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Calculate the number of unpaired electrons in gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution? - CBSE (Science) Class 12 - Chemistry

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Question

Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution?

Solution

  Gaseous ions Number of unpaired electrons
1 `Mn^(3+), [Ar]3d^4` 4
2 `Cr^(3+), [Ar]3d^3` 3
3 `V^(3+), [Ar]3d^2` 2
4 `Ti^(3+), [Ar]3d^1` 1

Cr3+ is the most stable in aqueous solutions owing to a `t_(2g)^3` configuration.

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APPEARS IN

 NCERT Solution for Chemistry Textbook for Class 12 (2018 to Current)
Chapter 8: The d-block and f-block Elements
Q: 24 | Page no. 235

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Solution Calculate the number of unpaired electrons in gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution? Concept: General Properties of the Transition Elements (D-block).
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