3.10 : How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer
The atomic radius increases down the group. This is because a new energy shell (i.e. principal quantum number increases by unity) is added at each succeeding element while the number of electrons in the valence shell remains the same. In other words, the electrons in the valence shell of each succeeding element lie farther and farther away from the nucleus. As a result, the force of attraction of the nucleus for the valence electrons decreases and hence the atomic size increases.
In contrast, the atomic size decreases as we move from left to right in a period. This is because within a period the outer electrons remain in the same shell but the nuclear charge increases by one unit at each succeeding element. Due to this increased nuclear charge, the attraction of the nucleus for the outer electrons increases and hence the atomic size decreases.
NCERT 3.11 : What do you understand by isoelectronic species? Name the species that will be isoelectronic with each of the following atoms or ions: (i) F– (ii) Ar (iii) Mg2+ (iv) Rb+
Answer :
Isoelectronic species are atoms, molecules, or ions that contain the exact same number of electrons. Even though they have a similar electronic configuration, they belong to different elements. Because they have different atomic numbers, their nuclear charges (the number of protons in the nucleus) are different, which causes them to differ in size and chemical behavior.
Here are the species that are isoelectronic with each of the given examples :
(i) Fluoride ion (F⁻)
A neutral fluorine atom has 9 electrons. Gaining one electron gives the F⁻ ion a total of 10 electrons. The following species also contain 10 electrons and are isoelectronic with it :
Nitride ion (N³⁻), Oxide ion (O²⁻), Neon atom (Ne), Sodium ion (Na⁺), Magnesium ion (Mg²⁺), Aluminum ion (Al³⁺)
(ii) Argon (Ar)
Argon is a noble gas that naturally contains 18 electrons. The following species also contain 18 electrons and are isoelectronic with it :
Phosphide ion (P³⁻), Sulfide ion (S²⁻), Chloride ion (Cl⁻), Potassium ion (K⁺), Calcium ion (Ca²⁺)
(iii) Magnesium ion (Mg²⁺)
A neutral magnesium atom has 12 electrons. Losing two electrons to form the Mg²⁺ ion leaves it with 10 electrons. Therefore, its isoelectronic species have exact same 10-electron species as follows :
Nitride ion (N³⁻), Oxide ion (O²⁻), Fluoride ion (F⁻), Neon atom (Ne), Sodium ion (Na⁺), and Aluminum ion (Al³⁺)
(iv) Rubidium ion (Rb⁺)
A neutral rubidium atom possesses 37 electrons. When it loses one electron to form the Rb⁺ ion, it is left with 36 electrons. The species that are isoelectronic with it and also contain 36 electrons are :
Selenide ion (Se²⁻), Bromide ion (Br⁻), Krypton atom (Kr), Strontium ion (Sr²⁺)
NCERT 3.12 : Consider the following species : N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺, and Al³⁺ (i) What is common in them? (ii) Arrange them in the order of increasing ionic radii
Answer :
All of these species N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺, and Al³⁺ have the exact same number of electrons. Each of them contains 10 electrons so they are called isoelectronic species. The correct order of increasing ionic radii (from smallest to largest) is :
Al3+ < Mg2+ < Na+ < F– < O2- < N3-
For a group of isoelectronic species, the number of electrons remains constant, but the nuclear charge (the number of protons in the nucleus) changes. As the nuclear charge increases, the nucleus exerts a much stronger electrostatic pull on the same 10 electrons, attracting them closer and reducing the size of the ion.
| Species | Number of Electrons | Nuclear Charge (Protons) | Effect on Size |
| Al³⁺ | 10 | +13 | Strongest nuclear pull, Smallest size |
| Mg²⁺ | 10 | +12 | Very strong nuclear pull |
| Na⁺ | 10 | +11 | Strong nuclear pull |
| F⁻ | 10 | +9 | Weaker nuclear pull |
| O²⁻ | 10 | +8 | Weak nuclear pull |
| N³⁻ | 10 | +7 | Weakest nuclear pull, Largest size |
Therefore, as you move from Al³⁺ to N³⁻, the nuclear charge decreases, causing the ionic radius increasing.
NCERT 3.13 : Explain why cations are smaller and anions are larger in radii than their parent atoms ?
Answer :
The size of an ion depends on the balance between the number of electrons and the positive charge of the nucleus (nuclear charge).
Cations (positively charged ions)
When an atom loses one or more electrons, it forms a cation. The loss of electrons reduces electron-electron repulsion in the atom. The number of protons in the nucleus remains the same, but there are now fewer electrons to balance the positive charge. As a result, the effective nuclear charge experienced by each electron increases. The electrons are pulled closer to the nucleus, so the ionic radius becomes smaller than that of the parent atom.
Example : Na to Na+
- Sodium loses one electron : 11 protons, 10 electrons.
- Increased nuclear pull causes the radius to decrease.
Anions (negatively charged ions)
When an atom gains one or more electrons, it forms an anion. The addition of electrons increases electron-electron repulsion in the atom. The number of protons in the nucleus remains the same, but there are now more electrons to attract. The effective nuclear charge per electron decreases. The electrons spread out more, so the ionic radius becomes larger than that of the parent atom.
Example: Cl to Cl–
- Chlorine gains one electron : 17 protons, 18 electrons.
- Reduced nuclear pull and increased repulsion cause the radius to increase.
Hence, Cations are smaller than parent atoms due to loss of electrons and stronger nuclear pull and Anions are larger than parent atoms due to gain of electrons and weaker nuclear pull.
NCERT 3.14 : What is the significance of the terms ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
Answer :
(i) Ionization Enthalpy
Ionization enthalpy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so as to convert it into a gaseous cation.
- The force with which an electron is attracted by the nucleus of an atom is significantly affected by the presence of other atoms within its molecule or in the neighborhood.
- For the purpose of determining ionization enthalpy, it is essential that these inter-atomic forces should be minimum.
- In the gaseous state, atoms are widely separated, so inter-atomic forces are minimal.
- Additionally, since it is not possible to isolate a single atom completely, the measurement is carried out at a low pressure, further reducing inter-atomic interactions.
Hence, the term “isolated gaseous atom” is included in the definition of ionization enthalpy.
(ii) Electron Gain Enthalpy
Electron gain enthalpy is the energy released when an isolated gaseous atom in the ground state accepts an extra electron to form a gaseous negative ion.
- The term isolated gaseous atom has already been explained above.
- The term ground state means that the atom must be present in its most stable state.
- If the atom is in an excited state, less energy will be released when it accepts an electron to form a gaseous anion.
- Therefore, for comparison purposes, the electron gain enthalpies of gaseous atoms must be determined in their respective most stable state, i.e., the ground state.
NCERT 3.15 : Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J/mol.
Answer :
The energy required to remove an electron from a hydrogen atom in the ground state is the negative of its ground state energy :
$$
E_{\text{electron}} = – (\text{energy in ground state}) = -(-2.18 \times 10^{-18}\ \text{J}) = 2.18 \times 10^{-18}\ \text{J}
$$
Use Avogadro’s number NA = 6.022 x 1023 mol-1 to convert energy for one atom to energy per mole :
$$
E_{\text{mole}} = E_{\text{electron}} \times N_A
= (2.18 \times 10^{-18}) \times (6.022 \times 10^{23})
$$
$$
E_{\text{mole}} \approx 13.13 \times 10^{5}\ \text{J mol}^{-1} = 1.313 \times 10^{6}\ \text{J mol}^{-1}
$$
$$
E_{\text{mole}} = 1.313 \times 10^{6}\ \text{J mol}^{-1} = 1313\ \text{kJ mol}^{-1}
$$
The ionization enthalpy of atomic hydrogen is approximately 1313 kJ/mol.
NCERT 3.16 : Among the second period elements, the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why (i) Be has higher first ionization enthalpy 
Δ1Hi than B (ii) O has lower Δ1Hi than N and F.
Answer :
(i)
The ionization enthalpy depends, among other factors, on the type of electron to be removed from the same principal shell.
- In the case of Be ($1s^2 2s^2$), the outermost electron is present in the 2s-orbital.
- In B ($1s^2 2s^2 2p^1$), the outermost electron is present in the 2p-orbital.
- Since 2s-electrons are more strongly attracted by the nucleus than 2p-electrons, less energy is required to remove a 2p-electron than a 2s-electron.
Consequently, the first ionization enthalpy (Δ1Hi) of Be is higher than that of B.
(II)
- The electronic configuration of N ($1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$), where the 2p-orbitals are exactly half-filled, is more stable than the electronic configuration of O ($1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1$), where the 2p-orbitals are neither half-filled nor completely filled.
- Therefore, it is more difficult to remove an electron from N than from O. As a result, Δ1Hi of N is higher than that of O.
- Further, the electronic configuration of F is $1s^2 2s^2 2p_x^2 2p_y^2 2p_z^1$. Because of the higher nuclear charge (+9), the first ionization enthalpy of F is higher than that of O.
NCERT 3.17 : Explain why the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium
Answer :
The electronic configurations of sodium and magnesium are :
$$
Na : 1s^2 2s^2 2p^6 3s^1
$$
$$
Mg : 1s^2 2s^2 2p^6 3s^2
$$
First Ionization Enthalpy
- In both Na and Mg, the first electron is removed from the 3s-orbital.
- The nuclear charge of Na (+11) is lower than that of Mg (+12), so the attraction of the nucleus on the 3s electron in Na is weaker.
- Hence, the first ionization enthalpy of sodium is lower than that of magnesium.
Second Ionization Enthalpy
After losing the first electron :
$$
Na^+ : 1s^2 2s^2 2p^6
$$
$$
Mg^+ : 1s^2 2s^2 2p^6 3s^1
$$
For Na+, the second electron must be removed from a stable inert gas configuration (Ne-like), which is very difficult. For Mg+, the second electron is removed from the 3s-orbital, which is easier than removing an electron from a stable configuration. Hence First ionization Δ1Hi of Na < Δ1Hi of Mg and Second ionization of Δ2Hi of Na > Δ2Hi of Mg.
NCERT 3.18 : Factors causing ionization enthalpy of main group elements to decrease down the group
Answer :
Within the main group elements, the ionization enthalpy decreases regularly as we move down the group due to the following factors:
(i) Atomic Size
- As we move down a group, a new principal energy shell is added for each succeeding element.
- This increases the distance of valence electrons from the nucleus.
- The force of attraction between the nucleus and the valence electrons decreases with distance.
- Result: Ionization enthalpy decreases down the group.
(ii) Screening (Shielding) Effect
- With the addition of new shells, the number of inner electron shells shielding the valence electrons also increases.
- This screening effect reduces the effective nuclear charge experienced by valence electrons.
- The nucleus attracts the valence electrons less strongly, leading to a further decrease in ionization enthalpy down the group.
Therefore, the combined effects of increasing atomic size and increasing shielding cause the ionization enthalpy of main group elements to decrease as we move down a group.
NCERT 3.19 : The first ionization enthalpy values (in kJ/mol) of group 13 elements are
| Element | B | Al | Ga | In | Tl |
|---|---|---|---|---|---|
| First ionization enthalpy | 801 | 577 | 579 | 558 | 589 |
How would you explain this deviation from the general trend ?
Answer :
In general moving down group 13 from B to Al, the ionization enthalpy decreases as expected due to increase in atomic size and screening effect, which outweigh the effect of increased nuclear charge. But the first ionization enthalpy of Ga (579 kJ/mol) is slightly higher than that of Al (577 kJ/mol).
Al follows immediately after s-block elements, while Ga follows after d-block elements. The extra d-electrons in Ga do not shield the outer electrons effectively. Therefore, the valence electrons are held more tightly, and more energy is required to remove them.
On Moving down to In, the increased shielding effect (due to 4d-electrons) outweighs the increased nuclear charge. Hence first ionization enthalpy of In (558 kJ/mol) is lower than that of Ga.
Moving further down, Tl has additional 4f and 5d electrons. Here, the increase in nuclear charge (+32 units from Ga to Tl) outweighs the shielding effect. Hence net result is that first ionization enthalpy of Tl (589 kJ/mol) is much higher than that of Al, Ga, and In.
NCERT 3.20 : Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl
Answer :
(i) Oxygen (O) vs Fluorine (F)
Both O and F are in the 2nd period. On moving from O to F, atomic size decreases and Nuclear charge increases. Both factors increase the attraction of the nucleus for the incoming electron, making the electron gain enthalpy more negative.
Also the electronic configuration after gaining one electron :
$F^- : 1s^2 2s^2 2p^6$ (stable inert gas configuration)
$O^- : 1s^2 2s^2 2p^5$ (not fully stable)
Therefore Fluorine has a more negative electron gain enthalpy than oxygen.
(ii) Fluorine (F) vs Chlorine (Cl)
Generally, electron gain enthalpy becomes less negative down a group. However, chlorine (Cl) has, more negative electron gain enthalpy than fluorine. Because Fluorine is very small, so electron–electron repulsions in its compact 2p-subshell are large. Incoming electron experiences more repulsion, making it harder to add compared to chlorine. Chlorine is larger, so electron-electron repulsion is smaller and the added electron is more easily accommodated. Hence Chlorine has a more negative electron gain enthalpy than fluorine due to reduced repulsion in its larger atomic orbitals.
NCERT 3.21 : Would you expect the second electron gain enthalpy of O to be positive, more negative, or less negative than the first? Justify your answer.
Answer :
The second electron gain enthalpy of O is positive. When an electron is added to an oxygen atom to form the O– ion, energy is released. This makes the first electron gain enthalpy negative.
$$
O (g) + e^- (g) \rightarrow O^- (g) \quad ; \quad \Delta_{eg}H_1 = -141\ \text{kJ mol}^{-1}
$$
When a second electron is added to O– to form O2-, energy must be absorbed to overcome the strong electrostatic repulsion between the negatively charged O– ion and the incoming electron. Therefore, the second electron gain enthalpy is positive.
$$
O^- (g) + e^- (g) \rightarrow O^{2-} (g) \quad ; \quad \Delta_{eg}H_2 = +78\ \text{kJ mol}^{-1}
$$
NCERT 3.22 : What is the basic difference between the terms electron gain enthalpy and electronegativity?
Both electron gain enthalpy and electronegativity describe the tendency of an atom to attract electrons, but they are different concepts :
Electron Gain Enthalpy :
It refers to the tendency of an isolated gaseous atom to accept an additional electron. The process leads to the formation of a negative ion. It is expressed in terms of energy change (kJ/mol).
Electronegativity (EN) :
It refers to the tendency of an atom of an element to attract the shared pair of electrons towards itself in a covalent bond. It is a relative number (dimensionless, no units).
Comparison Table of Electron Gain Enthalpy and Electronegativity is as follows :
| Aspect | Electron Gain Enthalpy ($\Delta_{eg}H$) | Electronegativity (EN) |
|---|---|---|
| Definition | Tendency of an isolated gaseous atom to accept an additional electron. | Tendency of an atom in a covalent bond to attract the shared pair of electrons. |
| Nature | Measured as an energy change when an electron is added. | A relative scale (dimensionless number). |
| Units | Expressed in kJ/mol | No units (relative number). |
| Condition | Applicable to isolated atoms in gaseous state. | Applicable to bonded atoms in a molecule. |
| Positive/Negative | Can be negative (energy released) or positive (energy absorbed). | Always a relative value; no positive or negative sign. |
| Example | For oxygen: $O(g) + e^- \rightarrow O^-(g); \, \Delta_{eg}H = -141 \, \text{kJ mol}^{-1}$ | On Pauling scale: EN of F = 4.0, O = 3.5, etc. |
NCERT 3.23 : How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Answer :
The electronegativity of any given atom is not constant. Therefore, the statement that the electronegativity of nitrogen on Pauling scale is 3.0 in all nitrogen compounds is incorrect. This is due to the following reasons :
Effect of Hybridization
The electronegativity increases as the percentage of s-character of a hybrid orbital increases i.e. the order is sp3 < sp2 < sp. Thus, nitrogen in an sp hybrid orbital has a higher electronegativity than in sp2, and in turn higher than in sp3.
Effect of Oxidation State
The electronegativity of nitrogen also increases with increasing oxidation state. For example, In NO (oxidation state of N = +2), electronegativity is lower. In NO2 (oxidation state of N = +4), electronegativity is higher.
Hence, electronegativity of nitrogen is variable, depending on its state of hybridization and oxidation state, and is not fixed at 3.0 in all its compounds.
NCERT 3.24 : Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron.
Answer :
(a) When an Atom Gains an Electron (Formation of Anion)
When a neutral atom gains an electron to form an anion, its radius increases because the nuclear charge remains the same but the number of electrons increases. The same positive charge now attracts a greater number of electrons, so the effective nuclear charge per electron decreases. As a result, the force of attraction of the nucleus on each electron decreases, and the electron cloud expands. Hence, the ionic radius of the anion is larger than that of the parent atom.
Example:
$$ \text{Cl} + e^- \;\;\longrightarrow \;\; \text{Cl}^- $$
Chlorine atom : 17 protons, 17 electrons, radius = 99 pm
Chloride ion : 17 protons, 18 electrons, radius = 181 pm
(b) When an Atom Loses an Electron (Formation of Cation)
When a neutral atom loses an electron to form a cation, its radius decreases. because the nuclear charge remains the same but the number of electrons decreases. The same positive charge is now shared by fewer electrons, so the effective nuclear charge per electron increases. As a result, the nucleus pulls the remaining electrons closer, and the size of the cation decreases.
Example :
$$ \text{Na} \;\;\longrightarrow \;\; \text{Na}^+ + e^- $$
- Sodium atom : 11 protons, 11 electrons, radius = 156 pm
- Sodium ion : 11 protons, 10 electrons, radius = 95 pm
NCERT 3.25 : Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Answer :
The first ionization enthalpies of two isotopes of the same element are expected to be the same. Ionization enthalpy depends mainly on :
- Nuclear charge (Z)
- Electronic configuration
- Atomic size (determined by number of protons and electron shells)
Since isotopes of the same element have the same atomic number (Z) and hence the same electronic configuration, their ionization enthalpies remain almost identical. The only difference between isotopes lies in their number of neutrons and hence their atomic mass, which has negligible effect on ionization enthalpy. Therefore, the first ionization enthalpies of two isotopes of the same element are practically the same.
NCERT 3.26 : What are the major differences between metals and non-metals?
Answer :
Elements which have a strong tendency to lose electrons to form cations are called metals, while those which have a strong tendency to accept electrons to form anions are called non-metals.
Generally metals are strong reducing agents, have low ionization enthalpies, less negative electron gain enthalpies and low electronegativity. They form basic oxides and generally form ionic compounds.
Non-metals are strong oxidizing agents, have high ionization enthalpies, highly negative electron gain enthalpies and high electronegativity. They form form acidic oxides and generally form covalent compounds
Comparison of Metals and Non-Metals
| Property | Metals | Non-Metals |
|---|---|---|
| Tendency | Strong tendency to lose electrons (form cations) | Strong tendency to gain electrons (form anions) |
| Redox nature | Strong reducing agents | Strong oxidizing agents |
| Ionization enthalpy | Low | High |
| Electron gain enthalpy | Less negative | Highly negative |
| Electronegativity | Low | High |
| Oxides | Form basic oxides | Form acidic oxides |
| Compounds | Generally form ionic compounds | Generally form covalent compounds |
NCERT 3.27 : Use the periodic table to answer the following questions
(a) Identify an element with five electrons in the outer subshell
(b) Identify an element that would tend to lose two electrons
(c) Identify an element that would tend to gain two electrons
(d) Identify the group having metal, non-metal, liquid as well as gas at room temperature
Answer :
(a) Identify an element with five electrons in the outer subshell
The general electronic configuration of the elements having five electrons in the outer subshell is ns2np5. This electronic configuration is characteristic of elements of Group 17, i.e., halogens (F, Cl, Br, I, At)
(b) Identify an element that would tend to lose two electrons
The elements which have a tendency to lose two electrons must have two electrons in the valence shell. Therefore, their general configuration should be ns2. This electronic configuration is characteristic of Group 2 elements, i.e., alkaline earth metals (Mg, Ca, Sr, Ba).
(c) Identify an element that would tend to gain two electrons
The elements which have a tendency to accept two electrons must have six electrons in the valence shell. Therefore, their general electronic configuration is ns2np4. This electronic configuration is characteristic of Group 16 elements (O, S, Se, Te)
(d) Identify the group having metal, non-metal, liquid as well as gas at room temperature
A metal which is liquid at room temperature is Mercury (Hg). It is a transition metal and belongs to Group 12.
A non–metal which is a gas at room temperature: Nitrogen (Group 15), Oxygen (Group 16), Fluorine, Chlorine (Group 17), Inert gases (Group 18).
A non–metal which is a liquid at room temperature: Bromine (Group 17).
NCERT 3.28 : The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Answer
The elements of Group 1 have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons, in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group, therefore, the reactivity of Group 1 elements increases in the same order :
Li < Na < K < Rb < Cs
In contrast, the elements of Group 17 have seven electrons in their respective valence shells and thus have a strong tendency to accept one more electron. The tendency to accept electrons, in turn, depends upon their electrode potentials. Since the electrode potentials of Group 17 elements decrease in the order :
F(+2.87V) > Cl(+1.36V) > Br(+1.08V) > I(+0.53V)
Therefore, their reactivity also decreases in the same order :
F > Cl > Br > I
NCERT 3.29 : Write the general outer electronic configuration of s –, p –, d – and f – block elements
Answer
(i) s–Block elements :
$ns^{1-2}, \; n = 2 \text{ to } 7$
(ii) p–Block elements :
$ns^2 \, np^{1-6}, \; n = 2 \text{ to } 6$
(iii) d–Block elements :
$(n-1)d^{1-10} \, ns^{0-2}, \; n = 4 \text{ to } 7$
(iv) f–Block elements :
$(n-2)f^{0-14} \, (n-1)d^{0-1} \, ns^2, \; n = 6 \text{ to } 7$
NCERT 3.30 : Assign the position of the element having outer electronic configuration
(i) $ns^2 \, np^4 \quad \text{for } n = 3$
(ii) $(n-1)d^2 \, ns^2 \quad \text{for } n = 4$
(iii) $(n-2)f^7 \, (n-1)d^1 \, ns^2 \quad \text{for } n = 6$
Answer :
(i) ns2 np4 for n = 3$
When n = 3, it suggests that the element belongs to the third period. Since the last electron enters the p–orbital, the element is a p–block element and its valence shell contains 2 + 4 = 6 electrons. Group number of the element is 10 + number of valence electrons = 10 + 6 = 16. Its complete electronic configuration is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^4$. Hence, the element is Sulphur (S).
(ii) (n – 1)d2 ns2 for n = 4
Here n = 4 suggests that the element lies in the 4th period. Since the d–orbitals are incomplete, it is a d–block element. Group number of the element is number of d–electrons + number of s–electrons = 2 + 2 = 4. Thus, the element lies in Group 4 and 4th period. Its complete electronic configuration is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^2 \, 4s^2$. Hence, the element is Titanium (Ti).
(iii) $(n-2)f^7 \, (n-1)d^1 \, ns^2$ for $n = 6$
Here n = 6 means the element lies in the 6th period. Since the last electron goes to the f–orbital, the element is an f–block element. All f–block elements lie in Group 3. Its complete electronic configuration is $[Xe] \, 4f^7 \, 5d^1 \, 6s^2$. Atomic number of the element is 54 + 7 + 1 + 2 = 64 and hence the element is Gadolinium (Gd).
NCERT 3.31 : The first (ΔiH₁) and the second (ΔiH₂) ionization enthalpies (in kJ mol⁻¹) and the (ΔegH) electron gain enthalpy (in kJ mol⁻¹) of a few elements are given below:
| Element | ΔiH₁ (kJ mol⁻¹) | ΔiH₂ (kJ mol⁻¹) | ΔegH (kJ mol⁻¹) |
|---|---|---|---|
| I | 520 | 7300 | −60 |
| II | 419 | 3051 | −48 |
| III | 1681 | 3374 | −328 |
| IV | 1008 | 1846 | −295 |
| V | 2372 | 5251 | +48 |
| VI | 738 | 1451 | −40 |
Which of the above elements is likely to be :
(a) the least reactive element
(b) the most reactive metal
(c) the most reactive non–metal
(d) the least reactive non–metal
(e) the metal which can form a stable binary halide of the formula MX₂ (X = halogen)
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen) ?
Answer
(a) Least reactive element : V.
The element V has very high ionization enthalpies (ΔiH₁ = 2372, ΔiH₂ = 5251) and a positive electron gain enthalpy (ΔegH = +48). These elements shows noble gas and are chemically unreactive.
(b) Most reactive metal : II.
The element II which has the lowest first ionization enthalpy (ΔiH₁ = 419) and a low negative electron gain enthalpy (ΔegH = -48) is the most reactive metal. This element easiest to lose an electron and most metallic/reactive metal among the list.
(c) Most reactive non–metal : III.
The element III has negative electron gain enthalpy (ΔegH = -328) and have strongest tendency to gain an electron. Thus it is most reactive non-metal (halogen).
(d) Least reactive non–metal : IV.
The element IV has a high negative electron gain enthalpy (ΔegH) but not so high first ionization enthalpy (ΔiH1) and hence least reactive non-metal (like iodine)
(e) Metal that can form a stable binary halide MX₂ : VI.
The element VI has relatively low ΔiH₁ and moderate ΔiH₂ (738 and 1451). The element is in +2 oxidation state is energetically feasible and form stable MX₂ (alkaline-earth-like behaviour).
(f) Metal that can form a predominantly covalent halide MX : I.
The element I has low ΔiH₁ (520) but very large ΔiH₂ (7300) therefore, it must be an alkali metal has stable +1 state; small highly polarising cation leads to covalent character in MX (Li-like behaviour).
NCERT 3.32 : Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Answer :
(a) Lithium and oxygen
Lithium is an alkali metal (Group 1). It has only one electron in the valence shell, therefore, its valency is 1. Oxygen is a group 16 element with a valence of 2. Therefore, formula of the compound formed would be Li2O (Lithium oxide).
(b) Magnesium and nitrogen
Magnesium is an alkaline earth metal (Group 2) and hence has a valence of 2. Nitrogen is a group 15 element with a valence of 8 – 5 = 3. Thus, the formula of the compound formed would be Mg3N2 (Magnesium nitride).
(c) Aluminium and iodine
Aluminium is a group 13 element with a valence of 3 while iodine is a halogen (group 17) with a valence of 1. Therefore, the formula of the compound formed would be AlI3 (Aluminium iodide).
(d) Silicon and oxygen
Silicon is a group 14 element with a valence of 4 while oxygen is a group 16 element with a valence of 2. Hence the formula of the compound formed is SiO2 (Silicon dioxide).
(e) Phosphorus and fluorine
Phosphorus is a group 15 element with a valence of 3 or 5 while fluorine is a group 17 element with a valence of 1. Hence the formula of the compound formed would be PF3 or PF5.
(f) Element 71 and fluorine
Element with atomic number 71 is a lanthanoid called lutetium (Lu). Its common valence is 3. Fluorine is a group 17 (halogen) element with a valence of 1. Therefore, the formula of the compound formed would be LuF3 (Lutetium fluoride).
NCERT 3.33 : In the modern periodic table, the period indicates the value of :
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number
Answer :
In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of the principal quantum number. Thus, the correct option is (c) principal quantum number
NCERT 3.34 : Which of the following statements related to the modern periodic table is incorrect?
(a) The p–block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p–subshell.
(b) The d–block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d–subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
Answer
Statement (b) is incorrect while the other statements are correct. The d–block has 10 columns, because a maximum of 10 electrons can occupy all the orbitals in a d–subshell.
NCERT 3.35 : Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons
Answer
Nuclear mass does not affect the valence shell because the nucleus consists of protons and neutrons. Where protons (i.e., nuclear charge) affect the valence shell but neutrons do not influence valence electrons. Thus, option (c) is wrong.
NCERT 3.36 : The size of isoelectronic species F⁻, Ne and Na⁺ is affected by
(a) nuclear charge (Z)
(b) valence principal quantum number (n)
(c) electron–electron interaction in the outer orbitals
(d) none of the factors because their size is the same.
Answer :
The size of isoelectronic ions depends upon the nuclear charge (Z). As the nuclear charge increases, the size decreases. For example : F⁻ (Z = 9) > Ne (Z = 10) and Na⁺(Z = 11). Therefore, statement (a) is correct, while all other statements are wrong.
NCERT 3.37 : Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Answer :
Statement (d) is incorrect. The correct statement is the removal of electron from orbitals bearing lower n value is more difficult than from orbitals having higher n value. All other statements are correct.
NCERT 3.38 : Considering the elements B, Al, Mg and K, the correct order of their metallic character is :
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer
In a period, metallic character increases as we move from right to left. Therefore, metallic character of K, Mg and Al decreases in the order :
K > Mg > Al
Within a group, metallic character increases from top to bottom. Thus, Al is more metallic than B. Hence, the correct sequence of decreasing metallic character is :
K > Mg > Al > B
Therefore, the correct option is (d).
NCERT 3.39 : Considering the elements B, C, N, F and Si, the correct order of their non–metallic character is
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Answer :
In a period, non–metallic character decreases from right to left. Thus, among B, C, N and F, the order is :
F > N > C > B
Within a group, non–metallic character decreases from top to bottom. Therefore, C is more non–metallic than Si. Hence, the correct sequence of decreasing non–metallic character is :
F > N > C > B > Si
Thus, the correct option is (c).
NCERT 3.40 : Considering the elements F, Cl, O and N the correct order of their chemical reactivity in terms of oxidizing property is :
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Answer :
Within a period, the oxidizing character increases from left to right. Therefore, among F, O and N, the oxidizing power decreases in the order :
F > O > N
Within a group, the oxidizing power decreases from top to bottom. Thus, F is a stronger oxidizing agent than Cl. Since O is more electronegative than Cl, oxygen is also a stronger oxidizing agent than chlorine. Hence, the overall decreasing order of oxidizing power is :
F > O > Cl > N
Thus, the correct option is (b).