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Chapter 22: Problem 51

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: $(\mathbf{a}) \mathrm{HNO}_{2},(\mathbf{b}) \mathrm{N}_{3}^{-},(\mathbf{c}) \mathrm{N}_{2} \mathrm{H}_{5}^{+},(\mathbf{d}) \mathrm{NO}_{3}^{-}$.

Short Answer

Expert verified
(a) HNO2: Lewis Structure: ``` O || N - O | H ``` Molecular Geometry: Trigonal Planar Oxidation State of Nitrogen: +2 (b) N3-: Lewis Structure: ``` N ≡ N - N ≡ N ↔ N ``` Molecular Geometry: Linear Oxidation State of Nitrogen: N1: 0, N2: +2, N3: -2 (c) N2H5+: Lewis Structure: ``` H | N - N - H | | H H ``` Molecular Geometry: Trigonal Pyramidal Oxidation State of Nitrogen: N1: -2, N2: +2 (d) NO3-: Lewis Structure: ``` O || N - O | O ↔ O ``` Molecular Geometry: Trigonal Planar Oxidation State of Nitrogen: +5

Step by step solution

01

(a) HNO2 Lewis structure

First, we need to determine the total number of valence electrons in the HNO2 molecule. Nitrogen has 5, Oxygen has 6 (x2 for two oxygen atoms), and Hydrogen has 1, for a total of 18 valence electrons. Next, we construct the skeleton of the molecule by connecting the central atom Nitrogen to each of the Hydrogen and Oxygen atoms with a single bond. The remaining electrons will be placed around the Oxygen atoms to satisfy their octet requirements. The HNO2 Lewis structure will be: ```O || N - O | H ```
02

(a) HNO2 molecular geometry and oxidation state of nitrogen

In HNO2, nitrogen has one double bond with one oxygen, one single bond with the other oxygen, and one single bond with hydrogen. It has 3 bonding domains, so the molecular geometry is trigonal planar. The oxidation state of nitrogen can be calculated as follows: Let x be the oxidation state of nitrogen, then x +2(1) + 2(-2) = 0 x -2 = 0 x = +2
03

(b) N3- Lewis structure

For N3-, we have 3 Nitrogen atoms with 5 valence electrons each, plus one extra electron due to the negative charge, for a total of 16 valence electrons. The Lewis structure of N3- is: ``` N ≡ N - N ≡ N ↔ N ```
04

(b) N3- molecular geometry and oxidation state of nitrogen

In N3-, two nitrogen atoms have two lone electron pairs and one single bond, and the central nitrogen atom has one triple bond and a lone pair. All three nitrogen atoms have a linear molecular geometry. The oxidation state of the left nitrogen atom (N1) is 0, the middle nitrogen atom (N2) is +2, and the right nitrogen atom (N3) is -2.
05

(c) N2H5+ Lewis structure

For N2H5+, we have 2 Nitrogen atoms with 5 valence electrons each and 5 Hydrogen atoms with 1 valence electron each, minus one electron due to the positive charge, for a total of 14 valence electrons. The Lewis structure of N2H5+ is: ``` H | N - N - H | | H H ```
06

(c) N2H5+ molecular geometry and oxidation state of nitrogen

In N2H5+, each nitrogen atom has three single bonds (one with each other and two with hydrogen atoms) and one lone pair of electrons. The molecular geometry of both nitrogen atoms is trigonal pyramidal. The oxidation state of the left nitrogen atom (N1) is -2, and the right nitrogen atom (N2) is +2.
07

(d) NO3- Lewis structure

For NO3-, we have 1 Nitrogen atom with 5 valence electrons and 3 Oxygen atoms with 6 valence electrons each, plus one extra electron due to the negative charge, for a total of 24 valence electrons. The Lewis structure of NO3- is: ``` O || N - O | O ↔ O ```
08

(d) NO3- molecular geometry and oxidation state of nitrogen

In NO3-, the nitrogen atom has three resonance structures with double bonds to oxygen atoms and no lone pairs. The molecular geometry for nitrogen is trigonal planar. The oxidation state of nitrogen in NO3- can be calculated as follows: Let x be the oxidation state of nitrogen, then x + 3(-2) = -1 x -6 = -1 x = +5

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Most popular questions from this chapter

Select the member of group 14 that best fits each description: \((\mathbf{a})\) has the greatest tendency to form multiple bonds with itself, (b) forms polymeric structures with oxygen, (c) is a metal with +2 and +4 oxidation states.

Borazine, \((\mathrm{BH})_{3}(\mathrm{NH})_{3},\) is an analog of \(\mathrm{C}_{6} \mathrm{H}_{6},\) benzene. It can be prepared from the reaction of diborane with ammonia, with hydrogen as another product; or from lithium borohydride and ammonium chloride, with lithium chloride and hydrogen as the other products. (a) Write balanced chemical equations for the production of borazine using both synthetic methods. (b) Draw the Lewis dot structure of borazine. (c) How many grams of borazine can be prepared from $2.00 \mathrm{~L}$ of ammonia at STP, assuming diborane is in excess?

A sulfuric acid plant produces a considerable amount of heat. This heat is used to generate electricity, which helps reduce operating costs. The synthesis of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) consists of three main chemical processes: (a) oxidation of \(S\) to \(\mathrm{SO}_{2}\), (b) oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3},\) (c) the dissolving of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and the subsequent reaction with water to form \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If the third process produces \(130 \mathrm{~kJ} / \mathrm{mol}\), how much heat is produced in preparing a mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from a mole of \(S\) ? How much heat is produced in preparing \(2000 \mathrm{~kg}\) of $\mathrm{H}_{2} \mathrm{SO}_{4} ?$

The standard heats of formation of $\mathrm{H}_{2} \mathrm{O}(g), \mathrm{H}_{2} \mathrm{~S}(g), \mathrm{H}_{2} \operatorname{Se}(g)$, and \(\mathrm{H}_{2} \mathrm{Te}(g)\) are \(-241.8,-20.17,+29.7,\) and $+99.6 \mathrm{~kJ} /$ mol, respectively. The enthalpies necessary to convert the elements in their standard states to one mole of gaseous atoms are \(248,277,227,\) and \(197 \mathrm{~kJ} / \mathrm{mol}\) of atoms for $\mathrm{O}, \mathrm{S}$, Se, and Te, respectively. The enthalpy for dissociation of \(\mathrm{H}_{2}\) is \(436 \mathrm{~kJ} / \mathrm{mol}\). Calculate the average \(\mathrm{H}-\mathrm{O}, \mathrm{H}-\mathrm{S}, \mathrm{H}-\mathrm{Se}\), and \(\mathrm{H}-\) Te bond enthalpies, and comment on their trend.

(a) How does the structure of diborane $\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\( differ from that of ethane \)\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) ?(\mathbf{b})$ Explain why diborane adopts the geometry that it does. (c) What is the significance of the statement that the hydrogen atoms in diborane are described as "hydridic"?
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