Chemical Bonding and Molecular Structure

A chemical bond is the attractive force that holds the constituent atoms, ions, etc. together in a chemical species. Bonds form because systems tend to become more stable by lowering their energy. Atoms combine either by transfer of electrons or by sharing of electrons so that they attain a stable electronic arrangement, usually an octet in the valence shell.

Lewis dot symbols show the valence electrons as dots around the element symbol.

- Mg: $\mathrm{Mg}\,\bullet\bullet$
- Na: $\mathrm{Na}\,\bullet$
- B: $\mathrm{B}\,\bullet\bullet\bullet$
- O: $\mathrm{O}\,\bullet\bullet\bullet\bullet\bullet\bullet$
- N: $\mathrm{N}\,\bullet\bullet\bullet\bullet\bullet$
- Br: $\mathrm{Br}\,\bullet\bullet\bullet\bullet\bullet\bullet\bullet$

These dots represent the number of valence electrons in each atom.

Lewis symbols:

- S has 6 valence electrons, so it is shown with 6 dots.
- **$\mathrm{S}^{2-}$ has gained 2 electrons, so it has 8 dots.
- Al has 3 valence electrons, so it is shown with 3 dots.
- $\mathrm{Al}^{3+}$ has lost 3 electrons, so it has no dots.
- H has 1 valence electron, so it has 1 dot.
- $\mathrm{H}^{-}$** has gained 1 electron, so it has 2 dots.

So:
- $\mathrm{S}\,\bullet\bullet\bullet\bullet\bullet\bullet$
- $[\mathrm{S}]^{2-}$ with 8 dots
- $\mathrm{Al}\,\bullet\bullet\bullet$
- $[\mathrm{Al}]^{3+}$ with 0 dots
- $\mathrm{H}\,\bullet$
- $[\mathrm{H}]^{-}\,\bullet\bullet$

Lewis structures:

- **$\mathrm{H_2S}$**: sulfur is central with two single bonds to H and two lone pairs on S.
- $\mathrm{H-S-H}$ with 2 lone pairs on sulfur.
- **$\mathrm{SiCl_4}$: silicon is central with four single bonds to four chlorine atoms; each Cl has three lone pairs.
- $\mathrm{BeF_2}$: beryllium is central with two single bonds to two fluorine atoms; Be has an incomplete octet.
- $\mathrm{CO_3^{2-}}$**: carbon is central; one double bond and two single bonds to oxygen appear in each canonical form, and the ion is written in brackets with overall $2-$ charge. It is best described by resonance.
- **$\mathrm{HCOOH}$**: $\mathrm{H-C(=O)-OH}$; the carbon has one double bond to one oxygen and single bonds to H and to the OH oxygen.

These structures satisfy the octet rule where possible, and the carbonate ion is represented by equivalent resonance forms.

The octet rule states that atoms can combine either by transfer of valence electrons or by sharing of valence electrons so that each atom attains an octet in its valence shell.

### Significance
- It explains the formation of many ionic and covalent bonds.
- It helps in writing Lewis structures of simple molecules and ions.
- It is especially useful for compounds of the second period elements.

### Limitations
The octet rule is not universal.
- Incomplete octet: some compounds have fewer than 8 electrons around the central atom, e.g. LiCl, BeH2, BCl3, BF3.
- Odd-electron molecules: molecules like NO and NO2 do not satisfy the octet rule for all atoms.
- Expanded octet: elements beyond the second period can have more than 8 valence electrons, e.g. PF5, SF6, H2SO4.
- It does not explain molecular shape.
- It does not explain relative stability or bond energy.

The favourable factors for the formation of an ionic bond are:

1. Low ionization enthalpy of the atom forming the cation, so electron loss is easy.
2. High negative electron gain enthalpy of the atom forming the anion, so electron gain is easy.
3. Large lattice enthalpy released when gaseous ions combine to form the crystal lattice, which stabilises the ionic solid.
4. Suitable arrangement and packing of ions in the crystal lattice.

Thus, ionic bonding is favoured between elements that can form ions easily and release enough energy on lattice formation to compensate for the energy absorbed in ion formation.

Using the VSEPR model, the shapes are:

- BeCl2: central Be has 2 bond pairs and no lone pair, so the arrangement is linear; bond angle $180^\circ$.
- BCl3: central B has 3 bond pairs and no lone pair, so the shape is trigonal planar; bond angle $120^\circ$.
- SiCl4: central Si has 4 bond pairs and no lone pair, so the shape is tetrahedral; bond angle $109.5^\circ$.
- AsF5: central As has 5 bond pairs and no lone pair, so the shape is trigonal bipyramidal.
- H2S: central S has 2 bond pairs and 2 lone pairs, so the shape is bent or V-shaped.
- PH3: central P has 3 bond pairs and 1 lone pair, so the shape is trigonal pyramidal.

The shapes follow from repulsion between electron pairs, with lone pairs causing greater distortion than bond pairs.

Both NH3 and H2O have a tetrahedral arrangement of electron pairs around the central atom, but their shapes differ because of lone pair repulsion.

- In NH3, nitrogen has 3 bond pairs and 1 lone pair. The lone pair repels the bond pairs more strongly than bond pair–bond pair repulsion, so the H–N–H angle is reduced from $109.5^\circ$ to about **$107^\circ$.
- In H2O, oxygen has 2 bond pairs and 2 lone pairs. Here there are two lone pairs**, so lone pair–lone pair and lone pair–bond pair repulsions are stronger. This compresses the H–O–H angle further, from $109.5^\circ$ to about **$104.5^\circ$.

So, the bond angle in water is less than in ammonia because H2O has more lone-pair repulsion**.

In the Lewis description, bond order is the number of bonds between two atoms.

The bond strength is expressed as follows:
- Higher bond order means greater bond enthalpy and therefore a stronger bond.
- Lower bond order means smaller bond enthalpy and a weaker bond.

So, as bond order increases, bond strength increases. For example, bond order in $\mathrm{N_2}$ is 3, so it is stronger than $\mathrm{O_2}$, where bond order is 2.

Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.

The **carbonate ion, $\mathrm{CO_3^{2-}}$, cannot be represented correctly by a single Lewis structure because all three C–O bonds are experimentally equivalent.

- A single Lewis structure with one double bond and two single bonds would show unequal bonds**, so it is inadequate.
- Therefore, $\mathrm{CO_3^{2-}}$ is represented by three canonical forms.
- In each canonical form, the position of the double bond changes, but the actual ion is a resonance hybrid of all three.
- Resonance stabilizes the ion because the energy of the resonance hybrid is lower than any one canonical structure.
- Resonance also averages the bond characteristics, so all three C–O bonds become identical in the real ion.

Thus, the true structure of carbonate is not any one Lewis structure, but the resonance hybrid of the three canonical forms.

No, these two structures of **$\mathrm{H_3PO_3}$ cannot be taken as the canonical forms of a resonance hybrid.

### Reasons
- In resonance, the positions of atoms must remain the same and only the distribution of electrons may change.
- The canonical forms must have similar energy, same skeletal arrangement, and same bonding/non-bonding pattern**.
- In the given case, the two structures of $\mathrm{H_3PO_3}$ differ in the position of hydrogen atoms attached to oxygen/phosphorus, so they are not just electron-shift forms.
- Therefore, they are not resonance structures; they represent different structural possibilities, not canonical forms of the same resonance hybrid.

The resonance structures are:

- **$\mathrm{SO_3}$**: three equivalent canonical forms with one $\mathrm{S=O}$ and two $\mathrm{S-O}$ bonds in each form, the double bond shifting among the three oxygen atoms.
- **$\mathrm{NO_2}$**: two canonical forms with one $\mathrm{N=O}$ bond and one $\mathrm{N-O}$ bond, and the unpaired electron distributed in the two forms.
- **$\mathrm{NO_3^-}$**: three equivalent canonical forms with one $\mathrm{N=O}$ bond and two $\mathrm{N-O^-}$ bonds in each form, the double bond shifting among the three oxygens.

These resonance forms show that the actual molecule/ion is a resonance hybrid with equivalent or delocalised bonding.

Using Lewis symbols, electron transfer occurs as follows:

### (a) K and S
- Potassium has 1 valence electron, so two K atoms each lose one electron.
- Sulfur has 6 valence electrons and gains 2 electrons.

So:
$$2\mathrm{K} \rightarrow 2\mathrm{K}^+ + 2e^-$$
$$\mathrm{S} + 2e^- \rightarrow \mathrm{S}^{2-}$$

This gives **$\mathrm{K_2S}$**.

### (b) Ca and O
- Calcium loses 2 electrons.
- Oxygen gains 2 electrons.

$$\mathrm{Ca} \rightarrow \mathrm{Ca}^{2+} + 2e^-$$
$$\mathrm{O} + 2e^- \rightarrow \mathrm{O}^{2-}$$

This gives **$\mathrm{CaO}$**.

### (c) Al and N
- Aluminium loses 3 electrons.
- Nitrogen gains 3 electrons.

$$\mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3e^-$$
$$\mathrm{N} + 3e^- \rightarrow \mathrm{N}^{3-}$$

This gives **$\mathrm{AlN}$**.

In each case, the cation and anion attain stable noble gas configurations.

Although both are triatomic molecules, **$\mathrm{H_2O}$ is bent while $\mathrm{CO_2}$ is linear because of the direction of bond dipoles and the overall molecular dipole moment.

- In $\mathrm{CO_2}$, the molecule is linear and the two equal C=O bond dipoles act in opposite directions. They cancel each other, so the net dipole moment is zero. A linear arrangement is therefore consistent.
- In $\mathrm{H_2O}$**, the molecule has a bent shape. The two O–H bond dipoles do not cancel because they are at an angle of $104.5^\circ$. Hence water has a net dipole moment of 1.85 D.

So, **$\mathrm{CO_2}$ is linear due to cancellation of dipoles, while $\mathrm{H_2O}$** is bent because the bond dipoles add up to give a resultant dipole moment.

Dipole moment is useful for several purposes:

- It helps in deciding whether a molecule is polar or non-polar.
- It gives information about the shape of molecules; for example, a zero dipole moment may suggest a symmetrical structure.
- It helps compare the polarity of bonds and molecules.
- It is useful in understanding molecular geometry in polyatomic molecules.
- It helps explain why molecules like **$\mathrm{NH_3}$ and $\mathrm{NF_3}$ have different dipole moments even though both are pyramidal.

Thus, dipole moment is an important measure of the polarity and structure** of molecules.

Electronegativity is the tendency of an atom in a molecule to attract the shared pair of electrons towards itself.

It differs from electron gain enthalpy as follows:
- Electronegativity is a relative tendency in a bonded state; it is not a directly measurable quantity.
- Electron gain enthalpy is the enthalpy change when a gaseous atom in the ground state gains an electron.
- Electron gain enthalpy has a definite thermodynamic meaning and can be positive or negative, whereas electronegativity is a scale-based property used for comparing atoms in molecules.

A polar covalent bond is a covalent bond in which the shared electron pair is not shared equally between the two atoms because one atom is more electronegative.

For example, in HF, fluorine is much more electronegative than hydrogen. So the shared pair of electrons is pulled more towards fluorine, producing partial charges:
- H becomes **$\delta^+$
- F becomes $\delta^-$

Thus HF contains a polar covalent bond. The molecule has a dipole moment** because of this unequal distribution of charge.

Ionic character increases with increasing difference in electronegativity.

- N₂: same atoms, so it is non-polar covalent and has the least ionic character.
- SO₂: polar covalent, but still largely covalent.
- ClF₃: more polar than SO₂ because of greater electronegativity difference.
- LiF: highly ionic.
- K₂O: also highly ionic; among the given compounds it is placed after LiF in increasing order.

So the order of increasing ionic character is:

**$\mathrm{N_2 < SO_2 < ClF_3 < LiF < K_2O}$**.

The correct Lewis structure of acetic acid is:

**$\mathrm{H_3C- C(=O)-OH}$

More explicitly:
- The first carbon is bonded to three H atoms and to the second carbon.
- The second carbon is double-bonded to one oxygen.
- The same carbon is single-bonded to another oxygen, which is bonded to hydrogen.

So the corrected structure is $\mathrm{CH_3COOH}$ with the carboxyl group written as $\mathrm{-C(=O)OH}$**.