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Chapter 9: Problem 111

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 \(\mathrm{PBrCl}_{4}\).

Short Answer

Expert verified
The X-P-X bond angle in a perfect tetrahedral electron-domain geometry is predicted to be 109.5° by the VSEPR model. As the electronegativity of the halide (X) increases, the X-P-X bond angle decreases due to increased lone pair-bonding electron repulsion. For PBrCl₄, the VSEPR model predicts a trigonal bipyramidal geometry with axial P-Cl bond angle of 180° and equatorial P-Br bond angles of approximately 120°.

Step by step solution

01

(a) Calculating the X-P-X angle using VSEPR model

For phosphorus trihalides (PX₃), the phosphorus (P) is the central atom, and the halides (X) surround it. The electron-domain geometry can be predicted using the VSEPR model. Here, phosphorus has 5 valence electrons and each halide X contributes one electron. The VSEPR model predicts a tetrahedral electron-domain geometry based on the sum of bonding pairs and lone pairs around the central atom. In PX₃ compounds, there are three bonding pairs (formed between the central atom P and halide atoms X) and one lone pair (on the central atom P). The VSEPR model predicts the X-P-X bond angle in a perfect tetrahedral electron-domain geometry to be 109.5°.
02

(b) General trend in the X-P-X angle with increasing halide electronegativity

As the electronegativity of the halide (X) increases, the X-P-X bond angle decreases. For example, the bond angle in PF₃ (96.3°) is smaller than that in PCl₃ (100.3°) and PBr₃ (101.0°). The trend continues (although less markedly) as we move from PBr₃ to PI₃ (102.0°).
03

(c) Using the VSEPR model to explain the observed trend in the X-P-X angle

The VSEPR model explains the observed trend in the X-P-X bond angle in phosphorus trihalides as a result of the interaction between lone pair electrons on the central phosphorus atom and bonding electrons. As the electronegativity of the halide (X) increases, the bonding electrons are drawn closer to the more electronegative halide atom. With the bonding electrons being closer to the halide atom (X), the repulsion between the lone pair and the bonding electrons on the central phosphorus (P) atom increases. Consequently, the central atom's lone pair pushes the bonding electron pairs slightly closer together. As a result, the X-P-X bond angle becomes smaller.
04

(d) Predicting the structure of PBrCl₄

PBrCl₄ is an exception as it contains four surrounding atoms. The phosphorus (P) is the central atom with three bromine (Br) atoms and one chlorine (Cl) atom surrounding it. Again, phosphorus has 5 valence electrons, with the three bromine atoms and the chlorine atom contributing one electron each. The VSEPR model predicts a trigonal bipyramidal electron-domain geometry for PBrCl₄ based on the five bonding pairs (formed between the central atom P and the surrounding atoms). There won't be any lone pair on the central atom P. Since the chlorine atom is more electronegative than the bromine atoms, it will take the axial position (due to its stronger repulsion with the equatorial atoms), while the bromine atoms occupy the equatorial plane. The axial P-Cl bond angle will be 180°, and the equatorial P-Br bond angles should be approximately 120°.

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Most popular questions from this chapter

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

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?

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?

Consider a molecule with formula \(\mathrm{AX}_{3}\). Supposing the \(\mathrm{A}-\mathrm{X}\) bond is polar, how would you expect the dipole moment of the \(\mathrm{AX}_{3}\) molecule to change as the \(\mathrm{X}-\mathrm{A}-\mathrm{X}\) bond angle increases from \(100^{\circ}\) to \(120^{\circ} ?\)

The orbital diagram that follows presents the final step in the formation of hybrid orbitals by a silicon atom. What type of hybrid orbital is produced in this hybridization? [Section 9.5\(]\)
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