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(a) What type of isomerism is shown by t...

(a) What type of isomerism is shown by the complex `[Co(NH_(3))_(5)(SCN)]^(2+)`?
(b) Why is `[NiCl_(4)]^(2-)` paramagnetic while `[Ni(CN)_(4)]^(2-)` is diamagnetic ? (Atomic number of `Ni = 28`)
(c) Why are low spin tetrahedral complexes rarely observed?

Text Solution

Verified by Experts

(a) Linkage isomerism is shown by complex `[Co(NH_(3))_(3)(SCN)]^(2+)`
(b) `[NiCl_(4)]^(2-)`
As `Cl^(-)` is weak ligand, hence, pairing of `e^(-)` does not take place. Hence, paramagnetic due to the presence of 2 unpaired `e^(-)s`.
ltBrgt as `CN^(-)` is strong ligand due to that, pairing of `e^(-)`s take place. Hence, no unpaired electrons due to that. It is diamagnetic in nature.
(c) In tetrahedral complexes, the splitting is much smaller than that is case of octahedral complexes, i.e., `Delta_(t)=(4)/(9)Delta_(o)`. This energy is so small, that it is unable to force the `e^(-)`s to pair up. Hence, low spin complexes of tetrahedral are rarely observed and they have high spin configuration.
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Knowledge Check

  • The complex ion [Ni(CN)_4]^(2-) is :

    A
    Square planar and diamagnetic
    B
    Tetrahedral and paramagnetic
    C
    Square planar and paramagnetic
    D
    Tetrahedral and diamagnetic

  • Give reason for the statement [Ni(CN)_(4)]^(2-) is diamagnetic while [NiCl_(4)]^(2-) is paramagnetic in nature .

    A
    In `[NiCl_(4)]^(2-)` no unpaired electrons are present while in `[Ni(CN)_(4)]^(2-)` two unpaired electrons are present
    B
    In `[Ni(CN)_(4)]^(2-)` , no unpaired electrons are present while in `[NiCl_(4)]^(2-)` two unpaired electrons are present .
    C
    `[NiCl_(4)]^(2-)` shows `dsp^(2)` hybridisation hence it is paramagnetic .
    D
    `[Ni(CN)_(4)]^(2-)` shows `sp^(3)` hybridisation hence it is diamagnetic .

  • The oxidation of central atom in the complex [ Co(NH_(3))_(4) CINO_(2)] is

    A
    ` +2`
    B
    `+3`
    C
    `+1`
    D
    zero

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