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(a) Methane \(\left(\mathrm{CH}_{4}\right)\) and the perchlorate ion \(\left(\mathrm{ClO}_{4}^{-}\right)\) 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
The bond angles in methane and perchlorate ion are close to or exactly 109.5°. The BF3 molecule is flat.

Step by step solution

01

Determine the shape and hybridization of the molecules

Methane (CH4) has a central carbon attached to four hydrogen atoms. The central carbon atom forms four sigma bonds with these hydrogen atoms and is sp3 hybridized. The perchlorate ion (ClO4-) has a central chlorine atom attached to four oxygen atoms. The central chlorine atom forms one sigma bond and three double bonds (one each with the oxygen atoms) and is also sp3 hybridized. Both CH4 and ClO4- have tetrahedral geometry because they have four electron domains around the central atom.
02

Determine the bond angles in a tetrahedral molecule

In a tetrahedral molecule, the angles between the bonds are determined by the equal repulsion between the four electron domains. This results in an ideal bond angle of 109.5°. In both the CH4 and ClO4- molecules, there should be bond angles close to, if not exactly, 109.5°. #a) Answer#: The bond angles in methane and perchlorate ion are close to or exactly 109.5°. #b) Flat molecule: NH3 or BF3#
03

Determine the shapes of NH3 and BF3

Ammonia (NH3) is composed of a nitrogen atom connected to three hydrogen atoms. It exhibits sp3 hybridization but with a lone pair of electrons on the central nitrogen atom results in a trigonal pyramidal geometry. Boron trifluoride (BF3) is composed of a boron atom connected to three fluorine atoms. It exhibits sp2 hybridization, and since there are no lone pairs of electrons on the boron atom, it results in a trigonal planar geometry.
04

Identify the flat molecule

The trigonal pyramidal geometry of NH3 means that it is not flat. On the other hand, BF3 has a trigonal planar geometry, which lacks a third dimension making it a flat molecule. #b) Answer#: The BF3 molecule is flat.

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Key Concepts

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

Tetrahedral Geometry
The concept of tetrahedral geometry is fundamental in understanding molecular shapes in chemistry. It refers to a molecule consisting of a central atom surrounded by four other atoms or groups that are equidistant from each other. This arrangement is analogous to a pyramid with a triangular base, hence the term 'tetrahedral'. Imagine placing a central atom in the middle of a ball, and then pushing four additional atoms towards the surface; they will naturally position themselves as far away from each other as possible.

In this configuration, the angles between any two bonds, which we call 'bond angles', are always approximately 109.5 degrees. This occurs because the electron pairs around the central atom tend to repel each other equally, creating a uniform distribution. Substances like methane \(\mathrm{CH}_{4}\) and the perchlorate ion \(\mathrm{ClO}_{4}^{-}\) adopt this geometry, signifying that all bond angles are nearly the same across these molecules.
Bond Angles
Bond angles are a pivotal aspect of molecular geometry, revealing the spatial arrangement of atoms within a molecule. The term specifically refers to the angle formed between two covalent bonds that share a common atom. For example, in tetrahedral molecules such as methane \(\mathrm{CH}_{4}\), the bond angles are crucial as they are indicative of the molecule's three-dimensional shape.

The ideal bond angle in a perfectly symmetrical tetrahedral molecule is 109.5 degrees, as seen in both methane and the perchlorate ion. However, it's important to understand that the presence of double bonds, lone pairs, or other factors can cause deviations from this ideal angle in other types of molecules.
Hybridization
Hybridization is the concept of merging atomic orbitals into new hybrid orbitals that can accommodate bonding pairs of electrons in molecules. It's an integral part of understanding how atoms bond in different geometric arrangements. For instance, a carbon atom in methane forms four equivalent sp3 hybrid orbitals from mixing one 's' and three 'p' orbitals. This hybridization results in four mutually perpendicular orbitals, shaping the tetrahedral geometry of methane.

Similarly, when a boron atom in boron trifluoride (BF3) hybridizes its one 's' and two 'p' orbitals, it forms three sp2 hybrid orbitals that lie in the same plane, which leads to the trigonal planar geometry of BF3.
Trigonal Planar
Molecules with trigonal planar geometry have three atoms bonded to a central atom—and this forms a plane. There are no atoms or lone pairs of electrons above or below this plane, making the molecule flat. Each bond angle in a perfectly trigonal planar molecule is 120 degrees. The central atom in such molecules is sp2 hybridized, which indicates the merging of three atomic orbitals (one 's' and two 'p' orbitals) to create three new equal sp2 hybrid orbitals.

An example of such a molecule is boron trifluoride (BF3), where the boron is at the center, and the three fluorine atoms are arranged equidistantly around it, creating a flat molecule with 120 degree bond angles.
Trigonal Pyramidal
Trigonal pyramidal geometry occurs in molecules with a central atom bonded to three other atoms, plus one lone pair of electrons. The presence of the lone pair affects the geometry, making it different from tetrahedral despite having similar sp3 hybridization. This lone pair exerts repulsive forces on the bonded electron pairs, causing them to move closer together and altering the bond angles slightly.

In ammonia (NH3), for instance, the nitrogen atom is the central atom, with three hydrogen atoms bonded to it and one lone pair of electrons. This setup results in bond angles slightly less than 109.5 degrees, making NH3 a molecule that is not flat but rather has a three-dimensional pyramidal shape.

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Most popular questions from this chapter

Dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right)\) has three forms (isomers), each of which is a different substance. (a) Draw Lewis structures of the three isomers, all of which have a carbon-carbon double bond. ( b) Which of these isomers has a zero dipole moment? (c) How many isomeric forms can chloroethylene, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\) have? Would they be expected to have dipole moments?

An AB \(_{3}\) molecules described as having a trigonal-bipyramidal electron- domain geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) trigonal planar, (ii) trigonal pyrametry of (iii) T-shaped, or (iv) tetrahedral?

Consider the \(\mathrm{H}_{2}^{+}\) ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}+\) ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+} ?\) (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higher-energy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\) ion to be stable or to fall apart? (f) Which of the following statements about part (e) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_ \(_{4} )\) according to the following unbalanced reaction: \begin{equation}\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) \end{equation} The O atom and the four \(\mathrm{F}\) atoms in OSF \(_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of OSF_ in which the formal charges of all atoms are zero.(c) Use average bond enthalpies (Table 8.3 ) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

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