To determine the geometry of molecules, we use the VSEPR (Valence Shell Electron Pair Repulsion) theory, which considers the number of bond pairs and lone pairs around the central atom.

(i) ${\mathrm{NH}}_3$ (Ammonia):

• Central atom: Nitrogen (N).
• Valence electrons of N = 5.
• Number of H atoms bonded = 3.
• Number of bond pairs = 3.
• Number of lone pairs = (5 - 3) / 2 = 1.
• Total electron pairs = 3 (bond pairs) + 1 (lone pair) = 4.
• Electron geometry: Tetrahedral.
• Molecular geometry: Due to the presence of one lone pair, the geometry is distorted from tetrahedral to pyramidal.

(ii) ${\mathrm{C}}_2{\mathrm{H}}_4$ (Ethene):

• Each carbon atom is bonded to two hydrogen atoms and one other carbon atom via a double bond.
• Consider one carbon atom: It forms 2 single bonds with H and 1 double bond with C.
• Number of electron domains around each carbon = 3 (2 single bonds + 1 double bond).
• Number of lone pairs = 0.
• Electron geometry: Trigonal planar.
• Molecular geometry: Planar (specifically, trigonal planar around each carbon, making the overall molecule planar).

(iii) ${\mathrm{ClO}}_3^{-}$ (Chlorate ion):

• Central atom: Chlorine (Cl).
• Valence electrons of Cl = 7.
• Number of O atoms bonded = 3.
• Charge = -1, so add 1 electron.
• Total electrons for bonding and lone pairs = 7 + 1 = 8.
• Number of bond pairs = 3 (Cl forms single bonds with O, with resonance).
• Number of lone pairs = (8 - 3) / 2 = 1. (Note: In VSEPR, double bonds are treated as a single electron domain for geometry prediction).
• Total electron pairs = 3 (bond pairs) + 1 (lone pair) = 4.
• Electron geometry: Tetrahedral.
• Molecular geometry: Due to the presence of one lone pair, the geometry is distorted from tetrahedral to pyramidal.

Therefore, the geometries are Pyramidal, Planar, Pyramidal.

Option Analysis:

• A) Pyramidal, pyramidal, tetrahedral: Incorrect, ${\mathrm{C}}_2{\mathrm{H}}_4$ is planar, and ${\mathrm{ClO}}_3^{-}$ is pyramidal.
• B) Pyramidal, tetrahedral, pyramidal: Incorrect, ${\mathrm{C}}_2{\mathrm{H}}_4$ is planar.
• C) Pyramidal, planar, tetrahedral: Incorrect, ${\mathrm{ClO}}_3^{-}$ is pyramidal.
• D) Pyramidal, planar, pyramidal: Correct.

Correct Answer: (D)