Inorganic Chemistry

PCl₅ has a trigonal bipyramidal geometry. In this geometry, there are two types of P-Cl bonds:

• Axial bonds: Two longer and weaker bonds positioned above and below the central phosphorus atom.
• Equatorial bonds: Three shorter and stronger bonds arranged in a trigonal plane around the central phosphorus atom.

Therefore, the statement "All five P - Cl bonds are identical in length" is incorrect. The other statements are correct: PCl₅ has two different bond angles (90° and 120°), exhibits sp³d hybridization, and consists of five sigma bonds.

Let’s match the processes with their respective catalysts:

• A. Haber process: This is the process for the industrial synthesis of ammonia (N₂ + 3H₂ → 2NH₃), and the catalyst used is Fe (iron).
  Thus, A-I.

• B. Wacker oxidation: This process involves the oxidation of alkenes to carbonyl compounds (like aldehydes or ketones), and the catalyst used is PdCl₂ (palladium chloride).
  Thus, B-II.

• C. Wilkinson catalyst: This catalyst is used in the hydrogenation of alkenes, and its composition is [(C₆H₅)₃RhCl] (tris(triphenylphosphine) rhodium chloride).
  Thus, C-III.

• D. Ziegler catalyst: This is used in the polymerization of alkenes, and the catalyst used is TiCl₄ with Al(C₂H₅)₃ (titanium tetrachloride and triethylaluminum).
  Thus, D-IV.

Thus, the correct answer is (2) A-I, B-II, C-III, D-IV.

To determine the correct geometry, we need to determine the number of electron pairs (bonding and lone pairs) around the central Xenon atom for each molecule.

A. XeO₃

• Xe has 8 valence electrons. Each O contributes 2 electrons for bonding. So Xe will have 3 double bonds with the 3 O atoms, leaving one lone pair. 3 bond pairs + 1 lone pair = 4 electron pairs around Xe.

• This gives sp³ hybridization. Because of the presence of a lone pair, the geometry is pyramidal.

B. XeF₂

• Xe has 8 valence electrons. Each F contributes 1 electron for bonding. So Xe will have 2 single bonds with the 2 F atoms, leaving three lone pairs. 2 bond pairs + 3 lone pairs = 5 electron pairs around Xe.

• This gives sp³d hybridization. The three lone pairs are arranged equatorially to minimize repulsion, resulting in a linear geometry.

C. XeOF₄

• Xe has 8 valence electrons. Each F contributes 1 electron for bonding, O contributes 2 electron bonding. So Xe will have 4 single bonds with the 4 F atoms, and 1 double bond with the 1 O atom, leaving one lone pair. 5 bond pairs + 1 lone pair = 6 electron pairs around Xe.

• This gives sp³d² hybridization. With one lone pair it makes Square pyramidal.

D. XeF₆

• Xe has 8 valence electrons. Each F contributes 1 electron for bonding. So Xe will have 6 single bonds with the 6 F atoms, leaving one lone pair. 6 bond pairs + 1 lone pair = 7 electron pairs around Xe.

• Due to steric congestion, the molecule does not exhibit octahedral geometry. Based on the VSEPR model, The molecule is predicted to have a distorted octahedral geometry. The central Xe atom has seven electron pairs around it (6 bonds and 1 lone pair), which cause its geometry to be distorted octahedral.

Matching the compounds to their geometries:

A (XeO₃) - II (sp³ pyramidal)

B (XeF₂) - I (sp³d linear)

C (XeOF₄) - IV (sp³d² square pyramidal)

D (XeF₆) - III (sp³d distorted octahedral)

The correct answer is (4) A-II, B-I, C-IV, D-III