Unit 3: Classification of Elements and Periodicity in Properties

---

The Big Picture: Why We Need to Classify Elements

In Unit 1 and Unit 2, we discovered that chemical matter is constructed from subatomic particles (electrons, protons, and neutrons) arranged in quantized quantum orbitals ($s, p, d, f$). However, when examining macroscopic chemistry, a formidable challenge arises:

• In the year 1800, only 31 chemical elements were known.
• By 1865, the number of discovered elements had more than doubled to 63.
• Today, 118 elements are experimentally confirmed, ranging from naturally occurring hydrogen ($Z = 1$) and uranium ($Z = 92$) to artificially synthesized superheavy transuranium elements such as oganesson ($Z = 118$).

If a chemist had to memorize the specific reactions, boiling points, densities, and chemical valencies of 118 separate elements and their millions of binary, ternary, and coordination compounds individually, the study of chemistry would dissolve into an unmanageable catalogue of disconnected facts.

The development of the Periodic Table is one of the grandest intellectual achievements in the history of science. It transforms what could have been an chaotic list of isolated facts into an elegant, predictive framework. Elements having similar valence shell electronic configurations are grouped together into families, exhibiting regular, predictable gradations in physical and chemical behavior across rows and columns.

Mastering this chapter equips you with the foundational intuition required to predict atomic size, ionization energetics, electron affinities, electronegativities, metallic character, and chemical reactivity across the entire periodic landscape.

---

1. Genesis of Periodic Classification

The development of the Modern Periodic Table was not an overnight discovery. It was the cumulative consequence of decades of systematizing empirical observations made by pioneering chemists seeking order within atomic masses and chemical affinities.

1.1 Dobereiner's Triads (1829)

In the early 1800s, the German chemist Johann Wolfgang Dobereiner was the first to identify systematic numerical relationships among the atomic weights and chemical properties of elements. By 1829, he grouped elements with similar chemical properties into clusters of three, termed Triads.

Limitations of Dobereiner's Triads

Dobereiner's classification could only be established for a very small number of elements known at the time. Several elements discovered subsequently did not fit into any triad (for instance, the halogen fluorine, with atomic weight $19.0$, could not form a valid triad with chlorine and bromine because $(19.0 + 80.0)/2 = 49.5$, far from chlorine's weight of $35.5$). Consequently, the relationship was dismissed by contemporaries as a numerical coincidence.

---

1.2 De Chancourtois's Telluric Helix (1862)

In 1862, the French geologist A.E. Beguyer de Chancourtois arranged the known chemical elements in order of increasing atomic weights along a spiral cylinder inscribed at an angle of $45^\circ$ to its base (the Telluric Helix or cylinder of elements).

He observed that elements possessing similar chemical properties aligned on the same vertical generating line of the cylinder at regular pitch intervals of 16 units of atomic weight. Although de Chancourtois was the first to observe periodic recurrence of properties, his paper included geological terminology rather than chemical concepts and lacked a clear visual chart, causing it to be overlooked by mainstream chemists.

---

1.3 Newlands' Law of Octaves (1865)

In 1865, the English chemist John Alexander Reina Newlands arranged all 56 known elements in increasing order of their atomic weights and noticed a striking pattern:

Limitations of Newlands' Law of Octaves

1. Valid Only Up to Calcium: The octave recurrence was strictly valid only for lighter elements up to calcium ($Z = 20$). Beyond calcium, every eighth element failed to show resemblance with the preceding one (for example, iron was placed in the same column as oxygen and sulfur despite having entirely different properties).
2. Assumption of Fixed Number of Elements: Newlands assumed that only 56 elements existed in nature and that no new elements would be discovered.
3. Improper Pairing: To fit elements into his table, Newlands placed dissimilar elements together in the same slot (such as placing cobalt and nickel together under the halogens fluorine and chlorine).

Although initially ridiculed by the Chemical Society of London, Newlands' fundamental insight regarding the periodicity of elements was later vindicated, and he was awarded the prestigious Davy Medal in 1887 by the Royal Society.

---

1.4 Lothar Meyer's Periodic Atomic Volume Curves (1869)

Working independently, the German chemist Julius Lothar Meyer plotted the physical properties of elements (such as atomic volume, melting point, and boiling point) against their increasing atomic weights.

Lothar Meyer obtained a periodically undulating curve:

• Sharp Peaks: Occupied by strongly electropositive alkali metals ($mathrm{Li}, mathrm{Na}, mathrm{K}, mathrm{Rb}, mathrm{Cs}$), indicating their exceptionally large atomic volumes.
• Descending Slopes: Occupied by alkaline earth metals ($mathrm{Be}, mathrm{Mg}, mathrm{Ca}, mathrm{Sr}, mathrm{Ba}$), which have smaller atomic volumes.
• Ascending Slopes: Occupied by electronegative halogens ($mathrm{F}, mathrm{Cl}, mathrm{Br}, mathrm{I}$).
• Troughs (Bottom Minima): Occupied by dense transition metals ($mathrm{Fe}, mathrm{Co}, mathrm{Ni}, mathrm{Cu}$, etc.) with high densities and melting points.

Unlike Newlands, Lothar Meyer observed that the length of the repeating periods varied (starting with short periods of 7 elements and expanding into longer periods). By 1868, he had drafted a comprehensive periodic table closely resembling the modern layout. However, his detailed work was published only after Dmitri Mendeleev's paper appeared in 1869.

---

1.5 Mendeleev's Periodic Table (1869 to 1905)

The Russian chemist Dmitri Ivanovich Mendeleev formulated the first comprehensive, universally accepted periodic system. Mendeleev went beyond physical properties, concentrating on the fundamental chemical behavior and empirical stoichiometries of the compounds formed by elements with oxygen (oxides) and hydrogen (hydrides).

Mendeleev arranged the known 63 elements in horizontal rows (series) and vertical columns (groups) in order of increasing atomic weights, ensuring that elements with identical chemical valence and empirical formula of their highest oxides ($R_2mathrm{O}, Rmathrm{O}, R_2mathrm{O}_3, Rmathrm{O}_2, R_2mathrm{O}_5, Rmathrm{O}_3, R_2mathrm{O}_7, Rmathrm{O}_4$) and hydrides ($Rmathrm{H}, Rmathrm{H}_2, Rmathrm{H}_3, Rmathrm{H}_4$) occupied the same vertical group.

Masterful Features of Mendeleev's Classification

1. Bold Gaps for Undiscovered Elements: Mendeleev recognized that the sequence of known elements was incomplete. Instead of forcing elements into ill-fitting slots, he left deliberate blanks, predicting not only the existence of undiscovered elements but also their quantitative atomic weights and physical properties.
   • For a gap under aluminium, he predicted Eka-Aluminium (discovered in 1875 by Lecoq de Boisbaudran as Gallium).
   • For a gap under silicon, he predicted Eka-Silicon (discovered in 1886 by Clemens Winkler as Germanium).
   • For a gap under boron, he predicted Eka-Boron (discovered in 1879 by Lars Nilson as Scandium).

The astounding precision of Mendeleev's quantitative forecasts established the worldwide authority of his Periodic Law:

1. Prioritizing Chemical Similarity over Strict Atomic Weight: Mendeleev realized that experimental atomic weights might contain measurement errors or that chemical periodicity superseded strict mass order. When a conflict arose, he placed elements where their chemical properties dictated:
   • Tellurium ($mathrm{Te} = 127.6$) and Iodine ($mathrm{I} = 126.9$): Strict atomic weight ordering would place iodine before tellurium. Mendeleev placed tellurium in Group VI (under sulfur and selenium) and iodine in Group VII (under chlorine and bromine) because iodine is chemically a halogen.
   • Cobalt ($mathrm{Co} = 58.9$) and Nickel ($mathrm{Ni} = 58.7$): Cobalt was placed before nickel to preserve the grouping of cobalt with rhodium and iridium.
   • Argon ($mathrm{Ar} = 39.9$) and Potassium ($mathrm{K} = 39.1$): Argon was placed before potassium in Group 0 when noble gases were discovered by Lord Rayleigh and William Ramsay in the 1890s.

Fundamental Defects of Mendeleev's Periodic Table

1. Anomalous Position of Hydrogen: Hydrogen exhibits dual characteristics: like alkali metals, it forms $+1$ ions ($H^+$) and halides ($Hmathrm{Cl}$), but like halogens, it is a diatomic gas ($H_2$) that gains an electron to form hydrides ($mathrm{Na}H$). Its placement in Group I was never fully justified.
2. Inverted Pairs of Atomic Weights: The pairs $mathrm{Ar}-mathrm{K}$, $mathrm{Co}-mathrm{Ni}$, and $mathrm{Te}-mathrm{I}$ violated the increasing order of atomic weights without a theoretical rationale under mass-based laws.
3. Placement of Isotopes: When isotopes were discovered (atoms of the same element having identical chemical properties but different atomic masses, such as $^{35}mathrm{Cl}$ and $^{37}mathrm{Cl}$), Mendeleev's law required them to occupy separate positions in the table, which was chemically untenable.
4. Dissimilar Elements Grouped Together: Chemically divergent elements were grouped together under sub-groups: highly reactive alkali metals ($mathrm{Li}, mathrm{Na}, mathrm{K}$) were placed in Group I alongside unreactive coinage metals ($mathrm{Cu}, mathrm{Ag}, mathrm{Au}$).

---

2. Modern Periodic Law and the Long Form of the Periodic Table

2.1 Moseley's Breakthrough: Atomic Number as the Fundamental Property (1913)

When Mendeleev published his table, subatomic particles were unknown. In 1913, the English physicist Henry Moseley bombarded various metallic targets with high-energy cathode ray electron beams and recorded the characteristic X-ray emission spectra produced.

Moseley found a profound mathematical relationship between the frequency ($\nu$) of the characteristic $K_\alpha$ X-ray lines and the atomic number ($Z$, representing the total nuclear positive charge):

$\sqrt{\nu} = a(Z - b)$

Where:

• $\nu$ is the frequency of the emitted characteristic X-ray line.
• $Z$ is the atomic number (number of protons in the atomic nucleus).
• $a$ and $b$ are empirical constants specific to the spectral series ($b = 1$ for the $K_\alpha$ series).

This proved conclusively that atomic number ($Z$), which corresponds to the nuclear charge, is the fundamental property of an element, not atomic weight ($A$).

Inverted pairs that troubled Mendeleev were resolved immediately:

• Argon ($Z = 18$, $A = 39.95$) naturally precedes Potassium ($Z = 19$, $A = 39.10$).
• Cobalt ($Z = 27$, $A = 58.93$) naturally precedes Nickel ($Z = 28$, $A = 58.69$).
• Tellurium ($Z = 52$, $A = 127.60$) naturally precedes Iodine ($Z = 53$, $A = 126.90$).

Because the atomic number ($Z$) in a neutral atom equals the number of electrons, the Modern Periodic Law reveals that chemical periodicity is the direct manifestation of the periodic variation in the ground-state electronic configurations of atoms.

---

2.2 Structural Architecture of the Long Form of the Periodic Table

The most widely adopted graphical representation of the Modern Periodic Law is the Long Form (Bohr's Table) of the Periodic Table, derived directly from the Aufbau principle and quantum mechanics:

1. Periods (Horizontal Rows):
   • There are 7 horizontal periods, numbered from 1 to 7.
   • The period number corresponds to the highest principal quantum number ($n$) of the electrons in that element's atoms.
2. Groups (Vertical Columns):
   • There are 18 vertical columns, designated as Groups 1 to 18 under the standardized IUPAC notation (replacing the older Roman numeral designations $mathrm{IA-VIIA}, mathrm{VIII}, mathrm{IB-VIIB}, 0$).
   • Elements within the same group share identical valence shell electronic configurations, having the same number and spatial symmetry of outer valence electrons.
3. Separate Inner-Transition Panels (f-Block):
   • Two separate 14-element series are placed at the bottom of the table to preserve a compact, manageable layout:
     • Lanthanoids: Elements from Cerium ($Z = 58$) to Lutetium ($Z = 71$) where the $4f$ subshell fills.
     • Actinoids: Elements from Thorium ($Z = 90$) to Lawrencium ($Z = 103$) where the $5f$ subshell fills.
   • Glenn T. Seaborg won the Nobel Prize in Chemistry in 1951 for synthesizing the transuranium elements ($Z = 94$ to $102$) and recognizing that actinoids represent a second $f$-electron inner transition series placed below the lanthanoids. In his honor, Element 106 was officially named Seaborgium (Sg).

---

3. IUPAC Nomenclature for Elements with Atomic Number $Z > 100$

When superheavy transuranic elements ($Z > 100$) are synthesized in particle accelerators, only a few fleeting atoms are produced per week or month. Historically, this led to priority disputes between competing American laboratories (such as Lawrence Berkeley National Laboratory) and Soviet laboratories (Joint Institute for Nuclear Research at Dubna):

• For Element 104, American scientists named it Rutherfordium (Rf), while Soviet scientists named it Kurchatovium (Ku).
• For Element 106, American scientists proposed Seaborgium (Sg), which was initially contested because Seaborg was still living.

To prevent nationalistic disputes and provide unambiguous terminology before official IUPAC ratification, IUPAC established a systematic nomenclature derived directly from the digits of the atomic number.

3.1 Numerical Roots and Suffix Rules

The systematic name is assembled by combining the numerical roots corresponding to the digits of the atomic number in sequential order, followed by the universal suffix "-ium":

Rules for Combining Roots:

1. If enn is followed by nil, the double 'n' is compressed into a single 'n' (e.g., enn + nil = ennil).
2. If bi or tri is followed by the suffix ium, the duplicate 'i' is dropped (e.g., bi + ium = bium, tri + ium = trium).
3. The three-letter systematic symbol is formed from the capitalized first letter of each numerical root (e.g., $104 \to$ u + n + q = Unq).

---

3.2 Complete IUPAC Superheavy Elements Registry ($Z = 101$ to $118$)

---

---

4. Electronic Configurations and the Periodic Table

The architecture of the Periodic Table is a physical consequence of the four quantum numbers ($n, l, m_l, m_s$) and the order of filling orbitals determined by the Aufbau principle ($(n + l)$ rule).

4.1 Electronic Configurations in Periods

The period number corresponds to the principal quantum number ($n$) of the valence shell. The total number of elements in any period equals twice the number of atomic orbitals available in that energy level (due to Pauli's Exclusion Principle allowing two electrons per orbital with opposite spins):

---

---

4.2 Groupwise Electronic Configurations

Elements belonging to the same vertical column share identical outer valence electron arrangements, accounting for their remarkably similar chemical properties:

---

5. Electronic Configurations and Division into Four Blocks ($s, p, d, f$)

Depending on the azimuthal quantum number ($l$) of the last subshell receiving the differentiating valence electron, all chemical elements are classified into four distinct structural blocks:

5.1 The s-Block Elements (Groups 1 and 2)

• Composition: Group 1 (Alkali Metals: $mathrm{Li}, mathrm{Na}, mathrm{K}, mathrm{Rb}, mathrm{Cs}, mathrm{Fr}$) and Group 2 (Alkaline Earth Metals: $mathrm{Be}, mathrm{Mg}, mathrm{Ca}, mathrm{Sr}, mathrm{Ba}, mathrm{Ra}$).
• Outer Valence Configuration: $ns^1$ for Group 1; $ns^2$ for Group 2.
• Physical Characteristics: Silvery-white, highly malleable, ductile, soft metals with low melting and boiling points.
• Energetics and Reactivity: Exceptionally low first ionization enthalpies that decrease down each group. They lose outer $s$-electrons readily to form univalent cations ($M^+$) and divalent cations ($M^{2+}$).
• Bonding Nature: Exclusively form ionic lattices (with notable covalent exceptions for lithium and beryllium salts due to high polarization power).
• Natural Abundance: Due to extreme chemical reactivity, they never occur uncombined in the free native state in nature.

---

5.2 The p-Block Elements (Groups 13 to 18)

• Composition: Groups 13 through 18. Together with the $s$-block elements, they are formally designated as the Representative Elements or Main Group Elements.
• Outer Valence Configuration: Ranges from $ns^2 np^1$ (Group 13: Boron Family) to $ns^2 np^6$ (Group 18: Noble Gases).
• Noble Gases (Group 18): Possess a completely closed octet configuration ($ns^2 np^6$, or $1s^2$ for Helium). Due to closed shell stabilization and large positive electron gain enthalpies, they exhibit negligible chemical reactivity.
• Halogens (Group 17) and Chalcogens (Group 16): Possess highly negative electron gain enthalpies, readily gaining one or two electrons to complete a stable noble gas configuration.
• Metallic to Non-Metallic Transition: Moving from left to right across a period, non-metallic character increases sharply; moving down a group, metallic character increases.

---

5.3 The d-Block Elements: Transition Elements (Groups 3 to 12)

• Composition: Groups 3 to 12 located in the central portion of the table. Characterized by the progressive filling of inner $(n-1)d$ orbitals.
• General Electronic Configuration: $(n - 1)d^{1-10} ns^{0-2}$.
  • Notable anomalies include Chromium ($[mathrm{Ar}] 3d^5 4s^1$), Copper ($[mathrm{Ar}] 3d^{10} 4s^1$), and Palladium ($[mathrm{Kr}] 4d^{10} 5s^0$) due to exchange energy stabilization of half-filled and completely filled subshells.
• Characteristic Transition Properties:
  1. Variable Oxidation States: Because $(n-1)d$ and $ns$ electrons have very close energies, both can participate in chemical bond formation (e.g., Manganese displays oxidation states from $+2$ to $+7$).
  2. Coloured Complex Ions: Unpaired $d$-electrons undergo $d-d$ electronic transitions upon absorbing specific wavelengths in the visible spectrum.
  3. Paramagnetism: Unpaired spins generate net permanent magnetic dipole moments.
  4. Catalytic Activity: Variable valence and available vacant $d$-orbitals provide reaction pathways with lower activation energies (e.g., Finely divided $mathrm{Fe}$ in the Haber process; $mathrm{V}_2mathrm{O}_5$ in the Contact process).
• The Pseudo-Transition Metals ($mathrm{Zn}, mathrm{Cd}, mathrm{Hg}$): Possess a completely filled $(n-1)d^{10} ns^2$ ground-state electronic configuration both as neutral atoms and in their common $+2$ oxidation states. They do not exhibit typical transition properties (they are softer with lower melting points: mercury is a liquid at room temperature).

---

5.4 The f-Block Elements: Inner-Transition Elements

• Composition: Two horizontal 14-element series placed beneath the main table:
  1. Lanthanoids ($4f$ series): Cerium ($Z = 58$) to Lutetium ($Z = 71$), outer configuration: $[\mathrm{Xe}] 4f^{1-14} 5d^{0-1} 6s^2$.
  2. Actinoids ($5f$ series): Thorium ($Z = 90$) to Lawrencium ($Z = 103$), outer configuration: $[\mathrm{Rn}] 5f^{1-14} 6d^{0-1} 7s^2$.
• Chemical Behavior: All are electropositive metals. Lanthanoids predominantly exhibit the stable $+3$ oxidation state. The chemistry of actinoids is far more intricate because $5f, 6d,$ and $7s$ energy levels have comparable energies, allowing oxidation states spanning from $+3$ to $+7$.
• Radioactivity: All actinoid elements are radioactive. Elements beyond uranium ($Z > 92$) do not occur naturally in appreciable quantities and are synthesized artificially through nuclear bombardment reactions (Transuranium Elements).

---

5.5 Classification into Metals, Non-Metals, and Metalloids

Beyond subshell configurations, elements are classified into three major physical categories:

1. Metals: Comprise more than $78\%$ of all known chemical elements, occupying the left, center, and bottom regions of the periodic table. They are characterized by high electrical and thermal conductivity, metallic luster, malleability, ductility, and high tensile strength. They have high melting and boiling points (with exceptions such as mercury which melts at $234\text{ K}$, and gallium and caesium which melt near human body temperature: $303\text{ K}$ and $302\text{ K}$).
2. Non-Metals: Less than 20 elements, concentrated in the upper right-hand corner of the table. Usually brittle solids or gases at room temperature (bromine is the sole liquid non-metal). They are poor electrical and thermal conductors with low melting and boiling points (carbon in diamond/graphite and boron are notable exceptions).
3. Metalloids (Semi-Metals): Elements running diagonally along the boundary separating metals from non-metals: Silicon ($mathrm{Si}$), Germanium ($mathrm{Ge}$), Arsenic ($mathrm{As}$), Antimony ($mathrm{Sb}$), and Tellurium ($mathrm{Te}$). They display dual physical and chemical characteristics, acting as intrinsic semiconductors foundational to solid-state microelectronics.

---

---

---

---

6. Periodic Trends in Physical Properties

Periodicity arises because atomic physical properties depend on two competing quantum-mechanical factors:

1. Attractive Electrostatic Pull of the Nucleus: Proportional to the effective nuclear charge ($Z_{\text{eff}}$).
2. Repulsive Screening / Shielding Effect ($sigma$): Inner core electron clouds screen the outermost valence electrons from the full positive nuclear charge: $Z_{\text{eff}} = Z - \sigma$

---

6.1 Atomic Radius

Unlike a macroscopic sphere, an isolated atom has no sharp, rigid physical boundary because the electron probability wave function $|\psi|^2$ decays asymptotically toward infinity. Therefore, atomic size is determined experimentally by measuring the internuclear equilibrium separation distance between bound atoms:

1. Covalent Radius ($r_{\text{cov}}$): Half of the equilibrium internuclear distance between two identical non-metallic atoms joined by a single covalent bond in a homonuclear diatomic molecule: $r_{\text{cov}} = \frac{1}{2} d_{A-A}$ Example: The internuclear bond length of a chlorine molecule ($mathrm{Cl}_2$) is $198\text{ pm}$. Thus, the covalent radius of chlorine is: $r_{\text{cov}}(\mathrm{Cl}) = \frac{198\text{ pm}}{2} = 99\text{ pm}$
2. Metallic Radius ($r_{\text{met}}$): Half of the internuclear distance separating two adjacent metal cation cores in a close-packed metallic crystal lattice: Example: The distance between two adjacent copper atoms in solid crystalline copper is $256\text{ pm}$. Thus, the metallic radius of copper is: $r_{\text{met}}(\mathrm{Cu}) = \frac{256\text{ pm}}{2} = 128\text{ pm}$
3. Van der Waals Radius ($r_{\text{vdw}}$): Half of the distance between the nuclei of two non-bonded adjacent atoms belonging to neighboring molecules in the solid state. Because Van der Waals forces are weak dispersion forces, atoms are held further apart: $r_{\text{vdw}} > r_{\text{met}} > r_{\text{cov}}$ Example: The Van der Waals radius of chlorine is $180\text{ pm}$, significantly larger than its covalent radius of $99\text{ pm}$.

Quantitative NCERT Atomic Radii Data ($pm$)

Periodic Trends and Atomic Mechanisms

1. Across a Period (Decreases from Left to Right): Successive electrons are added into the same principal quantum shell ($n$). However, each step increases the atomic number $Z$ by $+1$, increasing the positive nuclear charge. Because electrons in the same subshell provide poor mutual screening, effective nuclear charge ($Z_{\text{eff}}$) increases, pulling the valence electron clouds closer to the nucleus.
2. Down a Group (Increases from Top to Bottom): Moving down a group introduces an entirely new principal quantum shell ($n = 1 \to 2 \to 3 \to 4 \dots$). The additional intervening electron shells provide strong screening that offsets the increased nuclear charge. The valence electrons are situated farther from the nucleus, resulting in an expansion in atomic size.
3. Noble Gas Exception: Noble gases do not form single covalent homonuclear molecules under normal conditions. Their radii are measured as non-bonded Van der Waals radii, making their listed values appear larger than the preceding halogens (for example, Neon has an apparent radius of $160\text{ pm}$, compared to Fluorine's covalent radius of $64\text{ pm}$).

---

6.2 Ionic Radius

An ion is formed when an atom loses or gains electrons:

• Cations ($M^+$): Formed by the loss of one or more valence electrons. A cation is always smaller than its parent neutral atom because the removal of electrons often strips away the entire outermost valence shell. Furthermore, the remaining fewer electrons experience the same full positive nuclear charge, resulting in an increased $Z_{\text{eff}}$ that contracts the remaining electron cloud. Example: $\mathrm{Na} (186\text{ pm}) \xrightarrow{-e^-} \mathrm{Na}^+ (95\text{ pm})$.
• Anions ($X^-$): Formed by the gain of electrons. An anion is always larger than its parent neutral atom because the addition of electrons increases inter-electronic electrostatic repulsion within the valence shell without changing the nuclear charge, causing the electron cloud to expand outward. Example: $\mathrm{F} (64\text{ pm}) \xrightarrow{+e^-} \mathrm{F}^- (136\text{ pm})$.

Isoelectronic Series

Isoelectronic species are atoms and ions that contain the exact same total number of electrons and identical ground-state electronic configurations:

Consider the 10-electron isoelectronic series ($1s^2 2s^2 2p^6$): $\mathrm{N}^{3-}, \quad \mathrm{O}^{2-}, \quad \mathrm{F}^-, \quad \mathrm{Na}^+, \quad \mathrm{Mg}^{2+}, \quad \mathrm{Al}^{3+}$

---

---

6.3 Ionization Enthalpy ($\Delta_i H$)

$X(g) \xrightarrow{\Delta_i H_1} X^+(g) + e^-$

• Units: Expressed in kilojoules per mole ($ext{kJ/mol}$) or electron-volts per atom ($ext{eV/atom}$, where $1\text{ eV/atom} = 96.485\text{ kJ/mol}$).
• Thermodynamic Sign: Because work must always be done against the electrostatic attraction of the positive nucleus to overcome the binding potential, ionization enthalpies are always strictly positive (endothermic): $\Delta_i H > 0$.

Successive Ionization Enthalpies

Removing subsequent electrons from positive ions requires increasingly higher quantities of energy: $X^+(g) \xrightarrow{\Delta_i H_2} X^{2+}(g) + e^- \quad (\text{Second Ionization Enthalpy})$ $X^{2+}(g) \xrightarrow{\Delta_i H_3} X^{3+}(g) + e^- \quad (\text{Third Ionization Enthalpy})$

---

Quantitative First Ionization Enthalpies ($\Delta_i H_1$ in $\text{kJ/mol}$)

---

Fundamental Governing Factors Determining Ionization Enthalpy

1. Size of the Atom: Larger atomic radius $\implies$ valence electron is farther from the nucleus $\implies$ weaker electrostatic attraction $\implies$ lower $\Delta_i H$.
2. Nuclear Charge ($Z$): Higher positive nuclear charge $\implies$ stronger Coulomb attraction $\implies$ higher $\Delta_i H$.
3. Screening / Shielding Effect ($sigma$): Greater number of inner core electrons $\implies$ valence electron is shielded from the nuclear charge $\implies$ lower $\Delta_i H$.
4. Penetration Effect of Orbitals: For the same principal quantum shell ($n$), electrons in subshells that spend more time close to the nucleus penetrate inner shells more effectively: $\text{Penetration Power: } s > p > d > f$ Removing an $s$-electron requires more energy than removing a $p$-electron from the same shell.
5. Electronic Configuration Stability: Completely filled ($s^2, p^6, d^{10}$) and half-filled ($p^3, d^5$) subshells possess enhanced exchange energy stabilization and spherical symmetry, giving them unusually high ionization enthalpies.

---

Crucial NCERT Ionization Enthalpy Anomalies

Anomaly 1: Why is $\Delta_i H_1$ of Boron ($801\text{ kJ/mol}$) LOWER than Beryllium ($899\text{ kJ/mol}$)?

• Beryllium ($Z = 4$): Ground state configuration $[\mathrm{He}] 2s^2$. The electron to be removed resides in a fully filled, spherically symmetric $2s$ orbital.
• Boron ($Z = 5$): Ground state configuration $[\mathrm{He}] 2s^2 2p^1$. The electron to be removed resides in a $2p$ orbital.
• The $2s$ orbital penetrates closer to the nucleus than the $2p$ orbital. The $2p$ electron of boron is shielded from the nucleus by both the inner $1s^2$ core and the two $2s^2$ electrons. Consequently, the single $2p$ electron of boron experiences lower effective attraction and is removed more easily than the $2s$ electron of beryllium.

Anomaly 2: Why is $\Delta_i H_1$ of Oxygen ($1314\text{ kJ/mol}$) LOWER than Nitrogen ($1402\text{ kJ/mol}$)?

• Nitrogen ($Z = 7$): Electronic configuration $[\mathrm{He}] 2s^2 2p_x^1 2p_y^1 2p_z^1$. By Hund's rule, nitrogen possesses a symmetrically stable, half-filled $2p$ subshell, which maximizes quantum-mechanical exchange stabilization.
• Oxygen ($Z = 8$): Electronic configuration $[\mathrm{He}] 2s^2 2p_x^2 2p_y^1 2p_z^1$. In oxygen, two of the four $2p$ electrons must occupy the same $2p_x$ orbital, introducing electron-electron pairing repulsion. This repulsion destabilizes the electron pair, making it energetically easier to eject the fourth $2p$ electron compared to disrupting nitrogen's stable half-filled shell.

---

---

6.4 Electron Gain Enthalpy ($\Delta_{eg}H$)

$X(g) + e^- \xrightarrow{\Delta_{eg}H} X^-(g)$

Thermodynamic Sign and Conventions

• Exothermic (Negative $\Delta_{eg}H$): For most elements (especially non-metals), energy is released upon electron addition due to attractive nuclear pull. The more negative the value, the higher the electron affinity of the atom.
• Endothermic (Positive $\Delta_{eg}H$): For elements with exceptionally stable closed configurations (noble gases and alkaline earth metals), the added electron must enter an entirely new higher-energy subshell, requiring energy input ($Delta_{eg}H > 0$).
• Successive Electron Gain Enthalpy: While the first electron gain enthalpy can be negative, the second electron gain enthalpy ($\Delta_{eg}H_2$) is always strongly positive (endothermic) because an incoming negative electron experiences severe electrostatic repulsion from the existing negative anion: $\mathrm{O}(g) + e^- \to \mathrm{O}^-(g), \quad \Delta_{eg}H_1 = -141\text{ kJ/mol} \quad (\text{Exothermic})$ $\mathrm{O}^-(g) + e^- \to \mathrm{O}^{2-}(g), \quad \Delta_{eg}H_2 = +780\text{ kJ/mol} \quad (\text{Endothermic})$

Quantitative Electron Gain Enthalpies ($\Delta_{eg}H$ in $\text{kJ/mol}$)

Key Periodic Trends and The Second-Period Anomaly

1. Across a Period: Becomes generally more negative from left to right as $Z_{\text{eff}}$ increases and atomic radius decreases, reaching peak negative values in Group 17 halogens before jumping to positive values in noble gases.
2. Down a Group: Generally becomes less negative down a group because atomic radius expands, placing the incoming electron further from the nucleus.
3. The Crucial Anomaly (Chlorine vs Fluorine, Sulfur vs Oxygen):
   • Notice from the table that $\Delta_{eg}H$ of Chlorine ($-349\text{ kJ/mol}$) is more negative than Fluorine ($-328\text{ kJ/mol}$).
   • Similarly, $\Delta_{eg}H$ of Sulfur ($-200\text{ kJ/mol}$) is more negative than Oxygen ($-141\text{ kJ/mol}$).
   • Physical Reason: In Fluorine ($n = 2$), the added electron enters a extremely compact, dense $2p$ subshell where it experiences intense inter-electronic repulsion from the other seven valence electrons. In Chlorine ($n = 3$), the electron enters a larger, more diffuse $3p$ orbital where electron-electron repulsion is significantly lower. Consequently, Chlorine accepts an added electron with greater net energy release than Fluorine.

---

---

6.5 Electronegativity

Unlike ionization enthalpy or electron gain enthalpy, electronegativity is not a directly measurable thermodynamic property of an isolated atom. It is a dimensionless comparative property manifested only when an atom is bonded within a molecule.

Major Scales of Electronegativity

1. Pauling Scale (Linus Pauling, 1932): The most widely used scale, based on bond dissociation enthalpy differences: $\Delta = D_{A-B} - \sqrt{D_{A-A} \times D_{B-B}}$ $\chi_A - \chi_B = 0.208 \sqrt{\Delta} \quad (\text{where } \Delta \text{ is in kcal/mol})$ Pauling arbitrarily assigned the value of $4.0$ to Fluorine, the most electronegative element.
2. Mulliken-Jaffe Scale: Proposes that electronegativity is the average of an atom's ionization enthalpy and electron affinity: $\chi_M = \frac{\Delta_i H + A_e}{2}$
3. Allred-Rochow Scale: Evaluates the electrostatic force exerted by the effective nuclear charge on an electron at the covalent radius surface: $\chi_{AR} = 0.359 \frac{Z_{\text{eff}}}{r_{\text{cov}}^2} + 0.744$

Quantitative Electronegativity Values (Pauling Scale)

Correlation with Metallic and Non-Metallic Properties

• Increases Across a Period: As atomic radius decreases and $Z_{\text{eff}}$ increases, bonding electrons are attracted more strongly. Thus, non-metallic character increases across a period.
• Decreases Down a Group: As atomic radius expands, the positive nucleus is shielded from the bonding electron pair, diminishing electronegativity. Thus, metallic character increases down a group.

---

---

7. Periodic Trends in Chemical Properties

7.1 Periodicity of Valence and Oxidation States

The valence of an element represents its combining capacity with other atoms to form stable molecules. For representative elements, the valence is dictated by the outer shell electrons:

• For Groups 1, 2, 13, 14: Valence equals the number of valence electrons.
• For Groups 15, 16, 17, 18: Valence equals the number of valence electrons or $(8 - \text{valence electrons})$.

Modern Formulation: Oxidation State

Today, the broader term oxidation state is used. It represents the formal electric charge an atom would carry in a compound if all shared bonding electron pairs were assigned entirely to the more electronegative bonding partner.

Consider two compounds containing oxygen:

1. Oxygen Difluoride ($mathrm{OF}_2$):
   • Electronegativity order: $\mathrm{F} (4.0) > \mathrm{O} (3.5)$.
   • Fluorine is more electronegative than oxygen. Each fluorine atom is assigned an oxidation state of $-1$.
   • Oxygen shares two electrons with two fluorine atoms, giving oxygen an oxidation state of $+2$.
2. Sodium Oxide ($mathrm{Na}_2mathrm{O}$):
   • Electronegativity order: $\mathrm{O} (3.5) > \mathrm{Na} (0.9)$.
   • Oxygen is more electronegative than sodium, gaining two electrons to yield an oxidation state of $-2$.
   • Each sodium atom loses its $3s^1$ electron, giving sodium an oxidation state of $+1$.

Hydride and Oxide Stoichiometries Across Groups

---

---

7.2 Anomalous Properties of Second Period Elements and Diagonal Relationships

The first element of each group in the Second Period ($mathrm{Li}, mathrm{Be}, mathrm{B}, mathrm{C}, mathrm{N}, mathrm{O}, mathrm{F}$) exhibits distinct differences in physical and chemical behavior compared to the subsequent heavier congeners ($mathrm{Na}, mathrm{Mg}, mathrm{Al}, mathrm{Si}, mathrm{P}, mathrm{S}, mathrm{Cl}$) in their respective groups:

Three Fundamental Physical Origins of Anomalous Second-Period Behavior:

1. Exceptionally Small Atomic and Ionic Size: Yields a very high ionic charge-to-radius ratio (polarizing power).
2. Unusually High Electronegativity and Ionization Enthalpy.
3. Strict Absence of Low-Lying d-Orbitals: The valence shell of Period 2 elements has $n = 2$, which possesses only four quantum orbitals ($2s, 2p_x, 2p_y, 2p_z$). Consequently, the maximum covalency of second-period elements is strictly capped at 4.
   • Boron can only form $[\mathrm{BF}_4]^-$ (maximum 4 electron pairs), whereas Aluminium utilizes vacant $3d$ orbitals to form $[\mathrm{AlF}_6]^{3-}$.
   • Nitrogen can form only $\mathrm{NF}_3$ and cannot form $\mathrm{NCl}_5$, whereas Phosphorus readily forms $\mathrm{PCl}_5$ and $[\mathrm{PCl}_6]^-$.
4. Strong Capacity for $p\pi-p\pi$ Multiple Bonding: Due to compact atomic size, second period elements exhibit effective lateral orbital overlap, forming stable multiple bonds with themselves and adjacent small atoms: $\mathrm{C}=\mathrm{C}, \quad \mathrm{C}\equiv\mathrm{C}, \quad \mathrm{N}\equiv\mathrm{N}, \quad \mathrm{C}=\mathrm{O}, \quad \mathrm{C}\equiv\mathrm{N}, \quad \mathrm{N}=\mathrm{O}$ Heavier congeners have diffuse, larger $3p$ or $4p$ orbitals that cannot overlap laterally with sufficient energy to form stable $p\pi-p\pi$ bonds (for example, elemental nitrogen is a diatomic gas with a triple bond $\mathrm{N}\equiv\mathrm{N}$, whereas elemental phosphorus forms a tetrahedral $\mathrm{P}_4$ network with single $\mathrm{P}-\mathrm{P}$ bonds).

---

Diagonal Relationships in the Periodic Table

A striking observation is that the first member of a group in Period 2 displays strong chemical and structural resemblances to the second element of the neighboring group in Period 3:

• Lithium ($mathrm{Li}$) diagonally resembles Magnesium ($mathrm{Mg}$).
• Beryllium ($mathrm{Be}$) diagonally resembles Aluminium ($mathrm{Al}$).
• Boron ($mathrm{B}$) diagonally resembles Silicon ($mathrm{Si}$).

---

---

7.3 Periodic Trends and Chemical Reactivity

Chemical reactivity across a period displays a distinctive "V-shaped" or parabolic profile:

1. Extreme Left (Group 1 Alkali Metals): High chemical reactivity driven by low ionization enthalpy and high electropositivity (loss of valence electrons to form stable cations).
2. Extreme Right (Group 17 Halogens): High chemical reactivity driven by high electronegativity and strongly negative electron gain enthalpy (gain of electrons to form stable anions).
3. Center of Periodic Table (Group 14 and Transition Metals): Reactivity reaches a minimum.
4. Group 18 Noble Gases: Chemically inert due to completely closed octet valence configurations.

Periodicity in the Acid-Base Character of Oxides

The chemical character of normal oxides across a period displays a continuous gradation from strongly basic to strongly acidic:

• Basic Oxides: Formed by electropositive elements on the extreme left ($mathrm{Na}_2mathrm{O}, mathrm{CaO}, mathrm{BaO}$). Dissolve in water to yield basic hydroxides ($mathrm{OH}^-$ ions).
• Acidic Oxides: Formed by electronegative elements on the right ($mathrm{SO}_3, mathrm{CO}_2, mathrm{Cl}_2mathrm{O}_7$). React with water to form oxoacids ($mathrm{H}^+$ ions).
• Amphoteric Oxides: Display dual reactivity, reacting with strong acids as bases and with strong bases as acids ($mathrm{Al}_2mathrm{O}_3, mathrm{ZnO}, mathrm{As}_2mathrm{O}_3$): $\mathrm{Al}_2\mathrm{O}_3(s) + 6\mathrm{HCl}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \to 2[\mathrm{Al}(\mathrm{H}_2\mathrm{O})_6]^{3+}(aq) + 6\mathrm{Cl}^-(aq) \quad (\text{Acts as Base})$ $\mathrm{Al}_2\mathrm{O}_3(s) + 2\mathrm{NaOH}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \to 2\mathrm{Na}[\mathrm{Al}(\mathrm{OH})_4](aq) \quad (\text{Acts as Acid})$
• Neutral Oxides: Exhibit neither acidic nor basic properties, failing to form salts with either acids or bases: Carbon Monoxide ($mathrm{CO}$), Nitric Oxide ($mathrm{NO}$), and Nitrous Oxide ($mathrm{N}_2mathrm{O}$).

---

---

---

8. Comprehensive NCERT Chapter Exercises (3.1 to 3.40)

Exercise 3.1

Question: What is the basic theme of organisation in the periodic table?
Step-by-Step Solution:
The basic theme of organization in the periodic table is to classify chemical elements into periods (rows) and groups (columns) based on similarities in their physical and chemical properties. In the Modern Periodic Table, this classification is achieved by arranging elements in order of increasing atomic numbers ($Z$), which directly reflects the systematic filling of electron subshells ($s, p, d, f$) and aligns elements with identical valence shell electronic configurations into the same vertical group.

---

Exercise 3.2

Question: Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Step-by-Step Solution:
Mendeleev used increasing atomic weight (relative atomic mass) as the primary organizing criterion for his classification. However, he did not adhere to it rigidly. Whenever a conflict arose between strict atomic mass ordering and chemical similarity, Mendeleev gave priority to chemical properties and empirical stoichiometries of oxides and hydrides. For instance, he placed Tellurium (atomic weight $127.6$) before Iodine (atomic weight $126.9$), and Cobalt ($58.9$) before Nickel ($58.7$), correctly predicting that chemical periodicity was more fundamental than measured atomic weight.

---

Exercise 3.3

Question: What is the basic difference in approach between Mendeleev's Periodic Law and the Modern Periodic Law?
Step-by-Step Solution:

1. Mendeleev's Periodic Law: Posits that the physical and chemical properties of elements are periodic functions of their atomic weights. It lacks an underlying explanation for inverted pairs and the positioning of isotopes.
2. Modern Periodic Law: Posits that the physical and chemical properties of elements are periodic functions of their atomic numbers ($Z$), representing total nuclear positive charge (number of protons). Because chemical properties depend on outer electronic configurations, the atomic number provides a rigorous quantum-mechanical foundation for periodicity.

---

Exercise 3.4

Question: On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Step-by-Step Solution:

• For the sixth period, the principal quantum number of the valence shell is $n = 6$.
• According to the Aufbau $(n + l)$ rule, the subshells available for electron filling in Period 6 are:
  • $6s$: $1$ orbital ($n+l = 6+0 = 6$)
  • $4f$: $7$ orbitals ($n+l = 4+3 = 7$)
  • $5d$: $5$ orbitals ($n+l = 5+2 = 7$)
  • $6p$: $3$ orbitals ($n+l = 6+1 = 7$)
• Total number of available orbitals: $N_{\text{orbitals}} = 1 (6s) + 7 (4f) + 5 (5d) + 3 (6p) = 16\text{ orbitals}$
• Since each orbital accommodates at most 2 electrons with paired spins (Pauli Exclusion Principle): $N_{\text{elements}} = 16 \times 2 = 32\text{ elements}$ Therefore, the sixth period accommodates exactly 32 elements (from Caesium, $Z = 55$, to Radon, $Z = 86$).

---

Exercise 3.5

Question: In terms of period and group where would you locate the element with $Z = 114$?
Step-by-Step Solution:

• The nearest preceding noble gas is Radon ($mathrm{Rn}$, $Z = 86$).
• Number of electrons beyond Radon: $114 - 86 = 28$ electrons.
• Filling sequence beyond $[mathrm{Rn}]$ in Period 7: $7s^2 \to 5f^{14} \to 6d^{10} \to 7p^2$
• The valence shell has $n = 7 \implies$ Period 7.
• The differentiating valence electrons are in the $p$-subshell ($7p^2$). For $p$-block elements, Group Number = $12 + \text{number of } p\text{-electrons} = 12 + 2 = \mathbf{14}$.
• Location: Period 7, Group 14 (Carbon family; element named Flerovium, $mathrm{Fl}$).

---

Exercise 3.6

Question: Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Step-by-Step Solution:

• Period 3 corresponds to principal quantum number $n = 3$.
• Group 17 (Halogens) corresponds to the valence shell configuration $ns^2 np^5$.
• For $n = 3$, the valence configuration is $3s^2 3p^5$.
• Preceding noble gas core is Neon ($[mathrm{Ne}]$, $Z = 10$).
• Total atomic number: $Z = 10 (\mathrm{Ne}) + 2 (3s) + 5 (3p) = 17$
• The element is Chlorine ($mathrm{Cl}$, $Z = 17$).

---

Exercise 3.7

Question: Which element do you think would have been named by: (i) Lawrence Berkeley Laboratory, (ii) Seaborg's group?
Step-by-Step Solution:

• (i) Berkelium ($mathrm{Bk}$, $Z = 97$) and Californium ($mathrm{Cf}$, $Z = 98$), named after the City of Berkeley and the State of California where the Lawrence Berkeley National Laboratory is situated; also Lawrencium ($mathrm{Lr}$, $Z = 103$), named directly after Ernest O. Lawrence, the founder of the laboratory.
• (ii) Seaborgium ($mathrm{Sg}$, $Z = 106$), named by IUPAC in honor of Glenn T. Seaborg and his research team who discovered plutonium and several transuranium elements.

---

Exercise 3.8

Question: Why do elements in the same group have similar physical and chemical properties?
Step-by-Step Solution:
Chemical properties are determined primarily by the number and spatial symmetry of electrons in the outermost valence shell. Elements belonging to the same vertical group possess identical valence shell electronic configurations (for example, all Group 1 elements possess an $ns^1$ configuration; all Group 17 halogens possess an $ns^2 np^5$ configuration). Because they require the loss, gain, or sharing of the exact same number of electrons to achieve stable noble-gas configurations, they display matching chemical valencies, form analogous compounds, and follow regular gradations in physical properties down the group.

---

Exercise 3.9

Question: What does atomic radius and ionic radius really mean to you?
Step-by-Step Solution:

1. Atomic Radius: The effective distance from the center of the nucleus to the outermost boundary of the electron cloud in an atom. Because electron clouds have no sharp boundaries, it is operationally defined as half the internuclear distance between two identical bonded atoms: either as covalent radius (in non-metallic single covalent bonds) or as metallic radius (between adjacent metal cores in a solid metallic lattice).
2. Ionic Radius: The effective portion of the internuclear distance allocated to a spherical ion in an ionic crystal lattice. It represents the spatial distance from the nucleus up to which the ion exerts electrostatic influence over its surrounding counter-ions. Cations are smaller than their parent atoms ($r_{\text{cat}} < r_{\text{atom}}$), whereas anions are larger ($r_{\text{an}} > r_{\text{atom}}$).

---

Exercise 3.10

Question: How do atomic radius vary in a period and in a group? How do you explain the variation?
Step-by-Step Solution:

1. Variation Across a Period: Atomic radius decreases monotonically from left to right across a period.
   Explanation: Additional electrons enter the same principal quantum shell ($n$), while the positive nuclear charge ($Z$) increases by $+1$ at each successive step. Because electrons in the same subshell do not screen each other effectively, the effective nuclear charge ($Z_{\text{eff}}$) increases, drawing the valence electron cloud inward.
2. Variation Down a Group: Atomic radius increases regularly from top to bottom down a group.
   Explanation: At each successive period, a completely new principal energy level ($n$) is added. The shielding provided by the intervening core electrons outweighs the increase in nuclear charge, positioning the outermost valence electrons farther from the nucleus.

---

Exercise 3.11

Question: What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions: (i) $mathrm{F}^-$, (ii) $mathrm{Ar}$, (iii) $mathrm{Mg}^{2+}$, (iv) $mathrm{Rb}^+$.
Step-by-Step Solution:
Isoelectronic species are atoms, molecules, or ions that contain the identical number of electrons and identical ground-state electronic configurations.

• (i) $mathrm{F}^-$: Contains $9 + 1 = 10$ electrons. Isoelectronic species: $\mathbf{Ne}, \mathbf{Na}^+, \mathbf{O}^{2-}, \mathbf{Mg}^{2+}, \mathbf{Al}^{3+}, \mathbf{N}^{3-}$.
• (ii) $mathrm{Ar}$: Contains 18 electrons. Isoelectronic species: $\mathbf{K}^+, \mathbf{Ca}^{2+}, \mathbf{Cl}^-, \mathbf{S}^{2-}, \mathbf{P}^{3-}$.
• (iii) $mathrm{Mg}^{2+}$: Contains $12 - 2 = 10$ electrons. Isoelectronic species: $\mathbf{Ne}, \mathbf{Na}^+, \mathbf{F}^-, \mathbf{O}^{2-}$.
• (iv) $mathrm{Rb}^+$: Contains $37 - 1 = 36$ electrons. Isoelectronic species: $\mathbf{Kr}, \mathbf{Sr}^{2+}, \mathbf{Br}^-, \mathbf{Se}^{2-}$.

---

Exercise 3.12

Question: Consider the following species: $mathrm{N}^{3-}, mathrm{O}^{2-}, mathrm{F}^-, mathrm{Na}^+, mathrm{Mg}^{2+}$, and $mathrm{Al}^{3+}$.
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Step-by-Step Solution:
(a) Common Feature: All these species are isoelectronic. Each species contains exactly 10 electrons with the neon ground-state configuration $1s^2 2s^2 2p^6$.
(b) Order of Increasing Ionic Radii: For an isoelectronic series, ionic radius decreases as the positive nuclear charge ($Z$) increases.
Nuclear charges: $\mathrm{N} (Z = 7), \quad \mathrm{O} (Z = 8), \quad \mathrm{F} (Z = 9), \quad \mathrm{Na} (Z = 11), \quad \mathrm{Mg} (Z = 12), \quad \mathrm{Al} (Z = 13)$ Arranging in order of increasing ionic radii (from smallest to largest): $\mathbf{Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}}$ Quantitative values: $54\text{ pm} < 72\text{ pm} < 95\text{ pm} < 136\text{ pm} < 140\text{ pm} < 171\text{ pm}$.

---

Exercise 3.13

Question: Explain why cations are smaller and anions larger in radii than their parent atoms?
Step-by-Step Solution:

1. Cations are Smaller: Formed when a neutral atom loses one or more valence electrons. This often eliminates the outermost principal energy shell entirely. Furthermore, the remaining electrons experience an unchanged positive nuclear charge, increasing the effective nuclear charge per electron ($Z/e^-$). The valence cloud is drawn inward, decreasing ionic radius.
2. Anions are Larger: Formed when a neutral atom gains one or more electrons. The nuclear charge remains identical, but the addition of extra electrons increases inter-electronic electrostatic repulsion within the valence orbitals. The electron cloud expands outward to minimize potential energy, increasing ionic radius.

---

Exercise 3.14

Question: What is the significance of the terms "isolated gaseous atom" and "ground state" while defining the ionization enthalpy and electron gain enthalpy?
Step-by-Step Solution:

1. Isolated Gaseous Atom: In liquid or solid states, atoms interact strongly with neighboring atoms via metallic bonds, Van der Waals forces, or crystal lattice forces. To measure the intrinsic atomic binding energy of a single atom without interference from intermolecular interactions, the substance must be vaporized into widely separated, non-interacting gaseous atoms under low pressure.
2. Ground State: An atom can absorb energy to promote electrons into higher-energy excited orbitals. In excited states, electrons are situated further from the nucleus and require substantially less energy to remove. To establish a universal, reproducible thermodynamic baseline, all measurements must refer to the lowest, unexcited energy level (the electronic ground state).

---

Exercise 3.15

Question: Energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10^{-18}\text{ J}$. Calculate the ionization enthalpy of atomic hydrogen in terms of $\text{J/mol}$. (Hint: Apply the idea of mole concept).
Step-by-Step Solution:

---

Exercise 3.16

Question: Among the second period elements the actual ionization enthalpies are in the order:
$\mathrm{Li} < \mathrm{B} < \mathrm{Be} < \mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F} < \mathrm{Ne}$.
Explain why: (i) $\mathrm{Be}$ has higher $\Delta_i H$ than $\mathrm{B}$, (ii) $\mathrm{O}$ has lower $\Delta_i H$ than $\mathrm{N}$ and $\mathrm{F}$?
Step-by-Step Solution:

• (i) Beryllium versus Boron:
  • $\mathrm{Be}$ ($Z = 4$): $1s^2 2s^2$. The electron to be extracted is a $2s$ electron residing in a completely filled, penetrating subshell.
  • $\mathrm{B}$ ($Z = 5$): $1s^2 2s^2 2p^1$. The electron to be extracted resides in a $2p$ orbital that is screened by the $1s^2$ and $2s^2$ electrons. Because $2s$ electrons penetrate closer to the nucleus than $2p$ electrons, removing boron's $2p$ electron requires less energy than removing beryllium's $2s$ electron.
• (ii) Oxygen versus Nitrogen and Fluorine:
  • $\mathrm{N}$ ($Z = 7$): $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$. Nitrogen has an exceptionally stable, half-filled $2p^3$ subshell with three parallel spins, maximizing exchange energy.
  • $\mathrm{O}$ ($Z = 8$): $1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1$. In oxygen, the fourth $2p$ electron must pair up in the $2p_x$ orbital, introducing electron-electron repulsion that destabilizes the configuration. Consequently, removing this electron requires less energy than ionizing nitrogen ($1314\text{ kJ/mol}$ vs $1402\text{ kJ/mol}$).
  • Oxygen has lower $\Delta_i H$ than Fluorine ($1681\text{ kJ/mol}$) because Fluorine has a higher nuclear charge ($Z = 9$) and smaller atomic radius, substantially increasing $Z_{\text{eff}}$.

---

Exercise 3.17

Question: 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?
Step-by-Step Solution:

1. First Ionization Enthalpy ($\Delta_i H_1(\mathrm{Na}) < \Delta_i H_1(\mathrm{Mg})$):
   • Sodium ($mathrm{Na}$, $Z = 11$): $[mathrm{Ne}] 3s^1$. Removing the solitary $3s^1$ valence electron yields a stable noble-gas configuration ($[mathrm{Ne}]$) with low energy expenditure ($Delta_i H_1 = 496 ext{ kJ/mol}$).
   • Magnesium ($mathrm{Mg}$, $Z = 12$): $[mathrm{Ne}] 3s^2$. Removing an electron requires breaking into a stable, fully-filled $3s^2$ subshell under a higher nuclear charge ($Z = 12$), requiring higher energy ($Delta_i H_1 = 737 ext{ kJ/mol}$).
2. Second Ionization Enthalpy ($\Delta_i H_2(\mathrm{Na}) \gg \Delta_i H_2(\mathrm{Mg})$):
   • For Sodium: $mathrm{Na}^+$ has achieved the closed-shell noble gas configuration $[mathrm{Ne}] = 1s^2 2s^2 2p^6$. Removing a second electron requires pulling an electron from a deeply seated, core $2p$ orbital under an effective nuclear charge of 11 protons acting on 10 electrons: $\Delta_i H_2(\mathrm{Na}) = \mathbf{4562\text{ kJ/mol}}$
   • For Magnesium: $mathrm{Mg}^+$ has the configuration $[mathrm{Ne}] 3s^1$. Removing the second electron simply removes the remaining valence electron to reach the stable $[mathrm{Ne}]$ configuration: $\Delta_i H_2(\mathrm{Mg}) = \mathbf{1451\text{ kJ/mol}}$ Thus, $\Delta_i H_2(\mathrm{Na})$ is more than three times greater than $\Delta_i H_2(\mathrm{Mg})$.

---

Exercise 3.18

Question: What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Step-by-Step Solution:
Down any main group, two major factors influence ionization enthalpy:

1. Increase in Principal Quantum Shell ($n$): At each successive row, an additional electron shell is introduced. The outermost valence electrons reside farther from the nucleus, weakening the Coulomb electrostatic attraction.
2. Increased Shielding / Screening Effect: The number of intervening inner core electrons increases substantially down the group, effectively screening the outer valence electrons from the nuclear charge.
   Although the total nuclear charge ($Z$) increases down the group, the combined effect of increased atomic distance and inner electron screening outweighs the nuclear charge increase, resulting in a net decrease in ionization enthalpy down the group.

---

Exercise 3.19

Question: The first ionization enthalpy values (in $ext{kJ/mol}$) of group 13 elements are:
$\mathrm{B} (801), \mathrm{Al} (577), \mathrm{Ga} (579), \mathrm{In} (558), \mathrm{Tl} (589)$.
How would you explain this deviation from the general trend?
Step-by-Step Solution:
In a standard group, ionization enthalpy decreases regularly down the column. However, Group 13 displays anomalous behavior:

1. $\mathrm{Al} (577) \to \mathrm{Ga} (579)$: Gallium possesses an unexpectedly higher ionization enthalpy than aluminium. Gallium ($Z = 31$) is preceded by ten $3d$ transition elements ($3d^{10}$). Because $d$-orbitals are diffuse with poor screening capacity, they fail to shield the valence electrons from the $+10$ nuclear charge increase. This $d$-orbital contraction (transition contraction) causes gallium's valence electrons to be held more tightly than expected.
2. $\mathrm{In} (558) \to \mathrm{Tl} (589)$: Thallium ($Z = 81$) has a higher ionization enthalpy than Indium. Thallium is preceded by fourteen $4f$ inner-transition elements ($4f^{14}$). Due to their spatial shape, $4f$ orbitals have extremely poor screening power (Lanthanoid Contraction). The outer $6s^2$ electrons experience a strong effective nuclear charge and become chemically inert (the inert pair effect), requiring substantial energy to ionize.

---

Exercise 3.20

Question: Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) $mathrm{O}$ or $mathrm{F}$
(ii) $mathrm{F}$ or $mathrm{Cl}$
Step-by-Step Solution:

• (i) $mathrm{O}$ or $mathrm{F}$: Fluorine ($mathrm{F}$) has a much more negative electron gain enthalpy ($\Delta_{eg}H = -328\text{ kJ/mol}$) than Oxygen ($-141\text{ kJ/mol}$). Across Period 2 from oxygen to fluorine, nuclear charge increases ($Z = 8 \to 9$) and atomic size contracts. Adding one electron to fluorine completes a stable neon octet configuration ($2s^2 2p^6$).
• (ii) $mathrm{F}$ or $mathrm{Cl}$: Chlorine ($mathrm{Cl}$) has a more negative electron gain enthalpy ($\Delta_{eg}H = -349\text{ kJ/mol}$) than Fluorine ($-328\text{ kJ/mol}$). Fluorine's valence shell is a compact $2p$ orbital with a dense electron cloud, leading to substantial electron-electron repulsion upon adding an extra electron. In chlorine, the electron enters a larger $3p$ orbital with reduced repulsion.

---

Exercise 3.21

Question: Would you expect the second electron gain enthalpy of $mathrm{O}$ as positive, more negative or less negative than the first? Justify your answer.
Step-by-Step Solution:
The second electron gain enthalpy of oxygen is strongly positive (endothermic): $\mathrm{O}^-(g) + e^- \to \mathrm{O}^{2-}(g), \quad \Delta_{eg}H_2 = +780\text{ kJ/mol}$ Justification: The first electron addition is exothermic ($\Delta_{eg}H_1 = -141\text{ kJ/mol}$) because the neutral oxygen atom attracts an electron. However, adding a second electron to the negatively charged oxide ion ($mathrm{O}^-$) involves severe electrostatic repulsion between like charges. Energy must be supplied from external surroundings to force the second electron into the anion, making $\Delta_{eg}H_2$ strongly positive.

---

Exercise 3.22

Question: What is the basic difference between the terms electron gain enthalpy and electronegativity?
Step-by-Step Solution:

---

Exercise 3.23

Question: How would you react to the statement that the electronegativity of $mathrm{N}$ on Pauling scale is 3.0 in all the nitrogen compounds?
Step-by-Step Solution:
The statement is incorrect. Electronegativity is not an absolute constant for an element. It depends on several bonding variables:

1. Hybridization State: An orbital with higher $s$-character is held closer to the nucleus. The electronegativity of nitrogen increases with increasing $s$-character: $\mathrm{N}(sp^3) < \mathrm{N}(sp^2) < \mathrm{N}(sp)$
2. Oxidation State: Nitrogen in a higher oxidation state possesses greater positive charge and attracts shared electrons more strongly (e.g., Nitrogen in $+5$ oxidation state in $mathrm{NO}_3^-$ is significantly more electronegative than in $-3$ in $mathrm{NH}_3$).
3. Substituent Effects: Bonding with strongly electronegative atoms like fluorine increases the effective electronegativity of nitrogen.
   The Pauling value of $3.0$ is simply a standardized baseline reference for comparison.

---

Exercise 3.24

Question: Describe the theory associated with the radius of an atom as it: (a) gains an electron, (b) loses an electron.
Step-by-Step Solution:

• (a) When an atom gains an electron (Formation of Anion): The positive nuclear charge ($Z$) remains constant, but the number of valence electrons increases. This increases electron-electron electrostatic repulsion and mutual screening among valence electrons, decreasing the effective nuclear charge per electron ($Z_{\text{eff}}$). To minimize potential energy, the electron cloud expands outward, making the anion radius larger than the neutral atom.
• (b) When an atom loses an electron (Formation of Cation): The removal of valence electrons often completely empties the outermost quantum shell (as when $mathrm{Na}$ loses its $3s^1$ electron). The remaining electrons experience reduced mutual repulsion and are pulled inward by the unchanged positive nuclear charge, increasing $Z_{\text{eff}}$. Thus, the cation radius is smaller than the neutral atom.

---

Exercise 3.25

Question: Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Step-by-Step Solution:
The first ionization enthalpies for two isotopes of the same element are essentially identical.
Justification: Isotopes have different numbers of neutrons (differing in nuclear mass), but they have the exact same number of protons ($Z$) and the exact same electronic configuration. Because ionization enthalpy is dictated by the electrostatic Coulomb attraction between the positive nuclear charge ($Z$) and the valence electrons, nuclear mass has a negligible effect on outer electron binding energies.

---

Exercise 3.26

Question: What are the major differences between metals and non-metals?
Step-by-Step Solution:

---

Exercise 3.27

Question: Use the periodic table to answer the following questions:
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at room temperature.
Step-by-Step Solution:

• (a) An element with five electrons in the outer subshell has a valence configuration $ns^2 np^5$, which belongs to Group 17 (Halogens). Examples: Fluorine ($mathrm{F}$) or Chlorine ($mathrm{Cl}$).
• (b) An element tending to lose two electrons belongs to Group 2 (Alkaline Earth Metals) with valence configuration $ns^2$. Examples: Magnesium ($mathrm{Mg}$) or Calcium ($mathrm{Ca}$).
• (c) An element tending to gain two electrons belongs to Group 16 (Chalcogens) with valence configuration $ns^2 np^4$. Examples: Oxygen ($mathrm{O}$) or Sulfur ($mathrm{S}$).
• (d) Group 17 (Halogens):
  • Fluorine ($mathrm{F}_2$) and Chlorine ($mathrm{Cl}_2$): Non-metal gases.
  • Bromine ($mathrm{Br}_2$): Non-metal liquid.
  • Iodine ($mathrm{I}_2$): Non-metal solid.
  • Tennessine ($mathrm{Ts}$) / Astatine ($mathrm{At}$): Metalloid/metal solid.

---

Exercise 3.28

Question: The increasing order of reactivity among group 1 elements is $mathrm{Li} < mathrm{Na} < mathrm{K} < mathrm{Rb} < mathrm{Cs}$ whereas that among group 17 elements is $mathrm{F} > mathrm{Cl} > mathrm{Br} > mathrm{I}$. Explain.
Step-by-Step Solution:

• Group 1 Alkali Metals: Chemical reactivity depends on the ease of losing a valence electron to form univalent cations ($M \to M^+ + e^-$). Down Group 1, atomic radius increases and ionization enthalpy decreases. Because it becomes progressively easier to lose an electron, reactivity increases down the group: $\mathrm{Li} < \mathrm{Na} < \mathrm{K} < \mathrm{Rb} < \mathrm{Cs}$
• Group 17 Halogens: Chemical reactivity depends on the ease of gaining an electron to form halide anions ($X + e^- \to X^-$). Down Group 17, atomic radius increases and effective nuclear charge at the surface decreases, reducing electronegativity and oxidizing power. Because smaller halogens pull incoming electrons more strongly, reactivity decreases down the group: $\mathrm{F} > \mathrm{Cl} > \mathrm{Br} > \mathrm{I}$

---

Exercise 3.29

Question: Write the general outer electronic configuration of $s-, p-, d-$ and $f-$ block elements.
Step-by-Step Solution:

• $s$-Block Elements: $ns^{1-2}$ (where $n = 1$ to $7$)
• $p$-Block Elements: $ns^2 np^{1-6}$ (where $n = 2$ to $7$)
• $d$-Block Elements: $(n - 1)d^{1-10} ns^{0-2}$ (where $n = 4$ to $7$)
• $f$-Block Elements: $(n - 2)f^{1-14} (n - 1)d^{0-1} ns^2$ (where $n = 6, 7$)

---

Exercise 3.30

Question: Assign the position of the element having outer electronic configuration:
(i) $ns^2 np^4$ for $n = 3$
(ii) $(n - 1)d^2 ns^2$ for $n = 4$
(iii) $(n - 2)f^7 (n - 1)d^1 ns^2$ for $n = 6$, in the periodic table.
Step-by-Step Solution:

• (i) $ns^2 np^4$ for $n = 3$:
  Valence configuration: $3s^2 3p^4$. Period = $3$.
  Differentiating electron is in $p$-orbital: Group = $12 + 4 = \mathbf{16}$ (Chalcogens).
  The element is Sulfur ($mathrm{S}$, $Z = 16$).
• (ii) $(n - 1)d^2 ns^2$ for $n = 4$:
  Valence configuration: $3d^2 4s^2$. Period = $4$.
  Differentiating electron is in $d$-orbital: Group = number of $(n-1)d$ electrons + $ns$ electrons = $2 + 2 = \mathbf{4}$.
  The element is Titanium ($mathrm{Ti}$, $Z = 22$).
• (iii) $(n - 2)f^7 (n - 1)d^1 ns^2$ for $n = 6$:
  Valence configuration: $4f^7 5d^1 6s^2$. Period = $6$.
  Differentiating electron is in the $4f$ subshell, which belongs to the Lanthanoid series in Group 3.
  The element is Gadolinium ($mathrm{Gd}$, $Z = 64$).

---

Exercise 3.31

Question: The first ($\Delta_i H_1$) and second ($\Delta_i H_2$) ionization enthalpies and electron gain enthalpy ($\Delta_{eg}H$) in $ext{kJ/mol}$ of a few elements are given below:

Which of the above elements is likely to be:
(a) the least reactive element
(b) the most reactive metal
(c) the most reactive non-metal
(d) the least reactive non-metal
(e) the metal which can form a stable binary halide of the formula $M X_2$ ($X = \text{halogen}$)
(f) the metal which can form a predominantly stable covalent halide of the formula $M X$ ($X = \text{halogen}$)?

Step-by-Step Solution:

• (a) Least reactive element: Element V. It has an exceptionally high first ionization enthalpy ($2372\text{ kJ/mol}$) and a positive electron gain enthalpy ($+48\text{ kJ/mol}$), characteristic of a noble gas (Helium).
• (b) Most reactive metal: Element II. It has the lowest first ionization enthalpy ($419\text{ kJ/mol}$) and a low negative electron gain enthalpy ($-48\text{ kJ/mol}$), characteristic of a heavy alkali metal like Potassium ($mathrm{K}$).
• (c) Most reactive non-metal: Element III. It has a very high first ionization enthalpy ($1681\text{ kJ/mol}$) and a strongly negative electron gain enthalpy ($-328\text{ kJ/mol}$), matching Fluorine ($mathrm{F}$).
• (d) Least reactive non-metal: Element IV. It has a moderately high ionization enthalpy ($1008\text{ kJ/mol}$) and a high negative electron gain enthalpy ($-295\text{ kJ/mol}$), characteristic of Iodine ($mathrm{I}$).
• (e) Metal forming stable halide $M X_2$: Element VI. It has a moderate $\Delta_i H_1$ ($738\text{ kJ/mol}$) and a moderate $\Delta_i H_2$ ($1451\text{ kJ/mol}$), allowing removal of two valence electrons to form divalent cations, matching an alkaline earth metal like Magnesium ($mathrm{Mg}$).
• (f) Metal forming stable covalent halide $M X$: Element I. It has a low $\Delta_i H_1$ ($520\text{ kJ/mol}$) but a massive jump to $\Delta_i H_2$ ($7300\text{ kJ/mol}$), characteristic of Lithium ($mathrm{Li}$). Due to lithium's small size and high polarizing power (Fajans' rule), its halide $mathrm{LiCl}$ displays significant covalent character.

---

Exercise 3.32

Question: Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements:
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Step-by-Step Solution:

• (a) $\mathrm{Li}$ (Valence 1) and $\mathrm{O}$ (Valence 2) $\implies \mathbf{Li_2O}$ (Lithium Monoxide)
• (b) $\mathrm{Mg}$ (Valence 2) and $\mathrm{N}$ (Valence 3) $\implies \mathbf{Mg_3N_2}$ (Magnesium Nitride)
• (c) $\mathrm{Al}$ (Valence 3) and $\mathrm{I}$ (Valence 1) $\implies \mathbf{AlI_3}$ (Aluminium Triiodide)
• (d) $\mathrm{Si}$ (Valence 4) and $\mathrm{O}$ (Valence 2) $\implies \mathrm{Si}_2\mathrm{O}_4 = \mathbf{SiO_2}$ (Silicon Dioxide)
• (e) $\mathrm{P}$ (Valence 3 or 5) and $\mathrm{F}$ (Valence 1) $\implies \mathbf{PF_3}$ or $\mathbf{PF_5}$ (Phosphorus Trifluoride / Pentafluoride)
• (f) Element 71 is Lutetium ($mathrm{Lu}$, a lanthanoid with characteristic valence 3) and $\mathrm{F}$ (Valence 1) $\implies \mathbf{LuF_3}$ (Lutetium Trifluoride)

---

Exercise 3.33

Question: 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
Step-by-Step Solution:
Correct Answer: (c) principal quantum number.
The period number corresponds to the maximum principal quantum number ($n$) of the valence shell orbitals being occupied in that horizontal row.

---

Exercise 3.34

Question: Which of the following statements related to the modern periodic table is incorrect?
(a) The $p$-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a $p$-shell.
(b) The $d$-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a $d$-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number ($l$) for the last subshell that received electrons in building up the electronic configuration.
Step-by-Step Solution:
Correct Answer: (b).
Statement (b) is incorrect. A $d$-subshell contains 5 degenerate orbitals ($m_l = -2, -1, 0, +1, +2$). By Pauli's exclusion principle, each orbital accommodates 2 electrons with opposite spins, so a $d$-subshell holds a maximum of $5 \times 2 = 10$ electrons. Therefore, the $d$-block contains 10 columns (Groups 3 to 12), not 8 columns.

---

Exercise 3.35

Question: Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number ($n$)
(b) Nuclear charge ($Z$)
(c) Nuclear mass
(d) Number of core electrons
Step-by-Step Solution:
Correct Answer: (c) Nuclear mass.
Valence electrons are bound by Coulomb electrostatic attraction to the nuclear positive charge ($Z$) and screened by inner core electrons. While principal quantum number ($n$), nuclear charge ($Z$), and shielding by core electrons determine valence orbital energies, nuclear mass (the number of neutrons) exerts negligible electrostatic influence on valence electron behavior.

---

Exercise 3.36

Question: The size of isoelectronic species: $mathrm{F}^-, mathrm{Ne}$ and $mathrm{Na}^+$ is affected by:
(a) nuclear charge ($Z$)
(b) valence principal quantum number ($n$)
(c) electron-electron interaction in the outer orbitals
(d) none of the factors because their size is the same
Step-by-Step Solution:
Correct Answer: (a) nuclear charge ($Z$).
All three species have the identical number of electrons (10 electrons) and the same valence shell ($n = 2$). However, their nuclear charges differ: $mathrm{F}^-$ has $Z = 9$, $mathrm{Ne}$ has $Z = 10$, and $mathrm{Na}^+$ has $Z = 11$. As nuclear charge increases, the electron cloud is pulled inward with greater electrostatic force, decreasing atomic/ionic size.

---

Exercise 3.37

Question: Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower $n$ value is easier than from orbital having higher $n$ value.
Step-by-Step Solution:
Correct Answer: (d).
Statement (d) is incorrect. Electrons in orbitals with lower principal quantum numbers ($n$) reside closer to the nucleus, experience stronger electrostatic pull, and have lower screening. Therefore, removing an electron from an orbital with a lower $n$ value requires substantially more energy than removing an electron from an orbital with a higher $n$ value.

---

Exercise 3.38

Question: Considering the elements $mathrm{B}, mathrm{Al}, mathrm{Mg}$, and $mathrm{K}$, the correct order of their metallic character is:
(a) $mathrm{B} > mathrm{Al} > mathrm{Mg} > mathrm{K}$
(b) $mathrm{Al} > mathrm{Mg} > mathrm{B} > mathrm{K}$
(c) $mathrm{Mg} > mathrm{Al} > mathrm{K} > mathrm{B}$
(d) $mathrm{K} > mathrm{Mg} > mathrm{Al} > mathrm{B}$
Step-by-Step Solution:
Correct Answer: (d) $mathrm{K} > mathrm{Mg} > mathrm{Al} > mathrm{B}$.

• Metallic character increases down a group and decreases across a period from left to right.
• Potassium ($mathrm{K}$) is in Group 1, Period 4 (an electropositive alkali metal) and has the highest metallic character.
• Magnesium ($mathrm{Mg}$, Group 2, Period 3) is more metallic than Aluminium ($mathrm{Al}$, Group 13, Period 3).
• Boron ($mathrm{B}$, Group 13, Period 2) is a semi-metallic metalloid/non-metal with the lowest metallic character.
  Order: $\mathbf{K > Mg > Al > B}$.

---

Exercise 3.39

Question: Considering the elements $mathrm{B}, mathrm{C}, mathrm{N}, mathrm{F}$, and $mathrm{Si}$, the correct order of their non-metallic character is:
(a) $mathrm{B} > mathrm{C} > mathrm{Si} > mathrm{N} > mathrm{F}$
(b) $mathrm{Si} > mathrm{C} > mathrm{B} > mathrm{N} > mathrm{F}$
(c) $mathrm{F} > mathrm{N} > mathrm{C} > mathrm{B} > mathrm{Si}$
(d) $mathrm{F} > mathrm{N} > mathrm{C} > mathrm{Si} > mathrm{B}$
Step-by-Step Solution:
Correct Answer: (c) $mathrm{F} > mathrm{N} > mathrm{C} > mathrm{B} > mathrm{Si}$.

• Across Period 2 from left to right, non-metallic character increases: $\mathrm{B} < \mathrm{C} < \mathrm{N} < \mathrm{F}$.
• Carbon and Silicon belong to Group 14. Down the group, non-metallic character decreases, so Carbon is more non-metallic than Silicon ($mathrm{C} > mathrm{Si}$).
• Between Boron and Silicon: Boron is a metalloid in Period 2, whereas Silicon is in Period 3; Boron is more non-metallic than Silicon ($mathrm{B} > mathrm{Si}$).
  Order of decreasing non-metallic character: $\mathbf{F > N > C > B > Si}$.

---

Exercise 3.40

Question: Considering the elements $mathrm{F}, mathrm{Cl}, mathrm{O}$ and $mathrm{N}$, the correct order of their chemical reactivity in terms of oxidizing property is:
(a) $mathrm{F} > mathrm{Cl} > mathrm{O} > mathrm{N}$
(b) $mathrm{F} > mathrm{O} > mathrm{Cl} > mathrm{N}$
(c) $mathrm{Cl} > mathrm{F} > mathrm{O} > mathrm{N}$
(d) $mathrm{O} > mathrm{F} > mathrm{N} > mathrm{Cl}$
Step-by-Step Solution:
Correct Answer: (b) $mathrm{F} > mathrm{O} > mathrm{Cl} > mathrm{N}$.

• Oxidizing power measures the ability of an element to act as an electron acceptor. It depends on standard reduction potentials, high electronegativity, and high electron gain enthalpy.
• Fluorine ($mathrm{F}$) is the strongest oxidizing agent of all known chemical elements ($E^\circ = +2.87\text{ V}$).
• Oxygen ($mathrm{O}$) is the second most electronegative element ($chi = 3.5$) and a more powerful oxidizing agent than Chlorine ($chi = 3.0$).
• Nitrogen ($mathrm{N}$) possesses a stable half-filled $2p^3$ subshell and low electron affinity, making it the weakest oxidizer in the set.
  Order of decreasing oxidizing power: $\mathbf{F > O > Cl > N}$.