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Chapter 21: Problem 34

Give the Lewis structure of (a) an oxide of nitrogen in the \(+5\) state. (b) the strongest oxoacid of nitrogen. (c) a tetrahedral oxoanion of sulfur.

Short Answer

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Question: Identify the oxide of nitrogen in which nitrogen has an oxidation state of +5 and draw its Lewis structure. Also, determine the strongest oxoacid of nitrogen and draw its Lewis structure, and identify a tetrahedral oxoanion of sulfur and draw its Lewis structure. Answer: The oxide of nitrogen in which nitrogen has an oxidation state of +5 is dinitrogen pentoxide (N₂O₅). Its Lewis structure is: O = N - O - N = O | | O O The strongest oxoacid of nitrogen is nitric acid (HNO₃), and its Lewis structure is: O || O - N - O - H | O A tetrahedral oxoanion of sulfur is the sulfate ion (SO₄²⁻), and its Lewis structure is: O | O - S - O (--) | O

Step by step solution

01

(a) Finding the oxide of nitrogen in the +5 state

To find an oxide of nitrogen in which nitrogen is in the +5 oxidation state, consider the nitrogen and oxygen are combined in their usual oxidation states of -3 for nitrogen and -2 for oxygen. Let's set up an equation: $$ -3 \times n + (-2 \times m) = +5 $$ Where n is the number of nitrogen atoms and m is the number of oxygen atoms. The most common oxide that fits this description is dinitrogen pentoxide: N₂O₅, where nitrogen has a +5 oxidation state.
02

(a) Drawing Lewis structure of N₂O₅

Each nitrogen atom has 5 valence electrons, and each oxygen atom has 6 valence electrons. So there are a total of 10 electrons for 2 nitrogen atoms and 30 electrons for 5 oxygen atoms. It means we have 40 valence electrons to distribute in N₂O₅. The Lewis structure for N₂O₅ will look like this: O = N - O - N = O | | O O Where double bonds (N = O) have four shared electrons, and single bonds (N - O) have two shared electrons. There are 4 lone pairs on every terminal oxygen atom and 2 lone pairs on each bridging oxygen atom(vertical ones)
03

(b) Identifying the strongest oxoacid of nitrogen

The oxidation state of nitrogen in oxoacids increases with the acidity of the acid. The strongest oxoacid of nitrogen will have it in the highest oxidation state, which is +5. The strongest oxoacid of nitrogen is nitric acid, HNO₃.
04

(b) Drawing Lewis structure of HNO₃

The total number of valence electrons in HNO₃ is 1 (H) + 5 (N) + 18 (O) = 24 valence electrons. The Lewis structure for HNO₃ will look like this: O || O - N - O - H | O The nitrogen atom forms a double bond with the oxygen atom above it, a single bond with the oxygen atom on the left, and another single bond to the oxygen to the right, which is bonded to the hydrogen atom. The terminal oxygen atoms have two lone pairs, the left bridging oxygen atom has three lone pairs, and the right bridging oxygen atom has one lone pair.
05

(c) Identifying a tetrahedral oxoanion of sulfur

Sulfate ion (SO₄²⁻) is a common tetrahedral oxoanion of sulfur where sulfur is at the center surrounded by four oxygen atoms.
06

(c) Drawing Lewis structure of SO₄²⁻

The total number of valence electrons in SO₄²⁻ is 6 (S) + 4 x 6 (O) + 2 (for 2- charge) = 32 valence electrons. The Lewis structure for SO₄²⁻ ion will look like this: O | O - S - O | O Add (--) label on top of one of Oxygen atoms to represent the -2 charge. There are two double bonds between sulfur and two of the oxygen atoms. The other two oxygen atoms have single bonds to sulfur. All four oxygen atoms have two lone pairs, and sulfur has no lone pairs because it is using all available electrons to form bonds with oxygen atoms.

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Most popular questions from this chapter

The amount of sodium hypochlorite in a bleach solution can be determined by using a given volume of bleach to oxidize excess iodide ion to iodine; \(\mathrm{ClO}^{-}\) is reduced to \(\mathrm{Cl}^{-}\). The amount of iodine produced by the redox reaction is determined by titration with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} ; \mathrm{I}_{2}\) is reduced to \(\mathrm{I}^{-}\). The sodium thiosulfate is oxidized to sodium tetrathionate, \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\). In this analysis, potassium iodide was added in excess to \(5.00 \mathrm{~mL}\) of bleach \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\). If \(25.00 \mathrm{~mL}\) of \(0.0700 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) was required to reduce all the iodine produced by the bleach back to iodide, what is the mass percent of \(\mathrm{NaClO}\) in the bleach?

Give the formula of (a) an anion in which \(S\) has an oxidation number of \(-2\). (b) two anions in which \(\mathrm{S}\) has an oxidation number of \(+4\). (c) two different acids of sulfur.

Chlorine can remove the foul smell of \(\mathrm{H}_{2} \mathrm{~S}\) in water. The reaction is $$\mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)+\mathrm{S}(s)$$ If the contaminated water has \(5.0\) ppm hydrogen sulfide by mass, what volume of chlorine gas at STP is required to remove all the \(\mathrm{H}_{2} \mathrm{~S}\) from \(1.00 \times 10^{3}\) gallons of water \((d=1.00 \mathrm{~g} / \mathrm{mL}) ?\) What is the \(\mathrm{pH}\) of the solution after treatment with chlorine?

Give the Lewis structure of (a) \(\mathrm{Cl}_{2} \mathrm{O}\) (b) \(\mathrm{N}_{2} \mathrm{O}\) (c) \(\mathrm{P}_{4}\) (d) \(\mathbf{N}_{2}\)

Write a balanced net ionic equation for the disproportionation reaction of (a) hypochlorous acid to chlorine gas and chlorous acid in acidic solution. (b) chlorate ion to perchlorate and chlorite ions.
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