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Among the following pairs of ions the lo...

Among the following pairs of ions the lower oxidation state in aqueous solution is more stable than the other in

A

`Ti^(o+),Ti^(3+)`

B

`Cu^(o+),Cu^(2+)`

C

`Cr^(2+),Cr^(3+)`

D

`V^(2+),VO^(2+)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which ion among the given pairs has a more stable lower oxidation state in aqueous solution, we can analyze the stability of the oxidation states of the transition metals based on their electronic configurations and the inert pair effect. ### Step-by-Step Solution: 1. **Identify the Ions and Their Oxidation States**: - Let's consider the pairs of ions provided in the question. For example, Titanium (Ti) in +1 and +3 oxidation states, Copper (Cu) in +1 and +2 oxidation states, and Vanadium (V) in +2 and +4 oxidation states. 2. **Determine the Electronic Configuration**: - For Titanium (Ti): - Ti: [Ar] 3d² 4s² - Ti⁺: [Ar] 3d² 4s¹ (for +1 state) - Ti³⁺: [Ar] 3d¹ (for +3 state) - For Copper (Cu): - Cu: [Ar] 3d¹⁰ 4s¹ - Cu⁺: [Ar] 3d¹⁰ (for +1 state) - Cu²⁺: [Ar] 3d⁹ (for +2 state) - For Vanadium (V): - V: [Ar] 3d³ 4s² - V²⁺: [Ar] 3d³ (for +2 state) - V⁴⁺: [Ar] 3d² (for +4 state) 3. **Analyze Stability Using the Inert Pair Effect**: - The inert pair effect suggests that the lower oxidation states of heavier p-block elements are more stable due to the reluctance of the s-electrons to participate in bonding. - For Titanium, Ti³⁺ is more stable than Ti⁺ due to the inert pair effect. - For Copper, Cu²⁺ is more stable than Cu⁺, as the +2 oxidation state is favored in aqueous solution. - For Vanadium, V⁴⁺ is more stable than V²⁺, as the +4 oxidation state is more favorable. 4. **Conclusion**: - Based on the analysis, the lower oxidation state is more stable for Titanium (Ti⁺ is less stable than Ti³⁺), while for Copper and Vanadium, the higher oxidation states are more stable. ### Final Answer: The lower oxidation state is more stable for Titanium (Ti⁺ is more stable than Ti³⁺).

To determine which ion among the given pairs has a more stable lower oxidation state in aqueous solution, we can analyze the stability of the oxidation states of the transition metals based on their electronic configurations and the inert pair effect. ### Step-by-Step Solution: 1. **Identify the Ions and Their Oxidation States**: - Let's consider the pairs of ions provided in the question. For example, Titanium (Ti) in +1 and +3 oxidation states, Copper (Cu) in +1 and +2 oxidation states, and Vanadium (V) in +2 and +4 oxidation states. 2. **Determine the Electronic Configuration**: ...
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Knowledge Check

  • In which of the following pairs of ions, the higher oxidation state in aqueous solution is more stable than the other? I. TI^(+), TI^(3+) II. Cu^(+),Cu^(2+) III. Cr^(2+),Cr^(3+) IV. V^(2+),VO^(2+)(V^(4+))

    A
    I, II
    B
    II, III
    C
    II, IV
    D
    II, III, IV
  • Among the following which one is more stable ?

    A
    `PH_(3)`
    B
    `NH_(3)`
    C
    `AsH_(3)`
    D
    `SbH_(3)`
  • In which of the following pairs are both the ions coloured in aqueous solution-

    A
    ` Ni^(2+), Ti^(3+) `
    B
    `Sc^(3+),Ti^(3+)`
    C
    `Sc^(3+), Co^(2+)`
    D
    `Ni^(2+), Cu^(+)`
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