Inorganic Chemistry: Comprehensive Study Notes on Chemical Bonding

Fundamental Principles of Ionic and Covalent Bonding

Ionic bonds are characterized by being non-directional in nature, whereas covalent bonds are directional. In specific chemical species, such as Sodium Cyanide ($KCN$) or Ammonium Chloride ($NH_4Cl$), both ionic and covalent bonding can coexist within the same compound. For instance, in $KCN$, the bond between the potassium cation ($K^+$) and the cyanide anion ($CN^-$) is ionic, while the bond between carbon and nitrogen within the cyanide ion is covalent. A similar coexistence is found in Sodium Hydroxide ($NaOH$) and Sodium Tetrafluoroborate ($NaBF_4$).

The formation of an ionic bond between two species, such as $A$ and $B^-$, is most thermodynamically favorable when the ionization energy of the metal atom ($A$) is low, facilitating the removal of an electron, and the electron gain enthalpy of the non-metal atom ($B$) is high, indicating a strong affinity for that electron. In contrast, bonds formed between two identical atoms cannot be ionic, as there is no electronegativity difference to drive the transfer of electrons; such bonds are typically covalent, metallic, or coordinate.

Valence Bond Theory, Hybridization, and Orbital Characteristics

The strength of a chemical bond is determined by the extent of orbital overlap. When comparing different types of orbital overlaps, bond strength generally decreases in the following order: $2p-2p > 2p-2s > 2s-2s$. This is due to the directional nature of p-orbitals providing more effective overlapping compared to the spherical s-orbitals.

Hybridization explains the mixing of atomic orbitals of slightly different energies to form a new set of equivalent orbitals. For carbon, the electronegativity of different hybrid states follows the order $sp > sp^2 > sp^3$. This is because the s-character is highest in $sp$ ($50\%$), followed by $sp^2$ ($33.3\%$) and $sp^3$ ($25\%$), and electrons in s-orbitals are held more tightly by the nucleus. Correspondingly, the length of hybrid orbitals increases as the s-character decreases: $sp < sp^2 < sp^3$. As s-character decreases, bond angles decrease, bond lengths increase, and the physical size of the hybrid orbital increases.

In specific molecules, hybridization dictates geometry. Phosphorus pentachloride ($PCl_5$) features $sp^3d$ hybridization with angles of $120^\circ$ (equatorial) and $90^\circ$ (axial). Bromine trifluoride ($BrF_3$) utilizes $sp^3d$ hybridization and assumes a bent T-shape. Xenon difluoride ($XeF_2$) also involves $sp^3d$ hybridization. In the case of Xenon hexafluoride ($XeF_6$) in the vapor phase, the structure is described as a capped octahedron. For the methyl cation ($CH_3^+$), the geometry is trigonal planar due to $sp^2$ hybridization. In solid $Cl_2O_6$, the cationic part is $[ClO_2]^+$ which exhibits $sp^2$ hybridization.

VSEPR Theory and Molecular Geometry

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the shapes of molecules based on the repulsion between electron pairs. Species like Phosphorus pentachloride ($PCl_5$) and Xenon difluoride ($XeF_2$) are considered hypervalent because the central atom expands its octet to accommodate more than eight electrons. Conversely, some species like $BF_3$ are hypovalent, having fewer than eight electrons.

Geometry remains consistent among certain sets: for example, Sulphate ($SO_4^{2-}$), Tetrafluoroborate ($BF_4^-$), and Phosphonium ($PH_4^+$) all exhibit tetrahedral geometry. However, Xenon tetrafluoride ($XeF_4$) differs, possessing a square planar geometry despite its $sp^3d^2$ hybridization, due to the presence of two lone pairs. Molecules like Sulphur tetrafluoride ($SF_4$) adopt a see-saw shape, where the lone pair occupies an equatorial position to minimize lone pair-bond pair repulsions at $90^\circ$.

In the ethylene molecule ($C_2H_4$), if the molecule lies in the X-Y plane, the nodal plane of the $\pi$-bond is situated in the X-Y plane itself, as the p-orbitals involved in the $\pi$-bond are perpendicular to this plane. For coordinate bonds, species such as Carbon Monoxide ($CO$), Ozone ($O_3$), and Methyl Isocyanide ($CH_3-NC$) contain them, while the carbonate ion ($CO_3^{2-}$) is typically represented by resonance structures consisting of covalent bonds.

Bent’s Rule and Drago’s Rule

Bent's Rule suggests that more electronegative substituents prefer hybrid orbitals with less s-character, and lone pairs prefer orbitals with more s-character. This explains why the bond angle in dimethyl ether ($CH_3OCH_3$) is approximately $111.7^\circ$, which is larger than the tetrahedral angle of $109^\circ 28'$, due to the steric bulk of the methyl groups and the electronic properties of the oxygen hybrid orbitals which are $sp^3$ hybridized.

Drago's Rule explains why certain heavy p-block hydrides, like Phosphine ($PH_3$) and Hydrogen Sulphide ($H_2S$), do not show significant hybridization. In $NH_3$, the lone pair is in an $sp^3$ hybrid orbital and is highly directional, making it a better electron donor. In contrast, the lone pair in $PH_3$ remains in a nearly pure, spherical s-orbital, making it less directional and a poorer donor.

Dipole Moments and Ionic Character

The dipole moment ($\mu$) is a measure of molecular polarity. For alkyl halides, the decreasing order of dipole moment is $CH_3Cl > CH_3F > CH_3Br > CH_3I$. Although fluorine is more electronegative than chlorine, the larger bond length in $CH_3Cl$ results in a higher dipole moment ($\mu = q \times d$).

For a diatomic molecule like $HCl$, the percentage of ionic character can be calculated using the experimental dipole moment ($1.03\,D$) and the bond length (1.275\,\text{
}). $1.03\,D = 1.03 \times 10^{-18}\,esu\,cm$$d = 1.275 \times 10^{-8}\,cm$ Theoretical $μ = 4.8 \times 10^{-10}\,esu \times 1.275 \times 10^{-8}\,cm = 6.12 \times 10^{-18}\,esu\,cm = 6.12\,D$$\text{Percentage Ionic Character} = \frac{1.03}{6.12} \times 100 \approx 17\%$

Symmetric molecules like Carbon Tetrachloride ($CCl_4$), Benzene ($C_6H_6$), and Boron Trifluoride ($BF_3$) have zero net dipole moments due to their highly symmetric geometries where individual bond dipoles cancel out.

Hydrogen Bonding and Van der Waals Forces

Hydrogen bonding significantly influences physical properties. Ethanol ($C_2H_5OH$) has a higher boiling point than dimethyl ether ($CH_3OCH_3$) because ethanol molecules associate through intermolecular hydrogen bonding, while the ether does not. The strength of hydrogen bonds typically follows the order: $FH...F > OH...O > NH...N$, correlating with the electronegativity of the atoms involved. However, the exact order of the strength of hydrogen bonding for different substances is $HF > H_2O > H_2O_2 > H_2S$.

Intramolecular hydrogen bonding (within a single molecule) occurs in o-nitrophenol, o-boric acid, o-chlorophenol, chloral hydrate, and o-salicylaldehyde. This often results in a lower boiling point compared to their para-isomers, which exhibit intermolecular hydrogen bonding.

Van der Waals forces, such as London dispersion forces (instantaneous dipole-induced dipole), are responsible for the liquefaction of noble gases like Xenon. The melting points of noble gases increase with atomic mass because the increase in the number of electrons leads to greater polarizability and stronger dispersion forces. These forces depend on the distance ($r$) between particles: London forces and Debye forces vary as $r^{-6}$, while Keesom forces (dipole-dipole) vary as $r^{-3}$. The critical temperature of water is higher than that of oxygen ($O_2$) because of its significant dipole moment and hydrogen bonding.

Structural Silicates

Silicates are classified based on how the $SiO_4^{4-}$ tetrahedra are linked through shared oxygen atoms at the corners:

• Orthosilicates: No oxygen atoms are shared ($SiO_4^{4-}$).
• Pyrosilicates: One oxygen atom is shared per tetrahedron (e.g., Thortveitite $Sc_2Si_2O_7$).
• Chain/Cyclic Silicates: Two oxygen atoms are shared per tetrahedron. Pyroxene is a chain silicate, while cyclic silicates form rings.
• Amphiboles: These are double-chain silicates where tetrahedra share alternately two and three oxygen atoms, resulting in an average of $2.5$ shared oxygens per unit.
• Sheet Silicates: Three oxygen atoms are shared per tetrahedron (e.g., Alunite or Talc).
• Three-Dimensional Silicates: All four oxygen atoms are shared (e.g., Quartz, Zeolites, Feldspars).

In Silica ($SiO_2$), every silicon atom is tetrabonded to four oxygen atoms in a continuous lattice. The mineral Kinoite is represented by the formula $Ca_2Cu_2Si_3O_{10} \cdot 2H_2O$, reflecting a chain of three tetrahedra.

Oxoacids of Phosphorus and Sulphur

Phosphorus oxoacids exhibit varying oxidation states and bonding patterns. Pyrophosphoric acid ($H_4P_2O_7$) contains four $P-OH$ bonds and the oxidation state of Phosphorus is $+5$. White phosphorus ($P_4$) contains four triangles (trigons) and contains $6$ P-P bonds. In $P_4O_6$, the number of P-O bonds is $12$.

Sulphur oxoacids include Peroxodisulphuric acid ($H_2S_2O_8$), which features a peroxide ($O-O$) linkage and four $S=O$ bonds and two $S-OH$ bonds. Pyrosulphuric acid ($H_2S_2O_7$) contains a bridging oxo group ($S-O-S$). Sodium pentathionate ($Na_2S_5O_6$) contains chains of sulphur atoms, leading to an average oxidation number for sulphur of $2.0$. For Sodium tetrathionate ($Na_2S_4O_6$), the difference in oxidation numbers between the two types of sulphur atoms (terminal vs. central) is $5$.

Questions & Discussion

• In solid $\text{NaCl}$, how many directional bonds does each $\text{Na}^+$ form? Sodium chloride is purely ionic; therefore, there are no directional covalent bonds between the ions, making the total number of directional bonds zero.

• What is the number of $\text{p}π-\text{d}π$ bonds in $\text{SO}_2$? In sulphur dioxide ($SO_2$), there is one $\text{p}π-\text{p}π$ bond and one $\text{p}π-\text{d}π$ bond.

• Why does $\text{NCl}_5$ not exist while $\text{PCl}_5$ does? Nitrogen belongs to the second period and lacks vacant d-orbitals in its valence shell to expand its octet to five bonds, whereas Phosphorus has vacant 3d-orbitals.

• How do ice cubes unite when pressed? When two ice cubes are pressed together, the pressure melts the surface slightly; upon release, hydrogen bonds reform at the interface, uniting the cubes.

• Which non-metal does not exhibit a positive oxidation state? Fluorine is the most electronegative element and consistently maintains an oxidation state of $-1$ (or $0$ in $F_2$); it never exhibits a positive oxidation state.

• Compare Diamond and Graphite conduction. Diamond is an insulator because its carbon atoms are $sp^3$ hybridized and have no free electrons. Graphite is a conductor because its carbon atoms are $sp^2$ hybridized, leaving one free electron per atom to move between layers.

• What is the geometry of $\text{XeOF}_4$? The molecule has $sp^3d^2$ hybridization for Xenon, featuring one lone pair and five bond pairs, resulting in a square pyramidal shape.