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Chapter 22: Problem 1

What is the oxidation state of the central atom (in pink) in each of the following compounds or ions: \((a) S C l_{2} ;(b) P O C l_{3} ;(c)\left[X e F_{5}\right]^{-} ;(d)\left[S b_{2} F_{11}\right]^{-}\) \((\mathrm{c})\left[\mathrm{HPO}_{4}\right]^{2-} ;(\mathrm{f}) \mathrm{Al}_{2} \mathrm{O}_{3} ;(\mathrm{g}) \mathrm{Me}_{2} \operatorname{Sn} \mathrm{F}_{2} ;(\mathrm{h})\) \(\left[\mathrm{Cr}_{2} \mathrm{O}_{7}\right]^{2-} ;(\mathrm{i})\left[\mathrm{MnO}_{4}\right]^{-} ;(\mathrm{j})\left[\mathrm{IO}_{4}\right]^{-},(\mathrm{k})\left[\mathrm{S}_{2}\right]^{2-} ?\)

Short Answer

Expert verified
(a) +2, (b) +5, (c) +5, (d) +5, (e) +6, (f) +3, (g) +4, (h) +6, (i) +7, (j) +7, (k) -1.

Step by step solution

01

Determine Oxidation State for Each Atom

The oxidation state of elements in their pure form (like in molecules with the same atom such as O鈧�, N鈧�) is zero. For compounds, consider known oxidation states: Oxygen is typically -2, Hydrogen is typically +1, Fluorine is -1, and other halogens (like chlorine) are generally -1 unless they are bonded to a more electronegative element.
02

Solve for Compound (a) SC\(l_2\)

Assign the oxidation state for Cl as -1. Set up the equation: \(x + 2(-1) = 0\). Solve for \(x\): \(x = +2\). Sulfur (S) has an oxidation state of +2.
03

Solve for Compound (b) POCl\(_3\)

Assign oxidation states: O is -2, Cl is -1. Set up the equation: \(x + (-2) + 3(-1) = 0\). Solve for \(x\): \(x = +5\). Phosphorus (P) has an oxidation state of +5.
04

Solve for Ion (c) [XeF\(_5\)]^-

Assign oxidation state for F as -1. Set up the equation with the charge: \(x + 5(-1) = -1\). Solve for \(x\): \(x = +5\). Xenon (Xe) has an oxidation state of +5.
05

Solve for Ion (d) [Sb\(_2\)F\(_{11}\)]^-

Assign oxidation state for F as -1. For two Sb atoms: \(2x + 11(-1) = -1\). Solve for \(x\): \(2x = 10\), so \(x = +5\). Antimony (Sb) has an oxidation state of +5 per atom.
06

Solve for Ion (e) [HPO\(_4\)]^{2-}

Assign oxidation states: H is +1, O is -2. Set up the equation with the charge: \(+1 + x + 4(-2) = -2\). Solve for \(x\): \(x = +6\). Phosphorus (P) has an oxidation state of +6.
07

Solve for Compound (f) Al\(_2\)O\(_3\)

Assign oxidation state for O as -2. For 2 Al: \(2x + 3(-2) = 0\). Solve for \(x\): \(2x = +6\), \(x = +3\) per Al atom.
08

Solve for Compound (g) Me\(_2\)SnF\(_2\)

Assign Me as a neutral or absent (for simplification, consider Me as negligible in oxidation states). Assign Sn as \(+4\) typically in organotin compounds. Confirm using charge balance: no net charge should confirm this.
09

Solve for Ion (h) [Cr\(_2\)O\(_7\)]^{2-}

Assign oxidation state for O as -2. For 2 Cr: \(2x + 7(-2) = -2\). Solve for \(x\): \(2x = +12\), \(x = +6\) per Cr atom.
10

Solve for Ion (i) [MnO\(_4\)]^-

Assign oxidation state for O as -2. For Mn: \(x + 4(-2) = -1\). Solve for \(x\): \(x = +7\). Manganese (Mn) has an oxidation state of +7.
11

Solve for Ion (j) [IO\(_4\)]^-

Assign oxidation state for O as -2. For I: \(x + 4(-2) = -1\). Solve for \(x\): \(x = +7\). Iodine (I) has an oxidation state of +7.
12

Solve for Ion (k) [S\(_2\)]^{2-}

Assign S a neutral or natural state normally of 0 (it's elemental), but in this compound, set equation: \(2x = -2\). Solve for \(x\): \(x = -1\). Each Sulfur (S) has an oxidation state of -1.

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Key Concepts

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

Redox Chemistry
Redox Chemistry is crucial in understanding the transfer of electrons between species. Redox, short for reduction-oxidation reactions, involves two simultaneous processes:
  • Oxidation: This is when a species loses electrons. As a species loses electrons, its oxidation state increases. In the exercise, such processes are calculated for elements like sulfur in SC\(l_2\) where sulfur is oxidized by having an oxidation state of +2.
  • Reduction: Conversely, this is when a species gains electrons. The reduction process decreases its oxidation state.
The overall idea in redox chemistry is the balancing of these electron transfers between involved elements. For example, in the ion \([MnO_4]^鈭抃), manganese undergoes oxidation resulting in an oxidation state of +7, showcasing that complex ions often involve elements in varying oxidation states. Remember, calculating oxidation states is often a necessary first step in determining the direction and magnitude of electron transfer during a chemical reaction.
Chemical Bonding
Chemical bonding describes the forces holding atoms together in compounds. The main types of bonds include:
  • Ionic Bonds: Formed when electrons are transferred from one atom to another, creating ions which attract each other. Compounds like \[Sb_2F_{11}\]^- feature elements forming ionic bonds due to the electronegativity difference.
  • Covalent Bonds: Occur when atoms share electrons to achieve completing their outer shell of electrons. The molecular structure \([XeF_5]^鈭抃) exemplifies covalent bonding. Here, Xenon is the central atom sharing electrons with several fluorine atoms.
  • Metallic Bonds: Found in metals, where electrons are shared across a lattice of metal ions.
Understanding how different atoms bind helps in predicting behavior and reactivity of compounds. In redox reactions, as atoms shift between oxidized or reduced states, the chemical bonds may break and form differently, facilitating the reaction.
Ions and Compounds
Ions are charged species leading to the formation of compounds. Ions form when atoms gain or lose electrons. Let's consider:
  • Cations: Positively charged ions, formed by losing electrons. Example: Al in \(Al_2O_3\) having a +3 oxidation state.
  • Anions: Negatively charged ions, formed by gaining electrons. Example: \([Cr_2O_7]^{2-}\) where the complex ion has extra electrons giving it a negative charge.
Compounds are substances formed when two or more elements combine chemically, with their atoms bonded together. They can be molecular (like \(POCl_3\)) or ionic (like \(NaCl\)), depending on the elements and the type of bonding. Understanding ions and how they form compounds is vital in predicting the physical and chemical properties of materials in various states, as well as their reactivity in redox reactions.

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Most popular questions from this chapter

By considering the atomic orbitals that are available on carbon and silicon, suggest why two \(\mathrm{C}=\mathrm{O}\) double bonds are energetically preferred to four \(\mathrm{C}-\mathrm{O}\) single bonds, but four \(\mathrm{Si}-\mathrm{O}\) single bonds are energetically preferred to two \(\mathrm{Si}=\mathrm{O}\) double bonds.

Suggest why compounds of the noble gases appear to be essentially restricted to xenon. Your arguments might lead you to suggest that radon should form compounds similar to those of xenon why do you think such compounds have not been included in our discussions?

Write down possible products from the reaction of (a) \(\mathrm{I}_{2}\) with \(\mathrm{F}_{2}\) and (b) \(\mathrm{Cl}_{2}\) with \(\mathrm{Br}_{2}\), In each case, also give the expected reactions of the products you have written with water.

Discuss each of the following statements, giving examples (where appropriate) to support the statements. (a) Liquid \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is self-ionizing. (b) \(\mathrm{SbF}_{5}\) is a fluoride acceptor. (c) \(\mathrm{XeF}_{2}\) is used as a combined oxidizing and fluorinating agent. (d) \(\mathrm{Al}_{2} \mathrm{O}_{3}\) is amphoteric. (e) \(\mathrm{GeCl}_{4}\) is a Lewis acid. (f) \(\left[\mathrm{IF}_{5}\right]^{2-}\) and \(\left[\mathrm{XeF}_{5}\right]^{-}\) are isostructural. (g) \(\mathrm{At} 298 \mathrm{K}\) and 1 bar pressure, elemental nitrogen occurs as \(\mathrm{N}_{2}\) but elemental phosphorus occurs as \(\mathrm{P}_{4}\) molecules.

Suggest why some \(p\) -block elements occur in the elemental state while others occur only as compounds. Give examples.
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