How Do Metals and Non-Metals React? – Long Answer Questions
Medium Level (Application & Explanation)
Q1. Explain the process by which an ionic compound is formed when sodium reacts with chlorine. Describe each step in detail.
Answer:
Sodium (Na) is a metal with one electron in its outermost shell (configuration: 2,8,1).
It loses one electron to achieve a stable noble gas configuration, forming the cation Na⁺.
Chlorine (Cl) is a non-metal with seven electrons in its valence shell (configuration: 2,8,7).
It gains one electron to complete its octet, forming the anion Cl⁻.
The oppositely charged ions (Na⁺ and Cl⁻) are attracted to each other by strong electrostatic force.
This results in the formation of the ionic compound NaCl (sodium chloride).
Q2. Describe how magnesium reacts with oxygen to form magnesium oxide. What is the electron transfer involved?
Answer:
Magnesium (Mg) has two electrons in its outermost shell (2,8,2).
It loses both electrons to become Mg²⁺, reaching a stable electron configuration.
Oxygen (O₂) needs two electrons to complete its outer shell (2,6).
Each oxygen atom gains two electrons to become O²⁻.
Therefore, two Mg atoms give up a total of four electrons, and two O atoms (from O₂) each accept two electrons.
The resulting ions (Mg²⁺ and O²⁻) combine to form MgO, an ionic compound.
Q3. Explain why metals tend to lose electrons while non-metals tend to gain electrons during chemical reactions.
Answer:
Metals typically have 1-3 electrons in their outermost shell, so losing these few electrons helps them achieve a stable, noble gas configuration.
This process makes them cationic (positively charged).
Non-metals usually have 5-7 electrons in their valence shell, so they gain electrons to complete their octet.
Gaining electrons makes them anionic (negatively charged).
This transfer of electrons is energetically favorable and helps both achieve stability.
Thus, the difference in electron arrangement explains their typical behavior during chemical reactions.
Q4. What are the general properties of ionic compounds that result from the reaction between metals and non-metals? Explain each property with a reason.
Answer:
High melting and boiling points: Ionic bonds are very strong due to electrostatic attraction, so more heat is needed to break them.
Solubility in water: Ionic compounds dissociate into ions in water because water molecules can surround and separate the ions.
Electrical conductivity: In molten or dissolved states, the ions are free to move, allowing electrical current to pass.
Brittleness: When force is applied, electrostatic forces shift, causing like charges to come together and repel, breaking the crystal.
These properties arise because of the strong attractions and structured arrangements of ions in a crystal lattice.
Examples include NaCl, MgO, and CaCl₂.
Q5. Using the example of aluminum and oxygen, show how charges get balanced in the final ionic compound.
Answer:
Aluminum (Al) loses three electrons to form Al³⁺ ions.
Oxygen (O) gains two electrons to form O²⁻ ions.
To balance the charges, two Al atoms (total +6 charge) react with three O atoms (total -6 charge from three O²⁻ ions).
The reaction can be written as: 4Al+3O_2→2Al_2O_3.
Thus, the charges are balanced as 2×3+=6+ from aluminum and 3×2−=6− from oxygen.
The product is neutral Al₂O₃ (aluminum oxide), an ionic compound.
High Complexity (Analysis & Scenario-Based)
Q6. Suppose a metal M reacts with a non-metal X. M has an electronic configuration of 2,8,2 and X has 2,6. Predict the formula of the resulting compound and explain your reasoning.
Answer:
Metal M (configuration 2,8,2) is similar to magnesium (Mg), which loses 2 electrons to become M²⁺.
Non-metal X (configuration 2,6) is similar to oxygen (O), which gains 2 electrons to become X²⁻.
Both ions have charges of ±2, so one M ion combines with one X ion.
The resulting formula of the compound is MX (just like MgO).
The electron transfer is such that M loses 2 electrons (oxidation), and X gains 2 electrons (reduction), leading to a stable, neutral ionic compound.
The explanation is based on the need to balance the total positive and negative charges.
Q7. Analyze the structure of an ionic compound like NaCl and explain why it conducts electricity in solution but not in solid state.
Answer:
In solid NaCl, Na⁺ and Cl⁻ ions are fixed in a regular lattice arrangement and cannot move.
For electrical conduction, charge carriers must be mobile, but in solid NaCl, they are locked in place.
When NaCl is dissolved in water, the lattice breaks down, and the ions separate and become free to move.
These free ions act as charge carriers, allowing the flow of current through the solution.
Similarly, molten NaCl allows ions to move freely due to heat breaking the crystal lattice bonds.
Thus, mobility of ions is crucial for conduction, which only happens in the liquid or solution state.
Q8. If a student mixes potassium (K) and sulfur (S), predict the formula of the resulting ionic compound. Explain the transfer of electrons and charge balancing.
Answer:
Potassium (K) has 1 electron in its outermost shell and loses 1 electron to become K⁺ ion.
Sulfur (S) has 6 electrons in its valence shell and gains 2 electrons to become S²⁻ ion.
To balance charges, 2 K atoms are needed for every S atom (2×+1=+2 balances with −2 from S).
The resulting formula is K₂S.
Each K atom gives up 1 electron (total 2 electrons), and these are both taken up by S to complete its octet.
This electron transfer results in a stable ionic lattice in K₂S.
Q9. Consider the reaction between calcium and chlorine. Use electron dot structures to explain the formation of CaCl₂. Describe the transfer of electrons and combination of ions.
Answer:
Calcium (Ca) has 2 electrons in its outermost shell (electron dot: ..Ca).
It donates 1 electron to each chlorine atom, losing both electrons to become Ca²⁺.
Each chlorine (Cl) needs one electron to fill its octet (electron dot: ..Cl:).
After gaining an electron, each Cl atom becomes Cl⁻.
Thus, one Ca²⁺ ion combines with two Cl⁻ ions to balance the charge (2+ from Ca, and two 1- from Cl).
The resulting compound is CaCl₂, a neutral ionic compound.
Q10. Imagine two unknown elements, A and B. A has electronic configuration 2,8,7 and B has 2,8,1. Describe and justify the type of bond formed and write the formula of the compound.
Answer:
Element A (2,8,7) is similar to chlorine (Cl), a non-metal that needs 1 electron to complete its octet.
Element B (2,8,1) is like sodium (Na), a metal that loses 1 electron easily.
B donates 1 electron to A, so both achieve stable configurations (B becomes B⁺, A becomes A⁻).
The electrostatic attraction between B⁺ and A⁻ creates an ionic bond.
The chemical formula will be BA (direct 1:1 ratio, like NaCl).
This process is justified by the letter elements' tendency to lose (for metals) and gain (for non-metals) electrons.
