Answer :
To determine the oxidation numbers for nickel (Ni), sulfur (S), and oxygen (O) in the compound [tex]\( \text{Ni}_2(\text{SO}_4)_3 \)[/tex], we need to analyze the structure and charges of each component.
1. Identifying the components and their charges:
- The compound is [tex]\( \text{Ni}_2(\text{SO}_4)_3 \)[/tex].
- Sulfate ion ([tex]\( \text{SO}_4^{2-} \)[/tex]) has an overall charge of -2.
2. Determine the oxidation state of oxygen:
- In almost all compounds, oxygen has an oxidation state of -2.
- There are 4 oxygen atoms in [tex]\( \text{SO}_4 \)[/tex], so the total oxidation state for oxygen in one sulfate ion is [tex]\( 4 \times -2 = -8 \)[/tex].
3. Determine the oxidation state of sulfur:
- Let the oxidation state of sulfur be x.
- The sum of the oxidation states in the sulfate ion has to equal the charge of the sulfate ion, which is -2.
- Therefore, the equation is: [tex]\( x + (-8) = -2 \)[/tex].
- Solving for [tex]\( x \)[/tex] gives [tex]\( x - 8 = -2 \Rightarrow x = +6 \)[/tex].
- So, the oxidation state of sulfur in [tex]\( \text{SO}_4^{2-} \)[/tex] is +6.
4. Determine the oxidation state of nickel:
- We have two nickel atoms in [tex]\( \text{Ni}_2(\text{SO}_4)_3 \)[/tex], so let the oxidation state of each nickel be y.
- The total charge contribution from the three sulfate ions is [tex]\( 3 \times -2 = -6 \)[/tex].
- The compound [tex]\(\text{Ni}_2(\text{SO}_4)_3\)[/tex] is electrically neutral, so the total oxidation state for nickel must balance the negative charge from the sulfates: [tex]\( 2y + (-6) = 0 \)[/tex].
- Solving for y gives [tex]\( 2y = +6 \Rightarrow y = +3 \)[/tex].
- Therefore, the oxidation state of each nickel ion is +3.
Putting it all together, the oxidation numbers for nickel, sulfur, and oxygen in [tex]\( \text{Ni}_2(\text{SO}_4)_3 \)[/tex] are:
- Nickel: +3
- Sulfur: +6
- Oxygen: -2
So, the correct answer is:
a. [tex]\( \text{Ni} +3; \text{S} +6; \text{O} -2 \)[/tex]
1. Identifying the components and their charges:
- The compound is [tex]\( \text{Ni}_2(\text{SO}_4)_3 \)[/tex].
- Sulfate ion ([tex]\( \text{SO}_4^{2-} \)[/tex]) has an overall charge of -2.
2. Determine the oxidation state of oxygen:
- In almost all compounds, oxygen has an oxidation state of -2.
- There are 4 oxygen atoms in [tex]\( \text{SO}_4 \)[/tex], so the total oxidation state for oxygen in one sulfate ion is [tex]\( 4 \times -2 = -8 \)[/tex].
3. Determine the oxidation state of sulfur:
- Let the oxidation state of sulfur be x.
- The sum of the oxidation states in the sulfate ion has to equal the charge of the sulfate ion, which is -2.
- Therefore, the equation is: [tex]\( x + (-8) = -2 \)[/tex].
- Solving for [tex]\( x \)[/tex] gives [tex]\( x - 8 = -2 \Rightarrow x = +6 \)[/tex].
- So, the oxidation state of sulfur in [tex]\( \text{SO}_4^{2-} \)[/tex] is +6.
4. Determine the oxidation state of nickel:
- We have two nickel atoms in [tex]\( \text{Ni}_2(\text{SO}_4)_3 \)[/tex], so let the oxidation state of each nickel be y.
- The total charge contribution from the three sulfate ions is [tex]\( 3 \times -2 = -6 \)[/tex].
- The compound [tex]\(\text{Ni}_2(\text{SO}_4)_3\)[/tex] is electrically neutral, so the total oxidation state for nickel must balance the negative charge from the sulfates: [tex]\( 2y + (-6) = 0 \)[/tex].
- Solving for y gives [tex]\( 2y = +6 \Rightarrow y = +3 \)[/tex].
- Therefore, the oxidation state of each nickel ion is +3.
Putting it all together, the oxidation numbers for nickel, sulfur, and oxygen in [tex]\( \text{Ni}_2(\text{SO}_4)_3 \)[/tex] are:
- Nickel: +3
- Sulfur: +6
- Oxygen: -2
So, the correct answer is:
a. [tex]\( \text{Ni} +3; \text{S} +6; \text{O} -2 \)[/tex]
