Exercises
1. What is the basic theme of organisation in the periodic table?
Answer
The basic theme of organisation in the periodic table is the periodic recurrence of similar physical and chemical properties of elements when arranged in order of increasing atomic numbers. Elements with similar outer electronic configurations are placed in the same vertical column (group). The physical and chemical properties of elements are periodic functions of their atomic numbers (Modern Periodic Law).
2. Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Answer
Mendeleev used atomic mass as the basis for classifying elements. He arranged elements in the increasing order of their atomic weights. However, he did not strictly stick to this property. When elements with similar properties did not fall in the same group by atomic mass order, he ignored the atomic mass sequence. For example, iodine (atomic mass 127) was placed after tellurium (atomic mass 128) because iodine's properties resembled those of the halogens (F, Cl, Br) rather than Group VI elements. He also left gaps for undiscovered elements and predicted their properties.
3. What is the basic difference in approach between Mendeleev's Periodic Law and the Modern Periodic Law?
Answer
• Mendeleev's Periodic Law: "The properties of the elements are a periodic function of their atomic weights (masses)." Atomic mass was used as the basis.
• Modern Periodic Law: "The physical and chemical properties of the elements are periodic functions of their atomic numbers." Atomic number (number of protons) is the basis.
The fundamental difference is: Mendeleev used atomic mass while the Modern Periodic Law uses atomic number, which was shown to be a more fundamental property by Henry Moseley in 1913 through X-ray spectroscopy experiments.
4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer
The sixth period corresponds to the principal quantum number n = 6. The subshells filled in this period are:
• 6s subshell: l = 0, can hold 2 electrons
⇒ 2 elements
• 4f subshell: l = 3, can hold 14 electrons
⇒ 14 elements
• 5d subshell: l = 2, can hold 10 electrons
⇒ 10 elements
• 6p subshell: l = 1, can hold 6 electrons
⇒ 6 elements
Total = 2 + 14 + 10 + 6 = 32 elements
Hence, the sixth period contains 32 elements.
5. In terms of period and group where would you locate the element with Z = 114?
Answer
Z = 114: The electronic configuration of element 114 (Flerovium, Fl):
[Rn] 5f¹⁴ 6d¹⁰ 7s² 7p²
• The last electron enters the 7p subshell, so it belongs to Period 7.
• It has 2 electrons in the 7p subshell (configuration p²), placing it in Group 14.
Therefore, Z = 114 is located in Period 7, Group 14.
6. Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer
Group 17 elements are halogens. In the third period, the element in Group 17 is Chlorine (Cl) with atomic number Z = 17.
Its electronic configuration: 1s²2s²2p⁶3s²3p⁵ — 3rd period (n=3), 5 electrons in p-subshell
⇒ Group 17.
Q3.7. Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg's group?
Ans.
(i) Lawrence Berkeley Laboratory — Element 106 was named Seaborgium (Sg) by this laboratory in honour of Glenn T. Seaborg. Also, elements like Berkelium (Bk, Z=97) and Californium (Cf, Z=98) were named after Berkeley, California.
(ii) Seaborg's group — Glenn T. Seaborg's group discovered several transuranium elements (Z = 94 to 102). Among these, Americium (Am), Curium (Cm), Berkelium (Bk), Californium (Cf), Einsteinium (Es), Fermium (Fm), Mendelevium (Md), and Nobelium (No) were identified by his group. Seaborgium (Sg, Z=106) was named in his honour.
8. Why do elements in the same group have similar physical and chemical properties?
Answer
Elements in the same group have the same number of valence electrons and the same outer electronic configuration (same nsxnpy pattern, differing only in the value of n). Since chemical properties depend primarily on the number and arrangement of valence electrons, elements in the same group exhibit similar chemical behaviour. Their similar valence shell configurations lead to similar bonding patterns, oxidation states, and reactivity. Physical properties also show similarity because they are influenced by electronic configuration, atomic size, and metallic/non-metallic character.
9. What does atomic radius and ionic radius really mean to you?
Answer
Atomic radius: It is a measure of the size of an atom — the distance from the nucleus to the outermost shell of electrons. Since atoms do not have a sharp boundary (electron cloud), atomic radius is defined as half the distance between the centres of two identical atoms bonded together (covalent radius) or touching each other in a solid (van der Waals radius). For metals, it is the metallic radius (half the internuclear distance in a metallic crystal).
Ionic radius: It is the effective distance from the nucleus to the outermost electron in an ion. When an atom loses electrons to form a cation, it becomes smaller (ionic radius < atomic radius). When it gains electrons to form an anion, it becomes larger (ionic radius > atomic radius). Ionic radii are important in determining crystal structure and properties of ionic compounds.
10. How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer
Variation across a period: Atomic radius decreases from left to right across a period. As we move across a period, the number of protons (atomic number) increases while electrons are added to the same shell. The increased nuclear charge pulls the electron cloud closer to the nucleus, reducing atomic radius. E.g., Li (152 pm) > Be (111 pm) > B (88 pm) > C (77 pm) > N (74 pm) > O (73 pm) > F (72 pm).
Variation down a group: Atomic radius increases going down a group. As we descend a group, each successive element has one more principal shell. The addition of new electron shells increases the distance of the outermost electrons from the nucleus, despite the increase in nuclear charge. The shielding effect of inner electrons also reduces the effective nuclear charge on outer electrons. E.g., Li (152 pm) < Na (186 pm) < K (231 pm) < Rb (244 pm) < Cs (262 pm).
11. What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions:
Answer
(i) F⁻ (ii) Ar (iii) Mg²⁺ (iv) Rb⁺
Ans. Isoelectronic species are atoms/ions that have the same number of electrons but different nuclear charges (different elements).
(i) F⁻ has 10 electrons (Z=9, gains 1 electron). Isoelectronic species: Ne, Na⁺, Mg²⁺, Al³⁺, O²⁻
(ii) Ar has 18 electrons (Z=18). Isoelectronic species: Cl⁻, K⁺, Ca²⁺, S²⁻
(iii) Mg²⁺ has 10 electrons (Z=12, loses 2 electrons). Isoelectronic species: Ne, Na⁺, F⁻, Al³⁺, O²⁻
(iv) Rb⁺ has 36 electrons (Z=37, loses 1 electron). Isoelectronic species: Kr, Sr²⁺, Br⁻, Y³⁺
12. Consider the following species: N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Answer
(a) All these species are isoelectronic — they all have 10 electrons (same as Ne configuration: 1s²2s²2p⁶). Their atomic numbers differ: N(7), O(8), F(9), Na(11), Mg(12), Al(13).
(b) Among isoelectronic species, the one with higher nuclear charge has a smaller ionic radius (more protons attract the same number of electrons more strongly).
Increasing nuclear charge: N(7) < O(8) < F(9) < Na(11) < Mg(12) < Al(13)
Therefore, ionic radii increase in the reverse order:
Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
13. Explain why cations are smaller and anions larger in radii than their parent atoms?
Answer
Cations are smaller: When an atom loses one or more electrons to form a cation, the number of electrons decreases while the nuclear charge (number of protons) remains the same. The remaining electrons experience a greater effective nuclear charge (higher proton-to-electron ratio), pulling the electron cloud closer to the nucleus. Also, if the outermost shell is completely emptied, the cation has one fewer shell. Both factors reduce the radius. E.g., Na (186 pm)
⇒ Na⁺ (102 pm).
Anions are larger: When an atom gains one or more electrons to form an anion, the number of electrons increases while the nuclear charge remains constant. More electrons in the same shell increase electron-electron repulsion and reduce the effective nuclear charge felt by each electron, causing the electron cloud to expand. E.g., F (64 pm)
⇒ F⁻ (133 pm), Cl (99 pm)
⇒ Cl⁻ (181 pm).
14. What is the significance of the terms — 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?
Answer
'Isolated gaseous atom': The atom must be in the gaseous state to eliminate interatomic/intermolecular interactions. In solids or liquids, atoms interact with neighbours, which would affect the energy measurement. Isolation ensures that only the intrinsic property of the individual atom is measured, allowing valid comparisons between different elements.
'Ground state': The atom must be in its lowest energy electronic configuration (ground state) to ensure standardised, reproducible measurements. If the atom were in an excited state, it would require less energy to remove an electron (or gain one), giving a misleadingly low ionization enthalpy. Starting from the ground state provides a fixed, comparable reference point for all elements.
15. Energy of an electron in the ground state of the hydrogen atom is –2.18×10⁻¹⁸ J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol⁻¹.
Answer
Ionization enthalpy is the energy required to remove an electron from the ground state of a gaseous atom.
For hydrogen: Energy of electron = –2.18 × 10⁻¹⁸ J (negative sign indicates bound state)
Energy required to remove this electron = +2.18 × 10⁻¹⁸ J per atom
To find energy per mole, multiply by Avogadro's number (NA = 6.022 × 10²³ mol⁻¹):
Ionization enthalpy = 2.18 × 10⁻¹⁸ J × 6.022 × 10²³ mol⁻¹
= 2.18 × 6.022 × 10⁵ J mol⁻¹
= 13.128 × 10⁵ J mol⁻¹
= 1.312 × 10⁶ J mol⁻¹ = 1312 kJ mol⁻¹
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 ΔiH than B
(ii) O has lower ΔiH than N and F?
Answer
(i) Be (ΔiH = 900 kJ/mol) > B (ΔiH = 801 kJ/mol):
Be has electronic configuration 1s²2s². The 2s orbital is completely filled, giving extra stability to the fully filled subshell. B has configuration 1s²2s²2p¹. The electron to be removed from B is from the 2p orbital, which is higher in energy than the 2s orbital, so it requires less energy. Also, 2p electrons are better shielded by the 2s electrons. Therefore, Be has higher ionization enthalpy than B.
(ii) N (ΔiH = 1402 kJ/mol) > O (ΔiH = 1314 kJ/mol):
N has configuration 1s²2s²2p³ — a half-filled 2p subshell which is extra stable due to exchange energy. O has configuration 1s²2s²2p⁴. In oxygen, one 2p orbital has a pair of electrons; the electron-electron repulsion in the paired orbital makes it easier to remove one electron from O. Hence, O has lower ionization enthalpy than N. F > O because F has higher nuclear charge (Z=9 vs Z=8), making it harder to remove electrons.
17. How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer
Na: 1s²2s²2p⁶3s¹; Mg: 1s²2s²2p⁶3s²
First ionization enthalpy: Na (496 kJ/mol) < Mg (738 kJ/mol). Mg has higher nuclear charge (Z=12) than Na (Z=11), and both lose an electron from the 3s subshell. Greater nuclear charge in Mg means the 3s electron is held more tightly, so more energy is needed.
Second ionization enthalpy: Na²⁺ requires much more energy than Mg²⁺. After losing the first electron, Na⁺ has configuration 1s²2s²2p⁶ (noble gas configuration — very stable). Removing the second electron from Na⁺ requires breaking into the stable 2p⁶ core, which demands very high energy. For Mg⁺ (configuration 1s²2s²2p⁶3s¹), the second electron is still from the 3s orbital, so much less energy is needed. Hence, ΔiH₂ of Na >> ΔiH₂ of Mg.
18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer
Ionization enthalpy decreases down a group due to:
1. Increase in atomic radius: As we go down, new shells are added, increasing distance between the nucleus and valence electrons. Less electrostatic attraction
⇒ easier to remove electrons.
2. Increased shielding effect: More inner electron shells shield the valence electrons from the nucleus, reducing effective nuclear charge felt by the outermost electrons.
3. Lower effective nuclear charge (Zeff): Though nuclear charge increases down a group, the increase is largely offset by the shielding effect, resulting in a lower effective nuclear charge on valence electrons.
4. Increased number of inner electrons: More electrons in inner shells increase the shielding, making it easier to remove the outermost electron.
19. The first ionization enthalpy values (in kJ mol⁻¹) of group 13 elements are: B(801), Al(577), Ga(579), In(558), Tl(589). How would you explain this deviation from the general trend?
Answer
Expected trend: ionization enthalpy should decrease down Group 13 (as B > Al > Ga > In > Tl). However, the actual trend shows some irregularities:
• Ga (579) > Al (577): Gallium follows the 3d transition series. The 10 d-electrons of Ga are poor shielders of nuclear charge compared to p-electrons. The effective nuclear charge on Ga's 4p electron is higher than expected, making it slightly harder to remove than Al's 3p electron.
• Tl (589) > In (558): Thallium follows the 4f (lanthanide) series. The 14 f-electrons of Tl are even poorer shielders than d-electrons. This "lanthanide contraction" causes Tl to have a higher effective nuclear charge, resulting in higher ionization enthalpy than In despite being below it in the group.
These irregularities are attributed to the poor shielding by d- and f-electrons (d- and f-block contraction).
20. Which of the following pairs of elements would have a more negative electron gain enthalpy?
Answer
(i) O or F (ii) F or Cl
Ans.
(i) F has more negative electron gain enthalpy than O. F (Z=9) has higher nuclear charge than O (Z=8) in the same period. Higher nuclear charge means F has greater tendency to attract an incoming electron, releasing more energy. F has ΔegH = –328 kJ/mol; O has ΔegH = –141 kJ/mol.
(ii) Cl has more negative electron gain enthalpy than F. Although F is above Cl and has higher electronegativity, the very small size of F makes electron-electron repulsion very high when an extra electron is added to the compact 2p orbital. Cl (3p orbital) is larger, so the incoming electron experiences less repulsion and is accommodated more easily. Cl has ΔegH = –349 kJ/mol; F has ΔegH = –328 kJ/mol. This is an exception to the general trend in Group 17.
21. Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Answer
The second electron gain enthalpy of O would be positive.
First electron gain: O + e⁻
⇒ O⁻ (ΔegH₁ = –141 kJ/mol, exothermic — energy released)
For the second electron gain: O⁻ + e⁻
⇒ O²⁻, we are trying to add a negatively charged electron to an already negatively charged O⁻ ion. Significant electrostatic repulsion exists between the incoming electron and the negative charge already present. This process requires energy input, making the second electron gain enthalpy positive (endothermic). The actual value is about +780 kJ/mol. O²⁻ exists in compounds only because the lattice energy of ionic compounds (like MgO) compensates for this unfavourable energy.
22. What is the basic difference between the terms electron gain enthalpy and electronegativity?
Answer
Electron gain enthalpy: It is the enthalpy change when a neutral gaseous atom gains an electron to form a gaseous anion: X(g) + e⁻
⇒ X⁻(g). It is a measurable, quantitative thermodynamic property, expressed in kJ/mol. It applies to isolated gaseous atoms.
Electronegativity: It is a relative measure of the tendency of a bonded atom in a molecule to attract the shared pair of electrons towards itself. It is not an experimentally measurable property of an isolated atom but a property of an atom in a chemical bond. It is a dimensionless, relative scale (Pauling scale, Mulliken scale, etc.). It depends on the chemical environment and hybridisation state.
Key difference: Electron gain enthalpy refers to an isolated gaseous atom gaining an electron; electronegativity refers to an atom within a molecule attracting bonding electrons.
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
This statement is not entirely accurate. Electronegativity is not an absolute fixed property — it varies with the chemical environment, particularly with the hybridisation state of the atom. Nitrogen's electronegativity changes depending on its bonding:
• sp hybridised N (e.g., in HCN) is more electronegative than sp² hybridised N (e.g., pyridine), which is more electronegative than sp³ hybridised N (e.g., NH₃).
• Increased s-character increases electronegativity.
The value 3.0 is an average/typical value for nitrogen. While it serves as a useful approximation, stating it is constant for all nitrogen compounds is an oversimplification. Electronegativity is an empirical concept that depends on oxidation state, hybridisation, and the nature of adjacent atoms.
Q3.24. Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
Ans.
(a) Gains an electron (forms anion): When an atom gains an electron, the number of electrons increases while the nuclear charge (protons) stays the same. The electron-electron repulsions increase as more electrons are crammed into the same electron cloud. The effective nuclear charge per electron decreases (same nuclear charge distributed over more electrons), so the electrons spread out further. The electron cloud expands, and the ionic (anionic) radius becomes larger than the parent atom's radius. E.g., Cl (99 pm) → Cl⁻ (181 pm).
(b) Loses an electron (forms cation): When an atom loses an electron, the number of electrons decreases while the nuclear charge remains constant. The effective nuclear charge per electron increases (same nuclear charge for fewer electrons). The remaining electrons are attracted more strongly toward the nucleus, contracting the electron cloud. The cation's radius is smaller than the parent atom's radius. If all valence electrons are removed, the cation is an entirely inner-shell species with a much smaller radius. E.g., Na (186 pm) → Na⁺ (102 pm).
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 for two isotopes of the same element would be essentially the same.
Isotopes of an element have the same number of protons (same atomic number Z) and the same number of electrons. They differ only in the number of neutrons in the nucleus (different mass numbers). Since ionization enthalpy depends on the effective nuclear charge (number of protons), the electron configuration, atomic size, and electron-electron repulsions — all of which are determined by the number of protons and electrons, not neutrons — isotopes have virtually identical ionization enthalpies. Any tiny difference would be negligible due to the slight variation in nuclear mass affecting the position of the nucleus (very small isotope effect).
26. What are the major differences between metals and non-metals?
Answer
Property | Metals | Non-metals |
Physical state | Solids at room temperature (except Hg) | Solids, liquids, or gases |
Lustre | Metallic lustre, shiny | No lustre (except iodine, graphite) |
Conductivity | Good conductors of heat and electricity | Poor conductors (except graphite) |
Malleability/Ductility | Malleable and ductile | Brittle (if solid) |
Ionization enthalpy | Low (easily lose electrons) | High (tend to gain electrons) |
Electronegativity | Low | High |
Oxides | Form basic oxides | Form acidic oxides |
Ion formation | Form cations by losing electrons | Form anions by gaining electrons |
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 the room temperature.
Answer
(a) Five electrons in the outer subshell means np⁵ configuration
⇒ Group 17 (Halogens). Examples: F, Cl, Br, I.
(b) An element that tends to lose two electrons has two valence electrons
⇒ Group 2 (Alkaline earth metals). Example: Magnesium (Mg, Z=12) — loses 2 electrons to form Mg²⁺.
(c) An element that tends to gain two electrons needs two more electrons to complete its octet
⇒ configuration ns²np⁴
⇒ Group 16 (Chalcogens). Example: Oxygen (O, Z=8) or Sulphur (S, Z=16).
(d) Group 17 (Halogens): contains Astatine (metal/metalloid), Iodine (non-metal solid), Bromine (liquid), and Fluorine and Chlorine (gases).
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
Group 1 (Alkali metals): Reactivity increases down the group (Li < Na < K < Rb < Cs) because as we go down, atomic radius increases, ionization enthalpy decreases (outer electron is farther from nucleus and more shielded), and the ease of losing the valence electron increases. Greater ease of electron loss = greater reactivity for metals.
Group 17 (Halogens): Reactivity decreases down the group (F > Cl > Br > I) because as we go down, atomic radius increases and electron gain enthalpy becomes less negative (decreasing tendency to gain electrons). Halogens react by gaining electrons; the smaller and more electronegative atom gains electrons more readily. F is the most electronegative element with the strongest oxidising power. The reactivity of halogens as oxidising agents decreases going down the group.
29. Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Answer
• s-block elements: ns1–2 (Groups 1 and 2; alkali and alkaline earth metals)
• p-block elements: ns²np1–6 (Groups 13–18; non-metals, metalloids, noble gases)
• d-block elements: (n-1)d1–10ns0–2 (Groups 3–12; transition metals)
• f-block elements: (n-2)f1–14(n-1)d0–1ns² (Lanthanoids and Actinoids; inner transition metals)
30. Assign the position of the element having outer electronic configuration
(i) ns²np⁴ for n=3
(ii) (n-1)d²ns² for n=4
(iii) (n-2)f⁷(n-1)d¹ns² for n=6, in the periodic table.
Answer
(i) n=3: configuration 3s²3p⁴. p-block element, Period 3. p⁴
⇒ Group = 12+4 = 16. Answer: Period 3, Group 16
⇒ Sulfur (S, Z=16).
(ii) n=4: configuration 3d²4s². d-block element (transition metal), Period 4. 3d²
⇒ Group 4. Answer: Period 4, Group 4
⇒ Titanium (Ti, Z=22).
(iii) n=6: configuration 4f⁷5d¹6s². f-block element (lanthanoid), Period 6. Answer: Period 6, f-block (Lanthanoids)
⇒ Gadolinium (Gd, Z=64).
31. The first (ΔiH₁) and second (ΔiH₂) ionization enthalpies (in kJ mol⁻¹) and the electron gain enthalpy (ΔegH) of a few elements are given in the table. Which of the elements is likely to be:
Answer
(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 MX₂ (f) the metal which can form a predominantly stable covalent halide MX?
Answer
(a) Least reactive — Element V: Very high ΔiH₁ (2372) and positive ΔegH (+48)
⇒ noble gas. Neither loses nor gains electrons. Likely He or Ne.
(b) Most reactive metal — Element II: Lowest ΔiH₁ (419), negative ΔegH. Very low ionization enthalpy
⇒ easily loses one electron. Very high ΔH₂ shows it forms only +1 ion. Likely Cs or K (alkali metal).
(c) Most reactive non-metal — Element III: High ΔiH (1681) and most negative ΔegH (–328 kJ/mol)
⇒ strongly gains one electron. Likely F or Cl.
(d) Least reactive non-metal — Element IV: High ΔiH₁ (1008) and ΔegH = –295 kJ/mol. Moderate non-metal reactivity. Likely halogen (I).
(e) Metal forming MX₂ — Element VI: ΔiH₁ = 738 and ΔiH₂ = 1451. Both moderate
⇒ loses 2 electrons readily. Likely Ca or Mg (alkaline earth metal).
(f) Metal forming covalent MX — Element I: ΔiH₁ = 520, ΔH₂ = 7300 (very large jump). Loses only 1 electron. Likely Li — lithium halides have significant covalent character due to small size and high polarising power.
32. Predict the formulas of the stable binary compounds formed by the following pairs of elements:
Answer
(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
Ans.
(a) Li (+1) and O (–2): Li₂O
(b) Mg (+2) and N (–3): Mg₃N₂ [3×(+2) + 2×(–3) = 0]
(c) Al (+3) and I (–1): AlI₃
(d) Si (+4) and O (–2): SiO₂
(e) P and F: PF₅ (phosphorus pentafluoride, most stable; P can expand octet) or PF₃
(f) Element 71 is Lutetium (Lu), oxidation state +3; with F (–1): LuF₃
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
(c) principal quantum number. The period number corresponds to the highest principal quantum number (n) of the elements in that period.
34. Which of the following statements related to the modern periodic table is incorrect?
Answer
(b) is incorrect — The d-block has 10 columns (Groups 3–12), not 8 columns, because a d-subshell can hold a maximum of 10 electrons (5 orbitals × 2).
35. Which factor does not affect the valence shell? (a) n (b) Z (c) Nuclear mass (d) Number of core electrons
Answer
(c) Nuclear mass. Neutrons do not affect the electrostatic forces experienced by electrons. Chemical properties are determined by protons (Z), electron configuration (n), and core electron shielding — not by the number of neutrons.
36. The size of isoelectronic species F⁻, Ne and Na⁺ is affected by: (a) Z (b) n (c) electron-electron interaction (d) none
Answer
(a) nuclear charge (Z). All three have 10 electrons (same n=2 for outermost). Their sizes differ because of different Z: F(9), Ne(10), Na(11). Higher Z
⇒ smaller size. Order: F⁻ > Ne > Na⁺.
37. Which statement is incorrect about ionization enthalpy? (a) IE increases for each successive electron (b) Greatest increase on removal from noble gas core (c) Big jump marks end of valence electrons (d) Removal from lower n is easier than higher n
Answer
(d) is incorrect. Removal of an electron from an orbital with a higher n value (outer shell) is easier, not harder. Outer electrons are farther from the nucleus with more shielding, so they are removed more easily. Inner electrons (lower n) are tightly bound.
38. Correct order of metallic character for B, Al, Mg, K: (a) B>Al>Mg>K (b) Al>Mg>B>K (c) Mg>Al>K>B (d) K>Mg>Al>B
Answer
(d) K > Mg > Al > B. K has the lowest ionization enthalpy (most metallic); B is a metalloid (least metallic).
39. Correct order of non-metallic character for B, C, N, F, Si: (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
(c) F > N > C > B > Si. F is most electronegative; Si is the most metallic/metalloid in the group.
40. Correct order of oxidizing property for F, Cl, O, N: (a) F>Cl>O>N (b) F>O>Cl>N (c) Cl>F>O>N (d) O>F>N>Cl
Answer
(b) F > O > Cl > N. F is the strongest oxidising agent. O oxidises better than Cl due to its smaller size and higher effective nuclear charge. N has the weakest oxidising tendency among the four.
Important Extra Questions
What is the Aufbau principle and how does it relate to the arrangement of elements in the periodic table?
Answer
The Aufbau principle states that electrons fill orbitals in order of increasing energy (1s
⇒ 2s
⇒ 2p
⇒ 3s
⇒ 3p
⇒ 4s
⇒ 3d
⇒ 4p
⇒ 5s
⇒ 4d
⇒ 5p
⇒ 6s
⇒ 4f
⇒ 5d
⇒ 6p...). The periodic table is a graphical representation of this principle. The period number corresponds to the principal quantum number of the outermost electrons. The block (s, p, d, or f) indicates which subshell last received electrons. This filling order explains why Period 1 has 2 elements, Periods 2 and 3 have 8 each, Periods 4 and 5 have 18 each, and Period 6 has 32 elements.
Why does the first member of each group show anomalous properties compared to the rest?
Answer
The first member of each group shows anomalous properties due to: (1) Small size — leads to high charge density and strong polarising power; (2) High electronegativity; (3) No d-orbitals in the valence shell (2nd period), limiting coordination number to 4; (4) Ability to form strong π bonds (pπ-pπ): e.g., N₂ forms N≡N triple bond while P₄ has only single bonds; O forms O=O while S forms S₈ rings. These differences arise from strong orbital overlap in small first-row atoms and absence of d-orbitals for back-bonding or expanded octet.
What is shielding effect? How does it affect atomic radius and ionization enthalpy across a period and down a group?
Answer
The shielding (screening) effect is the reduction in effective nuclear charge (Zeff) experienced by outer electrons due to the presence of inner core electrons. Zeff = Z − σ (σ = shielding constant).
Across a period: Same-shell electrons shield each other poorly (σ ≈ 0.35 per same-shell electron). Z increases but shielding increases less, so Zeff increases
⇒ atomic radius decreases, ionization enthalpy increases.
Down a group: Each new period adds a full inner shell. Inner electrons shield effectively (σ ≈ 0.85 for n-1 shell, 1.00 for deeper shells). Despite increasing Z, Zeff increases only slightly while atomic radius increases significantly (new shell farther from nucleus). Ionization enthalpy decreases down a group.
Explain diagonal relationships in the periodic table with examples.
Answer
A diagonal relationship exists between certain Period 2 elements and their Period 3 diagonal neighbours: Li–Mg, Be–Al, B–Si. These diagonal pairs have similar charge-to-radius ratios (ionic potentials), leading to similar chemical behaviour. Examples:
• Li and Mg: Both form organometallic compounds, both form normal oxides (Li₂O, MgO) rather than peroxides, both react slowly with water, LiCl and MgCl₂ are slightly soluble.
• Be and Al: Both form amphoteric oxides (BeO, Al₂O₃ react with both acids and bases), both dissolve in NaOH, both form covalent chlorides that hydrolyse.
• B and Si: Both form covalent hydrides and halides, both form acidic oxides, borax and silicates have similar polymeric structures. Moving one period down (larger size) and one group right (higher charge) roughly compensates, maintaining similar polarising power.
Arrange the following in order of increasing first ionization enthalpy and explain: Na, Mg, Al, Si, P, S, Cl, Ar (Period 3).
Answer
General trend across Period 3: ionization enthalpy increases with increasing atomic number due to increasing nuclear charge.
However, there are two exceptions:
• Al (577 kJ/mol) < Mg (738 kJ/mol): Al (3s²3p¹) loses a 3p electron (higher energy, better shielded by 3s²) compared to Mg (3s²) losing a 3s electron.
• S (1000 kJ/mol) < P (1012 kJ/mol): P (3s²3p³) has a half-filled 3p subshell (extra stable due to exchange energy). S (3s²3p⁴) has paired electrons in one 3p orbital; repulsion between paired electrons makes it easier to remove one electron from S.
Increasing order: Na < Mg < Al < Si < S < P < Cl < Ar
Values (kJ/mol): Na(496) < Al(577) < Mg(738)
⇒ corrected order accounting for exceptions: Na < Mg < Al < Si < S < P < Cl < Ar.