IB DP Chemistry -Reactivity 3.4 Electron-pair sharing reactions - IB Style Questions For HL Paper 1A -FA 2025

IBDP Chemistry HL Paper 1A & 1B - All Chapters

Question 

What is the oxidation state of Pt in \( \mathrm{[Pt(NH_3)_3Cl]Cl_3} \)?

(A) \(+1\)
(B) \(+2\)
(C) \(+3\)
(D) \(+4\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{D}} \)

The three \( \mathrm{Cl^-} \) ions outside the complex ion give the complex ion an overall charge of \(+3\):

\( \mathrm{[Pt(NH_3)_3Cl]^{3+}} \)

The \( \mathrm{NH_3} \) ligands are neutral, while the coordinated \( \mathrm{Cl^-} \) has an oxidation state of \(-1\).

Let the oxidation state of Pt be \(x\):

\(x+3(0)+(-1)=+3\)

\(x-1=3\)

\(x=+4\)

Therefore, the correct answer is (D).

Question 

Which statements are correct for the complex ion \( \mathrm{[FeCl_4]^{2-}} \)?

I. Chloride ions are behaving as ligands.
II. The oxidation state of iron is \(+3\).
III. Iron forms coordination bonds with chloride ions.

(A) I and II only
(B) I and III only
(C) II and III only
(D) I, II and III
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{B}} \)

Statement I: Correct. The four \( \mathrm{Cl^-} \) ions donate lone pairs of electrons to the iron ion and therefore act as ligands.

Statement II: Incorrect. Let the oxidation state of iron be \(x\):

\(x+4(-1)=-2\)

\(x=+2\)

Therefore, the oxidation state of iron is \(+2\), not \(+3\).

Statement III: Correct. The chloride ions donate lone pairs to \( \mathrm{Fe^{2+}} \), forming coordinate bonds.

Therefore, statements I and III are correct, so the answer is (B).

Question

What is the mechanism for the reaction of propene with iodine in the dark?
A. electrophilic addition
B. electrophilic substitution
C. free radical substitution
D. nucleophilic substitution
▶️ Answer/Explanation
Detailed solution

Propene is an alkene, and alkenes typically react with halogens through electrophilic addition. This involves the π bond attacking a halogen molecule.

In the dark, iodine does not undergo photochemical homolysis, so no free radicals form. Therefore, free radical substitution does not occur.

The reaction proceeds by:

  • iodine acting as an electrophile (polarized by the alkene π electrons)
  • the π bond opening to form a di-iodo product

Thus, the correct mechanism is:

\(\boxed{\text{electrophilic addition}}\)

✅ Answer: (A)

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