IB DP Chemistry Reactivity 3.2 Electron transfer reactions IB Style Question Bank SL Paper 1

Question

Which change involves oxidation of N?
A. \(NH_3\) to \(N_2\)
B. \(NO_2\) to NO
C. \(N_2\) to AlN
D. \(NO_2\) to \(N_2O_4\)

▶️Answer/Explanation

Markscheme: A

\(\mathrm{NH}_3\) to \(\mathrm{N}_2\)

1. Oxidation State of Nitrogen in \(\mathrm{NH}_3\) : In ammonia \(\left(\mathrm{NH}_3\right)\), nitrogen has an oxidation state of -3. Each hydrogen has an oxidation state of +1 , and there are three hydrogen atoms.
2. Oxidation State of Nitrogen in \(N_2\) : In dinitrogen \(\left(N_2\right)\), nitrogen has an oxidation state of 0 . Each nitrogen atom contributes zero oxidation state.

So, the change involves nitrogen going from an oxidation state of – 3 in \(\mathrm{NH}_3\) to \(\mathrm{O}\) in \(\mathrm{N}_2\). Nitrogen is gaining more positive oxidation state, and this is characteristic of oxidation. The

 

Question

What is the formula of copper(I) sulfide?
A. CuS
B. \(Cu_2S\)
C. \(CuSO_3\)
D. \(Cu_2SO_3\)

▶️Answer/Explanation

Markscheme: B

The symbol of copper is ‘Cu’, having a valency of \(1^{+}\).
The symbol of sulphide is ‘ \(S\) ‘ with valency 2 ‘.

Step 1 –

Write the formula of copper [I] sulphide:

Symbol Valency

Copper Cu \(\Rightarrow 1^{+}\)
Sulphide \(\mathrm{S} \Rightarrow 2^{-}\)

Step 2 –

Write each symbol with its valency:
\[
\mathrm{Cu}^{1+} \mathrm{S}^{2-}
\]

Step 3 –

Interchanging the valencies:

The formula of the compound copper [I] sulphide is \(\mathrm{Cu}_2 \mathrm{~S}\).

Question

Which element has variable oxidation states in its compounds?
A. Potassium
B. Calcium
C. Fluorine
D. Bromine

▶️Answer/Explanation

Markscheme: D

Bromine is a halogen and belongs to Group 17 (also known as Group VIIA) of the periodic table. Elements in Group 17 are known as halogens, and they include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At).

Bromine, in particular, is interesting because it can exhibit a range of oxidation states in its compounds. The most common oxidation state of bromine is -1, where it gains one electron to achieve a stable electron configuration similar to that of noble gases. However, bromine can also exhibit positive oxidation states, such as +1, +3, +5, and +7, in certain compounds.

For example:

  • In hydrogen bromide (HBr), bromine has an oxidation state of -1.
  • In bromine pentafluoride (BrF5), bromine has an oxidation state of +5.
  • In bromine heptafluoride (BrF7), bromine has an oxidation state of +7.

The ability of bromine to exhibit various oxidation states is attributed to its position in the halogen group, which allows it to readily gain or lose electrons to achieve a stable electron configuration. This flexibility in oxidation states makes bromine versatile in forming a variety of compounds with different elements.

In contrast, elements like potassium (A), calcium (B), and fluorine (C) are less likely to exhibit a range of oxidation states in their compounds. Potassium and calcium are typically found in +1 and +2 oxidation states, respectively, while fluorine is usually found in the -1 oxidation state in its compounds.

Question

What happens to iodine when iodate ions, \({\text{IO}}_3^ – \), are converted to iodine molecules, \({{\text{I}}_{\text{2}}}\)?

A.     It undergoes reduction and its oxidation number changes from \( – 1\) to 0

B.     It undergoes oxidation and its oxidation number changes from \( – 1\) to 0

C.     It undergoes reduction and its oxidation number changes from \( + 5\) to 0

D.     It undergoes oxidation and its oxidation number changes from \( + 5\) to 0

Answer/Explanation

Markscheme

C

Examiners report

[N/A]

Question

Consider the following reactions of three unknown metals X, Y and Z.

\({\rm{2XN}}{{\rm{O}}_3}{\rm{(aq)}} + {\rm{Y(s)}} \to 2{\rm{X(s)}} + {\rm{Y(N}}{{\rm{O}}_3}{)_2}{\rm{(aq)}}\)

\({\rm{Y(N}}{{\rm{O}}_3}{)_2}{\rm{(aq)}} + {\rm{Z(s)}} \to\) No reaction

\(2{\rm{XN}}{{\rm{O}}_3}{\rm{(aq)}} + {\rm{Z(s)}} \to 2{\rm{X(s)}} + {\rm{Z(N}}{{\rm{O}}_3}{)_2}{\rm{(aq)}}\)

What is the order of increasing reactivity of the metals (least reactive first)?

A.     \({\text{X}} < {\text{Y}} < {\text{Z}}\)

B.     \({\text{X}} < {\text{Z}} < {\text{Y}}\)

C.     \({\text{Z}} < {\text{Y}} < {\text{X}}\)

D.     \({\text{Y}} < {\text{Z}} < {\text{X}}\)

Answer/Explanation

Markscheme

B

Examiners report

[N/A]
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