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Predict the molecular structure (including bond angles) for each of the following. a. \(\mathrm{PCl}_{3}\) b. \(\mathrm{SCl}_{2}\) c. \(\mathrm{SiF}_{4}\)

Short Answer

Expert verified
a. \(\mathrm{PCl}_{3}\) has a tetrahedral geometry with a bond angle of 109.5°. b. \(\mathrm{SCl}_{2}\) has a bent geometry with a bond angle of approximately 104° to 109.5°. c. \(\mathrm{SiF}_{4}\) has a tetrahedral geometry with a bond angle of 109.5°.

Step by step solution

01

a. \(\mathrm{PCl}_{3}\)

- Central atom: P (phosphorus) - Bonded atoms: 3 Cl (chlorine) - Lone pairs on P: 1 (5 valence electrons on P, 3 used for bonding, leaving 2) - Total electron pairs: 4 (3 bonded + 1 lone pair)
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b. \(\mathrm{SCl}_{2}\)

- Central atom: S (sulfur) - Bonded atoms: 2 Cl (chlorine) - Lone pairs on S: 2 (6 valence electrons on S, 4 used for bonding, leaving 2) - Total electron pairs: 4 (2 bonded + 2 lone pairs)
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c. \(\mathrm{SiF}_{4}\)

- Central atom: Si (silicon) - Bonded atoms: 4 F (fluorine) - Lone pairs on Si: 0 (4 valence electrons on Si, all used for bonding) - Total electron pairs: 4 (4 bonded + 0 lone pairs) #Step 2: Determine molecular geometry and bond angles# Now that we have the electron pair count, we can determine the molecular geometry and bond angles for each molecule, using VSEPR theory.
04

a. \(\mathrm{PCl}_{3}\)

Molecular geometry: tetrahedral (4 electron pairs) Bond angle: 109.5° (standard bond angle for tetrahedral geometry)
05

b. \(\mathrm{SCl}_{2}\)

Molecular geometry: bent (4 electron pairs, but only 2 form bonds) Bond angle: approximately 104° to 109.5° (angle varies due to electron repulsion from the 2 lone pairs)
06

c. \(\mathrm{SiF}_{4}\)

Molecular geometry: tetrahedral (4 bonded electron pairs) Bond angle: 109.5° (standard bond angle for tetrahedral geometry) In summary, the molecular structures and bond angles for the given molecules are: a. \(\mathrm{PCl}_{3}\) - Tetrahedral geometry with a bond angle of 109.5° b. \(\mathrm{SCl}_{2}\) - Bent geometry with a bond angle of approximately 104° to 109.5° c. \(\mathrm{SiF}_{4}\) - Tetrahedral geometry with a bond angle of 109.5°

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Key Concepts

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

VSEPR Theory
Understanding the shapes of molecules is crucial in the study of chemistry, and the Valence Shell Electron Pair Repulsion (VSEPR) theory plays a vital role in determining molecular geometry. The VSEPR theory posits that electron pairs located in the outermost shell of atoms, known as valence electrons, will position themselves as far apart as possible to minimize repulsion. This behavior affects the overall shape of the molecule.

VSEPR theory follows simple steps to predict molecular structure. First, we count the number of valence electrons around the central atom. Then, we determine the number of bonded atoms and lone pairs of electrons. The combination of these factors will lead to a specific molecular geometry, such as linear, bent, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral.

For example, consider the molecule PCl3 from the exercise, which has a central phosphorus atom bonded to three chlorine atoms with one lone pair. According to VSEPR theory, the best arrangement to reduce repulsion is a tetrahedral geometry, even though the bond angles are slightly squished from the ideal 109.5° due to the lone pair's presence.
Bond Angles

Standard Bond Angles and Deviations

Every molecular geometry has its standard bond angles; these are the angles one would expect if all positions around the central atom were occupied by bonded atoms with no lone pairs causing deviations. For a tetrahedral molecule like SiF4, the standard bond angle is 109.5°. However, when lone pairs are present, like in SCl2, the bond angles are generally less than the standard due to the higher repulsion lone pairs exert compared to bonded pairs of electrons.

The ability to predict these angles aids in understanding how molecules interact and react with one another, affecting the physical and chemical properties of substances. When tackling problems involving bond angles, considering the effects of electron pair repulsion on the geometry helps in making accurate predictions.
Chemical Bonding
Chemical bonding is the fundamental process that allows atoms to combine and form molecules. It involves the sharing or exchange of electrons among atoms to achieve a full outer shell, based on the octet rule. The type of bond—ionic, covalent, or metallic—depends on the elements involved and their electronegativity differences.

In molecules like PCl3, SCl2, and SiF4, discussed in the exercise, covalent bonding occurs as nonmetals are sharing electrons to obtain a stable electron configuration. The number of shared electron pairs determines whether a bond is single, double, or triple, with single bonds being the longest and weakest, and triple bonds being the shortest and strongest.

By understanding bonding concepts and how they relate to molecular geometry, one can predict properties such as the polarity of molecules, their reactivities, and interactions with other molecules, which is essential in fields like material science, biochemistry, and pharmacology.

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Most popular questions from this chapter

The most common type of exception to the octet rule are compounds or ions with central atoms having more than eight electrons around them. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3}\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

Which member of the following pairs would you expect to be more energetically stable? Justify each choice. a. \(\mathrm{NaBr}\) or \(\mathrm{NaBr}_{2}\) b. \(\mathrm{ClO}_{4}\) or \(\mathrm{ClO}_{4}^{-}\) c. \(\mathrm{SO}_{4}\) or \(\mathrm{XeO}_{4}\) d. \(\mathrm{OF}_{4}\) or \(\mathrm{SeF}_{4}\)

The compound hexaazaisowurtzitane is one of the highestenergy explosives known ( \(C\) \& E News, Jan. 17, 1994, p. 26). The compound, also known as CL-20, was first synthesized in 1987 . The method of synthesis and detailed performance data are still classified because of CL-20's potential military application in rocket boosters and in warheads of "smart" weapons. The structure of CL-20 is In such shorthand structures, each point where lines meet represents a carbon atom. In addition, the hydrogens attached to the carbon atoms are omitted; each of the six carbon atoms has one hydrogen atom attached. Finally, assume that the two \(\mathrm{O}\) atoms in the \(\mathrm{NO}_{2}\) groups are attached to \(\mathrm{N}\) with one single bond and one double bond. Three possible reactions for the explosive decomposition of \(\mathrm{CL}-20\) are i. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{~N}_{12} \mathrm{O}_{12}(s) \rightarrow 6 \mathrm{CO}(g)+6 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)+\frac{3}{2} \mathrm{O}_{2}(g)\) ii. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{~N}_{12} \mathrm{O}_{12}(s) \rightarrow 3 \mathrm{CO}(g)+3 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)\) iii. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{~N}_{12} \mathrm{O}_{12}(s) \rightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2}(g)\) a. Use bond energies to estimate \(\Delta H\) for these three reactions. b. Which of the above reactions releases the largest amount of energy per kilogram of CL-20?

Write Lewis structures that obey the octet rule for each of the following molecules. a. \(\mathrm{CCl}_{4}\) c. \(\mathrm{SeCl}_{2}\) b. \(\mathrm{NCl}_{3}\) d. ICl In each case, the atom listed first is the central atom.

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