Problems 1-10

1. Which halogen forms (X2+), (X3+)? Why more stable than (X+)?

 These cations are known for heavier halogens (especially Cl₂⁺, Br₂⁺, I₂⁺, and polyhalogen cations like (I3+), (Br3+)).

• A single (X+) cation would be a very high-energy species with the positive charge localized on one atom.

• In (X2+) or (X3+), the positive charge is delocalized over a multi-center bond, which is more stable.

Heavier halogens (Cl, Br, I) most readily form (X2+) and (X3+) cations. These species are more stable than (X+) because the charge and bonding are delocalized over several atoms, lowering Coulombic repulsion and stabilizing the cation.

 Additional Note: Compare the bonding in I3- and I3+

In the triiodide anion (I3- ), three-center four-electron (3c-4e) bonding is justified using Molecular Orbital (MO) Theory (LCAO), which explains how the ion can be stable without involving 𝑑-orbitals for hypervalency. 

1. Molecular Orbital (MO) Justification 

The 3c-4e bond is formed by the linear combination of one orbital from each of the three iodine atoms: 

• Bonding MO (𝜓1): An all-in-phase combination that is delocalized across all three centers. It is the lowest in energy.

• Non-bonding MO (𝜓2): An out-of-phase combination of the two-terminal iodine 𝑝𝑧

orbitals. The central atom does not contribute to this orbital (it has a node at the center), so the electron density resides primarily on the terminal atoms.

• Antibonding MO (𝜓3): An all-out-of-phase combination that remains unoccupied. 

Four electrons occupy these orbitals: two in the bonding MO and two in the non-bonding MO. Since only the bonding MO provides net stabilization, the total bond order is 1, which is shared across two bonds, giving each I–I interaction a bond order of 0.5.

Feature     I₃⁻     I₃⁺

Electron count in axial σ system                     4 e⁻                                                                2 e⁻

3-centre bond type                              3c–4e (hypervalent, electron-rich)                3c–2e (electron-deficient)

MO occupancy                                             σ₁² σ₂² σ₃⁰                                                       σ₁² σ₂⁰ σ₃⁰

Total I–I bond order                                    1 (≈½ per I–I)                                              1 (≈½ per I–I)

Geometry                                                 Linear, symmetric                                 Linear or nearly linear, symmetric

Relative I–I distance                             Longer, weaker bonds                                       Shorter, stronger bonds

Reactivity                                             Nucleophilic / reducing                                Strongly electrophilic/ oxidizing

Note: 1.  In I₃⁺ the three collinear iodine atoms contribute three p-orbitals which combine to give three σ-MOs. The molecule contains only two electrons in this σ-system, which occupy the lowest bonding MO (σ₁). This results in a delocalized 3-center–2-electron bond extending over all three iodine atoms, analogous to H₃⁺. By contrast, I₃⁻ contains four electrons filling σ₁ and σ₂, giving a 3-center–4-electron hypervalent bond.

2. VSEPR predicts a linear geometry for both I₃⁺ and I₃⁻. However, the bonding is not adequately explained by simple hybridisation models. Modern theory treats these ions as 3-centre–2-electron (I₃⁺) or 3-centre–4-electron (I₃⁻) systems formed from iodine p-orbitals, without the need to invoke d-hybridisation.

 Key conceptual difference:

• I₃⁻: bonding is “diluted” by occupation of a nonbonding MO → hypervalent, softer, more diffuse.

• I₃⁺: bonding involves only the strongly bonding MO, with no nonbonding electrons along the axis → more strongly bound but electron-poor.

Q2: Why can PF₆ have a higher EA than the F atom?

Answer:

·        Adding an electron to atomic F(328Kj/mol) gives F⁻, where the extra electron is confined to a small 2p orbital → strong e⁻–e⁻ repulsion.

·        In PF₆(772KJ/mol), an extra electron is delocalized over six F atoms and a P center, stabilised by strong P–F bonding and charge distribution over a large volume. This reduces electron–electron repulsion and strongly stabilizes the anion.

 Therefore, A molecule like PF₆ has more ways to delocalize and stabilize the added electron (bonding + charge distribution) than an isolated F atom, so its electron affinity can be higher even though fluorine atoms are very electronegative.

Q3: For each of the following pairs of elements, state which one has the higher electron affinity (EA). Briefly justify your choice using periodic trends.
Pairs:  1. Li or Cs   2. Li or F     3. Cs or F     4. F or Cl         5. Cl or Br    6. O or S        7.  S or Se

Answers:

Note: Electron affinity generally increases across a period and decreases down a group, but chlorine has a slightly higher EA than fluorine, and sulfur has a slightly higher EA than oxygen due to reduced electron–electron repulsion in the larger p-orbitals.

1. Li > Cs; EA decreases down Group 1, so lithium has the higher EA.

2. F > Li; F is a halogen with much higher EA than an alkali metal.

3. F > Cs; Fluorine, being a halogen, has far higher EA than cesium.

4. Cl > F; Chlorine has slightly higher EA than fluorine because the incoming electron experiences less electron–electron repulsion in the larger 3p orbital of Cl compared with the compact 2p orbital of F.

5. Cl > Br; EA decreases down Group 17, so chlorine has the higher EA.

6. S > O; Sulfur has a higher EA than oxygen for the same reason as F/Cl: the added electron experiences less repulsion in the larger 3p orbital of sulfur compared with oxygen’s smaller 2p orbital.

7. S > Se; EA decreases down the group, so sulfur has the higher EA.

Q4: For each of the following pairs, state which species is larger.   Explain your choice using the trends that:

Pairs:   1. K⁺ or Cs⁺ 2. La³⁺ or Lu³⁺  3. Cl⁻ or Br⁻   4. Ca²⁺ or Zn²⁺      5. Cs or Fr

 Answers:

• Cations become smaller as positive charge increases.

• Anions become larger as negative charge increases.

• Ionic radius increases down a group in the periodic table.

• Higher nuclear charge within the same shell reduces ionic size.

 Cs⁺ is larger than K⁺.
Both are Group-1 cations, but Cs⁺ lies further down the group, so it has more electron shells and a larger radius.

1. La³⁺ is larger than Lu³⁺.

Across the lanthanide series, there is lanthanide contraction, meaning Lu³⁺ has a higher effective nuclear charge and is therefore smaller than La³⁺.

3. Br⁻ is larger than Cl⁻.

Both are halide anions, but bromine is below chlorine in the periodic table, so Br⁻ has a larger ionic radius.

4. Ca²⁺ is larger than Zn²⁺.

Zn²⁺ has a higher nuclear charge while having electrons in the same principal shell as Ca²⁺, so Zn²⁺ is pulled in more strongly and is smaller.

5. Fr is larger than Cs.
   Atomic radius increases down Group 1, and francium is below cesium. (Although experimental data on Fr is limited, the trend predicts Fr to be larger.)