Problem 43 - Introductory Chemistry Atoms First

Question

Use the shortcut rules to assign an oxidation state to each atom in: (a) \(\mathrm{PCl}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{~S}\) (c) \(\mathrm{MnO}_{4}^{-}\) (d) \(\mathrm{HNO}_{3}\) (e) \(\mathrm{HCOOH}\) (f) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\)

Step-by-Step Solution

Answer

The oxidation states of the given compounds are: (a) PCl$_3$: P: +3, Cl: -1 (b) H$_2$S: H: +1, S: -2 (c) MnO$_4^{-}$: Mn: +7, O: -2 (d) HNO$_3$: H: +1, N: +5, O: -2 (e) HCOOH: H: +1, C: 0/+2, O: -2 (f) S$_2$O$_3^{2-}$: S: +2/+4, O: -2

Step 1: (a) PCl3

For PCl3 (phosphorus trichloride), let's assign the oxidation state of P as x and for Cl as -1 (as per the rules for halogens). Using rule 3: x + 3(-1)= 0, solving for x we get, x = +3. So, the oxidation states are P: +3, Cl: -1.

Step 2: (b) H2S

For H2S (hydrogen sulfide), let's assign the oxidation state of H as +1 (as per the rules for hydrogen) and for S as x. Using rule 3: 2(+1) + x = 0, solving for x we get, x = -2. So, the oxidation states are H: +1, S: -2.

Step 3: (c) MnO4-

For MnO4- (permanganate ion), let's assign the oxidation state of Mn as x and for O as -2 (as per the rules for oxygen). Using rule 3: x + 4(-2) = -1, solving for x we get, x = +7. So, the oxidation states are Mn: +7, O: -2.

Step 4: (d) HNO3

For HNO3 (nitric acid), let's assign the oxidation state of H as +1 (as per the rules for hydrogen), for N as x and for O as -2 (as per the rules for oxygen). Using rule 3: +1 + x + 3(-2) = 0, solving for x we get, x = +5. So, the oxidation states are H: +1, N: +5, O: -2.

Step 5: (e) HCOOH

For HCOOH (formic acid), let's assign the oxidation state of H as +1 (as per the rules for hydrogen), for C in both instances as x, and for O as -2 (as per the rules for oxygen). Using rule 3: +1 + x + 2(-2) + x + 1 = 0, solving for x we get, x = 0 (for the C bonded to the H), x = +2 (for the C bonded to the O). So, the oxidation states are H: +1, C: 0/+2, O: -2.

Step 6: (f) S2O3^2-

For S2O3^2- (thiosulfate ion), let's assign the oxidation state of S as x (for both sulfur atoms) and for O as -2 (as per the rules for oxygen). Using rule 3: 2x + 3(-2) = -2, solving for x we get, x = +2 (for the S bonded to the other S atom), x = +4 (for the S bonded to the three O atoms). So, the oxidation states are S: +2/+4, O: -2.

Key Concepts

PCl3 oxidation state

The compound phosphorus trichloride (\(\text{PCl}_3\)) includes phosphorus and chlorine atoms. When determining oxidation states, it is helpful to know that chlorine is a halogen. Halogens typically have an oxidation state of -1.
In \(\text{PCl}_3\), to find the oxidation state of phosphorus, assign it a symbol, like \(x\). Applying the neutral charge rule, we can write the equation: \[x + 3(-1) = 0\]. Solving this gives \(x = +3\). This means phosphorus has an oxidation state of +3, and each chlorine atom is -1.

• Phosphorus: +3
• Chlorine: -1

Understanding this calculation is crucial as it illustrates balancing of oxidation states to achieve neutrality in chemical compounds.

H2S oxidation state

Hydrogen sulfide (\(\text{H}_2\text{S}\)) is composed of hydrogen and sulfur atoms. For most compounds, hydrogen holds an oxidation state of +1.
In \(\text{H}_2\text{S}\), the oxidation state of sulfur needs to be determined assuming it is \(x\). By using the neutral compound rule, we set up the equation: \[2(+1) + x = 0\]. Solving this, we find \(x = -2\), indicating sulfur's oxidation state is -2.

• Hydrogen: +1
• Sulfur: -2

This calculation shows how hydrogen’s consistent behavior across different compounds simplifies the determination of oxidation states.

MnO4- oxidation state

In the permanganate ion (\(\text{MnO}_4^-\)), manganese (Mn) and oxygen are present. Oxygen usually carries an oxidation state of -2.
To find the oxidation state of manganese (\(x\)), we apply the charge rule for ions, leading to the equation: \[x + 4(-2) = -1\]. Solving it gives \(x = +7\), which means manganese's oxidation state is +7.

• Manganese: +7
• Oxygen: -2

This example highlights how ions carry a net charge, affecting the overall oxidation state determination.

HNO3 oxidation state

Nitric acid (\(\text{HNO}_3\)) contains hydrogen, nitrogen, and oxygen. As previously discussed, hydrogen is +1, while oxygen is -2.
To find the oxidation state of nitrogen (\(x\)), use the neutral compound rule with the equation: \[+1 + x + 3(-2) = 0\]. This calculation results in \(x = +5\), indicating nitrogen has an oxidation state of +5.

• Hydrogen: +1
• Nitrogen: +5
• Oxygen: -2

Awareness of typical oxidation states for common elements simplifies these calculations and is crucial in chemistry.

HCOOH oxidation state

Formic acid (\(\text{HCOOH}\)) encompasses hydrogen, carbon, and oxygen. Besides known values, carbon's oxidation state can vary.
In the calculation, assign two unknowns \(x\) (carbon bonded to hydrogen) and \(y\) (carbon bonded to oxygen). The equation appears as: \[+1 + x + 2(-2) + y + 1 = 0\]. Solving yields \(x = 0\) and \(y = +2\).

• Hydrogen: +1
• Carbon: 0 when bonded with \(\text{H}\), +2 when bonded with \(\text{O}\)
• Oxygen: -2

This showcases how the context and connection within molecules can change the carbon oxidation state indicator.

S2O3^2- oxidation state

In the thiosulfate ion (\(\text{S}_2\text{O}_3^{2-}\)), sulfur and oxygen exist. Oxygen's standard state is -2.
Let \(x\) and \(y\) represent the two S atoms' oxidation states. The ionic equation reads: \[2x + 3(-2) = -2\], resulting in a mixture of states, where one sulfur is +2 and another is +4.

• Sulfur: +2/+4 (depending on position in structure)
• Oxygen: -2

This ion illustrates how complex molecules can feature varying oxidation states even for atoms of the same element.