NCERT Solutions for Class 11 Chemistry Chapter 3: Classification of Elements and Periodicity in Properties

Concept. The periodic table is a chart in which all known elements are arranged so that elements with similar properties fall together and so that properties change in a regular, predictable way as we move through the chart. NCERT opens the unit by pointing out that in 1800 only 31 elements were known, by 1865 the number had more than doubled to 63, and today 114 (now 118) are known; studying each element and its countless compounds individually is impossible, so scientists searched for a systematic way to organise knowledge. The periodic table is the final answer to that search.

1. Elements are arranged in order of increasing atomic number. The modern periodic law states that the physical and chemical properties of elements are periodic functions of their atomic numbers. When elements are listed one after another by Z, similar properties reappear after fixed intervals — this “periodic recurrence” is the heart of the table.
2. Horizontal rows are called periods (there are seven). The period number equals the principal quantum number n of the valence shell, so every new period begins when a new shell starts filling, and the number of elements in a period (2, 8, 8, 18, 18, 32, 32) equals twice the number of orbitals being filled.
3. Vertical columns are called groups or families (numbered 1–18 by IUPAC). Elements in a group have the same valence-shell electronic configuration (for example all alkali metals are ns¹) and therefore similar chemical behaviour.
4. The table is divided into s-, p-, d- and f-blocks according to the kind of orbital that receives the last electron, which connects the table directly to the quantum-mechanical model of the atom from Chapter 2.
5. Periodic trends — atomic radius, ionization enthalpy, electron gain enthalpy, electronegativity, valence and metallic character — vary regularly across a period and down a group, so one can predict the properties of an element simply from its position.

Extra depth. Even the arrangement of the table itself follows from electronic structure: the 4f and 5f series are drawn separately below the table only to keep the table compact while keeping similar elements in one column. The table also “rationalizes known chemical facts and even predicts new ones” — Mendeleev’s prediction of gallium and germanium shows how powerful this organising theme is.

Conclusion: The basic theme of the periodic table is the periodic recurrence of properties when elements are arranged by increasing atomic number, so that elements with the same outer electronic configuration sit in the same vertical group.

Concept. Dmitri Mendeleev (1834–1907), a Russian chemist, was the first to publish the periodic law (1869). His law reads: “The properties of the elements are a periodic function of their atomic weights.” Working independently, the German chemist Lothar Meyer reached very similar conclusions, but Mendeleev’s system was more elaborate and he is credited with the first general periodic table.

• The property used: atomic weight (atomic mass). Mendeleev arranged the elements in horizontal rows and vertical columns in order of increasing atomic weight, in such a way that elements with similar properties occupied the same vertical column or “group”.
• He also relied on the similarity of empirical formulas and properties of the compounds formed by the elements — a broader range of physical and chemical properties than Lothar Meyer, who plotted mainly physical properties such as atomic volume, melting point and boiling point against atomic weight.
• Did he stick to atomic weight? No, not rigidly. He realised that some elements would not fit his scheme if the order of atomic weights was followed strictly. He ignored the order of atomic weights, thinking the measured values might be incorrect, and placed the elements with similar properties together.
• Example given in NCERT: iodine has a lower atomic weight than tellurium (Group VI), yet he placed iodine in Group VII with fluorine, chlorine and bromine because of the similarity of properties (Fig. 3.1). (Another well-known case, not in the text: argon, 39.9, would precede potassium, 39.1, if only mass were used.)
• He left gaps for undiscovered elements, keeping his primary aim of placing similar elements in the same group. Gallium and germanium were unknown then; he left a gap under aluminium and one under silicon and called them Eka-aluminium and Eka-silicon, predicting their properties.

Extra depth. The boldness of these quantitative predictions and their eventual success made Mendeleev and his table famous. The flaw that atomic weight alone cannot be the basis was removed only in 1913, when Henry Moseley showed from X-ray spectra that atomic number is the more fundamental property (see Ex 3.3). Mendeleev’s instinct to break the weight-order for chemical similarity was therefore right — the true order is by Z, and I (Z = 53) does come after Te (Z = 52).

Conclusion: Mendeleev classified elements mainly by increasing atomic weight together with similarity of properties and compound formulae, but he did not adhere to it strictly: he put iodine before tellurium and left gaps for undiscovered elements.

Concept. Both laws say that properties of elements are periodic; they differ in the variable on which the periodicity is based. Mendeleev’s law (1869) was framed when chemists knew nothing about the internal structure of the atom, so it used atomic weight. The modern law is the outcome of 20th-century discoveries about sub-atomic particles.

Extra depth. Henry Moseley, English physicist, studied characteristic X-ray spectra in 1913. A plot of √ν (where ν is the frequency of X-rays emitted) against atomic number Z gave a straight line, whereas a plot of √ν against atomic mass did not. He thereby showed that atomic number is a more fundamental property than atomic mass, and Mendeleev’s law was modified accordingly. Atomic number equals the nuclear charge, which equals the number of electrons in a neutral atom; hence the quantum numbers and electronic configurations that govern chemical behaviour are fixed by Z, and the Periodic Law is, in essence, the consequence of periodic variation in electronic configurations.

Conclusion: Mendeleev’s law arranges elements by atomic weight, whereas the modern law, following Moseley’s X-ray work, arranges them by atomic number and explains periodicity through electronic configuration.

Concept. The period number equals the principal quantum number n of the valence shell, and every period ends when the orbitals that are energetically accessible in that shell are completely filled. The number of elements in a period therefore equals the number of electrons that can be accommodated in the orbitals which fill during that period — two electrons per orbital (Pauli exclusion principle) — i.e. twice the number of orbitals.

1. For the sixth period n = 6. The allowed values of l are 0, 1, 2, 3, 4, 5 (l = 0 to n − 1), i.e. 6s, 6p, 6d, 6f, 6g and 6h subshells in principle.
2. By the (n + l) rule and the aufbau order, the orbitals that are actually filled in the sixth period, in order of increasing energy, are 6s < 4f < 5d < 6p. (6d, 6f etc. lie higher than 6p and are not filled until later.)
3. Number of orbitals: 6s = 1 (l = 0); 4f = 7 (l = 3); 5d = 5 (l = 2); 6p = 3 (l = 1). Total = 1 + 7 + 5 + 3 = 16 orbitals.
4. Each orbital holds a maximum of 2 electrons with opposite spins, so the maximum number of electrons = 16 × 2 = 32.
5. Since each successive element adds exactly one electron, 32 electrons means 32 elements — from caesium (Z = 55, 6s¹) to radon (Z = 86, 6s²4f¹⁴5d¹⁰6p⁶), which includes the fourteen lanthanoids Ce (58) – Lu (71).

Extra depth. Compare with the fifth period (Problem 3.2): orbitals 5s, 4d, 5p = 1 + 5 + 3 = 9 → 18 elements; and the fourth: 4s, 3d, 4p = 9 → 18 elements. The extra seven f-orbitals are what raise the sixth (and seventh) period to 32. The seventh period would also have 32 on the same reasoning (7s, 5f, 6d, 7p). The first period has just 1s (2 elements), and the second and third have 2s2p / 3s3p (4 orbitals, 8 elements).

Conclusion: Because the n = 6 period fills 6s, 4f, 5d and 6p — sixteen orbitals holding two electrons each — the sixth period contains exactly 32 elements.

Concept. To locate an element we write its electronic configuration: the period is the principal quantum number of the outermost shell, and the group follows from the number and type of valence electrons (for p-block elements, group = 10 + number of electrons in the valence shell).

1. Z = 114 means 114 electrons. The preceding noble gas is radon (Z = 86), so the configuration is [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p². (Check: 86 + 14 + 10 + 2 + 2 = 114.)
2. The highest principal quantum number is 7 → the element is in the 7th period.
3. The last electron enters the 7p subshell → it is a p-block element. The valence shell is 7s²7p² = 4 electrons, so group number = 10 + 4 = 14.
4. It therefore lies below lead (6s²6p²) in the carbon family, as the heaviest member of Group 14. Its IUPAC temporary name (Table 3.5) was ununquadium (Uuq); the official name is flerovium (Fl).

Extra depth. The same method works for undiscovered or newly named elements (compare Problem 3.3: Z = 117 → Group 17, Z = 120 → Group 2). Because the d- and f-orbitals of the preceding shells are completely filled for Z = 114, they do not change the group; only the s and p electrons of shell 7 matter.

Conclusion: The element with Z = 114 (flerovium, originally Uuq) is a p-block element placed in the 7th period and Group 14.

Concept. The periodic table is a map: the period (row) tells us the value of the principal quantum number n of the valence shell, and the group (column) tells us the valence-shell configuration. Group 17 is the halogen family, whose members all have the outer configuration ns²np⁵. The third period therefore corresponds to n = 3.

1. Combine the two pieces of information: n = 3 and configuration ns²np⁵ → valence-shell configuration 3s²3p⁵.
2. The inner shells must be completely filled, so the full configuration is 1s² 2s² 2p⁶ 3s² 3p⁵, i.e. [Ne] 3s²3p⁵.
3. Counting electrons: 2 + 2 + 6 + 2 + 5 = 17. For a neutral atom, number of electrons = atomic number, so Z = 17.
4. The element with Z = 17 is chlorine (Cl).

Extra depth. An alternative shortcut: third-period elements run from Z = 11 (Group 1) to Z = 18 (Group 18); the first two (Na, Mg) are Groups 1 and 2, and from Al onwards the group number is 10 more than the number of valence electrons. Group 17 is the second last, so Z = 18 − 1 = 17. Chlorine has the highly negative electron gain enthalpy (−349 kJ mol⁻¹, Table 3.7) typical of halogens and forms Cl⁻ by gaining one electron.

Conclusion: The element in the third period and Group 17 has the configuration [Ne]3s²3p⁵, so its atomic number is 17 (chlorine).

Concept. Names of the heaviest elements have traditionally been the privilege of the discoverer, ratified by IUPAC. Such elements are so unstable that only minute quantities, sometimes only a few atoms, are produced, and their synthesis is done at a handful of sophisticated laboratories. The question is based on the NCERT footnote that describes Glenn T. Seaborg’s work and on the fact that these laboratories honoured pioneers of nuclear science.

• (i) Lawrence Berkeley Laboratory → Lawrencium (Lr), Z = 103. The laboratory in Berkeley, California, is named after Ernest O. Lawrence, inventor of the cyclotron; the element lawrencium, which completes the actinoid series, commemorates him. (Other elements discovered at Berkeley and named for the place are berkelium, Bk, Z = 97, and californium, Cf, Z = 98.)
• (ii) Seaborg’s group → Seaborgium (Sg), Z = 106. NCERT states in the footnote that Seaborg’s work from 1940 (discovery of plutonium) and of all the transuranium elements from 94 to 102 led to the reconfiguration of the periodic table that places the actinoids below the lanthanoids; he received the 1951 Nobel Prize in chemistry. “Element 106 has been named Seaborgium (Sg) in his honour.”
• Note the pattern in Table 3.5: temporary IUPAC names derive from the digits of Z (un-nil-bi-…), and permanent names are later chosen by IUPAC vote — reflecting a country or state where discovery took place, or paying tribute to a notable scientist.

Extra depth. The controversy over element 104 shows why systematic names exist: American scientists called it rutherfordium and Soviet scientists kurchatovium, each claiming discovery. To avoid such disputes IUPAC recommended that, until discovery is proved and a name officially recognised, the systematic name be derived from the atomic number using the numerical roots for 0–9 (Table 3.4).

Conclusion: Lawrence Berkeley Laboratory would have named lawrencium (Lr, Z = 103), and Seaborg’s group the element seaborgium (Sg, Z = 106).

Concept. Chemical behaviour is decided by the valence electrons — their number, the orbital type they occupy, and their distance from the nucleus. The Periodic Law is, in essence, the consequence of the periodic variation of electronic configuration. Hence the answer lies in electronic configuration.

1. Same valence-shell electronic configuration. Elements in a vertical column have the same number of electrons in the outer orbitals, distributed in the same way. For Group 1 all members have ns¹ (Li 2s¹, Na 3s¹, K 4s¹, Rb 5s¹, Cs 6s¹, Fr 7s¹); Group 17 members all have ns²np⁵.
2. Same valence and same tendency to lose or gain electrons. Because the valence-shell configuration is identical, they have the same valence (all alkali metals form M⁺; all halogens form X⁻), and compounds of the same type (LiCl, NaCl, KCl; HF, HCl, HBr, HI).
3. Only the principal quantum number changes. Going down a group the shell number increases, which causes gradual, regular changes (size increases, ionization enthalpy decreases) rather than changes of kind — so properties show a similarity with a smooth trend.
4. Hence the properties are a periodic function of atomic number, not of atomic mass: elements of the same group reappear after fixed numbers of elements (8, 8, 18, 18, 32) when the same type of valence shell is repeated.

Extra depth. Similarity does not mean sameness. The first member of each group (Li, Be, B … F) differs somewhat from the others because of small size, high electronegativity and the absence of d-orbitals (section 3.7.2), and in transition-metal groups the trend can even be reversed. Physical properties (melting point, density, etc.) also vary gradually because the underlying cause — the same valence-shell structure with a steadily larger core — is the same.

Conclusion: Elements of the same group have the same number and distribution of electrons in their outermost orbitals, so they show similar valence, similar compounds and a regular gradation of physical and chemical properties.

Concept. An atom has no sharp boundary because the electron cloud around the nucleus thins out gradually (Chapter 2: probability density), and an atom is extremely small — about 1.2 Å = 1.2 × 10⁻¹⁰ m in radius. Therefore there is no practical way to measure the size of an isolated atom; “atomic radius” and “ionic radius” are operational definitions based on distances between nuclei in the combined state.

• Atomic radius is the effective size of an atom estimated from half the distance between the nuclei of two bonded atoms. In this book the term is used for both covalent and metallic radius.
• Covalent radius (non-metals): half the distance between two like atoms joined by a single covalent bond. Example: Cl–Cl distance in Cl₂ = 198 pm → radius of Cl = 99 pm.
• Metallic radius (metals): half the internuclear distance between neighbouring metal cores in a metallic crystal. Example: Cu–Cu = 256 pm → metallic radius of Cu = 128 pm.
• Ionic radius is the effective size of a cation or anion in an ionic crystal, estimated from the distances between cations and anions. Removing an electron gives a cation (smaller than the atom); adding one gives an anion (larger). Examples: F = 64 pm but F⁻ = 136 pm; Na = 186 pm but Na⁺ = 95 pm.
• Both radii are determined by X-ray or other spectroscopic methods. Noble gases (monoatomic) are compared via van der Waals radii, which are much larger than covalent radii.

Extra depth. Because values depend on the method (type of bond, coordination number, crystal), atomic radii are approximate and values from different tables may differ slightly. They are useful because they explain trends: radius controls ionization enthalpy, electron gain enthalpy, electronegativity, bond lengths and even metallic character — “ionization enthalpy and atomic radius are closely related properties”.

Conclusion: Atomic radius is half the distance between nuclei of bonded atoms (covalent or metallic radius), and ionic radius is the effective size of a cation or anion in a crystal; both are practical estimates because an isolated atom has no sharp boundary.

Concept. The two trends are best understood from two opposing factors: (i) effective nuclear charge (Z_eff, the nuclear attraction felt by a valence electron after shielding by inner electrons) and (ii) the number of electron shells (principal quantum number n).

1. Across a period (left to right): atomic radius generally decreases. NCERT Table 3.6(a): period 2 — Li 152, Be 111, B 88, C 77, N 74, O 66, F 64 pm; period 3 — Na 186, Mg 160, Al 143, Si 117, P 110, S 104, Cl 99 pm (Fig. 3.4a).
2. Explanation: across a period the outer electrons are added to the same valence shell, so shielding by electrons of the same shell is poor, but the nuclear charge increases by one unit per element. The effective nuclear charge therefore rises and pulls the electron cloud inward.
3. Down a group: atomic radius increases regularly with atomic number. Table 3.6(b): Li 152, Na 186, K 231, Rb 244, Cs 262 pm (alkali metals); F 64, Cl 99, Br 114, I 133, At 140 pm (halogens) (Fig. 3.4b).
4. Explanation: going down, the principal quantum number n increases, so the valence electrons are farther from the nucleus. The inner shells are filled with electrons and shield the outer electrons from the nuclear pull; this increased shielding outweighs the increased nuclear charge, so the atom grows.

Extra depth. Exceptions/notes. Noble gases are excluded: they are monoatomic, so their (non-bonded) radii are very large and must be compared with van der Waals radii, not covalent radii. In transition (3d) series the decrease across the period is much smaller, and still smaller across the 4f series, because added electrons go to inner d/f subshells, which add shielding that nearly compensates the extra nuclear charge. Atomic radius is the property that controls ionization enthalpy, electronegativity and metallic character.

Conclusion: Atomic radius decreases across a period because of increasing effective nuclear charge in the same shell, and increases down a group because of additional shells and greater shielding.

Concept. Isoelectronic species are atoms and ions that contain the same number of electrons. NCERT’s footnote adds that such species have the same number of atoms, same number of valence electrons and same structure, regardless of the nature of the elements involved. Their electronic configurations are identical, but they differ in nuclear charge, and therefore in size and properties. For example O²⁻, F⁻, Na⁺ and Mg²⁺ all contain 10 electrons.

1. Method: count the electrons in the given species, then find other atoms/ions with the same count by adding or removing electrons from neighbouring elements.
2. (i) F⁻: 9 + 1 = 10 electrons (1s²2s²2p⁶). Isoelectronic: Ne, Na⁺, Mg²⁺, Al³⁺, O²⁻, N³⁻. (Answer: e.g. Ne, or Na⁺.)
3. (ii) Ar: 18 electrons (1s²2s²2p⁶3s²3p⁶). Isoelectronic: K⁺, Ca²⁺, Sc³⁺, Cl⁻, S²⁻, P³⁻.
4. (iii) Mg²⁺: 12 − 2 = 10 electrons. Isoelectronic: Ne, Na⁺, Al³⁺, F⁻, O²⁻, N³⁻.
5. (iv) Rb⁺: 37 − 1 = 36 electrons ([Ar]3d¹⁰4s²4p⁶). Isoelectronic: Kr, Sr²⁺, Y³⁺, Br⁻, Se²⁻.

Extra depth. Because nuclear charge differs while electrons are the same, the ion with the larger nuclear charge is smaller: Al³⁺ < Mg²⁺ < Na⁺ < Ne < F⁻ < O²⁻ < N³⁻ (see Ex 3.12). Isoelectronic ions of s- and p-block elements usually have the noble-gas configuration, which explains why such ions are very common in ionic compounds.

Conclusion: Isoelectronic species have the same number of electrons; F⁻ ≈ Ne/Na⁺, Ar ≈ K⁺/Cl⁻, Mg²⁺ ≈ Ne/Al³⁺, Rb⁺ ≈ Kr/Sr²⁺.

Concept. This question applies the idea of isoelectronic species. The radius of an ion depends on the number of electrons (which determines how many shells and how much electron–electron repulsion) and on the nuclear charge (which determines the pull on those electrons).

1. (a) What is common: N³⁻ (7 + 3), O²⁻ (8 + 2), F⁻ (9 + 1), Na⁺ (11 − 1), Mg²⁺ (12 − 2) and Al³⁺ (13 − 3) all have 10 electrons with the identical configuration 1s²2s²2p⁶ (that of neon). They are isoelectronic.
2. (b) Why the radii differ: the number of electrons and shells is the same, so the only difference is nuclear charge: N (+7), O (+8), F (+9), Na (+11), Mg (+12), Al (+13). The greater the positive nuclear charge, the stronger the pull on the same 10 electrons and the smaller the ion.
3. Therefore the order of increasing ionic radii is: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻.

Extra depth. NCERT quotes: Na⁺ = 95 pm and F⁻ = 136 pm, consistent with the order. For isoelectronic cations the greater positive charge gives the smaller radius; for anions the greater negative charge means the electron repulsion outweighs nuclear charge and the ion expands. Values also depend on the data source (e.g. Mg²⁺ = 72 pm, Na⁺ = 102 pm in Table of section 3.7.2, 95 pm in section 3.7.1(b)), but the order never changes.

Conclusion: All six ions are isoelectronic with neon (10 electrons) and their radii increase as nuclear charge decreases: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻.

Concept. An atom is neutral: its Z protons are balanced by Z electrons. When it loses or gains electrons it becomes an ion, and the balance between nuclear attraction and electron repulsion changes, so the size changes.

1. Cation (positive ion) — smaller than the parent atom. A cation is formed by the loss of one or more electrons, so it has fewer electrons while its nuclear charge remains the same. The remaining electrons are therefore pulled in more strongly (effective nuclear charge per electron increases), and electron–electron repulsion decreases.
2. In many cases the entire outer shell is removed (Na: 2, 8, 1 → Na⁺: 2, 8), so the ion has one shell fewer than the atom. Example: Na = 186 pm, Na⁺ = 95 pm.
3. Anion (negative ion) — larger than the parent atom. An anion is formed by gaining one or more electrons while the nuclear charge is unchanged. More electrons mean increased repulsion among the electrons and a decrease in effective nuclear charge per electron, so the electron cloud expands. Example: F = 64 pm, F⁻ = 136 pm.
4. Further charge, further change: the greater the positive charge on a cation, the smaller its radius (Na⁺ > Mg²⁺ > Al³⁺); the greater the negative charge on an anion, the larger its radius (F⁻ < O²⁻ < N³⁻).

Extra depth. The same reasoning explains the sizes of isoelectronic series, and why ionic radii follow the same periodic trend as atomic radii. It also explains the properties of ionic compounds: a small, highly charged cation (high charge/radius ratio, e.g. Li⁺, Be²⁺) polarises the anion strongly, giving covalent character — the reason for the anomalous behaviour of lithium and beryllium (section 3.7.2).

Conclusion: A cation is smaller than its atom because fewer electrons are held by the same nuclear charge, while an anion is larger because extra electrons repel one another and reduce the effective nuclear charge per electron.

Concept. Ionization enthalpy and electron gain enthalpy are meant to be intrinsic, comparable properties of an element. The hint in the question says: “requirements for comparison purposes”. To compare elements fairly, the measurement conditions must be identical, and the two terms provide this standard.

• ‘Isolated’ and ‘gaseous’: in the gas phase the atoms are far apart, so there are no inter-atomic forces (no metallic bonding as in solids, no intermolecular attractions as in liquids). The energy measured is therefore only that required to remove (or add) an electron from the atom itself, and is not mixed with lattice, bonding or cohesion energies, which differ from element to element.
• ‘Ground state’: the lowest-energy state of the atom, with electrons in the aufbau configuration. If the atom were in an excited state, the electron to be removed would already be in a higher energy level and the ionization enthalpy would be smaller; the value would vary with the degree of excitation. Taking the ground state gives a unique reference value for every element.
• The same two conditions apply to the electron gain enthalpy: X(g) + e⁻ → X⁻(g), where X is an isolated gaseous atom in its ground state.
• Without these two specifications the values would depend on temperature, physical state and excitation, and trends such as those in Figs 3.5 and 3.6 could not be compared.

Extra depth. The same thermodynamic care appears in NCERT footnote (the footnote under Table 3.7 mentions that electron affinity is defined at absolute zero and must be corrected for heat capacities at other temperatures: Δ_egH = −A_e − 5/2 RT). Defining both quantities for gaseous ground-state atoms also lets us link them to the later study of bond energies and lattice enthalpies through Born–Haber-type cycles.

Conclusion: ‘Isolated gaseous atom’ removes inter-atomic influences and ‘ground state’ fixes the energy level, so ionization and electron gain enthalpies become intrinsic values that can be compared for all elements.

Concept. The ionization enthalpy is the energy needed to remove the electron from the ground state (n = 1) to infinity (n = ∞, E = 0): H(g) → H⁺(g) + e⁻. The value given is for one atom; ionization enthalpy is expressed per mole, so we multiply by Avogadro’s number (N_A = 6.022 × 10²³ mol⁻¹), as the hint says.

1. Energy of the electron in the ground state: E₁ = −2.18 × 10⁻¹⁸ J per atom. The energy of a free electron at rest is E_∞ = 0.
2. Energy required to ionise one atom: ΔE = E_∞ − E₁ = 0 − (−2.18 × 10⁻¹⁸ J) = +2.18 × 10⁻¹⁸ J per atom.
3. Convert to one mole: Δ_iH = 2.18 × 10⁻¹⁸ J atom⁻¹ × 6.022 × 10²³ atoms mol⁻¹.
4. = (2.18 × 6.022) × 10⁵ = 13.13 × 10⁵ J mol⁻¹ = 1.312 × 10⁶ J mol⁻¹ (= 1312 kJ mol⁻¹).

Δ_iH = 2.18 × 10⁻¹⁸ J × 6.022 × 10²³ mol⁻¹ = 1.312 × 10⁶ J mol⁻¹

Extra depth. The value 1312 kJ mol⁻¹ is for H; it is large because the single electron is held by a full unit of nuclear charge with no shielding. The sign is positive because energy is supplied (ionization enthalpies are always positive). Using the Bohr formula E_n = −2.18 × 10⁻¹⁸ (Z²/n²) J, the same energy can be obtained for He⁺ (Z = 2): 4 × 2.18 × 10⁻¹⁸ J per ion, i.e. 5248 kJ mol⁻¹. Compare with NCERT’s remark that second ionization enthalpy of helium is 5251 kJ mol⁻¹ (element V in Ex 3.31).

Conclusion: The ionization enthalpy of atomic hydrogen is 2.18 × 10⁻¹⁸ J × 6.022 × 10²³ mol⁻¹ = 1.312 × 10⁶ J mol⁻¹ (1312 kJ mol⁻¹).

Concept. Across the second period ionization enthalpy generally increases because effective nuclear charge rises while electrons are added to the same shell (Fig. 3.6a). Two “anomalies” — Be > B and N > O — interrupt this trend. They are explained by (a) the penetration of s versus p orbitals and (b) electron–electron repulsion in paired orbitals.

1. (i) Be (Z = 4, 1s²2s²) vs B (Z = 5, 1s²2s²2p¹). Although B has greater nuclear charge, its first ionization removes a 2p electron, whereas Be loses a 2s electron.
2. In the same principal quantum level an s electron is attracted to the nucleus more strongly than a p electron, because the 2s orbital penetrates closer to the nucleus (more probability near the nucleus). So the 2p electron of boron is more shielded by the inner core (1s² and the 2s² electrons) and is held less tightly.
3. It is therefore easier to remove the 2p electron from boron than the 2s electron from beryllium: Δ_iH (Be) > Δ_iH (B). (Be also has a completely filled, stable 2s² subshell.)
4. (ii-a) O vs N. N (1s²2s²2p³) has its three 2p electrons in different orbitals with parallel spins (Hund’s rule) — a half-filled, extra-stable arrangement. In O (2p⁴), two of the four 2p electrons must occupy the same 2p orbital, which raises electron–electron repulsion.
5. Hence it is easier to remove the fourth 2p electron from oxygen than one of the three 2p electrons from nitrogen: Δ_iH (O) < Δ_iH (N).
6. (ii-b) O vs F. F (Z = 9) has greater nuclear charge than O (Z = 8), and its electrons are in the same shell, so the outer electrons of F are held more tightly: Δ_iH (O) < Δ_iH (F). This is the normal trend. Together: O < N < F, as in the given order.

Extra depth. Order check: Li (520) < B (801) < Be (899) < C (1086) < O (1314) < N (1402) < F (1681) < Ne (2081) kJ mol⁻¹ (values from the standard data tables; NCERT gives only the order). Similar pairs appear in the third period (Mg > Al, P > S) and the answer to Problem 3.6 relies on the same Be/B logic.

Conclusion: Be exceeds B because a 2s electron is harder to remove than a 2p electron, and O is below N because of electron repulsion in the paired 2p orbital, while O is below F because F has the greater nuclear charge.

Concept. The first ionization enthalpy depends on the neutral atom’s valence electron, but the second depends on the electron structure of the cation that is left. A sudden jump occurs when an electron has to be removed from a stable, noble-gas-like core.

1. First ionization enthalpy: Na (1s²2s²2p⁶3s¹) loses its single 3s electron; Mg (…3s²) loses one of two 3s electrons. Mg has a larger nuclear charge (+12 vs +11) while the electrons are in the same shell with little extra shielding, so Mg holds its outer electron more tightly. Data: Δ_iH₁ (Na) = 496, (Mg) = 737 kJ mol⁻¹ (given in Problem 3.6).
2. Second ionization enthalpy of Na: after losing one electron Na becomes Na⁺ with the stable neon configuration 1s²2s²2p⁶. The next electron must be removed from the completely filled 2p subshell of an inner (n = 2) shell, which is closer to the nucleus and is held by a nuclear charge of +11, with no shielding from outside. This takes very much more energy (about 4562 kJ mol⁻¹).
3. Second ionization enthalpy of Mg: Mg⁺ has the configuration [Ne]3s¹; the second electron is a 3s valence electron, removed from an ion giving Mg²⁺ with the stable neon core. This requires about 1451 kJ mol⁻¹ — larger than the first, as expected for any cation, but much smaller than Na’s second.
4. Result: first: Na (496) < Mg (737); second: Na (4562) > Mg (1451).

Extra depth. This is the same logic as the large jump in ionization enthalpy at the end of valence electrons (Ex 3.37 (c)): a big jump marks the entry into the core. It is also why Na forms only Na⁺ and Mg forms Mg²⁺ (Ex 3.31: element I has Δ_iH₂ = 7300 against Δ_iH₁ = 520, an alkali metal; element VI with 738 and 1451 is an alkaline-earth metal).

Conclusion: Mg has the higher first ionization enthalpy because of higher nuclear charge, but Na has the higher second because Na⁺ has the stable neon core, from which an electron is very difficult to remove.

Concept. Ionization enthalpy measures how tightly the outermost electron is held. It depends on the balance of (i) the attraction of the electron to the nucleus and (ii) the repulsion of the electron by other electrons. Down a main group several factors change together, and all of them favour easier removal of the outer electron.

1. Increase in atomic size (distance from the nucleus). Down a group the principal quantum number n rises and atomic radius increases (Li 152 → Cs 262 pm). The outermost electron is farther from the nucleus; attraction falls with distance, so less energy is needed to remove it.
2. Increase in the shielding (screening) effect. More inner shells are filled down a group, and these intervening core electrons shield the valence electron from the nuclear charge. The effective nuclear charge felt by the valence electron is less than the actual charge on the nucleus.
3. Shielding outweighs the increase in nuclear charge. Nuclear charge also rises down the group, but NCERT states that “the increase in shielding outweighs the increasing nuclear charge”, so the net attraction on the outermost electron decreases.
4. Penetration and type of orbital. A higher-n electron occupies a larger, more diffuse orbital which penetrates less towards the nucleus than the corresponding lower-n orbital, giving weaker binding.
5. Result: Δ_iH decreases down the group, e.g. alkali metals (Li 520, Na 496, K 419 kJ mol⁻¹; compare Fig. 3.6b) — which is why reactivity of metals rises down a group.

Extra depth. In transition (d-block) elements the trend is not regular, because the added electrons go to inner (n−1)d orbitals and shield poorly (the same d-block contraction effect seen in Ex 3.19). For main-group elements, though, the trend is regular and is directly related to the metallic character (increasing down a group) and to the reducing power of the metals.

Conclusion: Down a group, larger atomic size and greater shielding by inner electrons outweigh the increase in nuclear charge, so the outer electron is held less tightly and the ionization enthalpy decreases.

Concept. The general trend is a regular decrease of ionization enthalpy down a group. Group 13 shows a decrease from B to Al, then a slight increase to Ga, a decrease to In and again an increase to Tl. The deviation arises from the filling of d- and f-subshells, which shield the nuclear charge poorly.

1. B (801) → Al (577): normal fall. Al has an extra shell and an electron configuration [Ne]3s²3p¹; the larger size and shielding by the neon core outweigh the extra nuclear charge.
2. Al (577) → Ga (579): slight rise. Between Al (Z = 13) and Ga (Z = 31) the ten 3d electrons are added (Ga: [Ar]3d¹⁰4s²4p¹), along with 10 extra protons. The d electrons are poor shielders (diffuse d orbitals shield the outer electron much less effectively than s and p electrons of the same shell). So the effective nuclear charge felt by the 4p electron of Ga is higher than expected, and the atom is smaller than expected — the ionization enthalpy does not fall.
3. Ga (579) → In (558): normal fall, because a new shell (n = 5) is added.
4. In (558) → Tl (589): rise. Between In (Z = 49) and Tl (Z = 81) both the 4f¹⁴ and 5d¹⁰ subshells are filled. The f electrons are even poorer shielders than d electrons, so the effective nuclear charge on the 6p electron is much larger — a combined effect of the d- and f-block (lanthanoid) contraction.

Extra depth. Poor shielding by d/f electrons is also the origin of the lanthanoid contraction (treated in later units) and of the stability of the lower oxidation state (+1) for Tl, the “inert-pair effect” — NCERT says these are dealt with in group-wise discussions. The same logic explains similar small irregularities in Groups 14–16 after the 3d and 4f series.

Conclusion: The slight rises at Ga and Tl are due to poor shielding of the nuclear charge by the intervening 3d and 4f/5d electrons, which increases the effective nuclear charge on the outer p electron.

Concept. Electron gain enthalpy (Δ_egH) is the enthalpy change when a neutral gaseous atom accepts an electron to form an anion; a more negative value means the atom takes up the electron more readily. Generally it becomes more negative across a period (higher Z_eff, smaller atom) and less negative down a group (larger atom) — but the second-period elements O and F are exceptions to the second rule.

1. (i) O or F → F is more negative. They are in the same period, and F has the higher nuclear charge and smaller size; adding an electron gives F⁻ with the stable neon configuration. Data (Table 3.7): O = −141, F = −328 kJ mol⁻¹.
2. (ii) F or Cl → Cl is more negative. By the group trend F should have the more negative value, but the exception applies: when an electron is added to F it goes into the small n = 2 shell, where it experiences significant repulsion from the seven electrons already present in this compact level. In Cl the electron enters the larger n = 3 shell (3p), where it occupies a bigger region of space and the inter-electron repulsion is much less. Data: F = −328, Cl = −349 kJ mol⁻¹.

Extra depth. The same anomaly is seen in Group 16 (O −141 vs S −200) and Group 1 is regular (Li −60, Na −53, K −48). Chlorine has the most negative electron gain enthalpy of all elements, which is why it is such a good oxidising agent among the halogens in terms of Δ_egH, although F is the stronger oxidising agent overall (see Ex 3.40) because of F₂’s low bond enthalpy and the high hydration enthalpy of F⁻.

Conclusion: F has the more negative electron gain enthalpy than O, but Cl has a more negative value than F because the extra electron in the compact n = 2 shell of F suffers greater repulsion.

Concept. The first electron gain enthalpy of oxygen, O(g) + e⁻ → O⁻(g), is negative (−141 kJ mol⁻¹, Table 3.7) because the neutral atom, which needs two electrons to complete its octet (2s²2p⁴ → 2p⁶), is attracted to the incoming electron. The second step is different in kind: O⁻(g) + e⁻ → O²⁻(g).

1. In the second step the incoming electron is added to an already negatively charged ion (O⁻). The negative ion repels the incoming electron through strong Coulombic repulsion; there is no net nuclear attraction advantage because the effective nuclear charge per electron is already reduced.
2. The electron also enters the compact n = 2 shell, where electron–electron repulsion is considerable (cf. the O/F anomaly in Ex 3.20).
3. Therefore energy must be supplied to force the second electron on to O⁻. The process is endothermic, so the second electron gain enthalpy of oxygen is positive (the commonly tabulated value is about +780 kJ mol⁻¹).
4. It is neither ‘more negative’ nor ‘less negative’ — it changes sign: first ≈ −141 kJ mol⁻¹, second ≈ +780 kJ mol⁻¹.

Extra depth. Then how does O²⁻ exist in oxides like MgO and Na₂O? Because the large positive second electron gain enthalpy is compensated by the very large lattice enthalpy released when the O²⁻ ion is surrounded by cations in the crystal — the energy bookkeeping you will study in Chapter 4 and Born–Haber cycles. Similarly S, Se have positive second gain enthalpies; and for N the first value itself is about zero or positive (half-filled 2p³).

Conclusion: The second electron gain enthalpy of O is positive because the incoming electron is repelled by the negatively charged O⁻ ion, so energy has to be supplied.

Concept. Both terms describe how an atom behaves towards electrons, which is why they are often confused. The difference lies in what is being described: an isolated atom in the gas phase, or an atom bonded in a molecule.

Extra depth. Electron gain enthalpy of fluorine is −328 kJ mol⁻¹ and chlorine −349 kJ mol⁻¹, but the electronegativity of fluorine (4.0) is higher than chlorine (3.0): the two properties do not always run parallel (the electron gain enthalpy anomaly of F is due to its compact n = 2 shell). Electronegativity scales include Pauling, Mulliken–Jaffe and Allred–Rochow. The Mulliken scale actually combines ionization enthalpy and electron gain enthalpy, showing that electronegativity depends partly on both.

Conclusion: Electron gain enthalpy is a measurable energy change for an isolated gaseous atom accepting an electron, whereas electronegativity is a non-measurable relative ability of a bonded atom to attract shared electrons.

Concept. The statement is incorrect. Electronegativity is not a fixed, measurable property of an isolated atom like atomic mass; NCERT explicitly states: “The electronegativity of any given element is not constant; it varies depending on the element to which it is bound.” The value 3.0 on the Pauling scale is only the approximate, average value of nitrogen.

1. Depends on the bonded partner. In N–F and N–O bonds nitrogen (3.0) is less electronegative than F (4.0) and O (3.5), so it behaves as the electropositive end; in N–H (H = 2.1) and N–C (C = 2.5) it is more electronegative. The “effective” tendency to attract electrons therefore differs.
2. Depends on oxidation state. An atom in a higher positive oxidation state attracts electrons more strongly: N in NO₂ (+4) or HNO₃ (+5) has greater electron-attracting power than N in NH₃ (−3), because its effective nuclear charge is greater and its valence electrons are already partly drawn away.
3. Depends on hybridisation. The more s-character in the hybrid orbital used (sp > sp² > sp³) the closer the electrons are to the nucleus, and the more electronegative the atom appears. N in a nitrile (–C≡N, sp) is more electronegative than N in an amine (sp³).
4. Conclusion: 3.0 is an approximate working value. Electronegativity values (Pauling, Mulliken–Jaffe, Allred–Rochow) are scales for comparison, not constants of nature.

Extra depth. The caveat is important for predicting bond polarity: a bond between N and an atom of similar electronegativity (e.g. C, 2.5 vs 3.0) is polar covalent, while with a large difference (> about 1.7) it is ionic. Because values change slightly with environment, textbooks give a single approximate number only for simplicity; this will be explored in the next chapter on chemical bonding.

Conclusion: The statement is not true: electronegativity of nitrogen is only about 3.0 on average and changes with the element it is bonded to, its oxidation state and its hybridisation.

Concept. The radius of an atom (or ion) is governed by the balance between the pull of the nucleus on the electron cloud and the repulsion among the electrons themselves. Changing the number of electrons while the nuclear charge stays the same upsets that balance and changes the size. This is the theory behind the difference between atomic and ionic radii (section 3.7.1(b)).

1. (a) Gaining an electron: radius increases. When an atom accepts one or more electrons it forms an anion. The nuclear charge (number of protons) is unchanged, but there are more electrons, so (i) electron–electron repulsion increases and (ii) the effective nuclear charge per electron decreases. The electron cloud expands and the anion is larger than the parent atom.
2. Example: F = 64 pm → F⁻ = 136 pm; Cl = 99 pm → Cl⁻ = 181 pm (standard data). The more electrons gained, the larger the ion: F⁻ < O²⁻ < N³⁻.
3. (b) Losing an electron: radius decreases. A cation has fewer electrons while its nuclear charge remains the same. The remaining electrons are attracted more strongly (higher effective nuclear charge per electron) and repulsion among them is reduced; often an entire outer shell is lost. So the cation is smaller than the parent atom.
4. Example: Na = 186 pm → Na⁺ = 95 pm; the greater the positive charge, the smaller the ion: Na⁺ > Mg²⁺ > Al³⁺.

Extra depth. The two ideas — effective nuclear charge and electron repulsion — are the same ones that explain periodic trends in atomic radius, ionization enthalpy and electron gain enthalpy. The size changes also influence properties of compounds: small, highly charged cations polarise large anions (Fajans’ rules, later), giving covalent character as in LiCl, BeCl₂.

Conclusion: On gaining electrons the atom expands because of increased electron repulsion and reduced effective nuclear charge; on losing electrons it shrinks because the same nuclear charge holds fewer electrons more tightly.

Concept. Isotopes are atoms of the same element (same atomic number Z) with different mass numbers because they have different numbers of neutrons (e.g. ³⁵Cl and ³⁷Cl, ¹²C and ¹⁴C). The ionization enthalpy is a property of the electron cloud, so we must ask whether neutrons affect it.

1. The first ionization enthalpy depends on (i) the nuclear charge (number of protons), (ii) the electronic configuration, (iii) the shielding of the outer electron by inner electrons and (iv) the atomic radius.
2. Isotopes of an element have the same number of protons and the same number of electrons, hence the identical electronic configuration and almost identical atomic size.
3. Neutrons carry no charge. They add to the nuclear mass but do not change the nuclear charge or the electrostatic attraction on the valence electron.
4. Therefore the first ionization enthalpies of two isotopes are the same (any difference is extremely small, of the order of the reduced-mass correction, and negligible for chemical purposes).

Extra depth. This is why isotopes of an element behave chemically alike and are hard to separate chemically; physical properties dependent on mass (rate of diffusion, vibrational frequency, spectral line positions via reduced mass) differ slightly. A related NCERT concept: the periodic law is based on atomic number, not atomic mass — isotopes (same Z, different mass) occupy the same place in the periodic table (Moseley’s result, Ex 3.3). Hydrogen isotopes protium, deuterium, tritium have the same first ionization enthalpy to a very good approximation.

Conclusion: Two isotopes of an element have the same first ionization enthalpy because they have the same nuclear charge and electronic configuration; the neutron number affects only the mass.

Concept. Elements are broadly divided into metals, non-metals and metalloids (Fig. 3.3). Metals comprise more than 78 % of all known elements and appear on the left side of the periodic table; non-metals (fewer than twenty in number) are located at the top right-hand side. The difference originates in electronic structure: metals have few valence electrons and low ionization enthalpy (they readily lose electrons), whereas non-metals have high electronegativity and high negative electron gain enthalpy (they readily gain electrons).

Extra depth. The change from metallic to non-metallic character is not abrupt; the thick zig-zag line in Fig. 3.3 separates the two, and elements bordering it — silicon, germanium, arsenic, antimony and tellurium — show properties of both and are called metalloids (semi-metals). In a horizontal row the property changes from metallic on the left to non-metallic on the right; down a group the elements become more metallic. Metallic character is directly related to the tendency to lose electrons (low ionization enthalpy, low electronegativity), whereas non-metallic character is directly related to electronegativity.

Conclusion: Metals are lustrous, malleable, ductile, good conductors with low ionization enthalpy forming cations and basic oxides, whereas non-metals are brittle, poor conductors with high electronegativity forming anions and acidic oxides.

Concept. These are application questions: the position of an element, read from the periodic table, tells us its valence-shell configuration and hence whether it loses or gains electrons. Metals on the left lose electrons to form cations; non-metals on the right gain electrons to form anions.

1. (a) Five electrons in the outer subshell: the outer subshell must be np⁵, i.e. a Group 17 element (halogen, ns²np⁵). Examples: fluorine (2p⁵), chlorine (3p⁵), bromine (4p⁵), iodine (5p⁵). (Group 15 elements have only three electrons in the np subshell, so they do not qualify.)
2. (b) Tends to lose two electrons: a Group 2 element (alkaline-earth metal, ns²) with low ionization enthalpy, e.g. magnesium or calcium: Mg → Mg²⁺ + 2e⁻ (giving the neon configuration).
3. (c) Tends to gain two electrons: a Group 16 element (chalcogen, ns²np⁴), e.g. oxygen or sulphur: O + 2e⁻ → O²⁻ (neon configuration).
4. (d) Group with a metal, a non-metal, a liquid and a gas at room temperature: Group 17 (halogens) — fluorine and chlorine are gases, bromine is a liquid, iodine is a non-metallic solid, and astatine (the radioactive heaviest member) shows metallic character (it lies at the metal–non-metal border). (Group 1 with hydrogen also contains a gas, a non-metal and metals, but no element in it is liquid at room temperature; so Group 17 is the standard answer.)

Extra depth. The questions test a skill needed throughout the chapter: connect valence configuration, position and reactivity. Alkali metals (ns¹) lose one electron, alkaline-earth (ns²) two, Group 13 three; halogens gain one, chalcogens two, Group 15 three. Because ions with noble-gas configuration are stable, elements near the noble gases react by gaining or losing as few electrons as needed.

Conclusion: (a) halogens such as F or Cl, (b) Group 2 elements such as Mg, (c) Group 16 elements such as O or S, and (d) Group 17.

Concept. Chemical reactivity is related to the tendency to lose electrons (metals) or to gain electrons (non-metals). Group 1 elements are metals, so their reactivity depends on the ease of losing the single ns¹ electron (ionization enthalpy); Group 17 elements are non-metals, so reactivity depends on the ease of gaining one electron (electron gain enthalpy / electronegativity).

1. Group 1 (alkali metals): reactivity = tendency to lose the outer s electron to give M⁺. Down the group the atomic size increases and the outer electron is farther from the nucleus and more shielded by the inner shells, so the ionization enthalpy decreases (Li 520, Na 496, K 419 kJ mol⁻¹). The electron is lost more easily, so reactivity increases: Li < Na < K < Rb < Cs.
2. Group 17 (halogens): reactivity = tendency to gain one electron to give X⁻ (they act as oxidising agents). Down the group the atom becomes bigger, so the added electron is farther from the nucleus and is less strongly attracted; electron gain enthalpy becomes less negative (apart from the F anomaly) and electronegativity decreases (F 4.0, Cl 3.0, Br 2.8, I 2.5). The tendency to accept an electron falls, so reactivity decreases: F > Cl > Br > I.
3. Fluorine is the most reactive despite a slightly less negative electron gain enthalpy than chlorine, because of its very high electronegativity and small size; reactivity is linked to oxidising power (Ex 3.40).

Extra depth. The two opposite trends within a family are summed up by NCERT: “within a group of representative metals (say alkali metals) reactivity increases on moving down the group, whereas within a group of non-metals (say halogens), reactivity decreases down the group”. The reason is the same property — increasing atomic size — acting on metals (making electron loss easier) and on non-metals (making electron gain harder). This also links to metallic character increasing down a group and the natural occurrence of reactive elements only in combined form.

Conclusion: Alkali metals become more reactive down the group because their ionization enthalpy falls and electron loss becomes easier, whereas halogens become less reactive because the larger atoms attract an added electron less strongly.

Concept. The block of an element is decided by the subshell that receives the last (differentiating) electron in the aufbau process. Because of this the general outer electronic configuration is characteristic for each block and can be used to place an element in the periodic table.

Extra depth. The group number follows from the outer configuration: s-block — group number = number of ns electrons (1 or 2); p-block — group number = 10 + number of (ns + np) electrons (13–18); d-block — group number = number of (n−1)d + ns electrons (3–12; for Zn, Cd, Hg with d¹⁰s² it is 12). Note that Zn, Cd and Hg, with (n−1)d¹⁰ns², are counted with the d-block but do not show most properties of transition elements. NCERT also lists the f-block outer configuration as (n−2)f^1–14(n−1)d^0–1ns² for the 14 + 14 elements.

Conclusion: The outer configurations are ns^1–2 for s-block, ns²np^1–6 for p-block, (n−1)d^1–10ns^0–2 for d-block and (n−2)f^1–14(n−1)d^0–1ns² for f-block elements.

Concept. To assign a position: (1) substitute n to get the actual configuration, (2) period = n of the valence shell, (3) block from the last subshell filled, (4) group from the number of valence electrons (p-block: 10 + ns + np; d-block: (n−1)d + ns), (5) add the core electrons to get Z and identify the element.

1. (i) 3s²3p⁴: full configuration 1s²2s²2p⁶3s²3p⁴; Z = 10 + 2 + 4 = 16. Period = 3; p-block; valence electrons = 6 → group = 10 + 6 = 16. The element is sulphur (S), a chalcogen, 3rd period, Group 16, p-block.
2. (ii) 3d²4s²: [Ar]3d²4s²; Z = 18 + 2 + 2 = 22. Period = 4; d-block (3d series); group = 2 + 2 = 4. The element is titanium (Ti), 4th period, Group 4, d-block (first transition series).
3. (iii) 4f⁷5d¹6s²: [Xe]4f⁷5d¹6s²; Z = 54 + 7 + 1 + 2 = 64. Period = 6; f-block (4f inner-transition series, lanthanoids). The element is gadolinium (Gd), 6th period, f-block (lanthanoid series, placed in the group-3 position but written in the lower panel). The half-filled 4f⁷ arrangement makes this configuration stable.

Extra depth. Check with the periodic table: Ti is the first-transition-series element after Sc (21); Gd is the 7th member of the lanthanoids counting from Ce (Ce 58, Pr 59, Nd 60, Pm 61, Sm 62, Eu 63, Gd 64), with half-filled 4f⁷ at Eu (4f⁷6s²) and Gd (4f⁷5d¹6s²). The f-block elements do not have a group number in the 1–18 notation: lanthanoids belong to the sixth period and are placed in the Group 3 position at La/Ac but displayed below the table.

Conclusion: The elements are sulphur (3rd period, Group 16, p-block), titanium (4th period, Group 4, d-block) and gadolinium (6th period, f-block lanthanoid).

Concept. Data-interpretation questions are solved by reading three clues from the numbers: (1) a very large jump from Δ_iH₁ to Δ_iH₂ means that after removing one electron a stable noble-gas core is reached — typical of a Group 1 metal; (2) moderate Δ_iH₁ and Δ_iH₂ both low suggests two electrons removed easily — Group 2; (3) very high Δ_iH and a large negative Δ_egH means a non-metal; a positive Δ_egH with the highest Δ_iH means a noble gas.

1. Element I (520, 7300, −60): Δ_iH₂ is 14 times Δ_iH₁ → one valence electron, then a noble-gas core; low first value; small negative Δ_egH → an alkali metal, lithium (Δ_iH₁ = 520).
2. Element II (419, 3051, −48): the same pattern but with a lower Δ_iH₁ (419) → the alkali metal further down, potassium; the most reactive metal.
3. Element III (1681, 3374, −328): very high Δ_iH₁ and a very large negative Δ_egH (−328) → a halogen, fluorine.
4. Element IV (1008, 1846, −295): high Δ_iH and a large negative Δ_egH (−295) but lower than III → another halogen further down, iodine.
5. Element V (2372, 5251, +48): the highest Δ_iH₁ and a positive Δ_egH → a noble gas with a closed shell, helium (Table 3.7: He = +48).
6. Element VI (738, 1451, −40): moderate and nearly proportional Δ_iH₁ and Δ_iH₂ (a jump of only 2×) → two valence electrons: Group 2 metal, magnesium (738, 1451).
7. Answers: (a) least reactive element — V (noble gas, He); (b) most reactive metal — II (K, lowest Δ_iH₁); (c) most reactive non-metal — III (F, most negative among F and I with highest electronegativity); (d) least reactive non-metal — IV (I; the noble gas V is already the least reactive element overall); (e) metal forming stable MX₂ — VI (Mg: both electrons removed fairly easily; its Δ_iH₃ would be very large, so it forms M²⁺); (f) metal forming predominantly covalent MX — I (Li: the small Li⁺ has very high charge/radius ratio, polarises the halide and gives covalent character, e.g. LiI; section 3.7.2(b)).

Extra depth. These six sets are real data (compare Table 3.7: Δ_egH of Li = −60, K = −48, F = −328, He = +48). The identification confirms that reactivity is extreme at the left (loss of electrons) and right (gain of electrons) of the periodic table and least in the noble gases (section 3.7.3). For (d) one may argue about III vs IV: both are non-metals; IV has the smaller (less negative) Δ_egH and lower Δ_iH, so it is the less reactive halogen.

Conclusion: (a) V (He); (b) II (K); (c) III (F); (d) IV (I); (e) VI (Mg); (f) I (Li), identified by the pattern of ionization and electron gain enthalpies.

Concept. The formula of a binary compound follows from the valences of the two elements. For representative elements the valence is the number of valence electrons or eight minus that number (Table 3.9). The compound is neutral, so the total positive and negative valence must balance: use the ‘criss-cross’ rule.

1. (a) Li (Group 1, valence 1) + O (Group 16, valence 2): 2 Li⁺ balance 1 O²⁻ → Li₂O.
2. (b) Mg (Group 2, valence 2) + N (Group 15, valence 3): 3 Mg²⁺ (+6) balance 2 N³⁻ (−6) → Mg₃N₂.
3. (c) Al (Group 13, valence 3) + I (Group 17, valence 1): Al³⁺ needs 3 I⁻ → AlI₃.
4. (d) Si (Group 14, valence 4) + O (valence 2): 1 Si (+4) balances 2 O (−4) → SiO₂.
5. (e) P (Group 15, valence 3 and 5) + F (valence 1): two compounds are stable: PF₃ (valence 3) and PF₅ (valence 5, since P can expand its octet using 3d orbitals; the 3rd-period element has nine valence orbitals).
6. (f) Element 71 (lutetium, Lu; a lanthanoid, 4f¹⁴5d¹6s²) + F: lutetium shows a stable +3 oxidation state (loses 5d¹ and 6s²), so LuF₃.

Extra depth. Formulas like Li₂O, AlI₃ and SiBr₄ (Problem 3.8) are all predicted this way. P shows variable valence (3, 5) because it is a Group 15 element (“3,5” in the valence row of section 3.7.2); its +5 state in PF₅ uses empty 3d orbitals, unavailable to nitrogen — hence NF₃ exists but NF₅ does not (first-element anomaly, section 3.7.2(b)). Element 71 is identified from the atomic number: Lu, the last lanthanoid (Z = 71).

Conclusion: (a) Li₂O, (b) Mg₃N₂, (c) AlI₃, (d) SiO₂, (e) PF₃ and PF₅, (f) LuF₃.

Concept. The modern periodic table is organised so that every property of its rows and columns can be traced to quantum numbers. NCERT section 3.3 states: “The period number corresponds to the highest principal quantum number (n) of the elements in the period” and section 3.5(a): “The period indicates the value of n for the outermost or valence shell.”

1. Correct option: (c) principal quantum number. Period 1 → n = 1 (K shell, 1s); period 2 → n = 2 (L shell, 2s2p); period 3 → n = 3; … period 7 → n = 7. Each new period begins when a new principal shell starts to fill (Li, Na, K, Rb, Cs, Fr).
2. Why (a) is wrong: atomic number increases continuously from 1 to 118 and does not repeat; it is the order of arrangement, not the period number. Each period contains a range of atomic numbers (period 3: 11–18).
3. Why (b) is wrong: atomic mass is not the basis of the modern table at all (that was Mendeleev’s basis); the modern law uses atomic number, and mass does not determine the period.
4. Why (d) is wrong: the azimuthal quantum number l (0, 1, 2, 3 → s, p, d, f) determines the block (s, p, d, f) — a column-wise classification — not the period. The block indicates the value of l for the last subshell that received electrons.

Extra depth. The number of elements per period is twice the number of orbitals filling in that shell (2, 8, 8, 18, 18, 32, 32). So the table turns the quantum-number rules from Chapter 2 into a chart: n sets the row, l the block, and the number of valence electrons the column. Beware the apparent exception in d- and f-blocks: for transition elements the filling d-subshell belongs to (n−1), but the period is still the n of the s-orbital being filled.

Conclusion: The period indicates the value of the principal quantum number n of the valence shell, so option (c) is correct.

Concept. The shape of the modern table follows from subshell capacities. A subshell with azimuthal quantum number l has (2l + 1) orbitals and can hold a maximum of 2(2l + 1) electrons. Since one column of the long form corresponds to one electron added, the number of columns in a block equals the maximum number of electrons in that subshell.

1. Subshell capacities: s (l = 0): 1 orbital, 2 electrons → 2 columns; p (l = 1): 3 orbitals, 6 electrons → 6 columns; d (l = 2): 5 orbitals, 10 electrons → 10 columns; f (l = 3): 7 orbitals, 14 electrons → 14 columns.
2. (a) correct: the p-block (Groups 13–18) has 6 columns since a p subshell holds a maximum of 6 electrons.
3. (b) INCORRECT: the d-subshell has five orbitals and holds a maximum of 10 electrons, so the d-block has 10 columns (Groups 3–12), not 8. The statement gives the wrong number.
4. (c) correct: it states the general rule: s → 2, p → 6, d → 10, f → 14 columns.
5. (d) correct: the block indicates l of the last subshell filled (l = 0 s-block, 1 p-block, 2 d-block, 3 f-block).
6. Answer: option (b).

Extra depth. Total columns: 2 + 6 + 10 = 18 in the long form (plus 14 for the f-series placed separately), matching the 18 IUPAC groups. This also explains why periods have 2, 8, 18, 32 elements: s, s+p, s+d+p, s+f+d+p. Another way to see the error in (b): d-block in the table spans Sc to Zn in the 4th period — ten elements, not eight.

Conclusion: Statement (b) is incorrect because the five d-orbitals hold 10 electrons, giving the d-block 10 columns, not 8.

Concept. The chemistry of an element depends on how tightly its valence electrons are held and where they are. The factors are: how far they are from the nucleus (n), how strongly the nucleus attracts them (Z), and how much that attraction is reduced by the core electrons (shielding).

1. (a) Valence principal quantum number (n) — affects the valence shell: larger n means electrons farther from the nucleus, so the atom is larger and ionization enthalpy is lower.
2. (b) Nuclear charge (Z) — affects it: a greater positive charge attracts the electrons more strongly (cf. radius decreasing across a period).
3. (c) Nuclear mass — does not affect the valence shell. Neutrons carry no charge; the electrostatic forces on the valence electrons depend on charge, not on mass. Isotopes with different mass numbers have identical chemistry (Ex 3.25).
4. (d) Number of core electrons — affects it by shielding: more core electrons reduce the effective nuclear charge felt by the valence electrons (as down a group).
5. Answer: option (c).

Extra depth. The three effective factors are combined in the concept of effective nuclear charge Z_eff = Z − shielding constant. Nuclear mass enters only in properties like rates of diffusion or vibrational frequency (isotope effects), not in the electronic energy that governs chemical reactivity. The periodic law itself is stated in terms of atomic number, not atomic mass, because Z, not mass, controls electronic configuration.

Conclusion: Nuclear mass does not influence the valence electrons, so option (c) is correct.

Concept. F⁻, Ne and Na⁺ are isoelectronic: each has 10 electrons in the configuration 1s²2s²2p⁶. For such a set, several of the factors that normally determine size are identical, and the one that changes must be identified.

1. (b) valence principal quantum number n: the same for all three (n = 2). So it cannot explain a difference in size.
2. (c) electron–electron interaction in the outer orbitals: the same number of electrons (10) in the same orbitals gives the same repulsion in all three. So it also does not change.
3. (a) nuclear charge (Z): F⁻ has Z = 9, Ne has Z = 10, Na⁺ has Z = 11. The same 10 electrons are pulled by a different number of protons: the greater the nuclear charge, the more strongly the cloud is pulled in, so the size decreases: Na⁺ < Ne < F⁻. Hence size is affected by nuclear charge.
4. (d) is wrong because isoelectronic does not mean ‘same size’; the sizes differ (F⁻ 136 pm, Na⁺ 95 pm).
5. Answer: option (a).

Extra depth. This is the rule from section 3.7.1(b): among isoelectronic species the cation with the greater positive charge has the smaller radius, and the anion with the greater negative charge has the larger radius. The radius of Ne here is a non-bonded (van der Waals) radius, so numerical comparison with ionic radii is only qualitative; the rule applies cleanly to ions (e.g. N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺).

Conclusion: Because F⁻, Ne and Na⁺ have the same n and same number of electrons, only the nuclear charge differs and so controls their size — option (a).

Concept. Ionization enthalpy (Δ_iH) is the energy needed to remove an electron from an isolated gaseous atom or ion in its ground state; successive ionization enthalpies are for successive electrons. We examine each statement against the NCERT rules.

1. (a) correct: Δ_iH₂ > Δ_iH₁, Δ_iH₃ > Δ_iH₂, … because removing an electron from a positive ion is harder than from a neutral atom (the same nuclear charge holds fewer electrons).
2. (b) correct: the largest jump occurs when an electron is removed from a stable, completely filled noble-gas core (e.g. Na⁺: 496 → 4562 kJ mol⁻¹).
3. (c) correct: once all valence electrons have gone, the next electron comes from the inner core, causing a big jump (e.g. Li: 520, 7300).
4. (d) INCORRECT: electrons in orbitals with lower n lie closer to the nucleus and are held more tightly, so their removal is more difficult, not easier. Electrons in higher n (outer) orbitals are farther from the nucleus and are shielded, so they are removed more easily.
5. Answer: option (d).

Extra depth. The big jump in successive ionization enthalpies is the experimental way to find the number of valence electrons: for element I in Ex 3.31 (520, 7300) there is a jump after the first, so it has one valence electron; for element VI (738, 1451) there will be a jump after the second. This reasoning is also behind the formation of ions with noble-gas configuration (M⁺, M²⁺) rather than higher charges.

Conclusion: Statement (d) is incorrect, because removal of an electron from an orbital of lower n is harder than from an orbital of higher n.

Concept. Metallic character is the tendency of an element to lose electrons and form cations. As NCERT states, metallic character increases down a group and decreases from left to right across a period. It is inversely related to electronegativity and ionization enthalpy.

1. Position of the elements: B (Group 13, period 2), Al (Group 13, period 3), Mg (Group 2, period 3), K (Group 1, period 4).
2. B vs Al (same group): Al is lower, so Al is more metallic than B. Al vs Mg (same period 3): Mg is to the left of Al, so Mg > Al. Mg vs K: K is the alkali metal at the extreme left (Group 1) and in a lower period, so K > Mg.
3. Combining: K > Mg > Al > B. Check with electronegativities: 0.8 (K), 1.2 (Mg), 1.5 (Al), 2.0 (B) — lower electronegativity means more metallic, which gives exactly this order. (First ionization enthalpies K 419, Mg 737, Al 577, B 801 kJ mol⁻¹: Al is slightly lower than Mg because of the Mg/Al 3s→3p anomaly, but overall metallic behaviour — formation of cations and basic oxides — still places Mg ahead of Al.)
4. Answer: option (d). Options (a) and (b) place B too high; (c) places K below Al and B.

Extra depth. Compare with Problem 3.4 (P < Si < Be < Mg < Na), which uses the same logic. Metallic character is highest at the bottom-left of the periodic table (Cs, Fr) and lowest at the top-right (F). B is a metalloid with mainly non-metallic behaviour; Al is a metal with amphoteric oxide Al₂O₃.

Conclusion: The correct order of metallic character is K > Mg > Al > B, so option (d) is correct.

Concept. Non-metallic character is the tendency of an element to gain electrons; it increases from left to right across a period and decreases down a group, and it runs parallel to electronegativity.

1. Second-period elements B, C, N, F (left to right: Group 13, 14, 15, 17): non-metallic character increases B < C < N < F. Electronegativities: B 2.0, C 2.5, N 3.0, F 4.0.
2. Si lies directly below C in Group 14 (period 3), so non-metallic character decreases from C to Si: C > Si. Si (electronegativity 1.8) is less non-metallic than C (2.5).
3. Compare B and Si: B is a second-period element with electronegativity 2.0 and Δ_iH₁ 801 kJ mol⁻¹, whereas Si is a third-period element with electronegativity 1.8 and Δ_iH₁ 786 kJ mol⁻¹. B (a small atom of period 2) has the stronger tendency to attract electrons, so B is more non-metallic than Si. (B and Si are a diagonal pair and both are metalloids, but B is the more non-metallic.)
4. Final order: F > N > C > B > Si. Options (a) and (b) place F or N too low; (d) puts Si above B, contradicting the electronegativity data.
5. Answer: option (c).

Extra depth. Electronegativity is the quickest guide: non-metallic character is directly related to electronegativity (section 3.7.1(e)): 4.0 > 3.0 > 2.5 > 2.0 > 1.8. The B/Si comparison is the tricky step: moving down a group lowers non-metallic character (B → Al) while moving right raises it (Al → Si); the two effects roughly cancel along this diagonal, so the numerical data decide.

Conclusion: The order of non-metallic character is F > N > C > B > Si, so option (c) is correct.

Concept. An oxidising agent is a species that accepts electrons. For non-metals, oxidising power therefore follows the tendency to gain electrons, i.e. it increases with electronegativity and non-metallic character, which increase across a period and decrease down a group.

1. Electronegativities (Pauling scale): F = 4.0, O = 3.5, Cl = 3.0, N = 3.0. In the second period, oxidising power increases N < O < F (left to right).
2. F is the strongest: highest electronegativity, smallest size and highest effective nuclear charge among the four; it is the most powerful oxidising agent of all elements.
3. O vs Cl: oxygen (3.5) is more electronegative than chlorine (3.0) and is the second best electron acceptor in this set; hence O > Cl.
4. Cl vs N: both have electronegativity 3.0, but Cl has a very negative electron gain enthalpy (−349 kJ mol⁻¹), whereas the half-filled 2p³ configuration of N makes its electron gain enthalpy much less favourable (practically zero or slightly positive). Thus Cl > N.
5. Final order: F > O > Cl > N.
6. Answer: option (b). Option (a) puts Cl above O; (c) puts Cl above F; (d) puts O above F and N above Cl.

Extra depth. This ranking follows the periodic-trend reasoning of NCERT section 3.7.3: reactivity of non-metals rises towards the top right (Ex 3.28, F > Cl > Br > I). It is the same logic that places the halogens and chalcogens as the typical oxidising agents. In real redox chemistry the oxidising power also depends on bond enthalpies and hydration enthalpies (for example F₂ has a weak F–F bond), which you will meet in Chapter 7 (Redox Reactions).

Conclusion: The oxidising power of the elements decreases in the order F > O > Cl > N, so option (b) is correct.