Answer :
Sure! Let's go through the solution step by step to answer each part of the question.
### (a) Electronic Arrangement for Ions
- Element D: With an atomic number of 14, this element is likely Silicon (Si). When it forms a 4+ ion, it loses 4 electrons. The electronic arrangement for the Si 4+ ion is:
- Before losing electrons: 2, 8, 4 (Silicon)
- After losing 4 electrons: 2, 8
- Element E: With an atomic number of 17, this element is likely Chlorine (Cl). When it forms a 1- ion, it gains 1 electron. The electronic arrangement for the Cl 1- ion is:
- Before gaining the electron: 2, 8, 7 (Chlorine)
- After gaining 1 electron: 2, 8, 8
### (b) Identification of Elements Based on Properties
- (i) Poor Conductor of Electricity: Element D, which is Silicon (a metalloid), is a poor conductor of electricity as compared to metals.
- (ii) Most Reactive Non-metal: Element E, which is Chlorine (Cl), is one of the most reactive non-metals.
### (c) Period of Element E
- Element E (Chlorine), with atomic number 17, belongs to the 3rd period of the periodic table.
### (d) Reactivity Explanation
- It was stated incorrectly as Element E losing its outermost electron more readily, but actually, Chlorine gains electrons to achieve a stable electronic configuration. In contrast, Element A (Sodium) loses its outer electron readily to achieve stability. Sodium (Na) easily loses an electron compared to Chlorine, which gains.
### (e) Bonding Representation
- Elements A and D: Here, Element A is Sodium (Na), and Element D is Silicon (Si). Typically, these do not form traditional ionic or covalent compounds, so representing their bonding with dots and crosses may not be applicable here since they are more likely to form different types of interactions, such as alloy structures when possible.
### (f) Melting Point Explanation
- Element C vs. Element A: Aluminum (Element C, with atomic number 13) has a higher melting point than Sodium (Element A, with atomic number 11). This is because Aluminum has stronger metallic bonds involving more electrons, whereas Sodium involves only one electron in its metallic bonding, resulting in its lower melting point.
### (g) Reaction with Water
- The reaction of Element A (Sodium) with water can be shown by the balanced chemical equation:
[tex]\[
2 \text{Na} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaOH} + \text{H}_2
\][/tex]
This equation demonstrates that Sodium reacts with water to form sodium hydroxide and release hydrogen gas.
### Separation of Potassium Sulfate and Lead (II) Sulfate
To separate a solid mixture of potassium sulfate (K₂SO₄) and lead (II) sulfate (PbSO₄):
1. Dissolve the Mixture: Add water to the mixture. Potassium sulfate will dissolve because it is soluble in water, while lead (II) sulfate will not dissolve due to its low solubility.
2. Filtration: Filter the mixture to separate the undissolved lead (II) sulfate from the solution.
3. Evaporation: Evaporate the water from the filtrate to recover the dissolved potassium sulfate as a solid.
These steps will effectively separate the solid components based on solubility differences.
### (a) Electronic Arrangement for Ions
- Element D: With an atomic number of 14, this element is likely Silicon (Si). When it forms a 4+ ion, it loses 4 electrons. The electronic arrangement for the Si 4+ ion is:
- Before losing electrons: 2, 8, 4 (Silicon)
- After losing 4 electrons: 2, 8
- Element E: With an atomic number of 17, this element is likely Chlorine (Cl). When it forms a 1- ion, it gains 1 electron. The electronic arrangement for the Cl 1- ion is:
- Before gaining the electron: 2, 8, 7 (Chlorine)
- After gaining 1 electron: 2, 8, 8
### (b) Identification of Elements Based on Properties
- (i) Poor Conductor of Electricity: Element D, which is Silicon (a metalloid), is a poor conductor of electricity as compared to metals.
- (ii) Most Reactive Non-metal: Element E, which is Chlorine (Cl), is one of the most reactive non-metals.
### (c) Period of Element E
- Element E (Chlorine), with atomic number 17, belongs to the 3rd period of the periodic table.
### (d) Reactivity Explanation
- It was stated incorrectly as Element E losing its outermost electron more readily, but actually, Chlorine gains electrons to achieve a stable electronic configuration. In contrast, Element A (Sodium) loses its outer electron readily to achieve stability. Sodium (Na) easily loses an electron compared to Chlorine, which gains.
### (e) Bonding Representation
- Elements A and D: Here, Element A is Sodium (Na), and Element D is Silicon (Si). Typically, these do not form traditional ionic or covalent compounds, so representing their bonding with dots and crosses may not be applicable here since they are more likely to form different types of interactions, such as alloy structures when possible.
### (f) Melting Point Explanation
- Element C vs. Element A: Aluminum (Element C, with atomic number 13) has a higher melting point than Sodium (Element A, with atomic number 11). This is because Aluminum has stronger metallic bonds involving more electrons, whereas Sodium involves only one electron in its metallic bonding, resulting in its lower melting point.
### (g) Reaction with Water
- The reaction of Element A (Sodium) with water can be shown by the balanced chemical equation:
[tex]\[
2 \text{Na} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaOH} + \text{H}_2
\][/tex]
This equation demonstrates that Sodium reacts with water to form sodium hydroxide and release hydrogen gas.
### Separation of Potassium Sulfate and Lead (II) Sulfate
To separate a solid mixture of potassium sulfate (K₂SO₄) and lead (II) sulfate (PbSO₄):
1. Dissolve the Mixture: Add water to the mixture. Potassium sulfate will dissolve because it is soluble in water, while lead (II) sulfate will not dissolve due to its low solubility.
2. Filtration: Filter the mixture to separate the undissolved lead (II) sulfate from the solution.
3. Evaporation: Evaporate the water from the filtrate to recover the dissolved potassium sulfate as a solid.
These steps will effectively separate the solid components based on solubility differences.