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Chapter 4: Problem 133

Assign the oxidation state for the element listed in each of the following compounds: \(\mathrm{S}\) in \(\mathrm{MgSO}_{4}\)_______ \(\mathrm{Pb}\) in \(\mathrm{PbSO}_{4}\)______ \(\mathrm{O}\) in \(\mathrm{O}_{2}\)___________ \(\mathrm{Ag}\) in Ag _________________________ \(\mathrm{Cu}\) in \(\mathrm{CuCl}_{2}\)_______

Short Answer

Expert verified
The oxidation states of the elements in the given compounds are: $\mathrm{S}$ in $\mathrm{MgSO}_{4}$: \(+6\) $\mathrm{Pb}$ in $\mathrm{PbSO}_{4}$: \(+2\) $\mathrm{O}$ in $\mathrm{O}_{2}$: \(0\) $\mathrm{Ag}$ in $\mathrm{Ag}$: \(0\) $\mathrm{Cu}$ in $\mathrm{CuCl}_{2}$: \(+2\)

Step by step solution

01

Rule 1: Oxidation State of a Pure Element

The oxidation state of a pure element is always zero. For example, O in O₂ will have an oxidation state of zero.
02

Rule 2: Oxidation State of Homoatomic Ions

The oxidation state of a monoatomic ion equals its charge. For example, Ag in Ag will have an oxidation state of zero since it's an uncharged, pure element.
03

Rule 3: The Sum of Oxidation States in a Compound

The sum of oxidation states of all elements in a compound is equal to the total charge of the compound. For electrically neutral compounds, the sum is zero. Now we will apply these rules to find oxidation states for the remaining elements in the given compounds.
04

Assigning Oxidation State of S in MgSO₄

We can use Rule 3 and the properties of the individual ions to deduce the oxidation state of S in MgSO₄. We know Mg is a group 2 element, so it forms an ion with a charge of +2. The charge on SO₄²⁻ is -2 (since it's a sulfate ion) and O has an oxidation state of -2. Let x be the oxidation state of S. Mg²⁺ + x + 4(-2) = 0 (as the compound is neutral) Solving for x: x = 2 - 4*(-2) = 2 + 8 = +6 The oxidation state of S in MgSO₄ is +6.
05

Assigning Oxidation State of Pb in PbSO₄

We can use Rule 3 and the properties of individual ions to deduce the oxidation state of Pb in PbSO₄. We know that SO₄²⁻ has a charge of -2 and O has an oxidation state of -2. Let x be the oxidation state of Pb; x + 4(-2) = 0 (as the compound is neutral) Solving for x: x = 4*(-2) = - 8 The oxidation state of Pb in PbSO₄ is +2.
06

Assigning Oxidation State of Cu in CuCl₂

We can use Rule 3 and the properties of individual ions to deduce the oxidation state of Cu in CuCl₂. We know that Cl⁻ has a charge of -1. Let x be the oxidation state of Cu; x + 2(-1) = 0 (as the compound is neutral) Solving for x: x = 2*(-1) = + 2 The oxidation state of Cu in CuCl₂ is +2. In conclusion, the oxidation states of the elements in the given compounds are: S in MgSO₄: +6 Pb in PbSO₄: +2 O in O₂: 0 Ag in Ag: 0 Cu in CuCl₂: +2

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Most popular questions from this chapter

Specify which of the following are oxidation–reduction reactions, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced. a. $\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)$ b. $\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)$ c. $\mathrm{SiCl}_{4}(i)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{HCl}(a q)+\mathrm{SiO}_{2}(s)$ d. $\mathrm{SiCl}_{4}(l)+2 \mathrm{Mg}(s) \rightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s)$ e. $\mathrm{Al}(\mathrm{OH})_{4}-(a q) \rightarrow \mathrm{AlO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$

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