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?
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 and Explanation
| Gaseous ions | Number of unpaired electrons | |
| (i) | Mn3+, [Ar]3d4 | 4 |
| (ii) | Cr3+, [Ar]3d3 | 3 |
| (iii) | V3+, [Ar]3d2 | 2 |
| (iv) | Ti3+, [Ar]3d1 | 1 |
Cr3+ is the most stable in aqueous solutions owing to a \(t^3_{2g}\) configuration
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Concepts Used:
D and F Block Elements
The d-block elements are placed in groups 3-12 and F-block elements with 4f and 5f orbital filled progressively. The general electronic configuration of d block elements and f- block elements are (n-1) d 1-10 ns 1-2 and (n-2) f 1-14 (n-1) d1 ns2 respectively. They are commonly known as transition elements because they exhibit multiple oxidation states because of the d-d transition which is possible by the availability of vacant d orbitals in these elements.
They have variable Oxidation States as well as are good catalysts because they provide a large surface area for the absorption of reaction. They show variable oxidation states to form intermediate with reactants easily. They are mostly lanthanoids and show lanthanoid contraction. Since differentiating electrons enter in an anti-penultimate f subshell. Therefore, these elements are also called inner transition elements.
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