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Chapter 9: Problem 85

Draw the Lewis structures for \(\mathrm{SeO}_{2}, \mathrm{PCl}_{3}, \mathrm{NNO}, \mathrm{COS},\) and \(\mathrm{PF}_{3} .\) Which of the compounds are polar? Which of the compounds exhibit at least one bond angle that is approximately \(120^{\circ}\) Which of the compounds exhibit \(s p^{3}\) hybridization by the central atom? Which of the compounds have a linear molecular structure?

Short Answer

Expert verified
SeO鈧�, PCl鈧�, NNO, COS, and PF鈧� are all polar compounds. SeO鈧� has a bond angle of approximately 120掳. PCl鈧� and PF鈧� exhibit sp鲁 hybridization by the central atom. COS has a linear molecular structure.

Step by step solution

01

Draw the Lewis structures for each compound

Using the element's group number, determine the number of valence electrons and draw the Lewis structures for each. Remember to satisfy the octet rule whenever possible. 1. 厂别翱鈧�: Selenium (Se) is in Group 16 and so has 6 valence electrons. Oxygen (O) is also in Group 16 and has 6 valence electrons. Draw a Lewis structure where Selenium is the central atom, connected to two Oxygen atoms, with double bonds and lone electron pairs completing the octet for each atom: O = Se = O Se: 6 valence electrons 2 Oxygens: 2 x 6 valence electrons = 12 valence electrons Total valence electrons = 18 2. 笔颁濒鈧�: Phosphorus (P) is in Group 15 with 5 valence electrons. Chlorine (Cl) is in Group 17 with 7 valence electrons. Draw a Lewis structure where Phosphorus is the central atom, connected to three Chlorine atoms, each by single bonds and lone electron pairs completing the octet for each atom: Cl - P - Cl | Cl P: 5 valence electrons 3 Chlorines: 3 x 7 valence electrons = 21 valence electrons Total valence electrons = 26 3. NNO: There are two possible structures for NNO, i.e., N-N=O or N=N=O. The second structure N=N=O is a correct representation after verifying the formal charges: N = N = O Each Nitrogen: 5 valence electrons Oxygen: 6 valence electrons Total valence electrons = 16 4. COS: Carbon (C) is in Group 14 with 4 valence electrons. Oxygen (O) and Sulfur (S) are in Group 16 with 6 valence electrons each. Draw a Lewis structure where Carbon is the central atom, connected to Oxygen by a double bond, connected to Sulfur by a double bond, and lone electron pairs completing the octet for Oxygen and Sulfur: O = C = S C: 4 valence electrons O and S: 2 x 6 valence electrons = 12 valence electrons Total valence electrons = 16 5. 笔贵鈧�: Phosphorus (P) is in Group 15 with 5 valence electrons. Fluorine (F) is in Group 17 with 7 valence electrons. Draw a Lewis structure with Phosphorus as the central atom, connected to three Fluorine atoms, each by single bonds, and lone electron pairs completing the octet for each atom: F - P - F | F P: 5 valence electrons 3 Fluorines: 3 x 7 valence electrons = 21 valence electrons Total valence electrons = 26
02

Determine the polarity of each compound

Determine the polarity of each compound by checking if the molecule is symmetrical or has a net dipole moment due to differences in electronegativity between the atoms: 1. 厂别翱鈧�: Polar, due to the bent structure and difference in electronegativity between Se and O atoms. 2. 笔颁濒鈧�: Polar, due to the trigonal pyramid structure and difference in electronegativity between P and Cl atoms. 3. NNO: Polar, due to the bent structure and difference in electronegativity between N and O atoms. 4. COS: Polar, due to the linear structure and difference in electronegativity between C, O, and S atoms. 5. 笔贵鈧�: Polar, due to the trigonal pyramid structure and difference in electronegativity between P and F atoms.
03

Identify which compounds exhibit a bond angle of approximately 120掳

Based on their Lewis structures, the following compounds have a bond angle of approximately 120掳: - 厂别翱鈧�: The bond angle is approximately 120掳 due to its bent structure. - COS: The bond angle is 180掳, so it does not meet the 120掳 criterion.
04

Identify compounds with sp鲁 hybridization by the central atom

Look at the molecular geometry of the central atom in each compound to determine if sp鲁 hybridization is present: - 笔颁濒鈧�: The central Phosphorus atom is sp鲁 hybridized, with a tetrahedral molecular geometry. - 笔贵鈧�: The central Phosphorus atom is sp鲁 hybridized, with a tetrahedral molecular geometry.
05

Identify linear molecular structures

Based on the Lewis structures, the compounds with linear molecular structures are: - COS: The molecular geometry is linear with 180掳 bond angle. In summary, SeO鈧�, PCl鈧�, NNO, COS, and PF鈧� are all polar compounds. SeO鈧� has a bond angle of approximately 120掳, while PCl鈧� and PF鈧� have sp鲁 hybridization by the central atom. COS has a linear molecular structure.

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Key Concepts

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

Molecular Polarity
Molecular polarity is a key characteristic of molecules that determines how they interact with electric fields and other molecules. It depends on both the symmetry of the molecule and the differences in electronegativity between the atoms involved. A polar molecule typically has an uneven distribution of electron density, leading to a dipole moment, meaning one part of the molecule is slightly positive, while another part is slightly negative.
In terms of the compounds discussed:
  • 厂别翱鈧�: This molecule is polar due to its bent shape. The difference in electronegativity between selenium and oxygen leads to an uneven distribution of charge.
  • 笔颁濒鈧�: Exhibits polarity because of its trigonal pyramidal shape and the difference in electronegativity between phosphorus and chlorine.
  • NNO: Also shows polarity as it has a straight structure with an electronegativity difference between the central nitrogen atoms and the oxygen.
  • COS: Despite its linearity, the unequal sharing of electrons causes a polar structure.
  • 笔贵鈧�: Due to its trigonal pyramidal shape and electronegativity difference, this molecule is also polar.
The polar nature of these molecules influences their solubility in various solvents and affects their biological and chemical activity.
Bond Angles
Bond angles are the angles between adjacent bonds in a molecule. They are essential in determining the molecule's overall shape. Certain bond angles are associated with specific geometries, frequently predicted by the VSEPR (Valence Shell Electron Pair Repulsion) theory.
For example:
  • 厂别翱鈧�: This compound features a bond angle close to 120掳 due to its bent geometric structure, resembling a V-shape.
  • PCl鈧� and 笔贵鈧�: Though they both have trigonal pyramidal shapes, their bond angles are slightly less than the typical tetrahedral angle of 109.5掳 due to the repulsion caused by lone pair electrons on the phosphorus atom.
  • COS: As it is linear, it has a bond angle of 180掳, not fitting the 120掳 angle criteria.
These bond angles contribute to molecular shape, thereby influencing reactivity and interactions with other molecules.
sp鲁 Hybridization
sp鲁 hybridization occurs when one s orbital and three p orbitals mix to form four equivalent orbitals, leading to a tetrahedral shape with bond angles around 109.5掳. This type of hybridization is common in molecules where the central atom makes four bonds or has a combination of bonds and lone pairs.
In relation to the compounds:
  • 笔颁濒鈧�: The phosphorus atom undergoes sp鲁 hybridization, forming a trigonal pyramidal molecular geometry with one lone pair affecting its shape.
  • 笔贵鈧�: Similarly, the phosphorus in this compound exhibits sp鲁 hybridization, resulting in a comparable geometry to PCl鈧� with a lone pair.
sp鲁 hybridization contributes to understanding molecular geometry, electron distribution, and potential reaction pathways in various contexts.
Linear Molecular Structure
A linear molecular structure is characterized by atoms connected in a straight line with a bond angle of 180掳. This shape arises when there are no lone pairs forcing deviations from linearity, typically in diatomic molecules or compounds bonded with double bonds.
Considering our specific compounds:
  • COS: This molecule is a classic example of a linear structure, where the atoms are aligned with a bond angle of 180掳. The double bonds between carbon, oxygen, and sulfur facilitate this alignment, ensuring the molecule remains straight.
A linear molecular structure affects how molecules stack or interact with others, influencing both physical properties and chemical reactivity.

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Most popular questions from this chapter

Bond energy has been defined in the text as the amount of energy required to break a chemical bond, so we have come to think of the addition of energy as breaking bonds. However, in some cases the addition of energy can cause the formation of bonds. For example, in a sample of helium gas subjected to a high-energy source, some Holecules exist momentarily and then dissociate. Use MO theory (and diagrams) to explain why He_ molecules can come to exist and why they dissociate.

Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as \(\sigma\) or \(\pi .\)

Describe the bonding in the \(\mathrm{O}_{3}\) molecule and the \(\mathrm{NO}_{2}^{-}\) ion using the localized electron model. How would the molecular orbital model describe the \(\pi\) bonding in these two species?

In Exercise 95 in Chapter 8 , the Lewis structures for benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) were drawn. Using one of the Lewis structures, estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) using bond energies and given that the standard enthalpy of formation of \(\mathrm{C}(g)\) is 717 \(\mathrm{kJ} / \mathrm{mol}\) . The experimental \(\Delta H_{\mathrm{f}}^{\circ}\) value of \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) is 83 \(\mathrm{kJ} / \mathrm{mol} .\) Explain the discrepancy between the experimental value and the calculated \(\Delta H_{\mathrm{f}}^{\circ}\) value for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Many important compounds in the chemical industry are derivatives of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right) .\) Two of them are acrylonitrile and methyl methacrylate. Complete the Lewis structures, showing all lone pairs. Give approximate values for bond angles \(a\) through \(f\) . Give the hybridization of all carbon atoms. In acrylonitrile, how many of the atoms in the molecule must lie in the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in methyl methacrylate and acrylonitrile?
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