Understanding Hypovalent and Hypervalent Compounds

Hypovalent compounds have atoms with fewer electrons than required by the octet rule, while hypervalent compounds have atoms with more electrons than the octet rule allows (typically in period 3+ elements). We analyze each compound by calculating the formal number of electrons around the central atom.

Step 1: Analyze XeF₄

Xenon (Xe) has 8 valence electrons. Each fluorine (F) contributes 1 bond (2 electrons shared), so 4 F atoms give 4 bonds. Total electrons around Xe: 8 (lone pairs) + 8 (bonding) = 16 electrons, which exceeds octet. Thus, it is hypervalent.

Calculation: $e-{\mathrm{count}}_{\mathrm{Xe}}=8+4\times 2=16$

Step 2: Analyze SO₂

Sulfur (S) has 6 valence electrons. Each oxygen (O) forms a double bond (contributing 4 electrons each to S). Total electrons around S: 6 (lone pairs) + 8 (bonding) = 14 electrons, exceeding octet. Thus, it is hypervalent.

Calculation: $e-{\mathrm{count}}_S=6+2\times 4=14$

Step 3: Analyze COCl₂

Carbon (C) has 4 valence electrons. Oxygen (O) forms a double bond (4 electrons), and each chlorine (Cl) forms a single bond (2 electrons each). Total electrons around C: 0 (lone pairs) + 8 (bonding) = 8 electrons, satisfying octet. Thus, it is neither hypovalent nor hypervalent.

Calculation:

Step 4: Analyze BF₄^–

Boron (B) has 3 valence electrons. Each fluorine (F) contributes 1 bond (2 electrons shared), so 4 F atoms give 4 bonds. The negative charge adds 1 extra electron. Total electrons around B: 0 (lone pairs) + 8 (bonding) + 1 (charge) = 9 electrons, exceeding octet. However, boron is period 2 and cannot expand octet; this is incorrect. Actually, BF₄^– has tetrahedral geometry with sp³ hybridization, and B has 8 electrons around it (4 bonds × 2 electrons = 8), satisfying octet due to the charge. Wait, let's recalculate properly: B original valence = 3, charge adds 1, so effective valence electrons = 4. With 4 single bonds, it shares 8 electrons, which is octet. So it is not hypervalent; it satisfies octet.

Revised calculation: Effective valence electrons of B = 3 + 1 (from charge) = 4. Bonding electrons = 4 × 2 = 8. Total electrons = 8, octet satisfied. Not hypervalent.

Final Answer

XeF₄ and SO₂ are hypervalent. COCl₂ and BF₄^– are neither hypovalent nor hypervalent (they satisfy the octet rule).

Related Topics and Formulae

Octet Rule: Atoms tend to have 8 electrons in their valence shell for stability.

Hypovalency: Central atom has fewer than 8 electrons (e.g., BF₃ with 6 electrons).

Hypervalency: Central atom has more than 8 electrons, common in elements from period 3 onwards (e.g., PCl₅, SF₆).

Formula for Electron Count: $\mathrm{Total} \mathrm{electrons}=\mathrm{valence} e-+(\mathrm{bonds}\times 2)+\mathrm{charge} \mathrm{adjustment}$

Theory: Hypervalent compounds often involve d-orbital participation in bonding, though this is debated; molecular orbital theory provides a better explanation for their stability.