Comprehensive Advanced Physical and General Chemistry Study Guide

Atomic Structure, Subatomic Particles, and Electron Configurations

  • An atom is the smallest constituent unit of an element that can participate in a chemical reaction.

  • Atoms consist of three primary subatomic particles: protons, neutrons, and electrons.

  • Fundamental Subatomic Properties:

    • Proton (pp): Relative mass = 11, Relative charge = +1+1.
    • Electron (e−e^-): Relative mass = 11840\frac{1}{1840}, Relative charge = −1-1.
    • Neutron (nn): Relative mass = 11, Relative charge = 00.
  • Atomic Relationships and Neutrality:

    • In an electrically neutral atom: Number of protons=Number of electrons\text{Number of protons} = \text{Number of electrons}.
    • The atomic number (ZZ) of an element is uniquely defined by its number of protons.
    • Mass number (nucleon number, AA): Mass number=Number of protons+Number of neutrons\text{Mass number} = \text{Number of protons} + \text{Number of neutrons}.
  • Behavior in an Electric Field:

    • When a collimated beam containing protons, neutrons, and electrons passes between oppositely charged plates:
    • Electrons deflect toward the positive (++ ) plate because they carry a negative charge. Their deflection angle is very large owing to their negligible mass.
    • Protons deflect toward the negative (−- ) plate because they carry a positive charge. Their deflection angle is smaller than that of electrons due to their significantly larger mass.
    • Neutrons continue straight without deflection because they carry no electrical charge (00).
    • Mathematical Relationship of Deflection:
    • Angle of deflection∝chargemass\text{Angle of deflection} \propto \frac{\text{charge}}{\text{mass}}
    • Angle of deflection=k×chargemass\text{Angle of deflection} = k \times \frac{\text{charge}}{\text{mass}}
    • The constant kk depends strictly on the potential difference (p.d.) applied across the positive and negative plates.

Deflection of subatomic particles in an electric field

  • Atomic Structural Architecture:

    • Nucleus:
    • Located at the center of the atom, containing protons and neutrons.
    • Carries a net positive charge.
    • Represents the heaviest part of the atom because it houses the high-mass nucleons.
    • An element's chemical identity is fundamentally defined by its nuclear proton count.
    • Quantum Shells:
    • Extranuclear regions housing orbiting electrons.
    • A shell represents the energetic pathway or orbital route taken by an electron as it circles the nucleus.
    • Electrons are constrained in their trajectories by continuous electrostatic forces of attraction exerted by the positively charged nucleus.
    • Each shell corresponds to a principal quantum number (nn), numbered sequentially starting from the shell nearest to the nucleus (n=1,2,3,4,…n = 1, 2, 3, 4, \dots).
    • The maximum electron capacity of any principal shell is governed by the formula 2n22n^2:
      • Shell n=1n = 1: 2(1)2=2 electrons2(1)^2 = 2\text{ electrons}
      • Shell n=2n = 2: 2(2)2=8 electrons2(2)^2 = 8\text{ electrons}
      • Shell n=3n = 3: 2(3)2=18 electrons2(3)^2 = 18\text{ electrons}
      • Shell n=4n = 4: 2(4)2=32 electrons2(4)^2 = 32\text{ electrons}
  • Subshells and Orbitals:

    • The number of subshells in any principal shell is equal to its shell number (nn).
    • Subshells differ in shape, spatial extension, and energy levels, and are categorized as s,p,d,s, p, d, and ff:
    • Shell 11: 11 subshell (1s1s)
    • Shell 22: 22 subshells (2s,2p2s, 2p)
    • Shell 33: 33 subshells (3s,3p,3d3s, 3p, 3d)
    • Shell 44: 44 subshells (4s,4p,4d,4f4s, 4p, 4d, 4f)
    • Orbitals:
    • An orbital is defined as the three-dimensional region of space around the nucleus where there is the highest probability of locating an electron.
    • Any single orbital can accommodate a maximum of two electrons.
    • The two electrons within an orbital must possess opposite spins (represented diagrammatically by antiparallel arrows in boxes: ↑↓\uparrow\downarrow) to minimize electrostatic repulsion between their identical negative charges, conferring energetic stability.
    • Orbital Capacities per Subshell:
      • ss subshell: 11 orbital, maximum 2 electrons2\text{ electrons}
      • pp subshell: 33 degenerate orbitals (px,py,pzp_x, p_y, p_z), maximum 6 electrons6\text{ electrons}
      • dd subshell: 55 orbitals, maximum 10 electrons10\text{ electrons}
      • ff subshell: 77 orbitals, maximum 14 electrons14\text{ electrons}
    • Orbital Geometry:
    • ss Orbitals: Completely spherical; non-directional. The diameter of the sphere increases proportionally with the principal quantum number (nn) as distance from the nucleus increases.
    • pp Orbitals: Dumbbell-shaped consisting of two distinct lobes separated by a nodal plane at the nucleus. The three degenerate orbitals lie mutually perpendicular along the Cartesian axes (px,py,pzp_x, p_y, p_z) and possess equal energy in an isolated atom.
  • Relative Energies of Orbitals and the Aufbau Principle:

    • Electrons within shells possess both potential and kinetic energies; larger principal quantum numbers correspond to higher energetic states.
    • As the distance from the nucleus increases, the energy gap between consecutive principal shells becomes progressively smaller.
    • This energy compression causes subshell overlap beginning at shell n=4n = 4, such that the 4s4s subshell possesses lower energy than the 3d3d subshell and is therefore populated first.

Relative energies of orbitals for the first four principal quantum shells

  • Rules for Constructing Electronic Configurations:

    • Aufbau Principle: Electrons fill orbitals of the lowest available energy levels before occupying higher levels (1s→2s→2p→3s→3p→4s→3d→4p…1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d \rightarrow 4p \dots).
    • Hund's Rule: When filling degenerate orbitals (such as the three pp orbitals), electrons occupy them singly with parallel spins before pairing occurs, minimizing inter-electronic repulsion.
    • Complete filling of a subshell must occur before proceeding to populate the next subshell.
  • Ground-State Configurations of the First 30 Elements:

    • 1H:1s1_1\text{H}: 1s^1
    • 2He:1s2_2\text{He}: 1s^2
    • 3Li:1s2 2s1_3\text{Li}: 1s^2\,2s^1
    • 4Be:1s2 2s2_4\text{Be}: 1s^2\,2s^2
    • 5B:1s2 2s2 2p1_5\text{B}: 1s^2\,2s^2\,2p^1
    • 6C:1s2 2s2 2p2_6\text{C}: 1s^2\,2s^2\,2p^2
    • 7N:1s2 2s2 2p3_7\text{N}: 1s^2\,2s^2\,2p^3
    • 8O:1s2 2s2 2p4_8\text{O}: 1s^2\,2s^2\,2p^4
    • 9F:1s2 2s2 2p5_9\text{F}: 1s^2\,2s^2\,2p^5
    • 10Ne:1s2 2s2 2p6_{10}\text{Ne}: 1s^2\,2s^2\,2p^6
    • 11Na:1s2 2s2 2p6 3s1_{11}\text{Na}: 1s^2\,2s^2\,2p^6\,3s^1
    • 12Mg:1s2 2s2 2p6 3s2_{12}\text{Mg}: 1s^2\,2s^2\,2p^6\,3s^2
    • 13Al:1s2 2s2 2p6 3s2 3p1_{13}\text{Al}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^1
    • 14Si:1s2 2s2 2p6 3s2 3p2_{14}\text{Si}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^2
    • 15P:1s2 2s2 2p6 3s2 3p3_{15}\text{P}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^3
    • 16S:1s2 2s2 2p6 3s2 3p4_{16}\text{S}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^4
    • 17Cl:1s2 2s2 2p6 3s2 3p5_{17}\text{Cl}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^5
    • 18Ar:1s2 2s2 2p6 3s2 3p6_{18}\text{Ar}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6
    • 19K:1s2 2s2 2p6 3s2 3p6 4s1_{19}\text{K}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1
    • 20Ca:1s2 2s2 2p6 3s2 3p6 4s2_{20}\text{Ca}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2
    • 21Sc:1s2 2s2 2p6 3s2 3p6 3d1 4s2_{21}\text{Sc}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^1\,4s^2
    • 22Ti:1s2 2s2 2p6 3s2 3p6 3d2 4s2_{22}\text{Ti}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^2\,4s^2
    • 23V:1s2 2s2 2p6 3s2 3p6 3d3 4s2_{23}\text{V}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^3\,4s^2
    • 24Cr:1s2 2s2 2p6 3s2 3p6 3d5 4s1_{24}\text{Cr}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1 (Anomalous: Promotion of one 4s4s electron to 3d3d provides a half-filled 3d53d^5 subshell, conferring extra stability)
    • 25Mn:1s2 2s2 2p6 3s2 3p6 3d5 4s2_{25}\text{Mn}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^2
    • 26Fe:1s2 2s2 2p6 3s2 3p6 3d6 4s2_{26}\text{Fe}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^6\,4s^2
    • 27Co:1s2 2s2 2p6 3s2 3p6 3d7 4s2_{27}\text{Co}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^7\,4s^2
    • 28Ni:1s2 2s2 2p6 3s2 3p6 3d8 4s2_{28}\text{Ni}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^8\,4s^2
    • 29Cu:1s2 2s2 2p6 3s2 3p6 3d10 4s1_{29}\text{Cu}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^1 (Anomalous: Promotion of one 4s4s electron to 3d3d produces a fully-filled 3d103d^{10} subshell, conferring extra stability)
    • 30Zn:1s2 2s2 2p6 3s2 3p6 3d10 4s2_{30}\text{Zn}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2
  • Electronic Configuration of Ions:

    • An ion is formed when an atom loses or gains electrons to achieve an energetically stable noble gas configuration.
    • Anions: Formed when non-metal atoms gain electrons. Because non-metals possess high electronegativities and multiple valence electrons, relatively little energy is required to complete their valence shell:
    • O:1s2 2s2 2p4→O2−:1s2 2s2 2p6\text{O}: 1s^2\,2s^2\,2p^4 \rightarrow \text{O}^{2-}: 1s^2\,2s^2\,2p^6
    • P:1s2 2s2 2p6 3s2 3p3→P3−:1s2 2s2 2p6 3s2 3p6\text{P}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^3 \rightarrow \text{P}^{3-}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6
    • S:1s2 2s2 2p6 3s2 3p4→S2−:1s2 2s2 2p6 3s2 3p6\text{S}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^4 \rightarrow \text{S}^{2-}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6
    • Cl:1s2 2s2 2p6 3s2 3p5→Cl−:1s2 2s2 2p6 3s2 3p6\text{Cl}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^5 \rightarrow \text{Cl}^-: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6
    • Cations: Formed when metal atoms lose electrons from the outermost principal quantum shell and outermost orbital:
    • Mg:1s2 2s2 2p6 3s2→Mg2+:1s2 2s2 2p6\text{Mg}: 1s^2\,2s^2\,2p^6\,3s^2 \rightarrow \text{Mg}^{2+}: 1s^2\,2s^2\,2p^6
    • Transition Metal Ion Formation Rule: Electrons in the 4s4s subshell are lost before any electrons from the 3d3d subshell:
    • 21Sc:1s2 2s2 2p6 3s2 3p6 3d1 4s2→Sc3+:1s2 2s2 2p6 3s2 3p6_{21}\text{Sc}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^1\,4s^2 \rightarrow \text{Sc}^{3+}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6
    • 24Cr:1s2 2s2 2p6 3s2 3p6 3d5 4s1→Cr3+:1s2 2s2 2p6 3s2 3p6 3d3_{24}\text{Cr}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1 \rightarrow \text{Cr}^{3+}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^3
    • 30Zn:1s2 2s2 2p6 3s2 3p6 3d10 4s2→Zn2+:1s2 2s2 2p6 3s2 3p6 3d10_{30}\text{Zn}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2 \rightarrow \text{Zn}^{2+}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}

Ionisation Energy and Periodic Trends

  • Definitions of Ionisation Energy (IE):

    • Ionisation Energy: The energy required to remove the most loosely bound electron (outermost electron) from a gaseous particle.
    • First Ionisation Energy (IE1IE_1): The energy required to remove one mole of electrons from one mole of gaseous atoms of an element to form one mole of gaseous unipositive ions (+1+1 charge):
    • Mg(g)→Mg+(g)+e−(1st IE of Mg)\text{Mg}(g) \rightarrow \text{Mg}^+(g) + e^- \quad (1\text{st } IE\text{ of Mg})
    • Ca(g)→Ca+(g)+e−(1st IE of Ca)\text{Ca}(g) \rightarrow \text{Ca}^+(g) + e^- \quad (1\text{st } IE\text{ of Ca})
    • F(g)→F+(g)+e−(1st IE of F)\text{F}(g) \rightarrow \text{F}^+(g) + e^- \quad (1\text{st } IE\text{ of F})
    • Na(g)→Na+(g)+e−(1st IE of Na)\text{Na}(g) \rightarrow \text{Na}^+(g) + e^- \quad (1\text{st } IE\text{ of Na})
    • Second Ionisation Energy (IE2IE_2): The energy required to remove one mole of electrons from one mole of gaseous unipositive ions to form one mole of gaseous dipositive ions (+2+2 charge):
    • Mg+(g)→Mg2+(g)+e−(2nd IE of Mg)\text{Mg}^+(g) \rightarrow \text{Mg}^{2+}(g) + e^- \quad (2\text{nd } IE\text{ of Mg})
    • Zn+(g)→Zn2+(g)+e−(2nd IE of Zn)\text{Zn}^+(g) \rightarrow \text{Zn}^{2+}(g) + e^- \quad (2\text{nd } IE\text{ of Zn})
    • Ca+(g)→Ca2+(g)+e−(2nd IE of Ca)\text{Ca}^+(g) \rightarrow \text{Ca}^{2+}(g) + e^- \quad (2\text{nd } IE\text{ of Ca})
    • Higher Ionisation Equations:
    • 1st IE of Chromium: Cr(g)→Cr+(g)+e−1\text{st } IE\text{ of Chromium: } \text{Cr}(g) \rightarrow \text{Cr}^+(g) + e^-
    • 3rd IE of Aluminium: Al2+(g)→Al3+(g)+e−3\text{rd } IE\text{ of Aluminium: } \text{Al}^{2+}(g) \rightarrow \text{Al}^{3+}(g) + e^-
    • 5th IE of Manganese: Mn4+(g)→Mn5+(g)+e−5\text{th } IE\text{ of Manganese: } \text{Mn}^{4+}(g) \rightarrow \text{Mn}^{5+}(g) + e^-
  • Endothermic Nature of Ionisation:

    • Removing an electron from a gaseous species is always endothermic (IE>0IE > 0) because electrostatic attractions between the positively charged nucleus and the negative electron must be overcome.
  • Successive Ionisation Energies:

    • Successive ionisation energies increase progressively (1st<2nd<3rd…1\text{st} < 2\text{nd} < 3\text{rd} \dots).
    • As each electron is removed, the ionic charge becomes increasingly positive while the nuclear charge remains constant.
    • The effective attraction experienced by the remaining electrons increases, pulling them closer to the nucleus and requiring substantially more energy for subsequent removals:
    • Mg(g)→Mg+(g)+e−ΔH=+736 kJ mol−1\text{Mg}(g) \rightarrow \text{Mg}^+(g) + e^- \quad \Delta H = +736\,kJ\,mol^{-1}
    • Mg+(g)→Mg2+(g)+e−ΔH=+1450 kJ mol−1\text{Mg}^+(g) \rightarrow \text{Mg}^{2+}(g) + e^- \quad \Delta H = +1450\,kJ\,mol^{-1}
    • Mg2+(g)→Mg3+(g)+e−ΔH=+7740 kJ mol−1\text{Mg}^{2+}(g) \rightarrow \text{Mg}^{3+}(g) + e^- \quad \Delta H = +7740\,kJ\,mol^{-1}
    • The sharp jump between the 2nd and 3rd ionisation energies of magnesium reflects the removal of an electron from an inner, complete principal quantum shell (2p2p), closer to the nucleus with reduced shielding.
  • Factors Affecting Ionisation Energy:

    • Atomic Radius (Size of the Atom):
    • Larger atoms exhibit weaker electrostatic attraction between the nucleus and outermost valence electrons.
    • IE∝1Distance between nucleus and outermost electronIE \propto \frac{1}{\text{Distance between nucleus and outermost electron}}
    • Magnitude of Nuclear Charge (Number of Protons):
    • Higher nuclear charge exerts a stronger electrostatic pull on valence electrons, demanding more energy for removal.
    • IE∝Nuclear chargeIE \propto \text{Nuclear charge}
    • Shielding Effect (Screening):
    • Inner shell electrons repel valence electrons, counteracting nuclear pull.
    • IE∝1Shielding effectIE \propto \frac{1}{\text{Shielding effect}}
  • Specific Comparative Examples:

    • Sodium (11Na_{11}\text{Na}) vs. Potassium (19K_{19}\text{K}):
    • Despite potassium possessing a higher nuclear charge (1919 protons vs. 1111 protons), its first IEIE is lower than that of sodium.
    • Potassium's valence electron resides in the 4s4s subshell, which is further from the nucleus and shielded by more inner shells than the 3s3s electron of sodium.
    • Magnesium (12Mg_{12}\text{Mg}) vs. Calcium (20Ca_{20}\text{Ca}):
    • Ca<MgCa < Mg. Calcium's valence electron is in the 4s4s subshell, experiencing greater distance and shielding compared to the 3s3s valence electrons in magnesium.
    • Lithium (3Li_3\text{Li}) vs. Sodium (11Na_{11}\text{Na}):
    • Na<LiNa < Li. Sodium's outermost electron occupies the 3s3s subshell, situated further from the nucleus and experiencing greater inner-shell shielding than lithium's 2s2s electron.
    • Nitrogen (7N_7\text{N}) vs. Oxygen (8O_8\text{O}):
    • Both elements have valence electrons in the 2p2p subshell experiencing identical shielding, and oxygen has a higher nuclear charge (88 protons vs. 77 protons).
    • However, oxygen exhibits a lower first IEIE than nitrogen.
    • In nitrogen, the 2p2p subshell contains three singly occupied orbitals (2px1 2py1 2pz12p_x^1\,2p_y^1\,2p_z^1). In oxygen, one orbital contains paired electrons (2px2 2py1 2pz12p_x^2\,2p_y^1\,2p_z^1).
    • The inter-electronic repulsion between paired electrons in the same orbital elevates their energy, allowing one of the paired electrons to be removed more easily.
    • Phosphorus (15P_{15}\text{P}) vs. Sulfur (16S_{16}\text{S}):
    • Sulfur has a lower first IEIE than phosphorus despite higher nuclear charge.
    • Inter-electronic spin-pair repulsion in the doubly occupied 3p3p orbital of sulfur lowers the energy required to remove the electron relative to the half-filled 3p33p^3 subshell of phosphorus.
  • Periodic Trends in First Ionisation Energy:

    • Down a Group: First IEIE decreases.
    • Principal quantum shells increase, increasing atomic radius.
    • Inner electron shells multiply, substantially increasing shielding.
    • The combined effects of increased distance and shielding completely outweigh the increase in nuclear charge, resulting in a reduced effective nuclear charge on valence electrons.
    • Across a Period (Left to Right): First IEIE generally increases.
    • Valence electrons enter the same principal shell, maintaining approximately constant shielding.
    • Nuclear charge increases progressively with atomic number, causing effective nuclear charge to rise and pulling valence electrons closer to the nucleus.
    • Exceptions Across a Period:
    • Group 2 to Group 13 (e.g., Mg\text{Mg} vs. Al\text{Al}): Aluminium's first IEIE is lower than magnesium's because aluminium's outermost electron occupies a 3p3p subshell, which is higher in energy, further from the nucleus, and shielded by the inner 3s23s^2 pair. Magnesium's 3s23s^2 subshell is completely filled, conferring extra stability.
    • Group 15 to Group 16 (e.g., P\text{P} vs. S\text{S}): Sulfur's first IEIE drops below phosphorus's due to spin-pair repulsion in the 3p43p^4 configuration.

Chemical Bonding and Molecular Structure

  • A chemical bond represents the electrostatic force of attraction holding particles together in a chemical substance.

  • Major classes include ionic, covalent, coordinate (dative), and metallic bonds.

  • Ionic Bonding:

    • The electrostatic force of attraction operating between oppositely charged ions.
    • Formed by the complete transfer of electrons from a metal atom to a non-metal atom.
    • Dot-and-Cross Formulations:
    • NaClNaCl: Metal Na (2.8.1)→[Na]+\text{Na}\,(2.8.1) \rightarrow [\text{Na}]^+; Non-metal Cl (2.8.7)→[Cl]−\text{Cl}\,(2.8.7) \rightarrow [\text{Cl}]^-.
    • MgOMgO: Metal Mg (2.8.2)→[Mg]2+\text{Mg}\,(2.8.2) \rightarrow [\text{Mg}]^{2+}; Non-metal O (2.6)→[O]2−\text{O}\,(2.6) \rightarrow [\text{O}]^{2-}.
    • Na2ONa_2O: Two [Na]+[\text{Na}]^+ cations donate to one [O]2−[\text{O}]^{2-} anion.
    • CaSCaS: Metal Ca (2.8.8.2)→[Ca]2+\text{Ca}\,(2.8.8.2) \rightarrow [\text{Ca}]^{2+}; Non-metal S (2.8.6)→[S]2−\text{S}\,(2.8.6) \rightarrow [\text{S}]^{2-}.
    • KFKF: Metal K (2.8.8.1)→[K]+\text{K}\,(2.8.8.1) \rightarrow [\text{K}]^+; Non-metal F (2.7)→[F]−\text{F}\,(2.7) \rightarrow [\text{F}]^-.
    • Ionic vs. Atomic Radii:
    • Cation Radius: A cation is smaller than its parent neutral atom (Na+<Na\text{Na}^+ < \text{Na}). Loss of valence electrons removes the outermost shell, reducing electron-electron repulsion and shielding; the unchanged nuclear charge pulls the remaining electron cloud closer to the nucleus.
    • Anion Radius: An anion is larger than its parent neutral atom (O2−>O\text{O}^{2-} > \text{O}). Gaining electrons increases inter-electronic repulsion without increasing nuclear charge, causing expansion of the electron cloud.
    • Factors Determining Ionic Bond Strength:
    • Ionic Radius: Smaller ions pack more closely together in the crystal lattice, leading to stronger electrostatic forces of attraction.
    • Ionic Charge Magnitude: Higher charges on the interacting ions generate exponentially stronger electrostatic attractions.
  • Covalent Bonding:

    • Formed by the sharing of electron pairs between atoms, typically between non-metals.
    • Can form between a metal and a non-metal if the difference in their electronegativities is very small.
    • Defined as the electrostatic attraction between the positively charged nuclei of the bonded atoms and the shared pair of electrons.
    • Dot-and-Cross Examples:
    • H2H_2: 1 bond pair, 0 lone pairs1\text{ bond pair, } 0\text{ lone pairs}
    • HClHCl: 1 bond pair, 3 lone pairs on Cl1\text{ bond pair, } 3\text{ lone pairs on Cl}
    • H2OH_2O: 2 bond pairs, 2 lone pairs on O2\text{ bond pairs, } 2\text{ lone pairs on O}
    • NH3NH_3: 3 bond pairs, 1 lone pair on N3\text{ bond pairs, } 1\text{ lone pair on N}
    • CO2CO_2: Two double bonds (4 shared pairs total4\text{ shared pairs total}, 2 lone pairs on each O2\text{ lone pairs on each O} )
    • CCl4CCl_4: 4 bond pairs, 3 lone pairs on each Cl4\text{ bond pairs, } 3\text{ lone pairs on each Cl}
    • Octet Deficiencies and Expansions:
    • Incomplete Octets: Compounds like AlCl3AlCl_3 (central aluminium surrounded by 66 valence electrons) and BeCl2BeCl_2 (central beryllium surrounded by 44 valence electrons) have incomplete octets; they are electron-deficient and reactive.
    • Expanded Octets: Phosphorus in PCl5PCl_5 possesses 1010 valence electrons; Sulfur in SF4SF_4 has 1010 electrons, and in SF6SF_6 has 1212 electrons. Elements in Period 3 and beyond can expand their octets because they possess accessible, empty 3d3d orbitals that can participate in bonding.
    • Molecular Existence Anomalies:
      • SF6 vs. OF6SF_6\text{ vs. } OF_6: SF6SF_6 exists because sulfur has empty 3d3d orbitals capable of accommodating 1212 bonding electrons. Oxygen (n=2n=2) lacks dd orbitals and cannot expand its octet beyond 88 electrons; hence OF6OF_6 cannot exist.
      • PCl5 vs. PI5PCl_5\text{ vs. } PI_5: Phosphorus has a small atomic radius and can accommodate five chlorine atoms. Iodine atoms have significantly larger atomic radii, making it impossible to fit five bulky iodine atoms around the small central phosphorus atom due to severe steric hindrance.
  • Coordinate (Dative) Covalent Bonding:

    • A covalent bond in which both shared electrons are provided by only one of the participating atoms.
    • Formation Conditions:
    • The donor particle must possess at least one lone pair of electrons.
    • The acceptor particle must possess at least one vacant orbital in its valence shell.
    • Examples:
    • Ammonium Ion (NH4+NH_4^+): Formed when ammonia reacts with an acid (H+H^+):       NH3+H+→NH4+NH_3 + H^+ \rightarrow NH_4^+       The nitrogen atom in ammonia donates its lone pair into the empty 1s1s orbital of the hydrogen ion.
    • Oxonium (Hydronium) Ion (H3O+H_3O^+): Formed when water coordinates with an acid (H+H^+):       H2O+H+→H3O+H_2O + H^+ \rightarrow H_3O^+
    • Aluminium Chloride - Ammonia Adduct (AlCl3⋅NH3AlCl_3 \cdot NH_3): The lone pair on ammonia's nitrogen coordinates into the empty orbital of the electron-deficient aluminium atom in AlCl3AlCl_3.
    • Aluminium Chloride Dimer (Al2Cl6Al_2Cl_6): In solid aluminium chloride, bonding is intermediate. At 178 ∘C178\,^\circ\text{C}, it sublimes to form vapor consisting of Al2Cl6Al_2Cl_6 dimers, where two bridging chlorine atoms each donate a lone pair to adjacent electron-deficient aluminium atoms. At high temperatures (800 ∘C800\,^\circ\text{C}), the dimer dissociates completely into trigonal planar AlCl3(g)AlCl_3(g).
  • Valence Bond Theory and Orbital Overlap:

    • Covalent bonds form when outer-shell atomic orbitals containing single unpaired electrons overlap, pairing the electrons.
    • Sigma (σ\sigma) Bonds:
    • Formed by direct head-on (axial) overlap of atomic orbitals (s−s,s−p, or p−ps-s, s-p,\text{ or } p-p).
    • All single covalent bonds are sigma bonds.
    • Sigma bonds are stronger than pi bonds because head-on overlap allows greater spatial overlap of atomic orbitals along the internuclear axis.
    • Pi (π\pi) Bonds:
    • Formed by the sideways (lateral) overlap of parallel pp orbitals above and below the internuclear plane.
    • A pi bond forms only when multiple bonds exist between two atoms: a double bond contains 1σ+1π1\sigma + 1\pi, and a triple bond contains 1σ+2π1\sigma + 2\pi.
    • Deducing Sigma and Pi Bond Counts:
    • H−C≡NH-C \equiv N: 2σ,2π2\sigma, 2\pi
    • Ethane (C2H6C_2H_6): 7σ,0π7\sigma, 0\pi
    • Propanoic acid (CH3CH2COOHCH_3CH_2COOH): 10σ,1π10\sigma, 1\pi
    • Sulfuric acid (H2SO4H_2SO_4 with two S=OS=O and two S−OHS-OH): 6σ,2π6\sigma, 2\pi
    • But-2-enal derivative (CH3−CH=CH−CHOCH_3-CH=CH-CHO): 10σ,2π10\sigma, 2\pi
  • Valence Shell Electron Pair Repulsion (VSEPR) Theory:

    • Electron pairs surrounding a central atom repel each other electrostatically and adopt a geometry that maximizes spatial separation, minimizing repulsion and giving the lowest energy.
    • Hierarchy of Repulsion:lone pair - lone pair>lone pair - bond pair>bond pair - bond pair\text{lone pair - lone pair} > \text{lone pair - bond pair} > \text{bond pair - bond pair}
    • Lone pairs are held closer to the central nucleus than bonding pairs, producing a more concentrated, localized electron cloud that exerts greater repulsive force, reducing adjacent bond angles.
    • Rules for Predicting Molecular Geometries:
    • Total electron pairs: Valence e− of central atom+electrons needed by peripheral atoms2\frac{\text{Valence } e^- \text{ of central atom} + \text{electrons needed by peripheral atoms}}{2}
    • Double bonds are treated as single bonding domains for geometry prediction.
    • For oxoanions/oxo-compounds (where terminal oxygen is double-bonded): Electron pairs=Valence e−+needed e−2−(Number of terminal oxygen atoms)\text{Electron pairs} = \frac{\text{Valence } e^- + \text{needed } e^-}{2} - (\text{Number of terminal oxygen atoms})
  • VSEPR Geometries and Bond Angles:

    • 2 Bond Pairs,0 Lone Pairs2\text{ Bond Pairs}, 0\text{ Lone Pairs}: Linear, bond angle = 180∘180^\circ. Examples: BeCl2,CO2,NO2+BeCl_2, CO_2, NO_2^+.
    • 3 Bond Pairs,0 Lone Pairs3\text{ Bond Pairs}, 0\text{ Lone Pairs}: Trigonal planar, bond angle = 120∘120^\circ. Examples: AlCl3,BH3,BCl3,CO32−,NO3−AlCl_3, BH_3, BCl_3, CO_3^{2-}, NO_3^-.
    • 4 Bond Pairs,0 Lone Pairs4\text{ Bond Pairs}, 0\text{ Lone Pairs}: Tetrahedral, bond angle = 109.5∘109.5^\circ. Examples: CH4,CCl4,SiCl4,SO42−,NH4+CH_4, CCl_4, SiCl_4, SO_4^{2-}, NH_4^+.
    • 3 Bond Pairs,1 Lone Pair3\text{ Bond Pairs}, 1\text{ Lone Pair}: Trigonal pyramidal, bond angle = 107∘107^\circ. The lone pair pushes the three bonding pairs closer together. Examples: NH3,PCl3,H3O+NH_3, PCl_3, H_3O^+.
    • 2 Bond Pairs,2 Lone Pairs2\text{ Bond Pairs}, 2\text{ Lone Pairs}: Bent / Non-linear / V-shaped, bond angle = 104.5∘104.5^\circ. Two lone pairs exert strong mutual repulsion, reducing the angle from tetrahedral. Examples: H2O,H2S,H2Se,F2OH_2O, H_2S, H_2Se, F_2O.
    • 5 Bond Pairs,0 Lone Pairs5\text{ Bond Pairs}, 0\text{ Lone Pairs}: Trigonal bipyramidal, bond angles = 90∘,120∘,180∘90^\circ, 120^\circ, 180^\circ. Example: PCl5PCl_5.
    • 6 Bond Pairs,0 Lone Pairs6\text{ Bond Pairs}, 0\text{ Lone Pairs}: Octahedral, bond angle = 90∘90^\circ. Example: SF6SF_6.
    • 4 Bond Pairs,2 Lone Pairs4\text{ Bond Pairs}, 2\text{ Lone Pairs}: Square planar, bond angle = 90∘90^\circ. The two lone pairs occupy opposite axial positions (180∘180^\circ apart) to minimize repulsion. Examples: XeF4,IF4−XeF_4, IF_4^-.
  • Worked Examples of Shape Determinations:

    • H2OH_2O: 2+62=4 pairs\frac{2 + 6}{2} = 4\text{ pairs} (2 bp,2 lp2\text{ bp}, 2\text{ lp}) →\rightarrow Bent (104.5∘104.5^\circ).
    • NH3NH_3: 5+32=4 pairs\frac{5 + 3}{2} = 4\text{ pairs} (3 bp,1 lp3\text{ bp}, 1\text{ lp}) →\rightarrow Trigonal pyramidal (107∘107^\circ).
    • NH4+NH_4^+: 5+4−12=4 pairs\frac{5 + 4 - 1}{2} = 4\text{ pairs} (4 bp4\text{ bp}) →\rightarrow Tetrahedral (109.5∘109.5^\circ).
    • SF6SF_6: 6+6(1)2=6 pairs\frac{6 + 6(1)}{2} = 6\text{ pairs} (6 bp6\text{ bp}) →\rightarrow Octahedral (90∘90^\circ).
    • CH3+CH_3^+: 4+3−12=3 pairs\frac{4 + 3 - 1}{2} = 3\text{ pairs} (3 bp3\text{ bp}) →\rightarrow Trigonal planar (120∘120^\circ).
    • F2OF_2O: 2+62=4 pairs\frac{2 + 6}{2} = 4\text{ pairs} (2 bp,2 lp2\text{ bp}, 2\text{ lp}) →\rightarrow Bent (104.5∘104.5^\circ).
    • CO2CO_2: 4+2(2)2−2=2 pairs\frac{4 + 2(2)}{2} - 2 = 2\text{ pairs} (2 bp2\text{ bp}) →\rightarrow Linear (180∘180^\circ).
    • CO32−CO_3^{2-}: 4+3(2)+22−3=3 pairs\frac{4 + 3(2) + 2}{2} - 3 = 3\text{ pairs} (3 bp3\text{ bp}) →\rightarrow Trigonal planar (120∘120^\circ).
    • SO2SO_2: 6+2(2)2−2=3 pairs\frac{6 + 2(2)}{2} - 2 = 3\text{ pairs} (2 bp,1 lp2\text{ bp}, 1\text{ lp}) →\rightarrow Bent.
    • SO32−SO_3^{2-}: 6+3(2)+22−3=4 pairs\frac{6 + 3(2) + 2}{2} - 3 = 4\text{ pairs} (3 bp,1 lp3\text{ bp}, 1\text{ lp}) →\rightarrow Trigonal pyramidal.

Electronegativity, Bond Polarity, and Intermolecular Forces

  • Electronegativity:

    • The power of an atom to attract the shared pair of electrons in a covalent bond toward itself.
    • Governing Factors:
    • Atomic Radius: Smaller atoms pull shared electrons closer to their nucleus, giving higher electronegativity.
    • Shielding: Fewer shielding shells permit a stronger effective nuclear pull on shared electrons.
    • Nuclear Charge: A higher positive charge in the nucleus increases the attraction for shared electron pairs.
    • Periodic Trends:
    • Down a group: Electronegativity decreases because increases in atomic radius and shielding outweigh the added nuclear charge.
    • Across a period: Electronegativity increases because atomic radius decreases, shielding remains constant, and nuclear charge increases.
    • Most electronegative elements in decreasing order: F>O>N≈Cl\text{F} > \text{O} > \text{N} \approx \text{Cl}.
  • Bond Polarity and Molecular Dipoles:

    • Differences in electronegativity generate bond dipoles: the more electronegative atom acquires a partial negative charge (δ−\delta-) and the less electronegative acquires a partial positive charge (δ+\delta+).
    • A large electronegativity difference leads to ionic character; a small or zero difference results in covalent character.
    • Pure non-polar bonds occur between identical elements or elements with nearly identical electronegativities (H−H,O=O,F−F,Cl−Cl,C−HH-H, O=O, F-F, Cl-Cl, C-H).
    • Molecular Polarity:
    • A molecule is polar only if it possesses polar bonds arranged asymmetrically such that individual bond dipoles do not cancel.
    • Highly symmetrical molecules with polar bonds are non-polar overall because their bond dipoles cancel out:
      • Linear Y−X−YY-X-Y (CO2,Cl−Be−ClCO_2, Cl-Be-Cl) →\rightarrow Non-polar.
      • Trigonal planar AY3AY_3 (BF3BF_3) →\rightarrow Non-polar; but AB2YAB_2Y is polar.
      • Tetrahedral AY4AY_4 (CCl4CCl_4) →\rightarrow Non-polar; but AY3XAY_3X (CHCl3CHCl_3) is polar.
      • Trans-1,2-dichloroethene →\rightarrow Non-polar; Cis-1,2-dichloroethene →\rightarrow Polar.
      • Trans-diamminedichloroplatinum(II) →\rightarrow Non-polar; Cisplatin →\rightarrow Polar.
      • 1,4-dichlorobenzene →\rightarrow Non-polar; 1,2-dichlorobenzene →\rightarrow Polar.
      • 1,3,5-trichlorobenzene →\rightarrow Non-polar; 1,2,3-trichlorobenzene →\rightarrow Polar.
  • Impact of Polarity on Bond Strength and Reactivity:

    • Bond Strength: Polar bonds are generally stronger than equivalent non-polar bonds because electrostatic attractions between dipoles reinforce the covalent sharing (C≡O>N≡NC \equiv O > N \equiv N).
    • Halogen Halide Trend: Along the series H−F,H−Cl,H−Br,H−IH-F, H-Cl, H-Br, H-I, the electronegativity of the halogen decreases, bonds become less polarized, and bond strength progressively decreases.
    • Reactivity: Polar molecules are generally more chemically reactive than non-polar molecules because their dipoles attract incoming charged species and electrophiles/nucleophiles. For example, carbon monoxide (C≡OC \equiv O) is more reactive than dinitrogen (N≡NN \equiv N) because of its dipole moment.
    • Factors Influencing Covalent Bond Strength:
    • Bond polarity (greater polarization increases electrostatic attraction, strengthening the bond).
    • Number of shared electron pairs (triple bonds are stronger than double bonds, which are stronger than single bonds).
    • Atomic radii of bonded atoms (smaller atoms allow closer orbital overlap, yielding stronger, shorter bonds).
  • Intermolecular Forces (Van der Waals Forces):

    • Forces of attraction operating between separate molecules (distinct from intramolecular covalent bonds within a molecule).
    • Permanent Dipole-Dipole (p.d.d.) Forces: Electrostatic attractions between molecules containing permanent dipoles (HCl,CHCl3HCl, CHCl_3).
    • Induced Dipole-Dipole (i.d.d.) Forces (London Dispersion Forces):
    • The primary intermolecular attraction in non-polar molecules.
    • Formed by temporary imbalances in electron distribution, generating instantaneous dipoles that induce complementary dipoles in neighboring molecules.
    • Factors Affecting i.d.d. Strength:
      • Electron Density / Number of Electrons: Molecules with larger electron clouds polarize more easily, producing stronger, more extensive i.d.d. forces.
      • Surface Area of Contact: Molecules with larger surface areas exhibit stronger i.d.d. forces. Unbranched alkanes pack more efficiently and have greater surface contact than branched isomers, giving them higher boiling points (e.g., butane boils higher than methylpropane).
    • Relative Strength of Forces:Hydrogen bonding>Permanent dipole-dipole>Induced dipole-dipole\text{Hydrogen bonding} > \text{Permanent dipole-dipole} > \text{Induced dipole-dipole}
  • Hydrogen Bonding:

    • A strong electrostatic attraction between a hydrogen atom covalently bound to a highly electronegative element (F,O,NF, O, N) and the lone pair of electrons on a neighboring electronegative atom (F,O,NF, O, N).
    • Criteria for Formation:
    • A hydrogen atom directly bonded to F,O,F, O, or NN.
    • An adjacent F,O,F, O, or NN atom carrying at least one available lone pair.
    • Comparative Physical Properties:
    • Noble Gases (Group 18): Boiling points increase from HeHe to XeXe due to increasing electron counts, yielding stronger and more extensive i.d.d. forces.
    • HClHCl vs. ArAr: HClHCl has a higher boiling point than argon because polar HClHCl molecules are held by stronger permanent dipole-dipole forces, whereas non-polar argon atoms rely on weaker i.d.d. forces.
    • Hexane vs. Propane: Hexane has a higher boiling point due to its larger electron cloud and larger surface area of contact.
    • H2OH_2O vs. HFHF vs. NH3NH_3:
      • Water has the highest boiling point (100 ∘C100\,^\circ\text{C}), followed by HFHF, with NH3NH_3 lowest.
      • Although fluorine is more electronegative than oxygen, each water molecule forms an average of two hydrogen bonds (possessing two hydrogen atoms and two lone pairs). Hydrogen fluoride can form only one hydrogen bond per molecule on average due to having only one hydrogen atom.
      • Ammonia has the lowest boiling point because nitrogen is the least electronegative of the three, forming weaker hydrogen bonds.
    • Ethanol vs. Ethylamine: Ethanol has a higher boiling point because oxygen is more electronegative than nitrogen, forming stronger hydrogen bonds.
    • 1,4-Benzenediol vs. 1,2-Benzenediol: 1,4-Benzenediol forms extensive intermolecular hydrogen bonds, leading to a higher melting/boiling point. In 1,2-benzenediol, adjacent hydroxyl groups form intramolecular hydrogen bonds, reducing intermolecular interactions.

Gas Laws, Ideal vs. Real Gases, and Physical States

  • Kinetic Theory of Gases:

    • Describes gas particles as being in continuous, rapid, random motion, colliding with each other and with the container walls.
  • Postulates of Ideal Gases:

    • Intermolecular forces of attraction between gas molecules are negligible (zero).
    • The actual volume of the gas molecules is negligible compared to the total volume occupied by the gas (∑Vmolecules≪Vcontainer\sum V_{\text{molecules}} \ll V_{\text{container}}).
    • Collisions between gas particles and with the container walls are perfectly elastic (no loss of kinetic energy).
    • The absolute temperature of a gas is directly proportional to the average kinetic energy of its molecules.
  • The Gas Laws:

    • Boyle's Law: At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure:     V∝1p  ⟹  p1V1=p2V2V \propto \frac{1}{p} \implies p_1 V_1 = p_2 V_2
    • Charles's Law: At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature (TT in Kelvin):     V∝T  ⟹  V1T1=V2T2V \propto T \implies \frac{V_1}{T_1} = \frac{V_2}{T_2}     Absolute zero corresponds to −273 ∘C-273\,^\circ\text{C} (0 K0\,K).
    • Avogadro's Law: At constant temperature and pressure, the volume occupied by a gas is directly proportional to the number of moles (nn) of gas:     V∝nV \propto n
  • The Ideal Gas Equation:

    • Combining the empirical gas laws yields:     pV=nRTpV = nRT
    • SI Units:
    • pp = Pressure in Pascals (PaPa), where 1 kPa=103 Pa1\,kPa = 10^3\,Pa
    • VV = Volume in cubic meters (m3m^3), where 1 dm3=10−3 m31\,dm^3 = 10^{-3}\,m^3 and 1 cm3=10−6 m31\,cm^3 = 10^{-6}\,m^3
    • nn = Moles of gas (molmol)
    • TT = Temperature in Kelvin (KK), where K=∘C+273K = ^\circ\text{C} + 273
    • RR = Molar gas constant = 8.31 J K−1 mol−18.31\,J\,K^{-1}\,mol^{-1}
    • Alternative Algebraic Forms:
    • Using molar mass (MrM_r): pV=mMrRTpV = \frac{m}{M_r}RT
    • Gas density (ρ=mV\rho = \frac{m}{V}): pMr=ρRTp M_r = \rho RT
    • Molar concentration (c=nVc = \frac{n}{V}): p=cRTp = cRT
  • Worked Ideal Gas Calculations:

    • Volume Calculation: Volume occupied by 0.500 mol0.500\,mol of CO2CO_2 at 150 kPa150\,kPa and 19 ∘C19\,^\circ\text{C}:     p=150×103 Pa,T=19+273=292 Kp = 150 \times 10^3\,Pa, \quad T = 19 + 273 = 292\,KV=nRTp=0.500×8.31×292150×103=8.09×10−3 m3V = \frac{nRT}{p} = \frac{0.500 \times 8.31 \times 292}{150 \times 10^3} = 8.09 \times 10^{-3}\,m^3
    • Pressure Calculation: Pressure of 4 g4\,g of oxygen (O2O_2, Mr=32M_r = 32) in a 5 dm35\,dm^3 flask at 127 ∘C127\,^\circ\text{C}:     V=5×10−3 m3,T=127+273=400 KV = 5 \times 10^{-3}\,m^3, \quad T = 127 + 273 = 400\,Kp=mRTMrV=4×8.31×40032×5×10−3=83100 Pa=83.1 kPap = \frac{m R T}{M_r V} = \frac{4 \times 8.31 \times 400}{32 \times 5 \times 10^{-3}} = 83100\,Pa = 83.1\,kPa
    • Volume of Methane: Volume of 272 g272\,g of CH4CH_4 (Mr=16M_r = 16) at 250 kPa250\,kPa and 54 ∘C54\,^\circ\text{C}:     p=250×103 Pa,T=54+273=327 Kp = 250 \times 10^3\,Pa, \quad T = 54 + 273 = 327\,K250×103×V=27216×8.31×327  ⟹  V=0.185 m3250 \times 10^3 \times V = \frac{272}{16} \times 8.31 \times 327 \implies V = 0.185\,m^3
    • Molar Mass Determination: A 2 dm32\,dm^3 flask holds 5.28 g5.28\,g of gas at 200 kPa200\,kPa and 20 ∘C20\,^\circ\text{C}:     V=2×10−3 m3,p=200×103 Pa,T=20+273=293 KV = 2 \times 10^{-3}\,m^3, \quad p = 200 \times 10^3\,Pa, \quad T = 20 + 273 = 293\,KMr=mRTpV=5.28×8.31×293(200×103)×(2×10−3)=32.1 g mol−1M_r = \frac{m R T}{p V} = \frac{5.28 \times 8.31 \times 293}{(200 \times 10^3) \times (2 \times 10^{-3})} = 32.1\,g\,mol^{-1}
    • Expansion and Compression Calculations (p1V1=p2V2p_1 V_1 = p_2 V_2):
    • 20 cm320\,cm^3 of H2H_2 at 200 kPa200\,kPa expands to 90 cm390\,cm^3 at constant temperature:       (200×103)×(2×10−5)=p2×(9×10−5)  ⟹  p2=44.4 kPa(200 \times 10^3) \times (2 \times 10^{-5}) = p_2 \times (9 \times 10^{-5}) \implies p_2 = 44.4\,kPa
    • A gas in a 2 dm32\,dm^3 flask at 2 atm2\,atm is transferred to a 1 dm31\,dm^3 flask:       2×2=p2×1  ⟹  p2=4 atm2 \times 2 = p_2 \times 1 \implies p_2 = 4\,atm
    • Mixture of Gases (p1V1+p2V2=p3V3p_1 V_1 + p_2 V_2 = p_3 V_3):
    • A 2 dm32\,dm^3 flask of CO2CO_2 at 10 kPa10\,kPa is connected to a 3 dm33\,dm^3 flask at 8 kPa8\,kPa (V3=5 dm3V_3 = 5\,dm^3):       (10×103)(2)+(8×103)(3)=p3×5(10 \times 10^3)(2) + (8 \times 10^3)(3) = p_3 \times 520000+24000=5p3  ⟹  p3=8800 Pa=8.8 kPa20000 + 24000 = 5 p_3 \implies p_3 = 8800\,Pa = 8.8\,kPa
  • Limitations of Ideal Gas Laws and Real Gas Deviations:

    • Real gases deviate from ideality because intermolecular forces exist between molecules, and gas molecules occupy a finite volume.
    • Conditions Causing Deviation:
    • High Pressure: Molecules are compressed close together. Their physical volume becomes a significant fraction of the container volume, and intermolecular attractions increase.
    • Low Temperature: Lower kinetic energy allows attractive forces to pull molecules together, reducing collisions with container walls.
    • Conditions Approaching Ideal Behavior: Low pressure and high temperature minimize intermolecular interactions and make molecular volume negligible.
    • Pressure Difference: Real gases exert a lower pressure than predicted by the ideal gas equation because intermolecular attractions pull molecules toward one another and away from container walls.
    • Most Ideal Gas: Helium deviates least because it is a single small atom with only two electrons, giving minimal van der Waals forces.
    • Relative Deviations:
    • CO2CO_2 deviates more than H2H_2 due to its larger electron cloud and stronger van der Waals forces.
    • NH3NH_3 deviates more than N2N_2 due to strong hydrogen bonding.
    • Argon deviates more than neon or helium due to its larger atomic size and stronger van der Waals forces.

Solid State Structures and Carbon Allotropes

  • Physical properties of substances depend on their internal structure and the nature of the chemical bonds present.

  • Four structural classes exist: Giant ionic, Giant molecular, Simple molecular, and Giant metallic.

  • Giant Ionic Structures:

    • Composed of a three-dimensional lattice of alternating positive and negative ions held by ionic bonds.
    • Each cation is surrounded by a set number of anions, and vice versa.
    • Coordination Number: The number of oppositely charged ions immediately surrounding a given ion.
    • Depends on the relative sizes and charges of the ions.
    • In NaClNaCl, each Na+\text{Na}^+ ion is coordinated by 6 Cl−6\text{ Cl}^- ions, and each Cl−\text{Cl}^- is surrounded by 6 Na+6\text{ Na}^+ ions (6:66:6 coordination).
    • Physical Properties:
    • High melting and boiling points due to strong electrostatic attractions throughout the lattice.
    • Soluble in polar solvents (e.g., water), which hydrate and separate ions, but insoluble in non-polar organic solvents.
    • Electrical non-conductors in the solid state because ions are fixed in the lattice; good electrical conductors when molten or dissolved in aqueous solution because ions become mobile.
    • Hard but brittle: mechanical stress can cause layers of ions to shift, placing like charges adjacent to one another; the resulting electrostatic repulsion cleaves the crystal.
  • Simple Molecular Structures:

    • Small covalent molecules held internally by strong covalent bonds, but mutually attracted by weak intermolecular forces.
    • Low melting and boiling points; electrical insulators in all states; generally insoluble in water but soluble in non-polar organic solvents.
  • Giant Molecular (Macromolecular) Structures:

    • High melting and boiling points; insoluble in all solvents; typically very hard and non-conductive (except graphite).
  • Allotropes of Carbon:

    • Allotropes are different structural forms of the same element existing in the same physical state.
    • Diamond:
    • Each carbon atom is covalently bonded to four other carbon atoms in an interlocking tetrahedral framework.
    • Hybridization: sp3sp^3 with bond angles of 109.5∘109.5^\circ.
    • One of the hardest known substances; high melting and boiling points; electrical non-conductor; insoluble in all solvents.
    • Uses: Jewelry, cutting tools, drill bits, and industrial wire-drawing dies.
    • Graphite:
    • Carbon atoms form flat, parallel layers of interconnected hexagonal rings.
    • Hybridization: sp2sp^2 with bond angles of 120∘120^\circ.
    • Each carbon bonds to three others; the fourth valence electron occupies an unhybridized 2p2p orbital.
    • Overlapping 2p2p orbitals form an extended delocalized π\pi electron cloud across each layer.
    • The carbon-carbon bonds within a layer are shorter and stronger than those in diamond due to partial double bond character.
    • Weak van der Waals forces hold adjacent layers together, allowing them to slide easily.
    • Good electrical conductor along its layers; soft and slippery; high melting and boiling points; chemically inert.
    • Uses: Dry lubricant, inert electrodes, and pencil leads.
    • Fullerenes (C60C_{60} Buckminsterfullerene):
    • Simple molecular nanoparticle (10−9 m10^{-9}\,m) shaped like a hollow sphere.
    • Consists of 6060 carbon atoms arranged in 2020 hexagons and 1212 pentagons.
    • Each carbon bonds to three others; hybridization is intermediate between sp2sp^2 and sp3sp^3, with bond angles of approximately 108∘108^\circ.
    • Possesses partially delocalized electrons, but delocalization is less extensive than in graphite; non-polar.
    • Low sublimation point and relatively soft due to weak van der Waals forces between molecules; poor electrical conductor; insoluble in water but slightly soluble in organic solvents.
    • More chemically reactive than diamond or graphite due to localized regions of higher electron density; reacts with H2,F2,Br2,H_2, F_2, Br_2, and O2O_2.
    • Uses: Targeted drug delivery cages, catalysts, and specialized lubricants.

Structure of buckminsterfullerene C60 molecule

  • Graphene:
    • A single two-dimensional layer of graphite, consisting of a hexagonal honeycomb lattice of carbon atoms.
    • Hybridization: sp2sp^2 with bond angles of 120∘120^\circ.
    • Carbon-carbon bonds are stronger than typical single bonds.
    • Nearly transparent; exceptionally strong and flexible; very low density; superior conductor of electricity and heat (outperforming copper and graphite); highly chemically reactive.
    • Uses: High-speed microcircuits, touchscreens, and solar cells.
  • Carbon Nanotubes:
    • Cylindrical structures formed by rolled graphene sheets.
    • Cylinder walls consist of hexagonal carbon rings; ends are capped by hemispherical fullerene-like structures containing both pentagons and hexagons.
    • Possesses high electrical conductivity, high melting points, and tensile strength up to 100100 times greater than steel of the same diameter.
    • Uses: Nanoscale electronics, reinforced composites, and sports equipment (e.g., tennis rackets).

Mass Spectrometry and Isotopic Analysis

  • Principles of Mass Spectrometry:

    • Vaporization: The sample is converted into gaseous atoms or molecules.
    • Ionization: High-energy electron bombardment knocks valence electrons out of the molecules, producing unipositive ions:     CH4+e−→CH4++2e−CH_4 + e^- \rightarrow CH_4^+ + 2e^-     This process deposits energy into the molecular ion, causing fragmentation along chemical bonds:     CH4+→CH3++H∙→CH2+→CH+→C+CH_4^+ \rightarrow CH_3^+ + H^\bullet \rightarrow CH_2^+ \rightarrow CH^+ \rightarrow C^+
    • Acceleration and Deflection: Gaseous cations are accelerated by an electric field and deflected by an external magnetic field according to their mass-to-charge (m/em/e) ratio.
    • Detection: An ion detector records the relative abundance of each arriving species.
  • Spectroscopic Interpretation:

    • Molecular Ion Peak (MM): The peak with the highest m/em/e ratio corresponds to the unfragmented ionized molecule, giving its relative molecular mass (MrM_r).
    • Fragmentation Pattern: Allows identification of specific structural groups and isomers.
    • Methane (CH4CH_4, Mr=16M_r = 16): Peaks at m/e=16 ([CH4]+),15 ([CH3]+),14 ([CH2]+),13 ([CH]+),12 (C+),1 (H+)m/e = 16\,([CH_4]^+), 15\,([CH_3]^+), 14\,([CH_2]^+), 13\,([CH]^+), 12\,(C^+), 1\,(H^+).
    • Propane (C3H8C_3H_8, Mr=44M_r = 44): Peaks at m/e=44 ([C3H8]+),43 ([C3H7]+),42 ([C3H6]+),29 ([C2H5]+),15 ([CH3]+),12 (C+),1 (H+)m/e = 44\,([C_3H_8]^+), 43\,([C_3H_7]^+), 42\,([C_3H_6]^+), 29\,([C_2H_5]^+), 15\,([CH_3]^+), 12\,(C^+), 1\,(H^+).
    • Butane Isomers: Unbranched butane can be distinguished from 2-methylpropane by the presence of a strong ethyl fragment ([C2H5]+[C_2H_5]^+) at m/e=29m/e = 29.
  • Halogen Isotopic Patterns:

    • Chlorine (Cl2Cl_2):
    • Naturally occurring isotopes: 35Cl^{35}Cl and 37Cl^{37}Cl in a 3:13:1 abundance ratio.
    • Molecular species (Cl2+Cl_2^+):
      • 35Cl−35Cl^{35}Cl-^{35}Cl (m/e=70m/e = 70): Abundance = 34×34=916\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}
      • 35Cl−37Cl^{35}Cl-^{37}Cl (m/e=72m/e = 72): Abundance = 2×(34×14)=6162 \times \left(\frac{3}{4} \times \frac{1}{4}\right) = \frac{6}{16}
      • 37Cl−37Cl^{37}Cl-^{37}Cl (m/e=74m/e = 74): Abundance = 14×14=116\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}
      • The molecular ion region displays peaks at 70,72,7470, 72, 74 in a 9:6:19:6:1 ratio.
    • Monatomic fragments (Cl+Cl^+): Peaks at m/e=35m/e = 35 (34\frac{3}{4}) and m/e=37m/e = 37 (14\frac{1}{4}).
    • Bromine (Br2Br_2):
    • Naturally occurring isotopes: 79Br^{79}Br and 81Br^{81}Br in an approximately 1:11:1 ratio.
    • Molecular species (Br2+Br_2^+):
      • 79Br−79Br^{79}Br-^{79}Br (m/e=158m/e = 158): Abundance = 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
      • 79Br−81Br^{79}Br-^{81}Br (m/e=160m/e = 160): Abundance = 2×(12×12)=12=242 \times \left(\frac{1}{2} \times \frac{1}{2}\right) = \frac{1}{2} = \frac{2}{4}
      • 81Br−81Br^{81}Br-^{81}Br (m/e=162m/e = 162): Abundance = 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
      • The molecular ion region displays peaks at 158,160,162158, 160, 162 in a 1:2:11:2:1 ratio.
    • Monatomic fragments (Br+Br^+): Peaks at m/e=79m/e = 79 (12\frac{1}{2}) and m/e=81m/e = 81 (12\frac{1}{2}).
  • The M+1M+1 Peak and Carbon Number Calculation:

    • The M+1M+1 peak is produced by the presence of the carbon-13 (13C^{13}C) isotope (natural abundance 1.1%1.1\% relative to 12C^{12}C at 98.9%98.9\%).
    • The ratio of peak heights gives the number of carbon atoms (nn) in the molecule:     AMAM+1=100n×1.1\frac{A_M}{A_{M+1}} = \frac{100}{n \times 1.1}n=AM+1×100AM×1.1n = \frac{A_{M+1} \times 100}{A_M \times 1.1}
    • Worked Calculations:
    • An organic compound has AM=48A_M = 48 and AM+1=1.7A_{M+1} = 1.7:       n=1.7×10048×1.1=17052.8=3.2≈3 carbon atomsn = \frac{1.7 \times 100}{48 \times 1.1} = \frac{170}{52.8} = 3.2 \approx 3\text{ carbon atoms}
    • An alkane has AM=51A_M = 51 and AM+1=0.55A_{M+1} = 0.55:       n=0.55×10051×1.1=0.98≈1  ⟹  CH4n = \frac{0.55 \times 100}{51 \times 1.1} = 0.98 \approx 1 \implies CH_4
    • Calculating M+1M+1 for butane (C4H10C_4H_{10}, n=4n=4) where AM=12%A_M = 12\%:       12AM+1=1004×1.1  ⟹  AM+1=12×4.4100=0.528%≈0.53%\frac{12}{A_{M+1}} = \frac{100}{4 \times 1.1} \implies A_{M+1} = \frac{12 \times 4.4}{100} = 0.528\% \approx 0.53\%
  • The M+2M+2 Peak:

    • A peak at m/e=M+2m/e = M+2 indicates the presence of chlorine or bromine in the molecule:
    • If M:M+2≈3:1M : M+2 \approx 3:1, the compound contains one chlorine atom.
    • If M:M+2≈1:1M : M+2 \approx 1:1, the compound contains one bromine atom.

Chemical Energetics, Calorimetry, and Standard Enthalpy Changes

  • Thermodynamic Principles:

    • Chemical energetics is the study of energy transformations occurring during chemical reactions.
    • Enthalpy Change (ΔH\Delta H): The heat energy transferred into or out of a system at constant pressure.
    • Standard Conditions: Temperature = 25 ∘C25\,^\circ\text{C} (298 K298\,K), Pressure = 1 atm1\,atm (101 kPa101\,kPa), with all participating substances in their standard physical states. Denoted by ΔH⊖\Delta H^\ominus (kJ mol−1kJ\,mol^{-1}).
  • Reaction Energetics:

    • Exothermic Reactions:
    • Heat energy is released to the surroundings (ΔH<0\Delta H < 0).
    • The temperature of the surroundings rises.
    • The internal energy of the system decreases; products are energetically more stable than reactants.
    • Examples: Combustion, neutralisation, dissolving NaOHNaOH in water, rusting.
    • Endothermic Reactions:
    • Heat energy is absorbed from the surroundings (ΔH>0\Delta H > 0).
    • The temperature of the surroundings falls.
    • The internal energy of the system increases; products are energetically less stable than reactants.
    • Examples: Photosynthesis, thermal decomposition, melting, boiling.
    • Isothermic Reactions: Processes in which no net energy exchange occurs (ΔH=0\Delta H = 0).
    • Activation Energy (EaE_a): The minimum energy colliding particles must possess for a reaction to occur. Low EaE_a corresponds to rapid reaction rates.
  • Laboratory Calorimetry:

    • Calorimetry experiments are typically conducted using a polystyrene cup to minimize heat exchange with the surroundings.

Simple polystyrene cup calorimeter setup

  • Procedure:

    1. Measure the initial and peak final temperatures to determine ΔT=Tfinal−Tinitial\Delta T = T_{\text{final}} - T_{\text{initial}}.
    2. Calculate heat exchanged: ΔE=−(mcΔT) J\Delta E = -(m c \Delta T)\,J, where mm is the mass of solution (density assumed 1.0 g cm−31.0\,g\,cm^{-3}) and c=4.18 J K−1 g−1c = 4.18\,J\,K^{-1}\,g^{-1}.
    3. Determine the moles of the limiting reactant.
    4. Calculate molar enthalpy: ΔH=−mcΔTnlimiting×1000 kJ mol−1\Delta H = \frac{-m c \Delta T}{n_{\text{limiting}} \times 1000}\,kJ\,mol^{-1}.
    • Worked Calorimetric Calculations:
  • Neutralisation of NaOHNaOH by HNO3HNO_3:20 cm3 of 0.2 mol dm−3 NaOH at 22 ∘C+40 cm3 of 0.2 mol dm−3 HNO3 at 22 ∘C→Tf=32 ∘C20\,cm^3\text{ of } 0.2\,mol\,dm^{-3}\,NaOH\text{ at } 22\,^\circ\text{C} + 40\,cm^3\text{ of } 0.2\,mol\,dm^{-3}\,HNO_3\text{ at } 22\,^\circ\text{C} \rightarrow T_f = 32\,^\circ\text{C}ΔT=32−22=10 ∘C\Delta T = 32 - 22 = 10\,^\circ\text{C}ΔE=−(20+40)(4.18)(10)=−2508 J\Delta E = -(20 + 40)(4.18)(10) = -2508\,Jn(NaOH)=0.2×201000=0.004 mol(limiting)n(NaOH) = \frac{0.2 \times 20}{1000} = 0.004\,mol \quad (\text{limiting})n(HNO3)=0.2×401000=0.008 moln(HNO_3) = \frac{0.2 \times 40}{1000} = 0.008\,molΔH=−25080.004×1000=−627 kJ mol−1\Delta H = \frac{-2508}{0.004 \times 1000} = -627\,kJ\,mol^{-1}

  • Reaction of MgMg with HClHCl:2.4 g of Mg reacts with 150 cm3 of 2 mol dm−3 HCl;Ti=25 ∘C,Tf=40 ∘C2.4\,g\text{ of } Mg\text{ reacts with } 150\,cm^3\text{ of } 2\,mol\,dm^{-3}\,HCl; \quad T_i = 25\,^\circ\text{C}, T_f = 40\,^\circ\text{C}ΔT=40−25=15 ∘C\Delta T = 40 - 25 = 15\,^\circ\text{C}ΔE=−(150)(4.18)(15)=−9405 J\Delta E = -(150)(4.18)(15) = -9405\,Jn(Mg)=2.424=0.1 mol(limiting),n(HCl)=2×1501000=0.3 moln(Mg) = \frac{2.4}{24} = 0.1\,mol \quad (\text{limiting}), \quad n(HCl) = \frac{2 \times 150}{1000} = 0.3\,molΔH=−94050.1×1000=−94.05 kJ mol−1\Delta H = \frac{-9405}{0.1 \times 1000} = -94.05\,kJ\,mol^{-1}

  • Reaction of H2SO4H_2SO_4 with Na2CO3Na_2CO_3:80 cm3 of 1 mol dm−3 H2SO4+60 cm3 of 1 mol dm−3 Na2CO3;ΔT=+4 ∘C80\,cm^3\text{ of } 1\,mol\,dm^{-3}\,H_2SO_4 + 60\,cm^3\text{ of } 1\,mol\,dm^{-3}\,Na_2CO_3; \quad \Delta T = +4\,^\circ\text{C}ΔE=−(80+60)(4.18)(4)=−2340.8 J\Delta E = -(80 + 60)(4.18)(4) = -2340.8\,Jn(Na2CO3)=1×601000=0.06 mol(limiting),n(H2SO4)=0.08 moln(Na_2CO_3) = \frac{1 \times 60}{1000} = 0.06\,mol \quad (\text{limiting}), \quad n(H_2SO_4) = 0.08\,molΔH=−2340.80.06×1000=−39.0 kJ mol−1\Delta H = \frac{-2340.8}{0.06 \times 1000} = -39.0\,kJ\,mol^{-1}

  • Combustion of Glucose:1.8 g of glucose (C6H12O6,Mr=180) warms 150 cm3 of water by 15 ∘C1.8\,g\text{ of glucose } (C_6H_{12}O_6, M_r = 180)\text{ warms } 150\,cm^3\text{ of water by } 15\,^\circ\text{C}ΔE=−(150)(4.18)(15)=−9405 J\Delta E = -(150)(4.18)(15) = -9405\,Jn=1.8180=0.01 moln = \frac{1.8}{180} = 0.01\,molΔHc=−94050.01×1000=−940.5 kJ mol−1\Delta H_c = \frac{-9405}{0.01 \times 1000} = -940.5\,kJ\,mol^{-1}

    • Standard Enthalpy Definitions:
  • Standard Enthalpy of Reaction (ΔHr⊖\Delta H_r^\ominus): Energy change when stoichiometric molar quantities react under standard conditions.

  • Standard Enthalpy of Formation (ΔHf⊖\Delta H_f^\ominus): Energy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.

    • ΔHf⊖\Delta H_f^\ominus for an element in its standard state is zero (O2(g)→O2(g),ΔHf⊖=0\text{O}_2(g) \rightarrow \text{O}_2(g), \Delta H_f^\ominus = 0).
  • Standard Enthalpy of Combustion (ΔHc⊖\Delta H_c^\ominus): Energy change when one mole of a substance burns completely in excess oxygen under standard conditions (always negative).

  • Standard Enthalpy of Atomisation (ΔHat⊖\Delta H_{at}^\ominus): Energy change when one mole of gaseous atoms is produced from an element in its standard state (always positive, involving only bond cleavage):     12Cl2(g)→Cl(g)(ΔHat⊖ of Cl2)\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g) \quad (\Delta H_{at}^\ominus\text{ of } Cl_2)Na(s)→Na(g)(ΔHat⊖ of Na)\text{Na}(s) \rightarrow \text{Na}(g) \quad (\Delta H_{at}^\ominus\text{ of } Na)

  • Standard Enthalpy of Neutralisation (ΔHneu⊖\Delta H_{neu}^\ominus): Energy change when an acid reacts with an alkali to form one mole of water under standard conditions (always negative):     H+(aq)+OH−(aq)→H2O(l)\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)

    • Strong acids (HCl,HNO3HCl, HNO_3) fully dissociate in water, yielding ΔHneu⊖=−57.3 kJ mol−1\Delta H_{neu}^\ominus = -57.3\,kJ\,mol^{-1}.
    • Weak acids (CH3COOHCH_3COOH) partially dissociate. Energy is absorbed to break O−HO-H bonds for complete ionization, resulting in a less exothermic value (ΔHneu⊖=−55.2 kJ mol−1\Delta H_{neu}^\ominus = -55.2\,kJ\,mol^{-1}).
  • Standard Enthalpy of Solution (ΔHsol⊖\Delta H_{sol}^\ominus): Energy change when one mole of an ionic compound dissolves in excess water to form an infinitely dilute solution under standard conditions.

  • Standard Enthalpy of Hydration (ΔHhyd⊖\Delta H_{hyd}^\ominus): Energy change when one mole of gaseous ions dissolves in water to form an infinitely dilute solution (always negative):     Na+(g)+aq→Na+(aq)\text{Na}^+(g) + aq \rightarrow \text{Na}^+(aq)

    • The oxide ion (O2−O^{2-}) has no hydration enthalpy because it reacts chemically with water to form hydroxide ions:       O2−(g)+H2O(l)→2OH−(aq)\text{O}^{2-}(g) + \text{H}_2\text{O}(l) \rightarrow 2\text{OH}^-(aq)

    • Hess's Law and Enthalpy Cycles:

  • The total enthalpy change of a chemical reaction is independent of the pathway taken, provided the initial and final states are identical.

  • Using Enthalpies of Formation:ΔHr⊖=∑ΔHf⊖(products)−∑ΔHf⊖(reactants)\Delta H_r^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})

    • Combustion of ethene: C2H4+3O2→2CO2+2H2O\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}ΔHr=2(−394)+2(−286)−(+52)=−1412 kJ mol−1\Delta H_r = 2(-394) + 2(-286) - (+52) = -1412\,kJ\,mol^{-1}
    • Decomposition of CaCO3CaCO_3: CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2ΔHr=(−635)+(−394)−(−1207)=+178 kJ mol−1\Delta H_r = (-635) + (-394) - (-1207) = +178\,kJ\,mol^{-1}
  • Using Enthalpies of Combustion:ΔHr⊖=∑ΔHc⊖(reactants)−∑ΔHc⊖(products)\Delta H_r^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})

    • Hydrogenation of ethene: C2H4(g)+H2(g)→C2H6(g)\text{C}_2\text{H}_4(g) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)ΔHr=(−1411)+(−286)−(−1560)=−137 kJ mol−1\Delta H_r = (-1411) + (-286) - (-1560) = -137\,kJ\,mol^{-1}

    • Bond Energies:

  • Bond energy is the energy required to break one mole of a specific covalent bond in the gas phase.

  • Bond breaking is endothermic; bond formation is exothermic.     ΔH=∑Ebonds broken−∑Ebonds formed\Delta H = \sum E_{\text{bonds broken}} - \sum E_{\text{bonds formed}}

Lattice Energy, Born-Haber Cycles, and Solution Energetics

  • Lattice Energy (ΔHlatt⊖\Delta H_{latt}^\ominus):

    • The enthalpy change when one mole of an ionic crystalline compound is formed from its isolated gaseous ions under standard conditions:     Na+(g)+Cl−(g)→NaCl(s)\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s)Mg2+(g)+2Cl−(g)→MgCl2(s)\text{Mg}^{2+}(g) + 2\text{Cl}^-(g) \rightarrow \text{MgCl}_2(s)Mg2+(g)+O2−(g)→MgO(s)\text{Mg}^{2+}(g) + \text{O}^{2-}(g) \rightarrow \text{MgO}(s)
    • Lattice energy is always exothermic (ΔHlatt<0\Delta H_{latt} < 0); a more negative value indicates stronger ionic bonding.
    • Factors Determining Lattice Energy:
    • Ionic Charge: Higher charges produce stronger electrostatic attraction, making lattice energy more exothermic.
    • Ionic Radius: Smaller ions pack closer together, making lattice energy more exothermic.
    • Examples:
      • LE(NaCl)LE(NaCl) is more exothermic than LE(KCl)LE(KCl) because Na+\text{Na}^+ has a smaller ionic radius than K+\text{K}^+.
      • LE(CaBr2)LE(CaBr_2) is more exothermic than LE(SrBr2)LE(SrBr_2) because Ca2+\text{Ca}^{2+} has a smaller radius than Sr2+\text{Sr}^{2+}.
      • LE(MgO)LE(MgO) is substantially more exothermic than LE(NaCl)LE(NaCl) because MgOMgO contains divalent ions (Mg2+,O2−\text{Mg}^{2+}, \text{O}^{2-}) with smaller radii than the monovalent ions in NaClNaCl.
  • Electron Affinity (ΔHEA⊖\Delta H_{EA}^\ominus):

    • First Electron Affinity (EA1EA_1): Energy change when one mole of gaseous atoms gains one mole of electrons to form one mole of uninegative gaseous ions (X(g)+e−→X−(g)X(g) + e^- \rightarrow X^-(g)). Exothermic for most elements because of the electrostatic attraction between the positive nucleus and incoming electron.
    • Second Electron Affinity (EA2EA_2): Energy change when one mole of gaseous uninegative ions gains one mole of electrons to form one mole of dinegative ions (O−(g)+e−→O2−(g)\text{O}^-(g) + e^- \rightarrow \text{O}^{2-}(g)). Always endothermic because energy is required to overcome electrostatic repulsion between the negative ion and incoming electron.
  • Born-Haber Cycles:

    • A thermodynamic cycle applying Hess's Law to calculate the lattice energy of an ionic solid.
    • Requires five key enthalpy terms:
    1. Standard enthalpy of formation of the ionic compound (ΔHf⊖\Delta H_f^\ominus )
    2. Enthalpies of atomisation of the elements (ΔHat⊖\Delta H_{at}^\ominus )
    3. Ionisation energies of the metal (IEIE )
    4. Electron affinities of the non-metal (EAEA )
    5. Lattice energy (ΔHlatt⊖\Delta H_{latt}^\ominus )
    • Energy Balance Equations:
    • For NaClNaCl:       ΔHf=ΔHat(Na)+1st IE(Na)+ΔHat(Cl)+1st EA(Cl)+LE\Delta H_f = \Delta H_{at}(\text{Na}) + 1\text{st } IE(\text{Na}) + \Delta H_{at}(\text{Cl}) + 1\text{st } EA(\text{Cl}) + LE
    • For KClKCl:       ΔHf=ΔHat(K)+1st IE(K)+ΔHat(Cl)+1st EA(Cl)+LE\Delta H_f = \Delta H_{at}(\text{K}) + 1\text{st } IE(\text{K}) + \Delta H_{at}(\text{Cl}) + 1\text{st } EA(\text{Cl}) + LE
    • For MgCl2MgCl_2:       ΔHf=ΔHat(Mg)+1st IE(Mg)+2nd IE(Mg)+2ΔHat(Cl)+2(1st EA(Cl))+LE\Delta H_f = \Delta H_{at}(\text{Mg}) + 1\text{st } IE(\text{Mg}) + 2\text{nd } IE(\text{Mg}) + 2\Delta H_{at}(\text{Cl}) + 2(1\text{st } EA(\text{Cl})) + LE
    • For MgOMgO:       ΔHf=ΔHat(Mg)+1st IE(Mg)+2nd IE(Mg)+ΔHat(O)+1st EA(O)+2nd EA(O)+LE\Delta H_f = \Delta H_{at}(\text{Mg}) + 1\text{st } IE(\text{Mg}) + 2\text{nd } IE(\text{Mg}) + \Delta H_{at}(\text{O}) + 1\text{st } EA(\text{O}) + 2\text{nd } EA(\text{O}) + LE
  • Energetics of Solution and Hydration:

    • Dissolving an ionic solid involves separating the lattice into gaseous ions and hydrating them:     ΔHhydration=ΔHsolution+LE\Delta H_{hydration} = \Delta H_{solution} + LEΔHsolution=ΔHhydration−LE\Delta H_{solution} = \Delta H_{hydration} - LE
    • A more negative enthalpy of solution (ΔHsol\Delta H_{sol}) generally correlates with greater solubility.
    • Hydration enthalpy is always exothermic because it involves forming ion-dipole attractions between ions and water molecules.
    • Magnitude increases with higher ionic charge and smaller ionic radius:
    • ΔHhyd(Li+)\Delta H_{hyd}(\text{Li}^+) is more exothermic than ΔHhyd(Na+)\Delta H_{hyd}(\text{Na}^+) because Li+\text{Li}^+ has a smaller ionic radius.
    • ΔHhyd(Mg2+)\Delta H_{hyd}(\text{Mg}^{2+}) is more exothermic than ΔHhyd(Na+)\Delta H_{hyd}(\text{Na}^+) due to its higher charge and smaller radius.

Entropy, Gibbs Free Energy, and Chemical Feasibility

  • Entropy (SS):

    • A measure of the disorder or randomness of a system, expressed in J K−1 mol−1J\,K^{-1}\,mol^{-1}.
    • Standard entropy (S⊖S^\ominus) is measured under standard conditions (25 ∘C,1 atm25\,^\circ\text{C}, 1\,atm).
    • All substances possess non-zero disorder; therefore, entropy is always positive (S⊖>0S^\ominus > 0).
    • Entropies of States:Sgas>Sliquid>SsolidS_{\text{gas}} > S_{\text{liquid}} > S_{\text{solid}}     Gas particles have high kinetic energy, large intermolecular spacing, and many more possible microstates.
    • Rigid structures have lower entropy (e.g., diamond has S⊖=2.4 J K−1 mol−1S^\ominus = 2.4\,J\,K^{-1}\,mol^{-1}, lower than graphite).
    • Representative Values: Diamond = 2.42.4, CaCO3(s)=69.9CaCO_3(s) = 69.9, H2O(l)=92.9H_2O(l) = 92.9, H2(g)=130.6H_2(g) = 130.6, CO2(g)=213.6 J K−1 mol−1CO_2(g) = 213.6\,J\,K^{-1}\,mol^{-1}.
  • Entropy Change of a Reaction (ΔSsystem⊖\Delta S^\ominus_{\text{system}}):

    • ΔS⊖>0\Delta S^\ominus > 0 indicates an increase in disorder (favorable for spontaneity), seen in melting, evaporation, sublimation, diffusion, and reactions producing more moles of gas.
    • Calculated as:     ΔSreaction⊖=∑S⊖(products)−∑S⊖(reactants)\Delta S^\ominus_{\text{reaction}} = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})
    • Worked Examples:
    • CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)ΔS⊖=[39.7+213.6]−[92.9]=+160.4 J K−1 mol−1\Delta S^\ominus = [39.7 + 213.6] - [92.9] = +160.4\,J\,K^{-1}\,mol^{-1}
    • 2NaHCO3(s)→Na2CO3(s)+CO2(g)+H2O(l)2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(l)ΔS⊖=[135.0+69.9+213.6]−[2(101.7)]=+215.1 J K−1 mol−1\Delta S^\ominus = [135.0 + 69.9 + 213.6] - [2(101.7)] = +215.1\,J\,K^{-1}\,mol^{-1}
    • 2KNO3(s)→2KNO2(s)+O2(g)2\text{KNO}_3(s) \rightarrow 2\text{KNO}_2(s) + \text{O}_2(g)ΔS⊖=[2(152.1)+205.0]−[2(133.1)]=+243.0 J K−1 mol−1\Delta S^\ominus = [2(152.1) + 205.0] - [2(133.1)] = +243.0\,J\,K^{-1}\,mol^{-1}
  • Total Entropy Change and Reaction Feasibility:

    • Feasibility cannot be determined using ΔSsystem\Delta S_{\text{system}} alone because chemical reactions exchange heat with their surroundings, altering the surrounding entropy.
    • ΔStotal⊖=ΔSreaction⊖+ΔSsurrounding⊖\Delta S_{\text{total}}^\ominus = \Delta S_{\text{reaction}}^\ominus + \Delta S_{\text{surrounding}}^\ominus
    • Surrounding entropy change: ΔSsurrounding⊖=−ΔH⊖T\Delta S_{\text{surrounding}}^\ominus = -\frac{\Delta H^\ominus}{T}
    • Exothermic reactions release heat, increasing surrounding entropy (ΔSsurr>0\Delta S_{\text{surr}} > 0).
    • Endothermic reactions absorb heat, decreasing surrounding entropy (ΔSsurr<0\Delta S_{\text{surr}} < 0).
    • A reaction is spontaneous if and only if ΔStotal⊖>0\Delta S_{\text{total}}^\ominus > 0.
    • Worked Feasibility Checks:
    • CaO+2NH4Cl→CaCl2+2NH3+H2O\text{CaO} + 2\text{NH}_4\text{Cl} \rightarrow \text{CaCl}_2 + 2\text{NH}_3 + \text{H}_2\text{O} at 298 K298\,K (ΔH⊖=+88.8 kJ mol−1\Delta H^\ominus = +88.8\,kJ\,mol^{-1}):       ΔSreaction=[104.6+2(192.3)+69.9]−[39.7+2(94.6)]=559.1−228.9=+330.2 J K−1 mol−1\Delta S_{\text{reaction}} = [104.6 + 2(192.3) + 69.9] - [39.7 + 2(94.6)] = 559.1 - 228.9 = +330.2\,J\,K^{-1}\,mol^{-1}ΔSsurrounding=−88.8×1000298=−297.99 J K−1 mol−1\Delta S_{\text{surrounding}} = -\frac{88.8 \times 1000}{298} = -297.99\,J\,K^{-1}\,mol^{-1}ΔStotal=+330.2−297.99=+32.3 J K−1 mol−1>0  ⟹  Feasible\Delta S_{\text{total}} = +330.2 - 297.99 = +32.3\,J\,K^{-1}\,mol^{-1} > 0 \implies \text{Feasible}
    • ZnCO3(s)→ZnO(s)+CO2(g)\text{ZnCO}_3(s) \rightarrow \text{ZnO}(s) + \text{CO}_2(g) (ΔH=+72 kJ mol−1,ΔS=+174.8 J K−1 mol−1\Delta H = +72\,kJ\,mol^{-1}, \Delta S = +174.8\,J\,K^{-1}\,mol^{-1}):
      • At 298 K298\,K:         ΔStotal=+174.8+(−72000298)=−66.8 J K−1 mol−1<0  ⟹  Not feasible\Delta S_{\text{total}} = +174.8 + \left(-\frac{72000}{298}\right) = -66.8\,J\,K^{-1}\,mol^{-1} < 0 \implies \text{Not feasible}
      • At 550 K550\,K:         ΔStotal=+174.8+(−72000550)=+43.9 J K−1 mol−1>0  ⟹  Feasible\Delta S_{\text{total}} = +174.8 + \left(-\frac{72000}{550}\right) = +43.9\,J\,K^{-1}\,mol^{-1} > 0 \implies \text{Feasible}
  • Gibbs Free Energy (ΔG⊖\Delta G^\ominus):

    • Represents the energy available to do useful work, combining enthalpy and entropy:     ΔG⊖=−TΔStotal⊖\Delta G^\ominus = -T \Delta S_{\text{total}}^\ominusΔG⊖=ΔH⊖−TΔSsystem⊖\Delta G^\ominus = \Delta H^\ominus - T \Delta S_{\text{system}}^\ominus
    • Criteria for Feasibility:
    • ΔG⊖<0\Delta G^\ominus < 0: Reaction is feasible/spontaneous. A more negative value indicates a greater thermodynamic driving force.
    • ΔG⊖>0\Delta G^\ominus > 0: Reaction is not feasible.
    • ΔG⊖=0\Delta G^\ominus = 0: System is at dynamic equilibrium (e.g., during melting and boiling).
  • Worked Gibbs Free Energy Problems:

    • Thermal Decomposition of Na2CO3Na_2CO_3:Na2CO3(s)→Na2O(s)+CO2(g)(ΔH=+323 kJ mol−1,ΔS=+153.7 J K−1 mol−1)\text{Na}_2\text{CO}_3(s) \rightarrow \text{Na}_2\text{O}(s) + \text{CO}_2(g) \quad (\Delta H = +323\,kJ\,mol^{-1}, \Delta S = +153.7\,J\,K^{-1}\,mol^{-1})
    • At 1200 K1200\,K:       ΔG=323000−1200(153.7)=+138.56 kJ mol−1>0  ⟹  Not feasible\Delta G = 323000 - 1200(153.7) = +138.56\,kJ\,mol^{-1} > 0 \implies \text{Not feasible}
    • Minimum temperature for feasibility (ΔG<0\Delta G < 0):       323−T(153.71000)<0  ⟹  T>2101.5 K323 - T\left(\frac{153.7}{1000}\right) < 0 \implies T > 2101.5\,K
    • Reduction of Iron(III) Oxide at 800 K800\,K:
    • Reaction (a): Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2ΔH=−25 kJ mol−1,ΔS=+15.2 J K−1 mol−1\Delta H = -25\,kJ\,mol^{-1}, \Delta S = +15.2\,J\,K^{-1}\,mol^{-1}ΔG=−25000−800(15.2)=−37.16 kJ mol−1  ⟹  Feasible\Delta G = -25000 - 800(15.2) = -37.16\,kJ\,mol^{-1} \implies \text{Feasible}
    • Reaction (b): ΔH=+491 kJ mol−1,ΔS=+542.9 J K−1 mol−1\Delta H = +491\,kJ\,mol^{-1}, \Delta S = +542.9\,J\,K^{-1}\,mol^{-1}ΔG=491−800(542.91000)=+56.68 kJ mol−1  ⟹  Not feasible\Delta G = 491 - 800\left(\frac{542.9}{1000}\right) = +56.68\,kJ\,mol^{-1} \implies \text{Not feasible}Tmin=4910.5429=904.4 KT_{\text{min}} = \frac{491}{0.5429} = 904.4\,K
    • Melting of Ice:ΔHfusion=+6 kJ mol−1 at 0 ∘C (273 K)\Delta H_{\text{fusion}} = +6\,kJ\,mol^{-1}\text{ at } 0\,^\circ\text{C } (273\,K)     At equilibrium (ΔG=0\Delta G = 0):     0=6000−273ΔS  ⟹  ΔS=+22.0 J K−1 mol−10 = 6000 - 273 \Delta S \implies \Delta S = +22.0\,J\,K^{-1}\,mol^{-1}
  • Effect of Temperature on Reaction Feasibility:

    • Endothermic Reaction (ΔH>0\Delta H > 0):
    • If ΔS<0\Delta S < 0: ΔG=(+ve)−(−TΔS)=+ve+TΔS\Delta G = (+ve) - (-T\Delta S) = +ve + T\Delta S. Always positive; never feasible at any temperature.
    • If ΔS>0\Delta S > 0: ΔG=(+ve)−(+TΔS)\Delta G = (+ve) - (+T\Delta S). Becomes feasible at high temperatures (T>ΔHΔST > \frac{\Delta H}{\Delta S}), as in the decomposition of CaCO3CaCO_3.
    • Exothermic Reaction (ΔH<0\Delta H < 0):
    • If ΔS<0\Delta S < 0: ΔG=(−ve)−(−TΔS)=−ve+TΔS\Delta G = (-ve) - (-T\Delta S) = -ve + T\Delta S. Feasible only at lower temperatures (T<ΔHΔST < \frac{\Delta H}{\Delta S}).
    • If ΔS>0\Delta S > 0: ΔG=(−ve)−(+TΔS)\Delta G = (-ve) - (+T\Delta S). Always negative; feasible at all temperatures.