Answer :
Certainly! Let's break down and answer each part of the question thoroughly:
### 1. Chromium, Cr, and its Reaction:
#### (a) Oxidation States:
(i) Oxidation state of chromium in [tex]\( \text{Cr}_2\text{O}_3 \)[/tex]:
- In [tex]\( \text{Cr}_2\text{O}_3 \)[/tex], oxygen has a known oxidation state of -2.
- Let the oxidation state of chromium be [tex]\( x \)[/tex].
- We know the sum of oxidation states in a neutral compound is zero.
- So, the equation is: [tex]\( 2x + 3(-2) = 0 \)[/tex].
- Simplifying: [tex]\( 2x - 6 = 0 \)[/tex], which gives [tex]\( 2x = 6 \)[/tex] and [tex]\( x = 3 \)[/tex].
The oxidation state of chromium in [tex]\( \text{Cr}_2\text{O}_3 \)[/tex] is +3. [tex]\(\quad [1]\)[/tex]
(ii) Oxidation state of aluminium in [tex]\( \text{Al}_2\text{O}_3 \)[/tex]:
- In [tex]\( \text{Al}_2\text{O}_3 \)[/tex], oxygen again has an oxidation state of -2.
- Let the oxidation state of aluminium be [tex]\( y \)[/tex].
- The equation is: [tex]\( 2y + 3(-2) = 0 \)[/tex].
- Simplifying: [tex]\( 2y - 6 = 0 \)[/tex], which gives [tex]\( 2y = 6 \)[/tex] and [tex]\( y = 3 \)[/tex].
The oxidation state of aluminium in [tex]\( \text{Al}_2\text{O}_3 \)[/tex] is +3. [tex]\(\quad [1]\)[/tex]
#### (b) Reducing Agent:
State the reducing agent in the reaction:
The reducing agent is the substance that donates electrons and gets oxidized in the reaction. In this case, Aluminium (Al) is the reducing agent as it gets oxidized from [tex]\( \text{Al} \)[/tex] (0 oxidation state) to [tex]\( \text{Al}_2\text{O}_3 \)[/tex] (+3 oxidation state).
The reducing agent is Aluminium (Al). [tex]\(\quad [1]\)[/tex]
#### (c) Test to Confirm Reducing Agent:
Describe a test to confirm Aluminium (Al) as a reducing agent:
- Test:
Use a solution of copper(II) sulfate ([tex]\(\text{CuSO}_4\)[/tex]) and add aluminium metal to it.
- Result:
The aluminium will reduce [tex]\(\text{Cu}^{2+}\)[/tex] ions to metallic copper, resulting in a reddish-brown precipitate of copper.
[tex]\[ \text{Test:} \quad \text{Use a solution of copper(II) sulfate (}\text{CuSO}_4\text{) and add aluminium metal to it.} \][/tex]
[tex]\[ \text{Result:} \quad \text{The aluminium will reduce \(\text{Cu}^{2+}\) ions to metallic copper, resulting in a reddish-brown precipitate.} \][/tex]
This confirms that Aluminium acts as a reducing agent. [tex]\(\quad [2]\)[/tex]
### 2. Borazine:
#### (a) Formulas:
(i) Molecular formula and empirical formula of borazine:
- The molecular formula represents the actual number of atoms of each element in a molecule. For borazine, it is [tex]\( \text{B}_3\text{N}_3\text{H}_6 \)[/tex].
- The empirical formula is the simplest whole-number ratio of atoms in the compound. For borazine, the empirical formula is also [tex]\( \text{B}_3\text{N}_3\text{H}_6 \)[/tex] as the ratio cannot be simplified further.
Molecular formula: [tex]\( \text{B}_3\text{N}_3\text{H}_6 \)[/tex] [tex]\(\quad [1]\)[/tex]
Empirical formula: [tex]\( \text{B}_3\text{N}_3\text{H}_6 \)[/tex] [tex]\(\quad [1]\)[/tex]
#### (ii) Bonding Electrons around Nitrogen Atom:
Number of bonding electrons around one nitrogen atom in borazine:
- Borazine has a structure similar to benzene with alternating boron and nitrogen atoms.
- Each nitrogen atom forms three bonds (one double bond and two single bonds) with neighboring atoms, and has one lone pair of electrons.
- Each bond has two electrons, so 3 bonds provide [tex]\(3 \times 2 = 6\)[/tex] bonding electrons.
- Additionally, there is one lone pair of electrons which accounts for 2 electrons.
Thus, the number of bonding electrons around one nitrogen atom in borazine is 8.
### Summary of Answers:
1. [tex]\[ (a) \quad \text{(i) Chromium in \( \text{Cr}_2\text{O}_3 \): } +3 \quad (ii) \text{Aluminium in \( \text{Al}_2\text{O}_3 \): } +3 \][/tex]
(b) \quad \text{Reducing agent: Aluminium (Al)}
\]
(c) \quad \text{Test: Use a solution of copper(II) sulfate ([tex]\( \text{CuSO}_4 \)[/tex]) and add aluminium metal to it.}
\]
\text{Result: The aluminium will reduce [tex]\( \text{Cu}^{2+} \)[/tex] ions to metallic copper, resulting in a reddish-brown precipitate.}
\]
2. [tex]\[ (a) \quad \text{(i) Molecular formula: \( \text{B}_3\text{N}_3\text{H}_6 \), Empirical formula: \( \text{B}_3\text{N}_3\text{H}_6 \)} \][/tex]
(ii) \quad \text{Number of bonding electrons around one nitrogen atom: 8}
\]
By following these steps, we ensure all parts of the questions are answered comprehensively.
### 1. Chromium, Cr, and its Reaction:
#### (a) Oxidation States:
(i) Oxidation state of chromium in [tex]\( \text{Cr}_2\text{O}_3 \)[/tex]:
- In [tex]\( \text{Cr}_2\text{O}_3 \)[/tex], oxygen has a known oxidation state of -2.
- Let the oxidation state of chromium be [tex]\( x \)[/tex].
- We know the sum of oxidation states in a neutral compound is zero.
- So, the equation is: [tex]\( 2x + 3(-2) = 0 \)[/tex].
- Simplifying: [tex]\( 2x - 6 = 0 \)[/tex], which gives [tex]\( 2x = 6 \)[/tex] and [tex]\( x = 3 \)[/tex].
The oxidation state of chromium in [tex]\( \text{Cr}_2\text{O}_3 \)[/tex] is +3. [tex]\(\quad [1]\)[/tex]
(ii) Oxidation state of aluminium in [tex]\( \text{Al}_2\text{O}_3 \)[/tex]:
- In [tex]\( \text{Al}_2\text{O}_3 \)[/tex], oxygen again has an oxidation state of -2.
- Let the oxidation state of aluminium be [tex]\( y \)[/tex].
- The equation is: [tex]\( 2y + 3(-2) = 0 \)[/tex].
- Simplifying: [tex]\( 2y - 6 = 0 \)[/tex], which gives [tex]\( 2y = 6 \)[/tex] and [tex]\( y = 3 \)[/tex].
The oxidation state of aluminium in [tex]\( \text{Al}_2\text{O}_3 \)[/tex] is +3. [tex]\(\quad [1]\)[/tex]
#### (b) Reducing Agent:
State the reducing agent in the reaction:
The reducing agent is the substance that donates electrons and gets oxidized in the reaction. In this case, Aluminium (Al) is the reducing agent as it gets oxidized from [tex]\( \text{Al} \)[/tex] (0 oxidation state) to [tex]\( \text{Al}_2\text{O}_3 \)[/tex] (+3 oxidation state).
The reducing agent is Aluminium (Al). [tex]\(\quad [1]\)[/tex]
#### (c) Test to Confirm Reducing Agent:
Describe a test to confirm Aluminium (Al) as a reducing agent:
- Test:
Use a solution of copper(II) sulfate ([tex]\(\text{CuSO}_4\)[/tex]) and add aluminium metal to it.
- Result:
The aluminium will reduce [tex]\(\text{Cu}^{2+}\)[/tex] ions to metallic copper, resulting in a reddish-brown precipitate of copper.
[tex]\[ \text{Test:} \quad \text{Use a solution of copper(II) sulfate (}\text{CuSO}_4\text{) and add aluminium metal to it.} \][/tex]
[tex]\[ \text{Result:} \quad \text{The aluminium will reduce \(\text{Cu}^{2+}\) ions to metallic copper, resulting in a reddish-brown precipitate.} \][/tex]
This confirms that Aluminium acts as a reducing agent. [tex]\(\quad [2]\)[/tex]
### 2. Borazine:
#### (a) Formulas:
(i) Molecular formula and empirical formula of borazine:
- The molecular formula represents the actual number of atoms of each element in a molecule. For borazine, it is [tex]\( \text{B}_3\text{N}_3\text{H}_6 \)[/tex].
- The empirical formula is the simplest whole-number ratio of atoms in the compound. For borazine, the empirical formula is also [tex]\( \text{B}_3\text{N}_3\text{H}_6 \)[/tex] as the ratio cannot be simplified further.
Molecular formula: [tex]\( \text{B}_3\text{N}_3\text{H}_6 \)[/tex] [tex]\(\quad [1]\)[/tex]
Empirical formula: [tex]\( \text{B}_3\text{N}_3\text{H}_6 \)[/tex] [tex]\(\quad [1]\)[/tex]
#### (ii) Bonding Electrons around Nitrogen Atom:
Number of bonding electrons around one nitrogen atom in borazine:
- Borazine has a structure similar to benzene with alternating boron and nitrogen atoms.
- Each nitrogen atom forms three bonds (one double bond and two single bonds) with neighboring atoms, and has one lone pair of electrons.
- Each bond has two electrons, so 3 bonds provide [tex]\(3 \times 2 = 6\)[/tex] bonding electrons.
- Additionally, there is one lone pair of electrons which accounts for 2 electrons.
Thus, the number of bonding electrons around one nitrogen atom in borazine is 8.
### Summary of Answers:
1. [tex]\[ (a) \quad \text{(i) Chromium in \( \text{Cr}_2\text{O}_3 \): } +3 \quad (ii) \text{Aluminium in \( \text{Al}_2\text{O}_3 \): } +3 \][/tex]
(b) \quad \text{Reducing agent: Aluminium (Al)}
\]
(c) \quad \text{Test: Use a solution of copper(II) sulfate ([tex]\( \text{CuSO}_4 \)[/tex]) and add aluminium metal to it.}
\]
\text{Result: The aluminium will reduce [tex]\( \text{Cu}^{2+} \)[/tex] ions to metallic copper, resulting in a reddish-brown precipitate.}
\]
2. [tex]\[ (a) \quad \text{(i) Molecular formula: \( \text{B}_3\text{N}_3\text{H}_6 \), Empirical formula: \( \text{B}_3\text{N}_3\text{H}_6 \)} \][/tex]
(ii) \quad \text{Number of bonding electrons around one nitrogen atom: 8}
\]
By following these steps, we ensure all parts of the questions are answered comprehensively.
