Question

Give the oxidation number of the underlined atoms in the following molecules and ions: (a) \(\underline{\mathrm{ClF}},\) (b) \(\underline{\mathrm{IF}}_{7}\), (c) \(\underline{\mathrm{CH}}_{4},\) (d) \(\underline{\mathrm{C}}_{2} \mathrm{H}_{2},\) (e) \(\underline{\mathrm{C}}_{2} \mathrm{H}_{4},\) (f) \(\mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4},(\mathrm{~g}) \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (h) \(\mathrm{KMnO}_{4},\) (i) \(\mathrm{NaHCO}_{3},(\mathrm{j}) \underline{\mathrm{L} i}_{2},(\mathrm{k}) \mathrm{NaIO}_{3},(\mathrm{l}) \mathrm{K} \underline{\mathrm{O}}_{2}\) \((\mathrm{m}) \underline{\mathrm{PF}}_{6}^{-},(\mathrm{n}) \mathrm{K} \underline{\mathrm{Au}} \mathrm{Cl}_{4}\)

Short answer

(a) Cl in ClF: +1, (b) I in IF7 : +7, (c) C in CH4: -4, (d) C in C2H2: -1, (e) C in C2H4: -2, (f) Cr in K2CrO4: +6, (g) Mn in KMnO4: +7, (h) C in NaHCO3: +4, (i) Li in Li2: 0, (j) I in NaIO3: +5, (k) K in KO2: +1, (l) P in PF6-: +5, (m) Au in KAuCl4 : +3

Step-by-step solution

Chlorine Monofluoride (ClF)

Chlorine and Fluorine are both halogens, but Fluorine is more electronegative. Therefore, the oxidation number of F is -1. Since the compound is neutral, the oxidation number of Cl must be +1 to balance the charge.

Iodine Heptafluoride (IF7)

Again, Fluorine is more electronegative and has an oxidation number of -1. This gives a total of -7 for seven F atoms. To balance the charge in a neutral molecule, the oxidation number of I is +7.

Methane (CH4)

The oxidation number of H is +1. So, the total for four H atoms is +4. For the molecule to be neutral, the oxidation of C should be -4.

Acetylene (C2H2)

For two H atoms, the total oxidation number would be +2. Therefore, for the molecule to be neutral, the total oxidation number for two C atoms should be -2, hence each C atom has an oxidation number of -1.

Ethylene (C2H4)

For four H atoms, the total oxidation number would be +4. Therefore, for the molecule to be neutral, the total oxidation number for two C atoms should be -4, hence each C atom has an oxidation number of -2.

Potassium Chromate (K2CrO4)

The oxidation number of K is +1, and that of O is -2. Two K atoms give a total of +2 and four O atoms give a total of -8. In order to balance the charges to zero, Cr should be +6.

Potassium Permanganate (KMnO4)

The oxidation number of K is +1 and that of O is -2. One K atom gives +1 and four O atoms give a total of -8. In order to balance the charges to zero, Mn should be +7.

Sodium Bicarbonate (NaHCO3)

The oxidation number of Na is +1, H is +1, and O is -2. One Na atom gives +1, one H atom gives +1, and three O atoms give a total of -6. Therefore, to balance the charges to zero, the oxidation number of C should be +4.

Lithium (Li2)

As Li is a free element here, its oxidation number is 0.

Sodium Iodate (NaIO3)

The oxidation number of Na is +1 and O is -2. One Na atom gives +1 and three O atoms give a total of -6. Therefore, to balance the charges to zero, the oxidation number of I should be +5.

Potassium Superoxide (KO2)

The oxidation number of K is +1. This is a peroxide, hence the oxidation number of O is -1. Two O atoms give a total of -2. To balance out these charges, the oxidation number of K is +1.

Hexafluorophosphate (PF6-)

The oxidation number of F is -1. Six F atoms give a total of -6. As this is an ion with a charge of -1, the charge contributed by P should be +5 to balance out.

Potassium Tetrachloroaurate (KAuCl4)

The oxidation number of K is +1 and Cl is -1. One K atom gives +1 and four Cl atoms give a total of -4. In order to balance the charges to zero, the oxidation number of Au should be +3.

Key concepts

Chemical Oxidation States

Understanding oxidation states in chemistry is crucial for studying various chemical processes, especially redox reactions, where the transfer of electrons between atoms plays a starring role. An oxidation number or state indicates the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic.

An oxidation number can help us deduce the distribution of electrons among atoms in a molecule. Consider a molecule of water, H₂O. We generally assign oxygen an oxidation number of -2, and since hydrogen is less electronegative, it receives a +1. This means for the two hydrogens, we have a total of +2, balancing the -2 from oxygen, leading to a neutral molecule.

In the case of the exercise, determining the oxidation state of a central atom helps us understand its electron configuration and its potential for chemical reactions. For instance, in methane (CH₄), the carbon atom is assigned an oxidation state of -4 since it is bonded to four hydrogen atoms, each contributing +1.

Redox Reactions

Redox reactions are chemical reactions that involve the change in oxidation states of atoms through the transfer of electrons. The term 'redox' is a portmanteau of reduction and oxidation. These two processes always occur together in a redox reaction.

Oxidation refers to the loss of electrons by an atom or molecule, resulting in an increase in its oxidation number. Conversely, reduction is the gain of electrons, leading to a decrease in oxidation number. For example, when sodium metal reacts with chlorine gas, sodium gets oxidized (loses an electron, oxidation number increases from 0 to +1), and chlorine gets reduced (gains an electron, oxidation number decreases from 0 to -1), forming sodium chloride (NaCl).

Understanding oxidation numbers allows us to identify which atoms get oxidized and which get reduced, providing a fuller picture of the reaction's mechanics.

Balancing Oxidation-Reduction Equations

Balancing oxidation-reduction (redox) equations is a method used in chemistry to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. This is important for the conservation of charge and mass in chemical reactions.

To balance these equations, we must first assign oxidation numbers to all atoms in the reaction. After determining which atoms change their oxidation states, we use the half-reaction method where oxidation and reduction processes are balanced separately and then combined. Here, electron transfers are explicitly shown, ensuring that the gain and loss of electrons are balanced.

For example, when balancing the equation for the reaction between magnesium and hydrochloric acid, we ensure that the increase in oxidation number of magnesium equals the decrease in oxidation number of hydrogen. This foundational concept in chemistry is not only crucial in laboratory settings but also underpins the mechanisms of batteries, corrosion, and even cellular respiration.