Mock Exam IB DP Chemistry HL Paper 1 Set 3

Question

Which will eventually yield the greatest mass of deposited copper in the electrolysis of a fixed volume of 1mol \(dm^{-3}\) \(CuSO_4 (aq)\)?

A. copper anode and inert cathode
B. inert anode and copper cathode
C. inert anode and inert cathode
D. zinc anode and zinc cathode

▶️Answer/Explanation

Markscheme: A

In the electrolysis of copper sulfate \(\left(\mathrm{CuSO}_4\right)\), copper ions \(\left(\mathrm{Cu}^{2+}\right)\) are reduced at the cathode and deposited as solid copper. The half-reaction at the cathode is:

\[
C u^{2+}+2 e^{-} \rightarrow C u
\]

At the anode, copper metal is oxidized to form copper ions. The half-reaction at the anode is:

\[
C u \rightarrow C u^{2+}+2 e^{-}
\]

Now, let’s consider the given options:

A. Copper anode and inert cathode: Copper would be oxidized at the anode, and copper ions would be reduced at the cathode.

B. Inert anode and copper cathode: Copper ions would be reduced at the cathode, and there would be no oxidation occurring at the inert anode.

C. Inert anode and inert cathode: There would be no copper ions present to be reduced or oxidized, so there would be no deposition of copper.

D. Zinc anode and zinc cathode: Zinc is not part of the copper sulfate electrolyte, so it won’t be involved in the deposition of copper.

The correct choice for the greatest mass of deposited copper is option A, where copper is oxidized at the anode and reduced at the cathode. This ensures a continuous supply of copper ions for deposition at the cathode, resulting in the maximum mass of deposited copper.

 

Question

Which compound is an aromatic ester?

▶️Answer/Explanation

Markscheme: B

An aromatic ester is a type of chemical compound that contains an aromatic (benzene-like) ring and an ester functional group. The ester functional group is characterized by the presence of a carbonyl group (C=O) bonded to an oxygen atom, and an alkyl (or aryl) group bonded to the oxygen. The general structure of an ester is R-COO-R’, where R and R’ are organic groups.

In the case of an aromatic ester, at least one of the organic groups attached to the carbonyl carbon is an aromatic ring. Aromatic rings are characterized by having a cyclic, planar structure with alternating single and double bonds, and they often exhibit a high degree of stability.

A.  This structure represents a carboxylic acid (RCOOH), not an ester.

B. This structure represents an aromatic ester. The carbonyl carbon is bonded to an oxygen atom and an aromatic ring.

C. This structure represents an aliphatic ester, not an aromatic ester.

D.  This structure represents a molecule with an aldehyde group and an alcohol group, not an ester.

So, the correct choice for an aromatic ester is option B.

Question

Which reaction mechanisms involve heterolytic fission of chlorine?

I. electrophilic addition
II. electrophilic substitution
III. nucleophilic substitution

A. I and II only
B. I and III only
C. II and III only
D. I, II and III

▶️Answer/Explanation

Markscheme: D

The heterolytic fission of chlorine involves the breaking of the chlorine molecule (\(Cl_2\)) into two ions, a chloride ion (\(Cl^-\)) and a positively charged species. Let’s analyze the given reaction mechanisms:

I. Electrophilic addition: This mechanism typically involves the addition of an electrophile to a double or triple bond. The addition step itself may not involve the heterolytic fission of chlorine.

II. Electrophilic substitution: In electrophilic aromatic substitution, chlorine can act as an electrophile, and the initial attack involves the heterolytic fission of chlorine, forming a chloride ion and a positive species.

III. Nucleophilic substitution: This mechanism involves the substitution of a nucleophile for a leaving group. Chlorine, when it leaves a molecule, may undergo heterolytic fission, producing a chloride ion.

Therefore, the correct answer is D: I, II, and III, as all three mechanisms can involve heterolytic fission of chlorine.

Question

 Which molecule has a tetrahedral molecular geometry?

A. $\mathrm{HNO}_3$
B. $\mathrm{SF}_4$
C. $\mathrm{XeF}_4$
D. $\mathrm{XeO}_4$

▶️Answer/Explanation

Solution:

Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. A tetrahedral molecular geometry is characterized by a central atom that is surrounded by four other atoms or electron pairs, arranged in a way that is roughly like the corners of a tetrahedron.

Out of the given options, only $\mathrm{XeO}_4$ has a tetrahedral molecular geometry. In this molecule, Xenon undergoes $\mathrm{SP}^3$ hybridization in the fourth excited in the formation of XeO 4 . In the fourth excited, xenon atom has 8 unpaired electrons. One “s” and three “p” – orbitals undergo $\mathrm{sp}^3$ hybridization. Since it has no lone pair of electrons, the shape of XeO 4 is tetrahedral with the bond angle of 109 degrees. It has four sigma and four pie bonds.

$\mathrm{HNO}_3$ (A) has a trigonal planar molecular geometry, while $\mathrm{SF}_4$ (B) has a seesaw molecular geometry and $\mathrm{XeF}_4$ (C) has a square planar molecular geometry.

$\colorbox{yellow}{ Correct Option: D}$

Question

 Alloying a metal with a metal of smaller atomic radius can disrupt the lattice and make it more difficult for atoms to slide over each other.

Which property will increase as a result?

A. Electrical conductivity
B. Ductility
C. Malleability
D. Strength

▶️Answer/Explanation

Solution:

Alloys are made by mixing two or more metals, or a metal and a non-metal, to create a new material with improved properties such as increased strength, hardness, and corrosion resistance. When a metal is alloyed with a metal of smaller atomic radius, the resulting alloy may have a distorted lattice structure, which makes it more difficult for the atoms to slide over each other. This increased resistance to deformation or movement of atoms leads to an increase in the strength of the alloy.

Therefore, the property that will increase as a result of alloying a metal with a metal of smaller atomic radius is strength (option D). Options A, B, and C are not directly affected by the disruption of the lattice structure, and may depend on other factors such as the composition of the alloy or the processing conditions.

$\colorbox{yellow}{Correct Option : D}$

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