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Pre AP Chemistry -2.3D Representations of Ionic and Covalent Compounds- MCQ Exam Style Questions -New Syllabus 2025-2026

Pre AP Chemistry -2.3D Representations of Ionic and Covalent Compounds- MCQ Exam Style Questions – New Syllabus 2025-2026

Pre AP Chemistry -2.3D Representations of Ionic and Covalent Compounds- MCQ Exam Style Questions – Pre AP Chemistry – per latest Pre AP Chemistry Syllabus.

Pre AP Chemistry – MCQ Exam Style Questions- All Topics

Question 

The electronic structures of two atoms, $\mathrm{P}$ and $\mathrm{Q}$, are shown.

P and Q combine together to form a compound.
What is the type of bonding in the compound and what is the formula of the compound?

▶️ Answer/Explanation
Solution

Ans: A

1. Electron Analysis: – P: 7 valence electrons (needs 1 to complete octet) – Q: 1 valence electron (can donate 1 electron)

2. Bond Formation: – Q donates its valence electron to P – Forms $\mathrm{Q}^+$ cation and $\mathrm{P}^-$ anion – Ionic bond results from this electron transfer

3. Compound Formula: – Charges balance directly (1:1 ratio) – Simplest formula is PQ

4. Why Other Options are Incorrect: – B: Wrong formula (P2Q doesn’t balance charges) – C: Covalent bonding doesn’t apply here – D: Wrong bonding type and formula

Question 

Which row describes the type of bonding present in substances 1 and 2?

▶️ Answer/Explanation
Solution

Ans: C

Analysis of bonding types:

  1. Substance 1 (Methane, CH4):
    • Covalent bonding between carbon and hydrogen atoms (electron sharing).
    • Typical of molecular compounds.
  2. Substance 2 (Graphite):
    • Covalent bonding within each carbon layer (strong C-C bonds).
    • Delocalized electrons between layers give conductivity.
  3. Potassium Chloride (KCl):
    • Ionic bonding (electron transfer from K to Cl).
    • Not relevant to this question but shown for comparison.

Thus, Row C correctly identifies both substances as having covalent bonding.

Question 

Which diagram correctly shows the ions present in the compound potassium fluoride?

▶️ Answer/Explanation
Solution

Ans: C

1. Ion Formation Process:

  • Potassium (K) donates its 4s¹ electron → K⁺ ion (loses 1 electron)
  • Fluorine (F) gains 1 electron → F⁻ ion (completes its octet)

2. Diagram Analysis:

  • Option C correctly shows:
    • K⁺ with 18 electrons (original 19 – 1 donated)
    • F⁻ with 10 electrons (original 9 + 1 gained)
    • Proper charge distribution (+1 and -1)

3. Why Other Options Are Incorrect:

  • A: Shows incorrect electron counts (neutral atoms)
  • B: Shows incorrect charges (reversed polarity)
  • D: Shows incorrect electron transfer (multiple electrons)

Chemical Insight: The 1:1 ratio of K⁺ to F⁻ ions in potassium fluoride forms a cubic crystal lattice structure, similar to NaCl, with strong electrostatic attractions giving it a high melting point (858°C).

4. Conclusion:

Diagram C accurately represents the ionic bonding in KF, showing the correct electron transfer and resulting ion charges.

Question 

Element X has six electrons in its outer shell.

How could the element react?

  1. by gaining two electrons to form a positive ion
  2. by losing six electrons to form a negative ion
  3. by sharing two electrons with two electrons from another element to form two covalent bonds
  4. by sharing two electrons with two electrons from another element to form four covalent bonds
▶️ Answer/Explanation
Solution

Ans: C

Analysis of element X with 6 valence electrons (Group 16 element like oxygen):

  1. Option A:
    • Gaining electrons would form a negative ion, not positive.
    • Incorrect charge direction.
  2. Option B:
    • Losing 6 electrons is energetically unfavorable.
    • Would form a positive ion, not negative.
    • Double incorrect.
  3. Option C:
    • Forms two covalent bonds by sharing its 2 unpaired electrons.
    • Example: H₂O (oxygen shares 2 electrons with 2 hydrogen atoms).
    • Correct and most likely behavior.
  4. Option D:
    • Would require promoting electrons to form 4 bonds (unlikely for Group 16).
    • Typical maximum is 2 covalent bonds for these elements.

Key concept: Group 16 elements typically complete their octet by forming two covalent bonds (e.g., in H₂O or CO₂).

Question 

The electronic structures of atoms P and Q are shown.

P and Q react to form an ionic compound.

What is the formula of this compound?

A. PQ2   B. P2Q   C. P2Q6   D. P6Q2

▶️ Answer/Explanation
Solution

Ans: B

1. Valence Electron Analysis:
– P: 2,8,1 configuration (Group 1 metal – e.g., Na)
– Q: 2,8,6 configuration (Group 16 nonmetal – e.g., O)

2. Ion Formation:
– Each P atom loses 1 electron to form P+
– Each Q atom gains 2 electrons to form Q2-

3. Compound Formation:
– Charge balance requires 2 P+ ions for every 1 Q2- ion
– Results in the formula P2Q (e.g., Na2O)

4. Why Other Options Are Incorrect:
– PQ2: Would create charge imbalance (+1 vs -4)
– P2Q6/P6Q2: Don’t satisfy octet rule or charge balance

The correct ionic formula is P2Q (Option B).

Question

A model of a molecule is shown.

Which row shows the formula of this molecule and describes the type of bonding between the atoms?

A 2BH3 / covalent
B 2BH3 / ionic
C B2H6 / covalent
D B2H6 / ionic

▶️ Answer / Explanation

The model represents diborane, \(B_2H_6\).

In this molecule:

• Two boron atoms are bonded to hydrogen atoms.
• Two hydrogen atoms form bridging bonds between the boron atoms.
• All bonds are covalent (electron sharing).

Therefore the molecule is \(B_2H_6\) and the bonding is covalent.

Answer: C

Question

Which dot-and-cross diagram represents a molecule of ammonia?

▶️ Answer / Explanation

Ammonia (\(NH_3\)) has:

• Three N–H covalent bonds.
• One lone pair of electrons on the nitrogen atom.
• Each bond contains one electron from nitrogen and one from hydrogen.

Only diagram C correctly shows:

• Three shared electron pairs between N and H.
• One lone pair on nitrogen.

Answer: C

Question

What is the dot-and-cross diagram for a water molecule?

▶️ Answer / Explanation

In a water molecule (\(H_2O\)):

• Oxygen has 6 valence electrons.
• Each hydrogen has 1 electron.
• Two O–H covalent bonds form by sharing one electron from H and one from O.

After bonding, oxygen still has two lone pairs of electrons.

The correct diagram must therefore show:

• Two shared electron pairs (O–H bonds).
• Two lone pairs on the oxygen atom.

Diagram A shows this correctly.

Answer: A

Question

In which molecule are all the outer electrons of the atoms used in covalent bonds?

A. CH4
B. HCl
C. H2O
D. NH3

▶️ Answer / Explanation

CH4: Carbon has 4 valence electrons and forms four C–H bonds, using all its outer electrons in bonding. ✔

HCl: Chlorine has three lone pairs of electrons remaining. ✘

H2O: Oxygen has two lone pairs remaining. ✘

NH3: Nitrogen has one lone pair remaining. ✘

Only methane has no lone pairs on the central atom, meaning all outer electrons are used in bonding.

Answer: A

Question

The diagram represents an ionic compound formed from three types of atom.

What is the chemical formula for this compound?

(A) \(Na_{2}S_{4}O\)
(B) \(NaO_{4}S_{2}\)
(C) \(Na_{2}SO_{4}\)
(D) \(S_{4}O_{2}Na\)

▶️ Answer / Explanation

The diagram shows:

• One sulfur atom at the center.
• Four oxygen atoms bonded around sulfur, forming a sulfate ion \(SO_4^{2-}\).
• Two sodium ions balancing the \(2-\) charge of the sulfate ion.

Therefore the compound contains:

2 sodium atoms
1 sulfur atom
4 oxygen atoms

This gives the formula:

\(Na_2SO_4\)

Answer: C

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