Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic. - Chemistry

Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic.

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Solution 1

Ni is in the +2 oxidation state i.e., in d8 configuration.

d8 Configuration 

There are 4 CN ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.

It now undergoes dsp2 hybridization. Since all electrons are paired, it is diamagnetic.

In case of [NiCl4]2−, Cl ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired 3delectrons. Therefore, it undergoes sp3 hybridization.

Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.

 

Solution 2

`[Ni(CN)_4]^(2-) : Ni(28): 1s^22s^2 2p^6 3s^2 3p^6 4s^2 3d^8`

or last shell EC is in `[Ni(CN)_4]^(2-)` , Ni is in +2 state

`:.Ni^2:3d^8 4s^0`

CN being strong field ligand pairs up the `e^(-1)`s thus

`[Ni(CN)_4]^(2-)`: 

No unpaired e-, hence `[Ni(CN)_4]^(2-)` is diamagnetic

`[NiCl_4]^(2-)`: Ni is in +2 state thus

`Ni^(2+): 3d^8 4s^2`

Cl- being a weak field ligand cannot pair up the `e^(-1)s`

`:.[NiCl_4]^(2-)` 

Due to unpaired e-1s, `[NiCl4]^(2-)` is paramagnetic

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 12 Chemistry Textbook
Chapter 9 Coordinate Compounds
Q 5 | Page 254
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