9.6 [NiCl₄ ] ^2– is paramagnetic while [Ni(CO)₄ ] is diamagnetic though both are tetrahedral. Why?

In [Ni (Cl)₄]^2- ion, Cl^- is a weak field ligand so it will not pair the unpaired electrons of Ni⁺² ion. Electronic configuration of Ni is: [Ar]3d⁸4s² where [Ar] = 1s²2s²2p⁶3s²3p⁶

Electronic configuration of Ni⁺² = [Ar]3d⁸ Outer electronic configuration of Ni⁺² = 3d⁸ Overall charge balance:

X + 4 (-1) = -2 X = + 2.

Therefore it undergoes sp³ hybridization. So it will have tetrahedral geometry.

Since there are 2 unpaired electrons in the d orbital so it is a paramagnetic compound. In [Ni (Co)₄]:

Overall charge is neutral and oxidation state of Ni can be calculated as:

X + 4 (0) = 0

x = 0

Ni is in zero oxidation state.

Co is a strong field ligand

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In [Ni (Cl)₄]^2- ion, Cl^- is a weak field ligand so it will not pair the unpaired electrons of Ni⁺² ion. Electronic configuration of Ni is: [Ar]3d⁸4s² where [Ar] = 1s²2s²2p⁶3s²3p⁶

Electronic configuration of Ni⁺² = [Ar]3d⁸ Outer electronic configuration of Ni⁺² = 3d⁸ Overall charge balance:

X + 4 (-1) = -2 X = + 2.

Therefore it undergoes sp³ hybridization. So it will have tetrahedral geometry.

Since there are 2 unpaired electrons in the d orbital so it is a paramagnetic compound. In [Ni (Co)₄]:

Overall charge is neutral and oxidation state of Ni can be calculated as:

X + 4 (0) = 0

x = 0

Ni is in zero oxidation state.

Co is a strong field ligand it causes pairing of the 4 unpaired electrons in d orbital. Also, it causes the 4s electrons to shift to the 3d orbital, thereby undergoes sp³ hybridisation forming tetrahedral geometry. Since it has no unpaired electron, therefore it is diamagnetic compound.

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<p>In [Ni (Cl)₄]^2- ion, Cl^- is a weak field ligand so it will not pair the unpaired electrons of Ni⁺² ion. Electronic configuration of Ni is: [Ar]3d⁸4s² where [Ar] = 1s²2s²2p⁶3s²3p⁶</p><p>Electronic configuration of Ni⁺² = [Ar]3d⁸ Outer electronic configuration of Ni⁺² = 3d⁸ Overall charge balance:</p><p>X + 4 (-1) = -2 X = + 2.</p><p>Therefore it undergoes sp³ hybridization. So it will have tetrahedral geometry.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1686713107phpKKyKKw_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1686713107phpKKyKKw.jpeg" alt="" width="352" height="177"></picture></div></div><p>Since there are 2 unpaired electrons in the d orbital so it is a paramagnetic compound. In [Ni (Co)₄]:</p><p>Overall charge is neutral and oxidation state of Ni can be calculated as:</p><p>X + 4 (0) = 0</p><p>x = 0</p><p>Ni is in zero oxidation state.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1686713198phpBZ0Obo_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1686713198phpBZ0Obo.jpeg" alt="" width="307" height="130"></picture></div></div><p>Co is a strong field ligand it causes pairing of the 4 unpaired electrons in d orbital. Also, it causes the 4s electrons to shift to the 3d orbital, thereby undergoes sp³ hybridisation forming tetrahedral geometry. Since it has no unpaired electron, therefore it is diamagnetic compound.</p>

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