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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掳 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 PBrCl鈧�, the VSEPR model predicts a trigonal bipyramidal geometry with axial P-Cl bond angle of 180掳 and equatorial P-Br bond angles of approximately 120掳.

Step by step solution

01

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

For phosphorus trihalides (PX鈧�), 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 PX鈧� 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掳.
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 PF鈧� (96.3掳) is smaller than that in PCl鈧� (100.3掳) and PBr鈧� (101.0掳). The trend continues (although less markedly) as we move from PBr鈧� to PI鈧� (102.0掳).
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 PBrCl鈧�

PBrCl鈧� 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 PBrCl鈧� 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掳, and the equatorial P-Br bond angles should be approximately 120掳.

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Key Concepts

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

Phosphorus Trihalides Explained
Phosphorus trihalides \(PX_3\) are chemical compounds in which a central phosphorus atom is bonded to three halogen atoms. These halogens can be fluorine (F), chlorine (Cl), bromine (Br), or iodine (I).
Phosphorus trihalides are recognized by their systematic structure, where phosphorus contributes five valence electrons, and each halogen adds one. This results in a total of eight electrons shared between phosphorus and the halogens, forming three covalent bonds.
The remaining electron pair on phosphorus forms a lone pair, which influences the molecule鈥檚 geometry significantly.
These compounds are interesting due to their varied applications and their specific geometric formation.
Electronegativity Trends and Their Impact
Electronegativity refers to the ability of an atom to attract shared electrons in a covalent bond. In the context of phosphorus trihalides:
  • Fluorine, being highly electronegative, attracts bonding electrons more strongly than other halogens.
  • As we progress down the halogen group from F to I, electronegativity decreases.
This electronegativity difference leads to varied bond angles among different trihalides.
For instance, in \(PF_3\), where fluorine is highly electronegative, the \(X-P-X\) bond angle is smallest at approximately 96.3掳, compared to the wider angles in other trihalides.
Understanding these trends helps predict changes in molecular structure and reactivity.
Predicting Molecular Geometry with VSEPR
The Valence Shell Electron Pair Repulsion (VSEPR) model is a tool used to predict the geometric arrangement of atoms in a molecule. For phosphorus trihalides, the VSEPR model helps explain their pyramidal shape.
With three bonding pairs and one lone pair around the phosphorus atom, the predicted geometry is a trigonal pyramid, fitting the electron-domain geometry of a tetrahedral.
This shape arises because electron pairs, whether bonding or lone pairs, repel each other and try to stay as far apart as possible.
The VSEPR model predicts a bond angle of 109.5掳 for a perfect tetrahedral geometry. However, the presence of a lone pair slightly squeezes the other bonds together, reducing the actual \(X-P-X\) bond angle.
Understanding Bond Angles in Phosphorus Trihalides
Bond angles in phosphorus trihalides are influenced by both the lone pair of electrons on phosphorus and the electronegativity of the halogens.
The presence of the phosphorus lone pair creates additional repulsion, reducing the bond angles slightly compared to a perfect tetrahedral angle of 109.5掳.
As electronegativity of the halogen increases, the bond pair electrons are pulled closer to the halogen, intensifying the repulsion from the lone pair on phosphorus.
This means higher electronegativity correlates with smaller \(X-P-X\) bond angles. For example, the slightly smaller bond angle in \(PF_3\) is attributed to fluorine's high electronegativity.
Understanding bond angles is crucial for predicting the geometry and behavior of molecules in chemical reactions.

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Most popular questions from this chapter

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{AsF}_{3},\) (b) \(\mathrm{CH}_{3}^{+},\) (c) \(\mathrm{BrF}_{3},\) (d) \(\mathrm{ClO}_{3}^{-},\) (e) \(\mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

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?

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{~N}_{2}\). A closely related substance is hydrazobenzene, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2} .\) The Lewis structures of these two substances are (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the \(\mathrm{N}\) atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the \(\mathrm{N}\) and the \(C\) atoms in each of the substances? (c) Predict the \(\mathrm{N}-\mathrm{N}-\mathrm{C}\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the \(2 s\) orbital on each atom, (b) the \(2 p_{z}\) orbital on each atom (assume both atoms are on the \(z\) -axis), \((\mathrm{c})\) the \(2 s\) orbital on one atom and the \(2 p_{z}\) orbital on the other atom.

(a) Which geometry and central atom hybridization would you expect in the series \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+} ?(\mathbf{b})\) What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?
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