(a) What conditions must be met if a molecule with polar bonds is nonpolar? (b) What geometries will signify nonpolar molecules for \(\mathrm{AB}_{2}, \mathrm{AB}_{3},\) and \(\mathrm{AB}_{4}\) geometries?

Short Answer

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A molecule with polar bonds can be nonpolar if: (1) the central atom has no lone pairs of electrons, (2) the surrounding atoms are identical or have the same electronegativity, and (3) the molecule has a symmetrical geometry, leading to the cancellation of the net electric dipole moment arising from the polar bonds. For the general formulas AB₂, AB₃, and AB₄, nonpolar geometries include linear for AB₂, trigonal planar for AB₃, and tetrahedral for AB₄, provided that the polar bonds' dipoles cancel out due to symmetry.

Step by step solution

01

Understanding Polarity

Polarity refers to the separation of electric charges within a molecule, leading to a net electric dipole moment. A molecule can have polar bonds, meaning that the atoms in those bonds have a notable difference in electronegativity – the ability of an atom to attract electrons in a bond towards itself. However, a molecule itself can still be nonpolar if the net electric dipole moment is zero because the polar bonds are arranged symmetrically, canceling out each other's effects.
02

Conditions for Nonpolarity in Molecules with Polar Bonds

A molecule with polar bonds can be nonpolar if the following conditions are met: 1. The central atom has no lone pairs of electrons. 2. The surrounding atoms are identical or have the same electronegativity. 3. The molecule has a symmetrical geometry, leading to the cancellation of the net electric dipole moment arising from the polar bonds. It's important to note that if these conditions are not satisfied, then the molecule will be polar.
03

Identifying Nonpolar Geometries for AB₂, AB₃, and AB₄

Now we will discuss different geometries of molecules with the general formulas AB₂, AB₃, and AB₄, and determine if they can form nonpolar molecules given the presence of polar bonds. 1. AB₂: The simplest geometry for AB₂ is linear, where the two B atoms are opposite each other (180° angle) with respect to the central A atom. With this configuration, the polar bonds' dipoles point in opposite directions and cancel each other, resulting in a nonpolar molecule. 2. AB₃: One possible geometry for AB₃ is trigonal planar, with bond angles of 120°. In this case, if all three B atoms are the same and no lone pairs are present on the central atom A, the dipoles of each polar bond would cancel out due to symmetry, and the molecule would be nonpolar. Another possible geometry is trigonal pyramidal, which will always result in a polar molecule. 3. AB₄: For AB₄, the geometry can either be tetrahedral or square planar. In the case of a tetrahedral geometry, if all four B atoms are the same and no lone pairs are present on the central atom A, the polar bonds' dipoles would cancel out each other due to symmetry, making the molecule nonpolar. In contrast, a molecule with square planar geometry will always result in a polar molecule. In conclusion, molecules with the general formulas AB₂, AB₃, and AB₄ can be nonpolar with the following geometries, provided that the polar bonds' dipoles cancel out due to symmetry: linear for AB₂, trigonal planar for AB₃, and tetrahedral for AB₄. However, other possible geometries, such as trigonal pyramidal and square planar, will result in polar molecules.

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Most popular questions from this chapter

Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the \(z\) -axis is defined as the metal-metal bond axis. (a) Which of the \(3 d\) orbitals (Figure 6.23\()\) can be used to make a \(\sigma\) bond between metal atoms? (b) Sketch the \(\sigma_{3 d}\) bonding and \(\sigma_{3 d}^{*}\) antibonding MOs. (c) With reference to the "Closer Look" box on the phases of orbitals, explain why a node is generated in the \(\sigma_{3 d}^{\star}\) MO. (d) Sketch the energy-level diagram for the \(\mathrm{Sc}_{2}\) molecule, assuming that only the \(3 d\) orbital from part (a) is important. (e) What is the bond order in \(\mathrm{Sc}_{2}\) ?

(a) What is meant by the term orbital overlap? (b) Describe what a chemical bond is in terms of electron density between two atoms.

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a sigma bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two \(s\) orbitals combine to form a \(\pi\) bond? Explain.

(a) What is the physical basis for the VSEPR model? (b) When applying the VSEPR model, we count a double or triple bond as a single electron domain. Why is this justified?

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X}: \mathrm{PF}_{3}, 96.3^{\circ} ; \mathrm{PCl}_{3}, 100.3^{\circ} ; \mathrm{PBr}_{3}\), \(101.0^{\circ} ; \mathrm{PI}_{3}, 102.0^{\circ} .\) The trend is generally attributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the halide electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCl}_{4}\).

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