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Form the following species, Ti^(3+),V^(...

Form the following species,
`Ti^(3+),V^(3+),Sc^(3+),Fe^(3+),Co^(3+), Cu^+` and`Mn^(3+)`, Which two will not be coloured in aqueous solution ?

A

`Mn^(3+)` and `Co^(3+)`

B

`V^3 ` and ` Co^(3+)`

C

`TI^(3+) `and `Sc^(3+)`

D

`Sc^(3+)`and `Cu^+`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which of the given species will not be colored in aqueous solution, we need to analyze the electronic configurations and the presence of unpaired electrons in each ion. The color in transition metal ions arises due to the presence of unpaired electrons in their d-orbitals. ### Step-by-Step Solution: 1. **Identify the Electronic Configurations**: - **Ti^(3+)**: Titanium has an atomic number of 22. The electronic configuration of neutral Ti is [Ar] 4s² 3d². For Ti^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in [Ar] 3d¹. - **V^(3+)**: Vanadium has an atomic number of 23. The electronic configuration of neutral V is [Ar] 4s² 3d³. For V^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in [Ar] 3d². - **Sc^(3+)**: Scandium has an atomic number of 21. The electronic configuration of neutral Sc is [Ar] 4s² 3d¹. For Sc^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in [Ar] (empty d-orbitals). - **Fe^(3+)**: Iron has an atomic number of 26. The electronic configuration of neutral Fe is [Ar] 4s² 3d⁶. For Fe^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in [Ar] 3d⁵. - **Co^(3+)**: Cobalt has an atomic number of 27. The electronic configuration of neutral Co is [Ar] 4s² 3d⁷. For Co^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in [Ar] 3d⁶. - **Cu^(+)**: Copper has an atomic number of 29. The electronic configuration of neutral Cu is [Ar] 4s² 3d¹⁰. For Cu^(+), we remove 1 electron (from 4s), resulting in [Ar] 3d¹⁰. - **Mn^(3+)**: Manganese has an atomic number of 25. The electronic configuration of neutral Mn is [Ar] 4s² 3d⁵. For Mn^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in [Ar] 3d⁴. 2. **Count Unpaired Electrons**: - **Ti^(3+)**: 1 unpaired electron (3d¹). - **V^(3+)**: 2 unpaired electrons (3d²). - **Sc^(3+)**: 0 unpaired electrons (empty d-orbitals). - **Fe^(3+)**: 5 unpaired electrons (3d⁵). - **Co^(3+)**: 4 unpaired electrons (3d⁶). - **Cu^(+)**: 0 unpaired electrons (3d¹⁰). - **Mn^(3+)**: 4 unpaired electrons (3d⁴). 3. **Determine Color**: - Ions with unpaired electrons typically exhibit color due to d-d transitions. Conversely, ions with all paired electrons do not exhibit color. - From our analysis: - **Sc^(3+)** has 0 unpaired electrons (colorless). - **Cu^(+)** has 0 unpaired electrons (colorless). - The other ions (Ti^(3+), V^(3+), Fe^(3+), Co^(3+), Mn^(3+)) all have unpaired electrons and will be colored. ### Conclusion: The two species that will not be colored in aqueous solution are **Sc^(3+)** and **Cu^(+)**.
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Knowledge Check

  • Out of the following ions Ti^(3+), V^(3+), Cu^(+), Sc^(3+), Mn^(3+) and Co^(2+) the colourless ions will be

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

  • Out of the following ions Ti^(3+), V^(3+), Cu^(+), Sc^(3+), Mn^(2+), Fe^(3+) and Co^(2+) the colourless ions will be

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

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