Question:
Which pair of ions act as strong reducing agents?
Which pair of ions act as strong reducing agents?
Show Hint
Lanthanides in the +2 state ($\text{Eu}^{2+}$, $\text{Yb}^{2+}$) act as reducing agents, while those in the +4 state ($\text{Ce}^{4+}$, $\text{Tb}^{4+}$) act as oxidizing agents to reach the stable +3 state.
Updated On: Jul 22, 2026
- $\text{Ce}^{4+}, \text{Tb}^{4+}$
- $\text{Eu}^{2+}, \text{Yb}^{2+}$
- $\text{Gd}^{3+}, \text{Lu}^{3+}$
- $\text{La}^{3+}, \text{Pm}^{3+}$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
This question is about the stability of oxidation states of lanthanides (f-block elements).
We need to identify which pair of lanthanide ions can act as strong reducing agents.
Step 2: Key Formula or Approach:
The most stable and common oxidation state of the lanthanides is +3.
A species acts as a reducing agent if it can easily lose electrons to be oxidized to a higher, more stable oxidation state.
Therefore, lanthanide ions in the +2 state will tend to get oxidized to the +3 state, acting as reducing agents.
Step 3: Detailed Explanation:
• Let us analyze the oxidation states of the options:
• 1. $\text{Eu}^{2+}$ and $\text{Yb}^{2+}$:
Europium (Eu) has the electronic configuration $[\text{Xe}] 4\text{f}^7 6\text{s}^2$. The ion $\text{Eu}^{2+}$ has the stable half-filled configuration $[\text{Xe}] 4\text{f}^7$.
Ytterbium (Yb) has the configuration $[\text{Xe}] 4\text{f}^{14} 6\text{s}^2$. The ion $\text{Yb}^{2+}$ has the stable fully-filled configuration $[\text{Xe}] 4\text{f}^{14}$.
Although $+2$ is a stable state for these ions, the $+3$ state remains the thermodynamic favorite in aqueous solutions.
Thus, $\text{Eu}^{2+}$ and $\text{Yb}^{2+}$ readily lose an electron to form $\text{Eu}^{3+}$ and $\text{Yb}^{3+}$, behaving as strong reducing agents.
• 2. $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$:
These ions are in the $+4$ oxidation state. To reach the stable $+3$ state, they have a strong tendency to gain an electron.
Therefore, they act as strong oxidizing agents, not reducing agents.
• 3. $\text{Gd}^{3+}$, $\text{Lu}^{3+}$, $\text{La}^{3+}$, and $\text{Pm}^{3+}$:
These ions are already in their most stable $+3$ oxidation state and do not show strong reducing or oxidizing behavior under normal conditions.
Step 4: Final Answer:
The pair of ions that act as strong reducing agents is $\text{Eu}^{2+}$ and $\text{Yb}^{2+}$.
This question is about the stability of oxidation states of lanthanides (f-block elements).
We need to identify which pair of lanthanide ions can act as strong reducing agents.
Step 2: Key Formula or Approach:
The most stable and common oxidation state of the lanthanides is +3.
A species acts as a reducing agent if it can easily lose electrons to be oxidized to a higher, more stable oxidation state.
Therefore, lanthanide ions in the +2 state will tend to get oxidized to the +3 state, acting as reducing agents.
Step 3: Detailed Explanation:
• Let us analyze the oxidation states of the options:
• 1. $\text{Eu}^{2+}$ and $\text{Yb}^{2+}$:
Europium (Eu) has the electronic configuration $[\text{Xe}] 4\text{f}^7 6\text{s}^2$. The ion $\text{Eu}^{2+}$ has the stable half-filled configuration $[\text{Xe}] 4\text{f}^7$.
Ytterbium (Yb) has the configuration $[\text{Xe}] 4\text{f}^{14} 6\text{s}^2$. The ion $\text{Yb}^{2+}$ has the stable fully-filled configuration $[\text{Xe}] 4\text{f}^{14}$.
Although $+2$ is a stable state for these ions, the $+3$ state remains the thermodynamic favorite in aqueous solutions.
Thus, $\text{Eu}^{2+}$ and $\text{Yb}^{2+}$ readily lose an electron to form $\text{Eu}^{3+}$ and $\text{Yb}^{3+}$, behaving as strong reducing agents.
• 2. $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$:
These ions are in the $+4$ oxidation state. To reach the stable $+3$ state, they have a strong tendency to gain an electron.
Therefore, they act as strong oxidizing agents, not reducing agents.
• 3. $\text{Gd}^{3+}$, $\text{Lu}^{3+}$, $\text{La}^{3+}$, and $\text{Pm}^{3+}$:
These ions are already in their most stable $+3$ oxidation state and do not show strong reducing or oxidizing behavior under normal conditions.
Step 4: Final Answer:
The pair of ions that act as strong reducing agents is $\text{Eu}^{2+}$ and $\text{Yb}^{2+}$.
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET The Lanthanoids Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions