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Chapter 10: Problem 53

Write plausible Lewis structures for the following odd-electron species: (a) \(\mathrm{CH}_{3} ;\) (b) \(\mathrm{ClO}_{2} ;\) (c) \(\mathrm{NO}_{3}\).

Short Answer

Expert verified
The Lewis structures for the odd-electron species CH3, ClO2 and NO3 have been drawn. For CH3, the odd electron is located on the Carbon atom. For ClO2, the odd electron is on the central Oxygen. For NO3, the Nitrogen atom carries the unpaired electron.

Step by step solution

01

Determine the total valence electrons for each molecule

For CH3, Carbon (C) has 4 valence electrons and Hydrogen (H) has 1, multiplied by 3 gives 3 electrons. Total is 4 + 3 = 7 valence electrons. For ClO2, Chlorine (Cl) has 7 valence electrons and Oxygen (O) has 6, multiplied by 2 gives 12 electrons. Total is 7 + 12 = 19 valence electrons. For NO3, Nitrogen (N) has 5 valence electrons and Oxygen (O) has 6, multiplied by 3 gives 18 electrons. Total is 5 + 18 = 23 valence electrons.
02

Assign the central atom and distribute the electrons

CH3: Carbon is the central atom with Hydrogen atoms around it. We put 2 electrons (a single bond) between the C and each H. This uses up 6 of our 7 electrons. The single remaining electron is placed on the central Carbon, making it an odd-electron species. ClO2: Oxygen is the central atom with Chlorine and another Oxygen around it. Single bond is made between Cl-O and O-O using 4 electrons. The rest are placed as lone pairs. The remaining electron (making it an odd-electron species) will be added to the central Oxygen. NO3: Nitrogen is the central atom with Oxygen atoms around it. Single bond is made among N-Ox3, using 6 electrons. Remaining ones are placed as lone pairs. The total gives 24 electrons, but only 23 are available so Nitrogen will carry one unpaired electron.
03

Draw the Lewis structure

Draw the molecules according to the previous steps, making sure to appropriately place the odd (unpaired) electron.

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Most popular questions from this chapter

Use the VSEPR theory to predict a probable shape of the molecule \(\mathrm{F}_{4} \mathrm{SCH}_{2}\), and explain the source of any ambiguities in your prediction.

The molecule \(\mathrm{H}_{2} \mathrm{O}_{2}\) has a resultant dipole moment of 2.2 D. Can this molecule be linear? If not, describe a shape that might account for this dipole moment.

Draw Lewis structures for two different molecules with the formula \(\mathrm{C}_{3} \mathrm{H}_{4}\). Is either of these molecules linear? Explain.

Without referring to tables in the text, indicate which of the following bonds you would expect to have the greatest bond length, and give your reasons. (a) \(\mathrm{O}_{2}\); (b) \(\mathrm{N}_{2} ;\) (c) \(\mathrm{Br}_{2} ;\) (d) \(\mathrm{BrCl}\).

Alternative strategies to the one used in this chapter have been proposed for applying the VSEPR theory to molecules or ions with a single central atom. In general, these strategies do not require writing Lewis structures. In one strategy, we write (1) the total number of electron pairs \(=[\) (number of valence electrons) \(\pm\) (electrons required for ionic charge) \(] / 2\) (2) the number of bonding electron pairs \(=\) (number of atoms) -1 (3) the number of electron pairs around central atom \(=\) total number of electron pairs \(-3 \times[\) number of terminal atoms (excluding \(\mathrm{H}\) )] (4) the number of lone-pair electrons = number of central atom pairs - number of bonding pairs After evaluating items \(2,3,\) and \(4,\) establish the VSEPR notation and determine the molecular shape. Use this method to predict the geometrical shapes of the following: (a) \(\mathrm{PCl}_{5} ;\) (b) \(\mathrm{NH}_{3} ;\) (c) \(\mathrm{ClF}_{3} ;\) (d) \(\mathrm{SO}_{2} ;\) (e) \(\mathrm{ClF}_{4}^{-}\); (f) \(\mathrm{PCl}_{4}^{+}\). Justify each of the steps in the strategy, and explain why it yields the same results as the VSEPR method based on Lewis structures. How does the strategy deal with multiple bonds?
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