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Chapter 10: Problem 65

Draw Lewis structures and give the other information requested for the following molecules: (a) \(\mathrm{BF}_{3}\). Shape: planar or nonplanar? (b) \(\mathrm{ClO}_{3}^{-}\). Shape: planar or nonplanar? (c) \(\mathrm{H}_{2} \mathrm{O}\). Show the direction of the resultant dipole moment. (d) \(\mathrm{OF}_{2}\). Polar or nonpolar molecule? (e) \(\mathrm{SeO}_{2}\). Estimate the OSeO bond angle.

Short Answer

Expert verified
The shape of \(\mathrm{BF}_{3}\) is planar, \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{H}_{2} \mathrm{O}\) are nonplanar with \(\mathrm{H}_{2} \mathrm{O}\) showing dipole moment towards O, \(\mathrm{OF}_{2}\) is a polar molecule, and the OSeO bond angle in \(\mathrm{SeO}_{2}\) is around 118°.

Step by step solution

01

Draw Lewis Structures and Determine Geometry for \(\mathrm{BF}_{3}\)

According to Lewis structure, B has 3 valence electrons and each F has 7 valence electrons. The B atom is connected to each F atom making a total of 3 covalent bonds and having no lone pairs left. This structure results in a trigonal planar shape, as the distribution of electrons is symmetric around the central atom B.
02

Draw Lewis Structures and Determine Geometry for \(\mathrm{ClO}_{3}^{-}\)

The Lewis diagram for \(\mathrm{ClO}_{3}^{-}\) has Cl (7 valence e⁻) as the central atom surrounded by three oxygens (each with 6 valence e⁻) and one extra electron provided by the ionic charge. There's a double bond between Cl and one of the oxygen atoms, and single bonds with the other two. Furthermore, there are 6 lone pairs of electrons. This distribution leads to a trigonal pyramidal shape, which is nonplanar.
03

Draw Lewis Structures and Determine Geometry for \(\mathrm{H}_{2} \mathrm{O}\)

In a water molecule, the central atom O (6 valence electrons) is bonded to two H atoms (1 valence electron each) to form two covalent bonds and is left with two lone pairs. This gives it a bent or V-shaped geometry which results in a resultant dipole moment towards the oxygen atom due to its higher electronegativity.
04

Draw Lewis Structures and Determine Geometry for \(\mathrm{OF}_{2}\)

The Lewis diagram for \(\mathrm{OF}_{2}\) has O (6 valence e⁻) as the central atom attached to two F atoms (7 valence e⁻ each) with a total of 2 lone pairs of electrons. Its molecular geometry is also bent, similar to water. The difference in electronegativity between F and O results in partial negative charges on the fluorine atoms, and with both dipoles additive along the same path, it became a polar molecule.
05

Draw Lewis Structure and Estimate Bond Angle for \(\mathrm{SeO}_{2}\)

In the Lewis diagram for \(\mathrm{SeO}_{2}\), Se (6 valence electrons) is the central atom having double bonds with both the O atoms (6 valence electrons each) and 1 lone pair. This results in a V-shaped or bent geometry (similar to \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{OF}_{2}\)). The bond angle in such cases is usually less than 120° due to presence of lone pair, let's say ~118°.

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Most popular questions from this chapter

Sketch the bond moments and resultant dipole moments for these molecules: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{PCl}_{3}, \mathrm{XeF}_{4}, \mathrm{PCl}_{5}\) \(\mathrm{SF}_{6}\)

The bonds in beryllium hydride \(\left(\mathrm{BeH}_{2}\right)\) molecules are polar, and yet the dipole moment of the molecule is zero. Explain.

Predict the geometry of sulfur dichloride \(\left(\mathrm{SCl}_{2}\right)\) and the hybridization of the sulfur atom.

How is the geometry of a molecule defined and why is the study of molecular geometry important?

The geometries discussed in this chapter all lend themselves to fairly straightforward elucidation of bond angles. The exception is the tetrahedron, because its bond angles are hard to visualize. Consider the \(\mathrm{CCl}_{4}\) molecule, which has a tetrahedral geometry and is nonpolar. By equating the bond moment of a particular \(\mathrm{C}-\mathrm{Cl}\) bond to the resultant bond moments of the other three \(\mathrm{C}-\mathrm{Cl}\) bonds in opposite directions, show that the bond angles are all equal to \(109.5^{\circ}\)
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