NCERT Class 11 Chemistry Chapter 3 Classification of Elements and periodicity in properties Solutions to each chapter is provided in the list so that you can easily browse through different chapters NCERT Class 11 Chemistry Chapter 3 Classification of Elements and periodicity in properties and select need one. NCERT Class 11 Chemistry Chapter 3 Classification of Elements and periodicity in properties Question Answers Download PDF. NCERT Chemistry Class 11 Solutions.
NCERT Class 11 Chemistry Chapter 3 Classification of Elements and periodicity in properties
Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. NCERT Class 11 Chemistry Chapter 3 Classification of Elements and periodicity in properties Solutions are part of All Subject Solutions. Here we have given NCERT Class 11 Chemistry Part: I, Part: II Notes. NCERT Class 11 Chemistry Chapter 3 Classification of Elements and periodicity in properties Notes, NCERT Class 11 Chemistry Textbook Solutions for All Chapters, You can practice these here.
Classification of Elements and periodicity in properties
Chapter: 3
| Part – I |
1. What is the basic theme of organisation in the periodic table?
Ans: The basic theme of organisation in the periodic table is based on atomic number, which arranges elements in order of increasing protons. Elements are grouped into periods (rows) and families (columns) based on similar properties and electron configurations, reflecting periodic trends in elemental behaviour. In fact, the valence shell electronic configuration of the elements placed in a group gets repeated after definite gaps of atomic numbers (8, 18, 18, 32).
2. Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Ans: The important property used by Mendeleev classified elements primarily by atomic mass and their chemical properties. He arranged elements in order of increasing atomic mass, placing those with similar properties into the same groups. Later, the periodic table was corrected by arranging elements according to atomic number instead of atomic mass.
3. What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
Ans: Mendeleev’s Periodic Law: Arranged elements based on increasing atomic weights.
Modern Periodic Law: Arranged elements based on increasing atomic numbers.
4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Ans: The sixth period has n=6 and includes the 6s, 4d, and 5f orbitals. Each can hold:
6s = 2 electrons
4d = 10 electrons
5f = 14 electrons
Totaling 2 + 10 + 14 = 32 electrons,
Therefore, based on the quantum numbers and the capacity of these orbitals, the sixth period can accommodate a maximum of 32 elements.
5. In terms of period and group where would you locate the element with Z = 114?
Ans: The element with atomic number 114, known as Flerovium (Fl). The gaps of atomic numbers in a group are 8, 8, 18, 18, 32. The element with atomic number Z = 114 in the same group must have atomic number equal to 114 – 32 = 82. This represent lead (pb) which is present in 6th period and group 14 of the p-block. This means that the group 14 of p-block and it must be a member of 7th period.and it is positioned in the same group as carbon (C) and silicon (Si), reflecting its similar chemical properties.
6. Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Ans: The element present in the third period and seventeenth group of the periodic table is Chlorine (Cl). Its atomic number is 17. Chlorine is a halogen and is located in the p-block of the periodic table, known for its high reactivity and presence in various compounds.
7. Which element do you think would have been named by:
(i) Lawrence Berkeley Laboratory.
Ans: Lawrencium (Lr), which has an atomic number of Z = 103
Berkelium (Bk), which has an atomic number of Z = 97
(ii) Seaborg’s group?
Ans: Seaborgium (Sg), Z = 106
8. Why do elements in the same group have similar physical and chemical properties?
Ans: Elements in the same group have similar physical and chemical properties due to their identical valence electron configurations, which leads to comparable bonding and reactivity patterns. Therefore, elements present in the same group have similar physical and chemical properties.
9. What does atomic radius and ionic radius really mean to you?
Ans: Atom: The atomic radius represents the distance from the nucleus of an atom to the outermost electron shell. It provides insight into the size of an atom, influencing properties like bond length, atomic interactions, and reactivity. Understanding atomic radius helps in predicting how atoms will behave in different chemical contexts. It can be measured either by X-ray 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 chlorine molecule is 198 pm. Therefore, the covalent radius of chlorine atom is 198/299 pm (0.99 Å). In case of metals, atomic radius is called metallic radius. It is defined as one-half the distance between the two adjacent atoms in the crystal lattice.
Ionic radius: The ionic radius refers to the size of an ion in a crystal lattice or molecule. An ion can be 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. As a result, force of attraction increases and hence the ionic size decreases. On the other hand, the size of the anion is larger than the parent atom because the addition of one or more electrons decreases the effective nuclear charge. As a result, the force of attraction decreases and hence the ionic size increases.
10. How do atomic radius vary in a period and in a group? How do you explain the Variation?
Ans: Atomic radius decreases across a period due to increasing nuclear charge and increases down a group due to added electron shells. Within a group, the atomic radius increases down the group. This is because a new energy shell is added at each succeeding element while the number of electrons in the valence shell remains to be 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 a group, on moving from top to bottom, the ionic radius increases as a new energy level is added at each succeeding element but the number of valence electrons remains same.
11. What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F–
(ii) Ar
(iii) Mg2+
(iv) Rb+
Ans: Ions of different elements which have the same number of electrons but different magnitude of the nuclear charge are called isoelectronic ions.
(i) F– has 10 (9+1) electrons. Therefore, the species nitride ion, N³– (7+3); oxide ion; 02- (8+2), neon, Ne (10 + 0); sodium ion, Na+ (11-1); magnesium ion, Mg2+ (12-2); aluminium ion Al3+ (13-3) etc. each one of which contains 10 electrons, are isoelectronic with it.
(ii) Ar has 18 electrons. Therefore, the species phosphide ion, P³– (15 + 3), sulphide ion; S2 (16 + 2); chloride ion, Cl– (17+1), potassium ion, K+ (19-1), calcium ion, Ca2+ (20-2), etc., each one of which contains 18 electrons, are isoelectronic with it.
(iii) Mg2+ has 10 (12-2) electrons, therefore, the species N³–, O2-, F–, Ne, Na+, Al3+, etc. each one of which contains 10 electrons, are isoelectronic with it.
(iv) Rb+ has 36 (37-1) electrons. Therefore, the species bromide ion, Br– (35 + 1), krypton Kr (36 + 0) and strontium Sr2+ (382) each one of which has 36 electrons, are isoelectronic with it.
12. Consider the following species: N3–, O2–, F– , Na+, Mg2+ and Al3+.
(a) What is common in them?
Ans: All the species listed are isoelectronic, meaning they all have the same number of electrons (10 electrons).
(b) Arrange them in the order of increasing ionic radii.
Ans: Increasing ionic radii order: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻. This order reflects the increasing ionic radii as the nuclear charge decreases, leading to less attraction between the nucleus and the electrons, thus increasing the size.
13. Explain why cation are smaller and anions larger in radii than their parent atoms?
Ans: Cations are smaller than their parent atoms because they lose electrons, reducing electron-electron repulsion and allowing the nucleus to pull the remaining electrons closer. Anions are larger than their parent atoms because they gain electrons, increasing electron-electron repulsion and causing the electron cloud to expand.
14. What is the significance of the terms ‘isolated gaseous atom’ and ‘ground state’ while defining the ionisation enthalpy and electron gain enthalpy? Hint: Requirements for comparison purposes.
Ans: Isolated gaseous atom: Ionization enthalpy is the minimum amount of energy required to remove the most loosely sound stat bound electron from an isolated gaseous atom so as to convert it into a gaseous reaction. The force with which an electron is attracted by the nucleus of an atom is appreciably affected by presence of other atoms within its molecule or in the neighbourhood. Ensures that the atom is free from intermolecular interactions and is in its pure state, enabling accurate measurement of energy changes.
Ground state: Guarantees the atom is in its lowest energy state, providing a consistent starting point. These conditions allow for accurate and comparable ionisation and electron gain enthalpy values Indicates that the atom is in its lowest energy state, ensuring consistency and comparability of ionisation and electron gain enthalpy values across different elements.
15. Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10–18J. Calculate the ionisation enthalpy of atomic hydrogen in terms of J mol–1. Hint: Apply the idea of mole concept to derive the answer.
Ans: Ionisation energy is the amount of energy required to remove the electron from the ground state to infinity.
The given energy of an electron in the ground state of a hydrogen atom is = -2.18 × 10-15J.
Energy of the electron at affinity = 0.
The energy required to remove an electron in the ground state of hydrogen atom = 0 – (its energy in the ground state)
= (-2.18 × 10-18J) = 2.18 × 10-18J.
∴ Ionisation enthalpy per mole of hydrogen atoms
= 1312.36 kj mol-1
= 1312.36 × 10-3 j mol-1
16. Among the second period elements the actual ionisation enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why:
(i) Be has higher ∆i H than B.
Ans: The ionisation enthalpy, among other things, depends upon the type of electron to be removed from the same principal shell. In case of Be (1s2 2s2) the outermost electron is present in 2s- orbital while in B (1s22s22p1) it is present in 2p- orbital. Since 2s-electrons are more strongly attracted by nucleus than 2p- electrons, therefore, lesser amount of energy is required to knock out a 2p- electron than a 2s-electron. Consequently, ΔiH of Be is higher than that Δi H of B.
(ii) O has lower ∆i H than N and F?
Ans: The electronic configuration of oxygen is 1s22s22p4. The 2p orbital contains 4 electrons out of which 2 are present in the same 2p-orbital. Due to this, the electron repulsion increases. N has stable half filled configuration. F has greater nuclear charge. Hence, the ionisation enthalpy of O is lower than that of N and F.
17. How would you explain the fact that the first ionisation enthalpy of sodium is lower than that of magnesium but its second ionisation enthalpy is higher than that of magnesium?
Ans: The first ionisation enthalpy of sodium (Na) is lower than that of magnesium (Mg) because sodium has a single electron in its outermost shell (3s¹), which is easier to remove compared to magnesium’s two outer electrons (3s²). Sodium’s outer electron is shielded by inner electrons and experiences less effective nuclear charge.
The electronic configurations of Na and Mg are:
Na: 1s2 2s2 2p6 3s¹
Mg: 1s2 2s2 2p6 3s2.
Thus, the first electron in both the cases has to be removed from the 3s-orbitals but the nuclear charge of Na (+11) is lower than that of Mg (+12), therefore, the first ionisation energy of sodium is lower than that of magnesium. After the loss of first electron, the electronic configuration of Nat is 1s22s22p6. Hence the electron is to be removal from inert (neon) gas configuration which is very stable and hence removal of second electron from sodium is very difficult. However, in case of magnesium, after the loss of first electron, the electronic configuration of Mg is 1s2 2s2 2p6 3s¹. Here, the electron is to be removed from a 3s orbital which is much faster than to remove an electron from inert gas configuration. Therefore, the second ionisation enthalpy of sodium is higher than that magnesium.
18. What are the various factors due to which the ionisation enthalpy of the main group elements tends to decrease down a group?
Ans: (i) Increase in the atomic size of elements: As you move down a group, the number of electron shells increases, causing the atomic radius to grow. The increased distance between the nucleus and the valence electrons reduces the electrostatic attraction between them, making it easier to remove an electron. Thus, they can be removed easily. Hence, on moving down a group, ionisation energy decreases.
(ii) Screening effect: With the addition of new shells, the number of inner electron shells which shield the valence electrons increases. In other words, the shielding effect or the screening effect increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionisation enthalpy decreases.
19. The first ionisation enthalpy values (in kJ mol–1) of group 13 elements are:
| B | AI | Ga | In | Ti |
| 801 | 577 | 579 | 558 | 589 |
How would you explain this deviation from the general trend?
Ans: On moving down a group, ionisation enthalpy generally decreases due to an increase in the atomic size and shielding. Thus, on moving down group 13, ionisation enthalpy decreases from B to Al. But, Ga has higher ionisation enthalpy than Al. The ionisation enthalpy of Ga is unexpectedly higher than that of Al. Ga contains 10d electrons in inner shell which are less penetrating. Their shielding is less effective than that of s and p electrons. The outer electron is held fairly strongly by the nucleus. The ionisation enthalpy increases slightly.
A similar increase is observed from In to Tl due to presence of 14f electrons in the inner shell of Tl which have poor shielding effect.The shielding provided by the electrons in both these orbitals is not very effective. Therefore, the valence electron is held quite strongly by the nucleus. Hence, the ionisation energy of Tl is on the higher side.
20. Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F.
Ans: O has more negative electron gain enthalpy than N. This is because nitrogen has stable configuration with half filled 2p orbitals. So, it will resist gain of electrons whereas oxygen has outer electronic configuration of 2p4. So, it will try to attain stability by gaining 2 electrons. Hence, the electron gain enthalpy of F is more negative than that of O.
(ii) F or Cl.
Ans: F and Cl belong to the same group of the periodic table. The electron gain enthalpy usually becomes less negative on moving down a group. However, in this case, the value of the electron gain enthalpy of Cl is more negative than that of F. This is because the atomic size of F is smaller than that of Cl. In F, the electron will be added to quantum level n = 2, but in Cl, the electron is added to quantum level n = 3.Therefore,Chlorine can accommodate an extra electron more easily, leading to a more negative electron gain enthalpy compared to Fluorine. Hence, the electron gain enthalpy of Cl is more negative than that of F.
21. Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Ans: The second electron gain enthalpy of O is positive as explained below:
When an electron is added to O atom to form O – ion, energy is released. Thus, first electron gain enthalpy of O is negative.
But when another electron is added to O – to form O 2– ion, energy is absorbed to overcome the strong electrostatic repulsion between the negatively charged O – ion and the second electron being added.
Thus, the second electron gain enthalpy of oxygen is positive.
22. What is the basic difference between the terms electron gain enthalpy and electronegativity?
Ans: Electron gain enthalpy is the tendency of an isolated gaseous atom to accept an additional electron to form a negative ion. Both electron gain enthalpy and electronegativity refer to the tendency of the atom of an element to attract electrons. Whereas electron gain enthalpy refers to the tendency of an isolated gaseous atom to accept an additional electron to form a negative ion, electronegativity refers to the tendency of the atom of an element to attract the shared pair of electrons towards it in a covalent bond.
23. How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Ans: The statement that the electronegativity of nitrogen is 3.0 in all its compounds on the Pauling scale is incorrect. Actually electronegativity varies with the state of hybridization and the oxidation state of the element. The electronegativity increases as the percentage of s- character of a hybrid orbital increases or the oxidation state of the element increases. For example, the electronegativity of N increases as we move from sp³-sp hybrid orbital. Similarly the electronegativity of N in NO2 where oxidation state of N is + 4 is higher than that in NO where the oxidation state of N is + 2. Therefore, assigning a fixed value to nitrogen’s electronegativity in all compounds is inaccurate.
24. Describe the theory associated with the radius of an atom as it
(a) gains an electron.
Ans: Gaining an Electron: When an atom gains an electron, it becomes an anion. The added electron increases electron-electron repulsion and expands the electron cloud, increasing the atomic radius. The added electron increases electron-electron repulsion and expands the electron cloud, increasing the atomic radius. Since the same nuclear charge now attracts greater number of electrons, therefore, force of attraction of the nucleus on the electrons of all the shells decreases (i.e., effective nuclear charge decreases) and hence the electron cloud expands. In other words, the distance between the centre of the nucleus and the last shell that contains electrons increases thereby increasing the ionic radius. Thus,
(b) loses an electron.
Ans: Losing an Electron: When an atom loses an electron, it forms a cation. The removal of an electron decreases electron-electron repulsion and increases the effective nuclear charge per electron, causing the electron cloud to contract. This leads to a smaller atomic radius, as the remaining electrons are pulled closer to the nucleus.
25. Would you expect the first ionisation enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Ans: Ionization enthalpy depends on factors such as electronic configuration (number of electrons) and nuclear charge (number of protons). Since the isotopes of an element have the same electronic configuration and same nuclear charge, they are expected to have same ionisation enthalpy.
26. What are the major differences between metals and nonmetals?
Ans: 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. Thus, metals are strong reducing agents, they have low ionisation enthalpies, have less negative electron gain enthalpies, low electronegativity, form basis oxides and ionic compounds. Metals are typically shiny, conductive, malleable, and ductile with high melting points.
Non-metals are strong oxidising agents with high ionisation enthalpies, high negative electron gain enthalpies and high electronegativities. They form acidic oxides and covalent compounds.
27. Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
Ans: The general electronic of the elements having five electrons in the outer subshell is ns2 nps. This electronic configuration characteristic of elements of group 17, i. e., halogen and their examples are F, Cl, Br, I, etc.
(b) Identify an element that would tend to lose two electrons.
Ans: The elements which have a tendency to lose two electrons must have two electrons in the valence shell. Therefore, their general electronic configuration should be ns2. This electronic configuration is characteristic of group 2 elements, i.e., alkaline earth metals and their examples are Mg Ca, Sr, Ba, etc.
(c) Identify an element that would tend to gain two electrons.
Ans: The elements which have a tendency to accept two electrons must have four electrons in the valence shell. Therefore, their general electronic configuration is ns2 np4. This electronic configuration is characteristic of group 16 elements and their examples are O and S.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
Ans: A metal which is liquid at room temperature is mercury. It is a transition metal and belongs to group 12.
A non-metal which is a gas at room temperature is nitrogen (group 15), oxygen (group 16), fluorine, chlorine (group 17) and inert gases (group 18).
A non-metal which is a liquid at room temperature is bromine (group 17).
28. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb Cs whereas that of group 17 is F > CI > Br > L. Explain.
Ans: The elements of group I 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 ionisation enthalpy. Since the ionisation 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 +1.36 V), Br (1.08V) and I (+0.53 V), therefore, their reactivities also decrease in the same order: F Cl > Br > 1
Alternatively, tendency to accept electrons can be linked to electron gain enthalpy. Since electron gain enthalpy becomes less and less negative as we move from Cl to I, therefore, reactivity increases from Cl to I, F, however, is the most reactive due to its low bond dissociation energy.
29. Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Ans: (i) s-Block elements: ns 1- 2
where n = 2 – 7
(ii) p-block elements: ns2 np1-6 np1- 6
where n = 2 – 6
(iii) d-block elements: (n – 1) d¹ – 10 ns0-2
where n = 4 – 7
(iv) f-block elements: (n – 2) f 0 – 4 (n – 1) d 0- 2 ns 2
where n = 6 – 7
30. Assign the position of the element having outer electronic configuration
(i) ns2np4 for n = 3
Ans: n = 3 suggests that the element belongs to third period. Since the last electron enters the p- orbital, therefore, the given element is a p- block element. Further since the valence shell contains 6 (2 + 4) electrons, therefore, group number of the element = 10 + no. of electrons in the valence shell 10 + 6 = 16.
The complete electronic configuration of the element is 1s 2 2s2 2p6 3s2 3p4 and the element is S (sulphur)
(ii) (n-1)d2ns2 for n=4, and
Ans: n = 4 suggests that the element lies in the 4th period.
Since the d-orbitals are incomplete, therefore, it is d-block element. The group number of the element no. of d-electrons + no. of s-electrons = 2 + 2 = 4. Thus, the elements lies in group 4 and 4th period. The complete electronic configuration of the
Elements is ls2 2s2 2p5 3p6 3d2 4s2 and the elements in Ti .
(iii) (n-2) f 7 (n-1)d1ns2 for n=6, in the periodic table.
Ans: Since n = 6, the element is present in the 6th period. It is an f –block element as the last electron occupies the f–orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe] 4f7 5d1 6s2.It means (6 – 2) f7 (6 -1)d1 6s2 or 4f75d16s2Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium.
31. The first (∆i H1) and the second (∆i H2) ionisation enthalpies (in kJ mol–1) and the (∆egH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
| Elements | ∆iH1 | ∆iH2 | ∆eqH |
| I | 520 | 7300 | –60 |
| II | 419 | 3051 | –48 |
| III | 1681 | 3374 | –328 |
| IV | 1008 | 1846 – | –295 |
| V | 2372 5 | 5251 | +48 |
| VI | 738 | 1451 | –40 |
Which of the above elements is likely to be:
(a) the least reactive element.
Ans: The element V has highest first ionisation enthalpy (ΔiH1) and positive electron gain enthalpy (Δeg H) and hence it is the least reactive element. Since inert gases have positive Delta (Δeg H, therefore, the element-V must be an inert gas. The values of Δi, H₁, Δi,H2 and Delta Δeg H match that of He.
(b) the most reactive metal.
Ans: The element II which has the least first ionisation enthalpy (ΔiH1) and a low negative electron gain enthalpy (Δeg H) is the most reactive metal. The values of Δi, H1, ΔiH2 and Delta Δeq H match that of K (potassium).
(c) the most reactive non-metal.
Ans: The element III which has high first ionisation enthalpy (Δi H₁) and a very high negative electron gain enthalpy (Δeg H) is the most reactive eg non-metal. The values of Δi H₁, Δi H₂ and Δeq H match that of F (fluorine).
(d) the least reactive non-metal.
Ans: The element IV has a high negative electron gain enthalpy (Δi, H) but not so high first ionisation enthalpy (Δ, H₁). Therefore, it is the least reactive non-metal. The values of Δi, H₁, Δi, H₂ and Δeg H match that of I (lodine).
(e) the metal which can form a stable binary halide of the formula MX2(X = halogen).
Ans: The element VI has low first ionisation a high negative electron gain enthalpy (A, H₁) but higher than that of alkali metals. Therefore, it appears that the element is an alkaline each metal and hence will form binary halide of the formula MX, (where X = halogen). The values of Δi, H₁, Δ, H₂ and Δeg H match that of Mg (magnesium). eq
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?
Ans: The element I has low first ionisation (Δi, H₁) but a very high second ionisation enthalpy (Δi, H₂), therefore, it must be an alkali metal. Since the metal forms a predominantly stable covalent halide of the formula MX (X = halogen), therefore, the alkali metal must be least reactive. The values of Δi, Η₁. Δ, Η₂ and Δeg H match that of Li (lithium).
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.
Ans: Lithium is an alkali metal (Group 1). It has only one electron in the valence shell, therefore, its valence 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.
Ans: 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.
Ans: Aluminium is 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 All3 (aluminium iodide).
(d) Silicon and oxygen.
Ans: 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.
Ans: 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.
Ans: Element with atomic number 71 is a anthanoide called lutetium (Lu). Its common valence 53. Fluorine is a group 17 (halogen) element with a valence of 1. Therefore, the formula of the compound formed would be LuF3 (lutetium fluoride).
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.
Ans: In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of principal quantum number. Thus, option. (c) principal quantum number.
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-shell.
(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.
Ans: Statement (b) is incorrect while other statement are correct. The correct statement (b) is: The d-block has 10 columns, because a maximum of 10 electrons can occupy all the orbitals in a d-subshell. Statements (a), (c), and (d) are correct.
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.
Ans: The nuclear mass does not affect the valence shell or the chemistry of the element. because the nucleus consists of protons and neutrons. Whereas protons, i.e., nuclear charge affects the valence shell but neutrons do not. Thus, option (c) is wrong.
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.
Ans: The size of the isoelectronic ions depends upon the nuclear charge (Z). As the nuclear charge increases, the effective nuclear attraction on the electrons. For example, F (+9) – Ne (+10) > Na+ (+11). Therefore, statement (a) correct while all other statements are wrong.
37. Which one of the following statements is incorrect in relation to ionisation enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionisation enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionisation enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Ans: Statement (d) is incorrect. The correct statement is Removal of electron from orbitals bearing lower n value is difficult than from orbital having higher n value. All other statements are correct.
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
Ans: In general, metallic character increases down a group and decreases across a period from left to right. Therefore, metallic character of K, Mg and Al decreases in the order: K > Mg > Al However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B. Therefore, the correct sequence of decreasing metallic character is: K > Mg > Al > B i.e., option (d) is correct.
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
Ans: Non-metallic character generally increases across a period from left to right and decreases down a group. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C > B However, within a group, non- metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F > N > C > B > Si i.e., option (c) is correct.
40. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is:
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Ans: Fluorine (F) is the most reactive as an oxidising agent due to its high electronegativity and ability to attract electrons. Within a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is: F > 0 > Cl > N i.e. option (b) is correct.
