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Among the following pairs of ions, lower oxidation state in aqueous solution is more stable than the other in
  • a)
    Ti+, Ti3+
  • b)
    Cu+, Cu2+
  • c)
    Cr2+, Cr3+
  • d)
    V2+,VO2+
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Among the following pairs of ions, lower oxidation state in aqueous s...
Ti+ does not exist, Cu+ does not dissolve in water, Cr2+ is less stable than Cr3+ (both do not exist in aqueous solution) O.S. of V in VO2+ is +6.
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Among the following pairs of ions, lower oxidation state in aqueous s...
Explanation:

The stability of an oxidation state in aqueous solution is determined by several factors, including the electronic configuration and the relative strengths of the oxidizing and reducing agents in the solution. In this case, we are comparing the stability of different oxidation states for the same element.

Let's examine each pair of ions and determine which oxidation state is more stable in aqueous solution:

a) Ti and Ti3
- Titanium (Ti) can exist in several oxidation states, including +2 and +3.
- In aqueous solution, Ti3+ is more stable than Ti2+. This is because Ti3+ has a half-filled d orbital (3d1) which provides greater stability.
- Ti2+ has a completely filled d orbital (3d2) which is less stable than the half-filled d orbital of Ti3+.

b) Cu and Cu2
- Copper (Cu) can also exist in several oxidation states, including +1 and +2.
- In aqueous solution, Cu2+ is more stable than Cu+. This is because Cu2+ has a completely filled d orbital (3d10) which provides greater stability.
- Cu+ has a partially filled d orbital (3d9) which is less stable than the completely filled d orbital of Cu2+.

c) Cr2 and Cr3
- Chromium (Cr) can exist in several oxidation states, including +2 and +3.
- In aqueous solution, Cr3+ is more stable than Cr2+. This is because Cr3+ has a half-filled d orbital (3d3) which provides greater stability.
- Cr2+ has a completely filled d orbital (3d4) which is less stable than the half-filled d orbital of Cr3+.

d) V2 and VO2
- Vanadium (V) can exist in several oxidation states, including +2 and +5.
- In aqueous solution, V2+ is more stable than VO2+. This is because V2+ has a completely filled d orbital (3d3) which provides greater stability.
- VO2+ has a partially filled d orbital (3d2) and also has the presence of oxygen, which makes it less stable than V2+.

Therefore, the correct answer is option D, V2 is more stable than VO2 in aqueous solution.
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Among the following pairs of ions, lower oxidation state in aqueous solution is more stable than the other ina)Ti+, Ti3+b)Cu+, Cu2+c)Cr2+, Cr3+d)V2+,VO2+Correct answer is option 'D'. Can you explain this answer?
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