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Arrange the oxides of nitrogen in increasing order of oxidation state of \(\mathrm{N}\) from \(+1\) to \(+5\). $$ \begin{aligned} &\mathrm{N}_{2} \mathrm{O}<\mathrm{N}_{2} \mathrm{O}_{3}<\mathrm{NO}_{2}<\mathrm{N}_{2} \mathrm{O}_{5}\\\ &\begin{aligned} &\text { (a) } \mathrm{N}_{2} \mathrm{O}<\mathrm{NO}<\mathrm{N}_{2} \mathrm{O}_{3}<\mathrm{NO}_{2}<\mathrm{N}_{2} \mathrm{O}_{5} \\ &\text { (b) } \mathrm{N}_{2} \mathrm{O}<\mathrm{NO}_{2}<\mathrm{N}_{2} \mathrm{O}_{3}<\mathrm{NO}<\mathrm{N}_{2} \mathrm{O} \\ &\text { (c) } \mathrm{N}_{2} \mathrm{O}_{5}<\mathrm{N}_{2} \mathrm{O}<\mathrm{NO}_{2}<\mathrm{N}_{2} \mathrm{O}_{3}<\mathrm{N}_{2} \mathrm{O}_{5} \end{aligned} \end{aligned} $$

Short Answer

Expert verified
The correct order of the oxides of nitrogen in increasing oxidation state of nitrogen is N2O, NO, N2O3, NO2, N2O5.

Step by step solution

01

Understanding the Oxidation State

The oxidation state of an element in a compound represents the hypothetical charge it would have if the compound was composed of ions. For the oxides of nitrogen, calculate the oxidation state of nitrogen by assuming that oxygen has an oxidation state of -2, since it's more electronegative.
02

Calculate the Oxidation States

To find the oxidation states of nitrogen in the given compounds, apply the formula: overall charge = (oxidation state of N)x(number of N atoms) + (-2)x(number of O atoms). Solve for the oxidation state of nitrogen in each compound: for N2O, the equation is 0 = (oxidation state of N)x(2) + (-2)x(1); for NO, it's 0 = (oxidation state of N)x(1) + (-2)x(1); for N2O3, 0 = (oxidation state of N)x(2) + (-2)x(3); for NO2, 0 = (oxidation state of N)x(1) + (-2)x(2); for N2O5, 0 = (oxidation state of N)x(2) + (-2)x(5).
03

Calculate the Oxidation Number for Each Compound

For N2O, N = +1; For NO, N = +2; For N2O3, N = +3; For NO2, N = +4; For N2O5, N = +5. These values are found by solving the equations for the oxidation state of N given in Step 2.
04

Arrange in Increasing Order

Arrange the compounds in order of increasing oxidation state of nitrogen: N2O (+1), NO (+2), N2O3 (+3), NO2 (+4), N2O5 (+5).

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Key Concepts

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

Chemical Bonding
Chemical bonding refers to the force that holds atoms together in compounds. It is a key concept in understanding the behavior of molecules, including the oxides of nitrogen. There are multiple types of chemical bonds, including ionic, covalent, and metallic bonds. In the oxides of nitrogen, the bonds are predominantly covalent, as they form between nonmetallic elements. Covalent bonds involve the sharing of electron pairs between atoms, contributing to the overall stability of the molecule.

For instance, in compounds such as

N2O and NO2, nitrogen shares electrons with oxygen to fulfill the octet rule, which states that atoms are most stable when they have eight electrons in their valence shell. However, the degree of sharing and the oxidation state of nitrogen varies among its different oxides, reflecting the diversity of chemical bonding scenarios. Understanding the nature of these bonds can be crucial in predicting the physical and chemical properties of these compounds.
Oxidation Number Calculation
The oxidation number, or oxidation state, provides insight into the electron distribution within compounds and is crucial for understanding redox reactions. To calculate the oxidation number of an element in a compound, we use a set of rules that prioritize elements based on their common oxidation states and other chemical properties.

Let's apply these rules to find the oxidation states of nitrogen in its oxides. Oxygen typically has an oxidation state of -2, except in peroxides. For a neutral compound, the sum of oxidation states of all elements must equal zero. Using the mathematical formula

0 = (oxidation state of N) × (number of N atoms) + (-2) × (number of O atoms), we can calculate the oxidation states of nitrogen in various oxides. Through simple algebra, we solve for the oxidation state of N, revealing the electron distribution within these compounds, and gain a deeper understanding of their electron configuration and bonding characteristics.
Properties of Oxides
Oxides are compounds formed by the combination of oxygen with another element. The properties of oxides can be predicted based on their oxidation states and the type of bonding present in the compound. These properties include acidity or basicity, volatility, and reactivity. Understanding the oxidation state of the central element is key to predicting such chemical behaviors.

For example, the oxides of nitrogen display a variety of properties. N2O, with the lowest oxidation state of nitrogen, is a neutral oxide and acts as a mild anesthetic. With increasing oxidation states, as in NO2 and N2O5, the oxides become more acidic and their reactivity also increases due to the greater number of bonds that can be formed or broken during chemical reactions. N2O5, being the oxide with the highest oxidation state of nitrogen, is the most acidic and oxidizing among them. By understanding the oxidation states and chemical bonding in these compounds, we can predict their behavior and potential applications in various chemical contexts.

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Most popular questions from this chapter

What is the correct representation of reaction occurring when \(\mathrm{HCl}\) is heated with \(\mathrm{MnO}_{2} ?\) (a) \(\mathrm{MnO}_{4}^{-}+5 \mathrm{Cl}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Cl}^{-}+5 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{MnO}_{2}+2 \mathrm{Cl}^{-}+4 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (c) \(2 \mathrm{MnO}_{2}+4 \mathrm{Cl}^{-}+8 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+2 \mathrm{Cl}_{2}+4 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{MnO}_{2}+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_{4}+\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Which of the following is not a rule for calculating oxidation number? (a) For ions, oxidation number is equal to the charge on the ion. (b) The oxidation number of oxygen is \(-2\) in all of its compounds. (c) The oxidation number of fluorine is \(-1\) in all of its compounds. (d) Oxidation number of hydrogen is \(+1\) except in binary hydrides of alkali metals and alkaline earth metals where it is \(-1\).

Which species is acting as a reducing agent in the following reaction? \(14 \mathrm{H}^{*}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3 \mathrm{Ni} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{Ni}^{2 *}\) (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) (b) Ni (c) \(\mathrm{H}^{*}\) (d) \(\mathrm{H}_{2} \mathrm{O}\)

The oxidation state of \(\mathrm{Fe}\) in \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) is (a) \(+2\) (b) \(+3\) (c) \(+4\) (d) \(+6\)

Given \(E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V} ; \quad E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V}\) \(E_{\mathrm{Ft}^{3}+i \mathrm{Fe}^{2}}=+0.76 \mathrm{~V} ; E^{\circ} \mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=+1.60 \mathrm{~V}\) Which of the following statements is not correct? (a) \(\mathrm{Fe}^{3+}\) does not oxidise \(\mathrm{Ce}^{3+}\). (b) Cu reduces \(\mathrm{Ag}^{+}\)to \(\mathrm{Ag}\). (c) Ag will reduce \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\). (d) \(\mathrm{Fe}^{3+}\) reduces \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\).

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