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An ion M^(2+), forms the complexes [M(H(...

An ion `M^(2+)`, forms the complexes `[M(H_(2)O)_(6)]^(2+),[M(en)_(3)]^(2+) and [MBr_(6)]^(4-)`, match the complex with the appropriate colour:

A

`[MBr_(6)]^(4-) blue, [M(H_(2)O)_(6)]^(2+) green, [M(en)_(3)]^(2+) red `

B

`[MBr_(6)]^(4-)green, [M(H_(2)O)_(6)]^(2+) blue , [M(en)_(3)]^(2+) red `

C

`[MBr_(6)]^(4-) green , [M(H_(2)O)_(6)]^(2+) red, [M(en)_(3)]^(2+) blue`

D

`[MBr_(6)]^(4-) red , [M(H_(2)O)_(6)]^(2+) green , [M(en)_(3)]^(2+) blue`

Text Solution

Verified by Experts

The correct Answer is:
A
Order of crystal field splitting `en gt H_(2)O gt Br^(-)` so, more stronger ligand will absorb lower wavelength light and reflect back higher wavelength light.
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Given that the energy of the photons of different colours decreases in the order of VIBGYOR (Violet gt Indigo gt Blue gt Green gt Yellow gt Red ) and that if complex absorbs a photon of low energy then it shows colour of high energy photon. If an ion , M^(2+) , forms the complexes [M(H_(2)O)_(6)]^(2+),[MBr_(6)]^(4-), and [M(en)_(3)]^(2+) . The colours of the complexes, though not necessarily in order, are green, red, and blue. Match the complexes with the appropriate colour. ( Do not use the table of complementry colours for this question )

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Knowledge Check

  • Using valence bond theory , the complex [Cr(H_(2)O)_(6)]^(3+) can be described as

    A
    `sp^(3) d^(2)` , outer orbital complex , paramagnetic
    B
    `dsp^(2)` , inner orbital complex , diamagnetic
    C
    `d^(2) sp^(3)` , inner orbital complex , paramagnetic
    D
    `d^(2) sp^(3)` , outer orbital complex , diamagnetic .