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Predict the number of unpaired electrons...

Predict the number of unpaired electrons in the square planar `[Pt(CN)_4]^(2-)` ion.

Text Solution

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`._(78)Pt` lies in Group 10 with the configuration `5d^(9)6s^(1)`. Hence, `Pt^(2+)` has the configuration `d^(8)`

For square planar shape, the hybridisation is `dsp^(2)`. Hence, the unpaired electrons in 5d pair up to make one d-orbital empty for `dsp^(2)` hybridization. Thus, there is no unpaired electron.
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Knowledge Check

  • What is/are number(s) of unpaired electrons in the square planar [Pt(CN)_(4)]^(2-) ion?

    A
    Zero
    B
    1
    C
    4
    D
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  • What is (are) number (s) of unpaired electron(s) in the square planar [Pt(CN)_4]^(2-) ion ?

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    0
    B
    1
    C
    4
    D
    6

  • The number of unpaired electrons in Zn^(2+) is

    A
    2
    B
    3
    C
    4
    D
    0

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