Copper (Cu) Chemistry: Practice Questions

Multiple Choice Questions (MCQs)

Question 1

Which of the following statements about copper is incorrect? a) Cu$^+$ is diamagnetic. b) Cu$^{2+}$ compounds are generally coloured. c) Copper reacts with dilute hydrochloric acid to produce hydrogen gas. d) Cu$^+$ disproportionates in aqueous solution.

Solution

The correct answer is c).

• Explanation:
  • a) Cu$^+$ has an electronic configuration of [Ar]3d$^{10}$. All electrons are paired, making it diamagnetic. This statement is correct.
  • b) Cu$^{2+}$ has an electronic configuration of [Ar]3d$^9$. It has one unpaired electron, leading to d-d transitions which absorb certain wavelengths of light and transmit others, resulting in colour (typically blue or green in aqueous solution). This statement is correct.
  • c) Copper is less reactive than hydrogen and does not react with non-oxidizing acids like dilute hydrochloric acid or dilute sulfuric acid to produce hydrogen gas. It reacts with oxidizing acids (e.g., HNO$_3$, concentrated H$_2$SO$_4$). This statement is incorrect.
  • d) Cu$^+$ is unstable in aqueous solution and undergoes disproportionation to Cu$^{2+}$ and Cu(s). The standard electrode potential for 2Cu$^+$(aq) → Cu$^{2+}$(aq) + Cu(s) is positive (+0.37 V), indicating spontaneity. This statement is correct.

Question 2

Copper turnings are added to concentrated nitric acid. The gas evolved is: a) NO b) NO$_2$ c) N$_2$O d) NH$_3$

Solution

The correct answer is b).

• Explanation:
  • Copper reacts with concentrated nitric acid to produce nitrogen dioxide (NO$_2$), a reddish-brown gas, along with copper nitrate and water.
  • The reaction is: Cu(s) + 4HNO$_3$(conc) → Cu(NO$_3$)$_2$(aq) + 2NO$_2$(g) + 2H$_2$O(l)
  • With dilute nitric acid, copper produces nitric oxide (NO), which is colourless but turns reddish-brown on exposure to air (due to oxidation to NO$_2$).
  • The reaction with dilute HNO$_3$ is: 3Cu(s) + 8HNO$_3$(dil) → 3Cu(NO$_3$)$_2$(aq) + 2NO(g) + 4H$_2$O(l)

Question 3

An aqueous solution of copper sulphate is treated with excess aqueous ammonia. The colour of the resulting solution will be: a) Blue b) Deep blue c) Green d) Colourless

Solution

The correct answer is b).

• Explanation:
  • When aqueous ammonia is added to copper sulphate solution, a pale blue precipitate of copper(II) hydroxide, Cu(OH)$_2$, is initially formed: CuSO$_4$(aq) + 2NH$_4$OH(aq) → Cu(OH)$_2$(s) + (NH$_4$)$_2$SO$_4$(aq)
  • Upon adding excess aqueous ammonia, the precipitate dissolves to form a deep blue soluble complex, tetraamminecopper(II) ion, [Cu(NH$_3$)$_4$]$^{2+}$: Cu(OH)$_2$(s) + 4NH$_3$(aq) → [Cu(NH$_3$)$_4$]$^{2+}$(aq) + 2OH$^-$(aq)
  • This deep blue colour is characteristic and is often used as a test for Cu$^{2+}$ ions.

Assertion-Reason Questions

The following questions consist of two statements, one labelled as Assertion (A) and the other as Reason (R). Examine both statements carefully and select the correct option from the following: a) Both A and R are true, and R is the correct explanation of A. b) Both A and R are true, but R is not the correct explanation of A. c) A is true, but R is false. d) A is false, but R is true.

Question 1

Assertion (A): Cu$^+$ ion is diamagnetic. Reason (R): Cu$^+$ ion has a completely filled d-orbital (3d$^{10}$).

Solution

The correct answer is a).

• Explanation:
  • Assertion (A) states that Cu$^+$ is diamagnetic. This is true. Diamagnetism arises from the absence of unpaired electrons.
  • Reason (R) states that Cu$^+$ has a completely filled d-orbital (3d$^{10}$). This is also true. The electronic configuration of Cu is [Ar]3d$^{10}$4s$^1$. When it loses one electron to form Cu$^+$, its configuration becomes [Ar]3d$^{10}$, meaning all its 3d electrons are paired.
  • Since a completely filled d-orbital (3d$^{10}$) implies all electrons are paired, it directly explains why Cu$^+$ is diamagnetic. Hence, R is the correct explanation for A.

Question 2

Assertion (A): Copper is an excellent conductor of electricity. Reason (R): Copper has a low first ionization enthalpy.

Solution

The correct answer is b).

• Explanation:
  • Assertion (A) states that copper is an excellent conductor of electricity. This is true. Copper is one of the best electrical conductors due to its metallic structure.
  • Reason (R) states that copper has a low first ionization enthalpy. This is also true relative to many non-metals, allowing it to readily lose its valence electron.
  • However, the direct reason for copper’s high electrical conductivity is the presence of delocalized valence electrons (often called a “sea of electrons”) that are free to move throughout the metallic lattice, rather than specifically its low ionization enthalpy. While low ionization enthalpy contributes to the metallic bonding and the availability of these mobile electrons, it is not the direct and sole explanation for conductivity in the context of the electron sea model. Therefore, both A and R are true, but R is not the correct explanation of A.

Short Answer Questions

Question 1

Explain why Cu$^+$ ion is unstable in aqueous solution while Cu$^{2+}$ is stable.

Model Answer

The Cu$^+$ ion (electronic configuration [Ar]3d$^{10}$) is unstable in aqueous solution and undergoes disproportionation, forming more stable Cu$^{2+}$ ions and solid copper: 2Cu$^+$(aq) → Cu$^{2+}$(aq) + Cu(s)

This disproportionation is spontaneous in aqueous medium because:

1. Hydration Enthalpy: Cu$^{2+}$ has a much higher (more negative) hydration enthalpy compared to Cu$^+$. This is due to the smaller size and higher charge density of Cu$^{2+}$, which allows for stronger interaction with water molecules. This large hydration energy compensates for the higher second ionization enthalpy required to form Cu$^{2+}$ from Cu$^+$.
2. Lattice Enthalpy: For solid compounds, the stability of Cu$^{2+}$ compounds is generally higher than Cu$^+$ compounds due to stronger ionic bonding arising from the higher charge of Cu$^{2+}$.
3. Standard Electrode Potentials: The standard electrode potential (E°) for the disproportionation reaction (E°cell = +0.37 V) is positive, indicating that the reaction is thermodynamically favourable.
   • Cu$^+$(aq) + e$^-$ → Cu(s) ; E° = +0.52 V
   • Cu$^{2+}$(aq) + e$^-$ → Cu$^+$(aq) ; E° = +0.15 V
   • The spontaneous nature of this reaction signifies the instability of Cu$^+$ in water.

Conversely, Cu$^{2+}$ ion (electronic configuration [Ar]3d$^9$) is stable in aqueous solutions due to its high hydration enthalpy which significantly offsets the energy required for its formation.

Question 2

Give two reactions to illustrate the oxidizing nature of concentrated sulfuric acid with copper.

Model Answer

Concentrated sulfuric acid (H$_2$SO$_4$) acts as a strong oxidizing agent, especially when heated, reducing itself while oxidizing less reactive metals like copper. Here are two reactions:

1. Reaction with Copper (producing sulfur dioxide): When copper is heated with concentrated sulfuric acid, copper is oxidized to copper(II) sulfate, and sulfuric acid is reduced to sulfur dioxide gas. Cu(s) + 2H$_2$SO$_4$(conc) $\xrightarrow{\text{heat}}$ CuSO$_4$(aq) + SO$_2$(g) + 2H$_2$O(l) In this reaction, the oxidation state of sulfur in H$_2$SO$_4$ changes from +6 to +4 in SO$_2$, indicating reduction. Copper’s oxidation state changes from 0 to +2, indicating oxidation.

2. Another representation focusing on redox changes: This is essentially the same reaction but highlights the redox half-reactions more clearly: Oxidation half-reaction: Cu(s) → Cu$^{2+}$(aq) + 2e$^-$ Reduction half-reaction: SO$_4^{2-}$(aq) + 4H$^+$(aq) + 2e$^-$ → SO$_2$(g) + 2H$_2$O(l) Combining these gives the overall reaction as shown above. This clearly demonstrates H$_2$SO$_4$ as the oxidizing agent.

High-Order Thinking Skills (HOTS) Question

Copper exhibits +1 and +2 oxidation states. Which of these is more stable in aqueous solution and why? Explain using relevant thermodynamic parameters.

Detailed Chemical Explanation

In aqueous solutions, the +2 oxidation state (Cu$^{2+}$) of copper is significantly more stable than the +1 oxidation state (Cu$^+$). This stability difference can be explained by considering the interplay of ionization enthalpies and hydration enthalpies.

1. Electronic Configuration:

   • Cu: [Ar]3d$^{10}$4s$^1$
   • Cu$^+$: [Ar]3d$^{10}$ (stable, completely filled d-subshell)
   • Cu$^{2+}$: [Ar]3d$^9$ (unstable, incomplete d-subshell, but not necessarily unstable in solution)

2. Ionization Enthalpies:

   • The formation of Cu$^+$ requires the first ionization enthalpy (Δ$_i$H$_1$).
   • The formation of Cu$^{2+}$ requires the sum of the first and second ionization enthalpies (Δ$_i$H$_1$ + Δ$_i$H$_2$).
   • Since the second ionization enthalpy (Δ$_i$H$_2$) is always greater than the first (Δ$_i$H$_1$), forming Cu$^{2+}$ requires significantly more energy than forming Cu$^+$. From an ionization energy perspective alone, Cu$^+$ appears more favorable.

3. Hydration Enthalpies:

   • When an ion is placed in water, it gets hydrated, releasing energy known as hydration enthalpy (Δ$_{hyd}$H). This energy stabilizes the ion.
   • Cu$^{2+}$ has a much higher (more negative, i.e., more energy released) hydration enthalpy compared to Cu$^+$. This is because:
     • Higher Charge: Cu$^{2+}$ has twice the charge of Cu$^+$.
     • Smaller Ionic Radius: Cu$^{2+}$ has a smaller ionic radius than Cu$^+$ (due to higher effective nuclear charge).
   • These factors lead to a stronger electrostatic attraction between Cu$^{2+}$ ions and the polar water molecules, resulting in substantially greater hydration energy.

4. Overall Stability and Disproportionation: The large hydration enthalpy of Cu$^{2+}$ effectively compensates for the high second ionization enthalpy required to remove the second electron from Cu$^+$. The instability of Cu$^+$ in aqueous solution is evident from its tendency to undergo disproportionation: 2Cu$^+$(aq) → Cu$^{2+}$(aq) + Cu(s)

   To evaluate the spontaneity of this reaction, we look at standard electrode potentials:

   • Reduction of Cu$^+$: Cu$^+$(aq) + e$^-$ → Cu(s); E° = +0.52 V
   • Reduction of Cu$^{2+}$: Cu$^{2+}$(aq) + e$^-$ → Cu$^+$(aq); E° = +0.15 V

   For the disproportionation reaction 2Cu$^+$(aq) → Cu$^{2+}$(aq) + Cu(s), we can consider it as:

   • Cu$^+$(aq) → Cu$^{2+}$(aq) + e$^-$ (Oxidation)
   • Cu$^+$(aq) + e$^-$ → Cu(s) (Reduction)

   The standard cell potential (E°cell) for this disproportionation is: E°cell = E°(reduction of Cu$^+$) - E°(reduction of Cu$^{2+}$ to Cu$^+$) E°cell = (+0.52 V) - (+0.15 V) = +0.37 V

   Since the E°cell is positive (+0.37 V), the disproportionation of Cu$^+$ in aqueous solution is a spontaneous process (ΔG° = -nFE°cell, so ΔG° will be negative). This means Cu$^+$ is thermodynamically unstable in water and will readily convert to the more stable Cu$^{2+}$ and elemental copper.

Conclusion: Despite Cu$^+$ having a stable 3d$^{10}$ configuration and requiring less energy for its formation, the significantly larger hydration enthalpy of the Cu$^{2+}$ ion, coupled with the positive standard electrode potential for the disproportionation of Cu$^+$, makes the Cu$^{2+}$ oxidation state more stable in aqueous solutions.