Question

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?

Short answer

In summary, \(\mathrm{PCl}_{5}\) has a bond angle of approximately \(120^{\circ}\), \(\mathrm{XeCl}_{2}\) exhibits \(d^{2}sp^{3}\) hybridization, \(\mathrm{ICl}_{5}\) has a square planar molecular structure, and the polar compounds are \(\mathrm{TeCl}_{4}\) and \(\mathrm{ICl}_{5}\).

Step-by-step solution

Drawing Lewis structures for given compounds

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 = \((8 + (2 \times 7)) = 22\)

Determining bond angles

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 \(120^{\circ}\).

Checking for \(d^{2}sp^{3}\) hybridization

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.

Identifying square planar molecular structures

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.

Identifying polar compounds

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 \(\mathrm{ICl}_{5}\).

Key concepts

Bond Angles

Bond angles describe the angles between adjacent bonds originating from the same atom. They play a crucial role in shaping a molecule's physical and chemical properties. In the context of the compounds discussed, only \(\mathrm{PCl}_{5}\) features a bond angle of approximately \(120^{\circ}\). This occurs in its trigonal bipyramidal geometry, where the three equatorial chlorine atoms form angles of \(120^{\circ}\) with each other. These angles are significant because they impact how molecules interact with each other and dictate the spatial distribution of atoms in a molecule. Understanding bond angles can reveal insights into molecular conformations and reactivity patterns.
Knowing which molecules have specific bond angles helps predict physical properties like melting and boiling points, due to molecular packing efficiency. It also aids in understanding the potential for molecular interactions and reactions.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It can be predicted using the

• VSEPR theory (Valence Shell Electron Pair Repulsion theory), which explains the shape of a molecule based on its electron pair repulsion.

For example, \(\mathrm{TeCl}_{4}\) adopts a seesaw shape due to one lone pair influencing the spatial arrangement of bonded atoms, while \(\mathrm{ICl}_{5}\) has a square pyramidal structure with a central iodine atom at the apex and five chlorine atoms forming the base.
Meanwhile, \(\mathrm{PCl}_{5}\) shows a trigonal bipyramidal form, with three equatorial chlorine atoms and two axial positions. Understanding the geometry provides insight into potential chemical behavior, like reaction sites or intermolecular interactions.

• For practice, visualize these structures and use molecular models to get a clear idea of the spatial arrangements.

Hybridization

Hybridization explains how atomic orbitals mix to form new hybrid orbitals, which determine bonding and molecular shape. For the compounds in question, different types of hybridizations influence their geometries; these are key in molecular structure predictions.

• \(\mathrm{TeCl}_{4}\): Exhibits \(sp^{3}\) hybridization, accommodating its electron pair and four bonded chlorines in a seesaw shape.
• \(\mathrm{PCl}_{5}\): Shows \(sp^{3}d\) hybridization, which allows it to expand its valency to hold five chlorines in a trigonal bipyramidal arrangement.
• \(\mathrm{XeCl}_{2}\): Uniquely presents \(d^{2}sp^{3}\) hybridization, leading to a linear structure facilitated by the inclusion of d-orbitals.

Hybridization helps in understanding how molecules form and break bonds by sharing or distributing electrons effectively, thereby dictating their geometry.

Polarity of Compounds

Polarity in molecules results from differences in electronegativity between atoms, leading to uneven electron distribution, which creates dipoles. Compounds such as \(\mathrm{TeCl}_{4}\) and \(\mathrm{ICl}_{5}\) are polar because their asymmetrical shapes lead to a non-cancelling dipole moment, with electronegativity differences contributing to this asymmetry. Compounds are polar when

• they feature significant differences in electronegativity.
• their shapes do not symmetrically cancel dipole moments.

This polarity affects physical properties like solubility and boiling/melting points. It also plays a role in chemical reactivity and intermolecular attractions, influencing how substances dissolve or interact with others. Recognizing compound polarity assists in predicting interactions and behavior in different chemical environments.