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(a) Methane \(\left(\mathrm{CH}_{4}\right)\) and the perchlorate ion \(\left(\mathrm{ClO}_{6}\right)\) are both described as tetrahedral. What does this indicate about their bond angles? (b) The \(\mathrm{NH}_{3}\) molecule is trigonal pyramidal, while \(\mathrm{BF}_{3}\) is trigonal planar. Which of these molecules is flat?

Short Answer

Expert verified
(a) In tetrahedral molecules like methane (\(\mathrm{CH}_4\)) and the perchlorate ion (\(\mathrm{ClO}_4^-\)), the bond angles between any two bonds are 109.5 degrees. (b) \(\mathrm{NH}_3\) has a trigonal pyramidal geometry and is not flat, while \(\mathrm{BF}_3\) is trigonal planar and has a flat shape.

Step by step solution

01

Answer(a): Bond Angles in Tetrahedral Molecules

For both methane (\(\mathrm{CH}_4\)) and the perchlorate ion (\(\mathrm{ClO}_4^-\)), their tetrahedral geometry indicates that the bond angles between any two bonds are 109.5 degrees. In a tetrahedral molecule, the central atom is surrounded by four other atoms, which are located at the vertices of a regular tetrahedron. The bond angles (angle between any two bonds) in such a geometry are all equal to 109.5 degrees.
02

Answer(b): Flat Molecule Between \(\mathrm{NH}_3\) and \(\mathrm{BF}_3\)

The \(\mathrm{NH}_3\) molecule has a trigonal pyramidal geometry, while the \(\mathrm{BF}_3\) molecule is trigonal planar. In a trigonal pyramidal molecule, the central atom is surrounded by three other atoms and has a lone pair of electrons. The presence of a lone pair introduces a 3-dimensional shape and distortion from the ideal planar geometry. Because of this, the \(\mathrm{NH}_3\) molecule is not flat. On the other hand, in a trigonal planar molecule such as \(\mathrm{BF}_3\), the central atom is surrounded by three other atoms and has no lone pairs of electrons. As a result, the molecule exhibits a flat or planar shape with bond angles of 120 degrees. Accordingly, the \(\mathrm{BF}_3\) molecule is flat.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tetrahedral Molecule
Understanding the structure of a tetrahedral molecule is fundamental in chemistry as it affects the molecule's properties and reactivity.

Imagine the molecule like a tripod; one that is perfectly balanced with a central atom resting in the center, where the legs (bonds) are heading off into the corners of an imaginary tetrahedron. This shape is one of the most common and important molecular geometries in chemistry. The central atom is bonded to four other atoms, placed at equal distance from each other and arranged at the vertices of a hypothetical tetrahedron.

As a result, all the bond angles in a perfect tetrahedral molecule are the same, measuring about \b109.5°\b. This happens because the shape allows for maximum spatial distribution, minimizing the repulsion between the electron pairs located in the bonds. This four-sided, three-dimensional geometric form is not just a theoretical construct; countless molecules adopt this structure, including methane (\bCH\(_4\)\b) and the perchlorate ion (\bClO\(_4^-\)\b), providing a stable, energetically favorable configuration.
Trigonal Pyramidal Geometry
Shifting our focus to molecules with a trigonal pyramidal geometry, we delve into a slightly less symmetric world where lone pairs come into play.

When a central atom is bonded to three other atoms but also possesses a lone pair of electrons, such as in ammonia (\bNH\(_3\)\b), it creates what's known as trigonal pyramidal geometry. Unlike tetrahedral geometry, the presence of this lone pair creates an asymmetry, pushing the bonded atoms slightly closer together and causing a distortion. As a result, the bond angles in a trigonal pyramidal molecule are typically less than the ideal 109.5° seen in tetrahedral molecules, leaning closer to \b107°\b.

The best way to visualize this is to imagine a pyramid with a very wide base. If you were to remove one of the corners of a tetrahedron (representing the lone pair), you'd end up with the trigonal pyramidal shape. This shape is important for understanding the molecular geometry of many compounds that contain a central atom with a lone pair, as it can significantly influence the molecule's chemical behavior and physical properties.
Bond Angles
Bond angles are the cornerstone of understanding molecular geometry. They can determine the shape and polarity of the molecule and thus its interactions with other substances.

The angle created by the intersection of two bonds, originating from the same central atom, is known as the bond angle. These angles are a direct consequence of the electron pair repulsion theory, which dictates that electron pairs (bonding or lone pairs) will arrange themselves as far apart as possible to minimize repulsion.

For example, in a tetrahedral molecule, the optimal bond angle of \b109.5°\b ensures the least amount of repulsion among bonding electron pairs. In contrast, a molecule with trigonal planar geometry, such as \bBF\(_3\)\b, exhibits a bond angle of \b120°\b due to its flat, three-sided shape with no lone electron pairs, while a trigonal pyramidal molecule like \bNH\(_3\)\b has bond angles slightly less than 109.5° because of the lone pair.

By comprehending the intricacies of bond angles, students can not only predict the shapes of molecules but also foresee the potential interactions they may have in a complex chemical environment.

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Most popular questions from this chapter

There are two compounds of the formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) : The compound on the right, cisplatin, is used in cancer therapy. The compound on the left, transplatin, is ineffective for cancer therapy. Both compounds have a square-planar geometry. (a) Which compound has a nonzero dipole moment? (b) The reason cisplatin is a good anticancer drug is that it binds tightly to DNA. Cancer cells are rapidly dividing, producing a lot of DNA. Consequently cisplatin kills cancer cells at a faster rate than normal cells. However, since normal cells also are making DNA, cisplatin also attacks healthy cells, which leads to unpleasant side effects. The way both molecules bind to DNA involves the \(\mathrm{Cl}^{-}\) ions leaving the Pt ion, to be replaced by two nitrogens in DNA. Draw a picture in which a long vertical line represents a piece of DNA. Draw the \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\) fragments of cisplatin and transplatin with the proper shape. Also draw them attaching to your DNA line. Can you explain from your drawing why the shape of the cisplatin causes it to bind to DNA more effectively than transplatin?

(a) Starting with the orbital diagram of a sulfur atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(S F_{2}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Would the hybridization scheme in part (a) be appropriate for \(\mathrm{SF}_{4} ?\) Explain.

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion, and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\) (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? Explain.

Ethyl acetate, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\), is a fragrant substance used bothas a solvent and as an aroma enhancer. Its Lewis structure is (a) What is the hybridization at each of the carbon atoms of the molecule? (b) What is the total number of valence electrons in ethyl acetate? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

(a) Does \(\mathrm{SCl}_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does \(\mathrm{BeCl}_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

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