Draw the Lewis structures for $\mathrm{TeCl}_{4}, \mathrm{ICl}_{5}, \mathrm{PCl}_{5}, \mathrm{KrCl}_{4},\( and \)\mathrm{XeCl}_{2}$ . Which of the compounds exhibit at least one bond angle that is approximately $120^{\circ} ?\( Which of the compounds exhibit \)d^{2} s p^{3}$ hybridization? Which of the compounds have a square planar molecular structure? Which of the compounds are polar?

1. \(\mathrm{TeCl}_{4}\): Te = Group 16 element (6 valence electrons) Cl = Group 17 element (7 valence electrons) Total valence electrons = \((6 + (4 \times 7)) = 34\) 2. \(\mathrm{ICl}_{5}\): I = Group 17 element (7 valence electrons) Cl = Group 17 element (7 valence electrons) Total valence electrons = \((7 + (5 \times 7)) = 42\) 3. \(\mathrm{PCl}_{5}\): P = Group 15 element (5 valence electrons) Cl = Group 17 element (7 valence electrons) Total valence electrons = \((5 + (5 \times 7)) = 40\) 4. \(\mathrm{KrCl}_{4}\): Kr = Group 18 element (8 valence electrons in its outermost shell) Cl = Group 17 element (7 valence electrons) Total valence electrons = \((8 + (4 \times 7)) = 36\) 5. \(\mathrm{XeCl}_{2}\): Xe = Group 18 element (8 valence electrons in its outermost shell) Cl = Group 17 element (7 valence electrons) Total valence electrons = \((8 + (2 \times 7)) = 22\)

1. \(\mathrm{TeCl}_{4}\): Tetrahedral (4 electron domains) - No bond angle of approximately \(120^{\circ}\) 2. \(\mathrm{ICl}_{5}\): Square pyramidal (6 electron domains) - No bond angle of approximately \(120^{\circ}\) 3. \(\mathrm{PCl}_{5}\): Trigonal bipyramidal (5 electron domains) - Has a bond angle of approximately \(120^{\circ}\) 4. \(\mathrm{KrCl}_{4}\): Tetrahedral (4 electron domains) - No bond angle of approximately \(120^{\circ}\) 5. \(\mathrm{XeCl}_{2}\): Linear (3 electron domains) - No bond angle of approximately \(120^{\circ}\) Only \(\mathrm{PCl}_{5}\) has a bond angle of approximately \(120^{\circ}\).

1. \(\mathrm{TeCl}_{4}\): \(sp^{3}\) hybridization - No 2. \(\mathrm{ICl}_{5}\): \(sp^{3}d\) hybridization - No 3. \(\mathrm{PCl}_{5}\): \(sp^{3}d\) hybridization - No 4. \(\mathrm{KrCl}_{4}\): \(sp^{3}\) hybridization - No 5. \(\mathrm{XeCl}_{2}\): \(d^{2}sp^{3}\) hybridization - Yes Only \(\mathrm{XeCl}_{2}\) exhibits \(d^{2}sp^{3}\) hybridization.

Only \(\mathrm{ICl}_{5}\) has a square planar molecular structure, since it has a central Iodine atom surrounded by 5 Chlorine atoms and has a square pyramidal structure. All the corner Chlorine atoms form a square base surrounding the Iodine atom, which sits on the pyramid's apex.

Polar compounds have asymmetrical charge distributions due to differences in electronegativity between bonded atoms. In this case: 1. \(\mathrm{TeCl}_{4}\): Polar - the electronegativity difference between Te and Cl atoms leads to polar bonds, and the tetrahedral shape distributes these dipoles asymmetrically. 2. \(\mathrm{ICl}_{5}\): Polar - the electronegativity difference between I and Cl atoms leads to polar bonds, and the square pyramidal shape distributes these dipoles asymmetrically. 3. \(\mathrm{PCl}_{5}\): Non-polar - the electronegativity difference between P and Cl atoms leads to polar bonds, but the trigonal bipyramidal shape distributes these dipoles symmetrically, which cancels them out. 4. \(\mathrm{KrCl}_{4}\): Non-polar - the electronegativity difference between Kr and Cl atoms is negligible, leading to non-polar bonds. 5. \(\mathrm{XeCl}_{2}\): Non-polar - the electronegativity difference between Xe and Cl atoms leads to polar bonds, but the linear shape distributes these dipoles symmetrically, which cancels them out. Polar compounds are \(\mathrm{TeCl}_{4}\) and \(\mathrm{ICl}_{5}\).