Question

Incorrect statement is :

• A. MgO > AlF3 > MgF2 : Lattice energy
• B. Li > Na > Al > Mg : Electron affinity
• C. SF6> PF5 > SiF4 : Lewis acidic character
• D. SiCl4 > SiBr4 > Sil4 : Decreasing order of electronegativity of Si

Answer 1

Answer: (C)

SF6> PF5 > SiF4 : Lewis acidic character

Answer 2

Let's analyze each statement:

(a) MgO > AlF₃ > MgF₂ : Lattice energy

Lattice energy is directly proportional to the product of the charges of the ions ($Z_1 Z_2$) and inversely proportional to the sum of their ionic radii ($r_1 + r_2$). The formula for lattice energy is approximately given by $U \propto \frac{Z_1 Z_2}{r_1 + r_2}$.

• MgO: Mg²⁺ (Z=2), O²⁻ (Z=2). Product of charges = 2 × 2 = 4. Ionic radii: r(Mg²⁺) ≈ 72 pm, r(O²⁻) ≈ 140 pm. Sum of radii ≈ 212 pm. Relative value ≈ 4/212 ≈ 0.0188.
• AlF₃: Al³⁺ (Z=3), F⁻ (Z=1). Product of charges = 3 × 1 = 3. Ionic radii: r(Al³⁺) ≈ 53.5 pm, r(F⁻) ≈ 133 pm. Sum of radii ≈ 186.5 pm. Relative value ≈ 3/186.5 ≈ 0.0161.
• MgF₂: Mg²⁺ (Z=2), F⁻ (Z=1). Product of charges = 2 × 1 = 2. Ionic radii: r(Mg²⁺) ≈ 72 pm, r(F⁻) ≈ 133 pm. Sum of radii ≈ 205 pm. Relative value ≈ 2/205 ≈ 0.0097.

Comparing the relative values: 0.0188 (MgO) > 0.0161 (AlF₃) > 0.0097 (MgF₂). Thus, the order MgO > AlF₃ > MgF₂ for lattice energy is correct.

(b) Li > Na > Al > Mg : Electron affinity

Electron affinity (EA) is the energy change when an electron is added to a neutral gaseous atom. A higher positive value (more exothermic) indicates a greater electron affinity.

• Li: Group 1, Period 2. EA = 59.6 kJ/mol.
• Na: Group 1, Period 3. EA = 52.8 kJ/mol. (EA decreases down a group, so Li > Na).
• Mg: Group 2, Period 3. Has a stable s² configuration. Adding an electron requires energy (endothermic). EA = -21 kJ/mol.
• Al: Group 13, Period 3. Has an empty p-orbital. EA = 42.5 kJ/mol.

Comparing the values: 59.6 (Li) > 52.8 (Na) > 42.5 (Al) > -21 (Mg). The order Li > Na > Al > Mg is correct.

(c) SF₆ > PF₅ > SiF₄ : Lewis acidic character

A Lewis acid is an electron pair acceptor. Lewis acidity depends on the electron deficiency of the central atom and the availability of vacant orbitals.

• SiF₄: Silicon is in Group 14, oxidation state +4. It has vacant 3d orbitals and is electron deficient. It is a strong Lewis acid, forming [SiF₅]⁻ and [SiF₆]²⁻.
• PF₅: Phosphorus is in Group 15, oxidation state +5. It has vacant 3d orbitals and is electron deficient. It is a Lewis acid, forming [PF₆]⁻.
• SF₆: Sulfur is in Group 16, oxidation state +6. While it has vacant 3d orbitals, the sulfur atom is coordinatively saturated (surrounded by 6 fluorine atoms, resulting in 12 valence electrons) and sterically hindered. This makes SF₆ exceptionally stable and chemically inert, acting as a very poor Lewis acid.

Comparing SiF₄ and PF₅: Si is less electronegative than P. In SiF₄, the Si-F bonds are more polar, leading to a greater partial positive charge on Si, making it more electron deficient and a stronger Lewis acid than PF₅.

Therefore, the correct order of Lewis acidity is SiF₄ > PF₅ > SF₆. The given statement SF₆ > PF₅ > SiF₄ is the reverse of the actual order and is incorrect.

(d) SiCl₄ > SiBr₄ > SiI₄ : Decreasing order of electronegativity of Si

Electronegativity is an intrinsic atomic property and does not change for the same atom in different compounds. The electronegativity of Silicon (Si) is constant, regardless of the halogen it is bonded to.

What changes in these compounds is the partial positive charge on the Si atom, which is influenced by the electronegativity of the bonded halogens.

Electronegativity of halogens: Cl (3.16) > Br (2.96) > I (2.66).

In SiCl₄, Si is bonded to highly electronegative Cl atoms, pulling electron density away from Si, resulting in the largest partial positive charge on Si. In SiBr₄, the partial positive charge on Si is less than in SiCl₄. In SiI₄, the partial positive charge on Si is the least.

So, the partial positive charge on Si decreases in the order: SiCl₄ > SiBr₄ > SiI₄.

However, the statement explicitly refers to the "electronegativity of Si", which is an atomic constant. Therefore, the statement is fundamentally incorrect in its premise.

Comparing (c) and (d): Statement (c) presents a chemical trend that is factually reversed, which is a clear chemical error. Statement (d) uses the term "electronegativity of Si" incorrectly, implying it changes, which is a definitional error. In the context of competitive exams, a reversed trend is often the intended incorrect statement. Both statements are incorrect, but (c) represents a more direct factual error about a chemical property trend.

Final decision: Statement (c) is definitively incorrect as it reverses the established trend of Lewis acidity.