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प्रश्न
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 an aqueous solution?
उत्तर
| Gaseous ions | Electronic configurations | Number of unpaired electrons |
| (1) Mn3+ | 3d44s0 | 4 |
| (2) Cr3+ | 3d34s0 | 3 |
| (3) V3+ | 3d24s0 | 2 |
| (4) Ti3+ | 3d14s0 | 1 |
Of these, Cr3+ is most stable in aqueous solution because it has a half-filled t2g level.
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संबंधित प्रश्न
Complete the following chemical equations:
`(i) Cr_2O_7^(2-)+6Fe^(2+)+14H^+ ->`
`(ii) 2CrO_4^(2-)+2H^+ ->`
`(iii) 2MnO_4^-+5C_2O_4^(2-)+16H^+ ->`
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Ti4+, V2+, Mn3+, Cr3+
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
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1) Which ion is most stable in an aqueous solution and why?
2) Which ion is a strong oxidising agent and why?
3) Which ion is colourless and why?
Write balanced chemical equations for the conversion of `CrO_4^(2-)` to `Cr_2O_7^(2-)` in acidic medium and `Cr_2O_7^(2-)` to `CrO_4^(2-)`
in basic medium.
Account for the following :
Ti4+ is colourless whereas V4+ is coloured in an aqueous solutions.
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- Write the IUPAC name of the complex compound.
- Draw the geometrical isomers of the complex compound.
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