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Chapter 4: Problem 49

Give oxidation number for the underlined atoms in the following molecules and ions: (a) \(\underline{C S}_{2} \mathrm{O}\) (b) \(\mathrm{CaI}_{2},\) (c) \(\underline{\mathrm{Al}_{2} \mathrm{O}_{3}},\) (d) \(\mathrm{H}_{3} \mathrm{As} \mathrm{O}_{3}\) (e) \(\underline{\mathrm{Ti}} \mathrm{O}_{2}\) (f) \(\underline{M}_{0} \mathrm{O}_{4}^{2-}\) \((\mathrm{g}) \underline{\mathrm{Pt}} \mathrm{Cl}_{4}^{2-},(\mathrm{h}) \underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\) (i) \(\underline{\operatorname{Sn}} \mathrm{F}_{2}\) (j) \(\underline{C l F}_{3}\) (k) \(\underline{\mathrm{Sb}} \mathrm{F}_{6}^{-}\)

Short Answer

Expert verified
(a) 1, (b) 1.5, (c) 4, (d) 6, (e) 2 and 4 for \(PtCl_{4}^{2-}\) and \(PtCl_{6}^{2-}\) respectively, (f) 2, (g) 3, (h) 5

Step by step solution

01

Assign Oxidation State for \(CS_2O\)

Since the oxidation number of Oxygen (\(O\)) is usually -2 and there are two Oxygen atoms, the total is -4. Being a neutral molecule the sum of oxidation states should be zero. Therefore the oxidation of the entire molecule is \(2x + (-4) = 0\), solving for \(x\), which represents the combined oxidation state of the two Sulphur (\(S\)) atoms, we get \(x = 2\), and Single sulfur atom will have oxidation number 1.
02

Assign Oxidation State for \(Al_2O_3\)

The oxidation number of Oxygen (\(O\)) is usually -2 and there are three Oxygen atoms, so the total is -6. Being a neutral molecule the sum of oxidation states should be zero. Therefore the oxidation of the entire molecule is \(2x + (-6) = 0\), solving for \(x\), it represents the combined oxidation state of the two aluminum (\(Al\)) atoms, we get \(x = 3\), so single Aluminium atom will have an oxidation number of 1.5.
03

Assign Oxidation State for \(TiO_2\)

The oxidation number of Oxygen (\(O\)) is usually -2, and there are two Oxygen atoms, so the total is -4. Being a neutral molecule the sum of oxidation states should be zero, hence the oxidation state of Titanium (\(Ti\)) is 4.
04

Assign Oxidation State for \(M_0O_{4}^{2-}\)

Here, the oxidation number of Oxygen (\(O\)) is -2, and there are four Oxygen atoms, so the total is -8. As per rule, the sum of the oxidation states equals to the charge of the ion, thus oxidation state of \(M_0\) is 6.
05

Assign Oxidation State for \(PtCl_{4}^{2-}\) and \(PtCl_{6}^{2-}\)

Here, the oxidation number of \(Cl\) is -1, and as per rule, the sum of oxidation states equals to the charge of the ion. So for \(PtCl_{4}^{2-}\), with 4 \(Cl\) atoms the total is -4, thus oxidation number of \(Pt\) is 2. For \(PtCl_{6}^{2-}\), with 6 \(Cl\) atoms the total is -6, hence oxidation state of \(Pt\) is 4.
06

Assign Oxidation State for \(SnF_{2}\)

The oxidation number of Fluorine (\(F\)) is usually -1 and there are two fluorine atoms so the total is -2. Being a neutral molecule the sum of oxidation states should be zero, thus the oxidation state of Tin (\(Sn\)) is 2.
07

Assign Oxidation State for \(ClF_{3}\)

The oxidation number of Fluorine (\(F\)) is usually -1 and there are three fluorine atoms so the total is -3. Being a neutral molecule the sum of oxidation states should be zero, hence the oxidation state of Chlorine (\(Cl\)) is 3.
08

Assign Oxidation State for \(SbF_{6}^{-}\)

Here, the oxidation number of Fluorine (\(F\)) is -1, and there are six Fluorine atoms, so the total is -6. As per rule, the sum of oxidation states equals to the charge of the ion, thus oxidation state of Antimony (\(Sb\)) is 5.

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Most popular questions from this chapter

Write the equation for calculating molarity. Why is molarity a convenient concentration unit in chemistry?

Phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) is an important industrial chemical used in fertilizers, in detergents, and in the food industry. It is produced by two different methods. In the electric furnace method, elemental phosphorus \(\left(\mathrm{P}_{4}\right)\) is burned in air to form \(\mathrm{P}_{4} \mathrm{O}_{10},\) which is then reacted with water to give \(\mathrm{H}_{3} \mathrm{PO}_{4} .\) In the wet process, the mineral phosphate rock fluorapatite \(\left[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\right]\) is reacted with sulfuric acid to give \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (and \(\mathrm{HF}\) and \(\mathrm{CaSO}_{4}\) ). Write equations for these processes and classify each step as precipitation, acid-base, or redox reaction.

A student carried out two titrations using a \(\mathrm{NaOH}\) solution of unknown concentration in the buret. In one titration she weighed out \(0.2458 \mathrm{~g}\) of KHP (see Section 4.7 ) and transferred it to an Erlenmeyer flask. She then added \(20.00 \mathrm{~mL}\) of distilled water to dissolve the acid. In the other titration she weighed out \(0.2507 \mathrm{~g}\) of KHP but added \(40.00 \mathrm{~mL}\) of distilled water to dissolve the acid. Assuming no experimental error, would she obtain the same result for the concentration of the \(\mathrm{NaOH}\) solution?

Predict the outcome of the reactions represented by the following equations by using the activity series, and balance the equations. (a) \(\mathrm{Cu}(s)+\mathrm{HCl}(a q) \longrightarrow\) (b) \(\mathrm{I}_{2}(s)+\operatorname{NaBr}(a q) \longrightarrow\) (c) \(\mathrm{Mg}(s)+\mathrm{CuSO}_{4}(a q) \longrightarrow\) (d) \(\mathrm{Cl}_{2}(g)+\mathrm{KBr}(a q) \longrightarrow\)

Can the following decomposition reaction be characterized as an acid-base reaction? Explain. \(\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\)
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