How Do Metals and Non-Metals React? – Long Answer Questions (CBSE Class 10 Science)
Medium Level (Application & Explanation)
Q1. Explain why metals tend to lose electrons and form cations. Support your answer with electron configurations and equations for Na, Mg, and K.
Answer: Metals generally have 1, 2, or 3 electrons in their outermost shell, which they can lose easily to achieve a stable noble gas configuration. This process of losing electrons is called oxidation, and it leads to the formation of positively charged ions (cations). The energy required to remove these few electrons is relatively low for metals, so they prefer electron loss over gain. For example:
- Sodium (Na): 2,8,1 → loses 1 e⁻ to form Na⁺
Na → Na⁺ + e⁻ - Magnesium (Mg): 2,8,2 → loses 2 e⁻ to form Mg²⁺
Mg → Mg²⁺ + 2e⁻ - Potassium (K): 2,8,8,1 → loses 1 e⁻ to form K⁺
K → K⁺ + e⁻
Thus, by oxidation, metals become stable and form cations that can combine with non-metal anions to make compounds.
Q2. Why do non-metals gain electrons to form anions? Explain using chlorine, oxygen, and bromine as examples.
Answer: Non-metals usually possess 5, 6, or 7 valence electrons. To complete their octet (8 electrons) and become stable like noble gases, they tend to gain electrons. This process is called reduction and results in the formation of negatively charged ions (anions). Gaining electrons is easier for non-metals due to their higher electron affinity and non-metallic character. Examples:
- Chlorine (Cl): 2,8,7 → gains 1 e⁻ to form Cl⁻
Cl + e⁻ → Cl⁻ - Oxygen (O): 2,6 → gains 2 e⁻ to form O²⁻
O + 2e⁻ → O²⁻ - Bromine (Br): 2,8,18,7 → gains 1 e⁻ to form Br⁻
Br + e⁻ → Br⁻
In all cases, reduction helps non-metals achieve a stable electron configuration, which is why they readily form anions.
Q3. Describe how an ionic (electrovalent) compound forms between a metal and a non-metal. Illustrate using NaCl, MgO, and CaCl₂.
Answer: An ionic compound forms when a metal atom donates electrons to a non-metal atom, resulting in the formation of cations and anions. These oppositely charged ions are then held together by strong electrostatic forces of attraction, forming an ionic bond.
- Sodium chloride (NaCl): Sodium loses 1 electron and chlorine gains 1 electron.
Na → Na⁺ + e⁻; Cl + e⁻ → Cl⁻; hence, Na⁺ + Cl⁻ → NaCl
Overall: 2Na + Cl₂ → 2NaCl - Magnesium oxide (MgO): Magnesium loses 2 electrons and oxygen gains 2 electrons.
Mg → Mg²⁺ + 2e⁻; O + 2e⁻ → O²⁻; hence, Mg²⁺ + O²⁻ → MgO
Overall: 2Mg + O₂ → 2MgO - Calcium chloride (CaCl₂): Calcium loses 2 electrons; each chlorine gains 1 electron.
Ca²⁺ + 2Cl⁻ → CaCl₂; Overall: Ca + Cl₂ → CaCl₂
Thus, electron transfer leads to ionic bonds and stable compounds.
Q4. State the key physical properties of ionic compounds and explain why they show these properties with examples.
Answer: Ionic compounds have distinct properties due to strong electrostatic forces between ions:
- High melting and boiling points: A lot of energy is needed to overcome the strong ionic bonds.
Example: NaCl melts at 801°C; MgO has an even higher melting point (~2852°C). - Solubility in water: Water molecules are polar and can surround and separate ions, making compounds like NaCl and KBr soluble.
- Electrical conductivity: Ionic compounds do not conduct electricity in solid state because ions are fixed. However, when molten or dissolved in water, free ions carry charge and conduct electricity.
- Hard and brittle nature: A slight shift in the lattice brings like charges together, causing repulsion and the crystal to shatter.
Overall, these properties arise from the ionic lattice and ion–dipole interactions with water.
Q5. Describe an activity that demonstrates ionic bond formation using sodium, water, and a model for sodium chloride. Include observations and safety points.
Answer:
- Materials: Small piece of sodium (Na), water, beaker, universal indicator, model kit (colored balls/sticks). Safety: Sodium must be handled only by the teacher with tongs; use a safety screen and dry apparatus.
- Step 1 (Reaction with water): Place a tiny piece of sodium on water. Observe a rapid reaction forming sodium hydroxide and hydrogen gas.
2Na + 2H₂O → 2NaOH + H₂↑
The indicator turns blue, showing an alkaline solution due to Na⁺ and OH⁻ formation. - Link to chlorine: Explain that sodium reacts vigorously with chlorine gas to form sodium chloride (ionic), showing electron transfer.
2Na + Cl₂ → 2NaCl - Static model: Remove one “electron” from Na and give it to Cl. Now Na⁺ and Cl⁻ attract and stick together, representing an ionic bond.
This activity visualizes electron transfer, ion formation, and electrostatic attraction in ionic compounds.
High Complexity (Analytical & Scenario-Based)
Q6. “Solid sodium chloride does not conduct electricity, but its aqueous solution does.” Analyze and explain this observation in detail.
Answer:
- In solid NaCl, ions are locked in a rigid ionic lattice. The Na⁺ and Cl⁻ ions cannot move freely, so there are no mobile charge carriers, and electrical conduction does not occur.
- When NaCl is dissolved in water, the polar water molecules surround and separate the ions (hydration). The lattice breaks down, producing free Na⁺ and Cl⁻ ions in the solution.
- These free ions can move under an applied potential, acting as charge carriers, so the solution conducts electricity. Such solutions are called electrolytes.
- In the molten state (liquid NaCl), the lattice is broken by heat, again freeing ions to move, so molten NaCl also conducts.
- Conclusion: Conductivity in ionic compounds depends on ion mobility. Solid state lacks mobility; aqueous and molten states provide free-moving ions, enabling conduction.
Q7. You observe a white powder after burning magnesium ribbon in air. Explain what the product is and how it forms at the atomic level.
Answer: The white powder is magnesium oxide (MgO). Its formation can be understood by electron transfer and ionic bonding:
- When heated, magnesium atoms undergo oxidation, losing two electrons:
Mg → Mg²⁺ + 2e⁻ - Oxygen atoms undergo reduction, gaining two electrons to complete their octet:
O + 2e⁻ → O²⁻ - The oppositely charged ions Mg²⁺ and O²⁻ attract strongly due to electrostatic force, forming the ionic compound MgO.
- Overall reaction: 2Mg + O₂ → 2MgO
- The product is a white, high-melting solid, consistent with ionic compounds’ properties. The strong ionic bonds in its crystal lattice account for its stability, hardness, and very high melting point.
Q8. Element A has electronic configuration 2,8,1 and element B has 2,7. Predict the ions formed, the formula of the compound, and explain the steps using the octet rule.
Answer:
- Element A (2,8,1) resembles sodium-like behavior; it will lose 1 electron to achieve a noble gas configuration, forming A⁺. This is oxidation.
- Element B (2,7) resembles chlorine-like behavior; it will gain 1 electron to complete its octet, forming B⁻. This is reduction.
- The electron transfer from A to B leads to the formation of oppositely charged ions that attract each other by electrostatic force.
- The simplest ratio that neutralizes charge is 1:1, so the formula of the compound is AB.
- Symbolically: A → A⁺ + e⁻; B + e⁻ → B⁻; then A⁺ + B⁻ → AB.
This satisfies the octet rule for both atoms. If B had configuration 2,6 (oxygen-like), two A atoms would be needed, giving A₂B, similar to MgO vs Na₂O analogies.
Q9. Design a simple circuit experiment to test the conductivity of NaCl in solid, molten, and aqueous states. Predict observations and explain scientifically.
Answer:
- Setup: A battery, bulb/LED, switch, and electrodes (graphite or inert). Prepare three samples: solid NaCl, molten NaCl (teacher demonstration), and NaCl solution in water.
- Procedure:
- Place electrodes on a solid NaCl crystal. Close the circuit—observe that the bulb does not glow.
- In a controlled setup, test molten NaCl—the bulb glows, showing conduction.
- Immerse electrodes in aqueous NaCl—the bulb glows here as well.
- Explanation:
- In the solid state, ions are fixed in a lattice—no mobile charge carriers.
- In molten and aqueous states, the lattice breaks; free Na⁺ and Cl⁻ ions move under potential, carrying current.
- Conclusion: Conduction in ionic substances requires free ions, which are present in molten and aqueous states but not in solids.
Q10. A friend argues that both metals and non-metals can either gain or lose electrons while reacting. Use evidence and reasoning to correct this misconception.
Answer:
- In general, metals lose electrons because they have few valence electrons (1–3) and relatively low ionization energy. Losing these electrons gives them a stable noble gas configuration. Examples:
Na → Na⁺ + e⁻; Mg → Mg²⁺ + 2e⁻ - Non-metals gain electrons because they have 5–7 valence electrons and higher electron affinity. Gaining electrons allows them to complete the octet. Examples:
Cl + e⁻ → Cl⁻; O + 2e⁻ → O²⁻ - In ionic bonding between a metal and a non-metal, electron transfer is directional: from metal (oxidation) to non-metal (reduction).
- This consistent behavior explains why compounds like NaCl, MgO, CaCl₂ form with cations (Na⁺, Mg²⁺, Ca²⁺) and anions (Cl⁻, O²⁻).
Thus, the usual rule is: metals lose, non-metals gain—not interchangeably.
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