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Write the chemical formula for each of the following compounds, and indicate the oxidation state of nitrogen in each: (a) sodium azide, (b) ammonium ion, (c) nitrous acid, \((\mathbf{d})\) magnesium nitride, \((\mathbf{e})\) diazene, \((\mathbf{f})\) sodium nitrate, (g) nitrogen trifluoride, (h) nitric acid.

Short Answer

Expert verified
\( (a) \: NaN_3 \: (N: -\frac{1}{3}) \newline (b) \: NH_4^+ \: (N: -3) \newline (c) \: HNO_2 \: (N: +3) \newline (d) \: Mg_3N_2 \: (N: -3) \newline (e) \: N_2H_2 \: (N: -1) \newline (f) \: NaNO_3 \: (N: +5) \newline (g) \: NF_3 \: (N: +3) \newline (h) \: HNO_3 \: (N: +5) \)

Step by step solution

01

(a) Sodium Azide

Sodium azide has the chemical formula NaN鈧. The azide ion (N鈧冣伝) has an overall charge of -1, and the sodium ion (Na鈦) has a charge of +1. Therefore, the oxidation state of nitrogen in sodium azide is -1/3 since there are three nitrogen atoms, and the overall charge of the azide ion is -1.
02

(b) Ammonium Ion

The ammonium ion has the chemical formula NH鈧勨伜. In this ion, nitrogen is bonded to four hydrogen atoms, each having an oxidation state of +1. The nitrogen atom has an oxidation state of -3 because the total charge of the ammonium ion is +1, and there are four hydrogen atoms with a combined oxidation state of +4.
03

(c) Nitrous Acid

Nitrous acid has the chemical formula HNO鈧. In this molecule, the nitrogen atom is bonded to one hydrogen atom (oxidation state +1) and two oxygen atoms (oxidation state -2 each). The oxidation state of nitrogen in nitrous acid is +3 because the total charge of the molecule is 0, and the nitrogen atom balances the charges of the surrounding atoms.
04

(d) Magnesium Nitride

Magnesium nitride has the chemical formula Mg鈧僋鈧. In this compound, three magnesium atoms (oxidation state +2 each) are bonded to two nitrogen atoms (oxidation state -3 each). Since there are two nitrogen atoms and the overall charge of the molecule is 0, the oxidation state of each nitrogen atom in magnesium nitride is -3.
05

(e) Diazene

Diazene has the chemical formula N鈧侶鈧. In diazene, two nitrogen atoms (each sharing a double bond with each other) are bonded to one hydrogen atom each. The oxidation state of each nitrogen atom in diazene is -1 because each hydrogen atom has an oxidation state of +1 and the overall charge of the molecule is 0.
06

(f) Sodium Nitrate

Sodium nitrate has the chemical formula NaNO鈧. In this compound, one sodium atom (oxidation state +1) is bonded to a nitrate ion (NO鈧冣伝). The nitrate ion (NO鈧冣伝) consists of one nitrogen atom and three oxygen atoms (each with an oxidation state of -2). The oxidation state of nitrogen in sodium nitrate is +5 because the total charge of the nitrate ion is -1, and the nitrogen atom balances the charges of the surrounding oxygen atoms.
07

(g) Nitrogen Trifluoride

Nitrogen trifluoride has the chemical formula NF鈧. In this molecule, one nitrogen atom is bonded to three fluorine atoms, each of which has an oxidation state of -1. The oxidation state of nitrogen in nitrogen trifluoride is +3 because the total charge of the molecule is 0 and the nitrogen atom balances the charges of the surrounding fluorine atoms.
08

(h) Nitric Acid

Nitric acid has the chemical formula HNO鈧. In this molecule, the nitrogen atom is bonded to one hydrogen atom (oxidation state +1) and three oxygen atoms (each with an oxidation state of -2). The oxidation state of nitrogen in nitric acid is +5 because the total charge of the molecule is 0, and the nitrogen atom balances the charges of the surrounding atoms.

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Key Concepts

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

Oxidation States
Understanding oxidation states is crucial for determining how many electrons an atom either gains or loses during a chemical reaction. This concept helps us understand the electron balance in chemical compounds. For example, when examining sodium azide (NaN鈧), we find that the azide ion (N鈧冣伝) brings the overall charge to -1. To determine the oxidation state of nitrogen here, divide the charge equally among the three nitrogen atoms, resulting in an oxidation state of -1/3 per nitrogen atom. Remember, oxidation states can vary across different compounds for the same element, depending on its chemical environment.
Nitrogen Compounds
Nitrogen is a versatile element that forms various compounds with different elements. Each nitrogen compound exhibits unique properties and uses. Consider ammonium ion (NH鈧勨伜), where nitrogen forms bonds with four hydrogen atoms. Here, hydrogen typically has an oxidation state of +1 which, combined with the overall +1 charge of the ion, results in nitrogen having an oxidation state of -3. Another interesting compound is nitrogen trifluoride (NF鈧), where nitrogen is connected with three highly electronegative fluorine atoms, leading to nitrogen's oxidation state being +3. Understanding these typical oxidation states helps in predicting the behavior of nitrogen in chemical reactions.
Molecular Structure
The molecular structure indicates how atoms are arranged within a molecule, affecting the molecule's properties and reactivity. Take magnesium nitride (Mg鈧僋鈧) as an example. In this compound, three magnesium atoms (each with a +2 oxidation state) bond with two nitrogen atoms (each with a -3 oxidation state). The structured balance between these ions results in a stable, consistently neutral compound. Another example is diazene (N鈧侶鈧), where two nitrogen atoms are connected by a double bond. This structure leads to nitrogen atoms sharing electrons more equally, which impacts the overall energy stability of the molecule.
Inorganic Chemistry
Inorganic chemistry is the branch that deals with inorganic compounds, including those without carbon-hydrogen bonds like the nitrogen compounds discussed here. It is a broad field covering various elements and compounds, including metals, minerals, and organometallic compounds. Nitric acid (HNO鈧) is an important inorganic compound where one nitrogen atom is bonded to three oxygen atoms and one hydrogen atom. The knowledge of inorganic chemistry helps us understand compounds like nitric acid, which is essential in chemical manufacturing, agriculture, and even in some cleaning products. This field provides the foundation for manipulating and utilizing compounds in various industrial and research processes.

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Most popular questions from this chapter

Both dimethylhydrazine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}\), and methylhydrazine, \(\mathrm{CH}_{3} \mathrm{NHNH}_{2}\), have been used as rocket fuels. When dinitrogen tetroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) is used as the oxidizer, the products are \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2}\), and \(\mathrm{N}_{2}\). If the thrust of the rocket depends on the volume of the products produced, which of the substituted hydrazines produces a greater thrust per gram total mass of oxidizer plus fuel? (Assume that both fuels generate the same temperature and that \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed.)

Although the \(\mathrm{ClO}_{4}^{-}\) and \(\mathrm{IO}_{4}^{-}\) ions have been known for a long time, \(\mathrm{BrO}_{4}^{-}\) was not synthesized until \(1965 .\) The ion was synthesized by oxidizing the bromate ion with xenon difluoride, producing xenon, hydrofluoric acid, and the perbromate ion. (a) Write the balanced equation for this reaction. (b) What are the oxidation states of Br in the Br-containing species in this reaction?

Write the formulas for the following compounds, and indicate the oxidation state of the group 14 element or of boron in each: (a) stannous fluoride, (b) germane, (c) diborane, (e) tin selenide, (d) tin(II) sulfate, (f) zinc carbonate.

Hydrogen peroxide is capable of oxidizing (a) hydrazine to \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O},(\mathbf{b}) \mathrm{SO}_{2}\) to \(\mathrm{SO}_{4}^{2-},(\mathbf{c}) \mathrm{NO}_{2}^{-}\) to \(\mathrm{NO}_{3}^{-},(\mathbf{d}) \mathrm{H}_{2} \mathrm{~S}(g)\) to \(\mathrm{S}(s),(\mathbf{e}) \mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\). Write a balanced net ionic equation for each of these redox reactions.

The physical properties of \(\mathrm{D}_{2} \mathrm{O}\) differ from those of \(\mathrm{H}_{2} \mathrm{O}\) because (a) D has a different electron configuration than O. (b) \(\mathrm{D}\) is radioactive. (c) D forms stronger bonds with \(\mathrm{O}\) than \(\mathrm{H}\) does. (d) D is much more massive than H.

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