9.5 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.

According to the valence band theory, the central metal atom or ion under the influence of ligands can use its (n-1)d, ns, np (inner orbital complex) or ns, np, and (outer orbital complex)orbitals for hybridisation to form equivalent set of orbitals of definite geometry.

In [Ni (CN)₄]^2-, oxidation state of Ni can be calculated as :

Using overall charge balance as the whole ion has overall -2 charge:

x + 4 (-1) = -2 (? CN^- has -1 negative charge) x = + 2

Ni is in + 2 oxidation state.

Electronic configuration of Ni is: [Ar]3d⁸4s² Where, [Ar] = 1s²2s²2p⁶3s²3p⁶

Electronic configuration of Ni⁺² = [Ar]3d⁸ Outer electronic configuration of Ni⁺² = 3

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According to the valence band theory, the central metal atom or ion under the influence of ligands can use its (n-1)d, ns, np (inner orbital complex) or ns, np, and (outer orbital complex)orbitals for hybridisation to form equivalent set of orbitals of definite geometry.

In [Ni (CN)₄]^2-, oxidation state of Ni can be calculated as :

Using overall charge balance as the whole ion has overall -2 charge:

x + 4 (-1) = -2 (? CN^- has -1 negative charge) x = + 2

Ni is in + 2 oxidation state.

Electronic configuration of Ni is: [Ar]3d⁸4s² Where, [Ar] = 1s²2s²2p⁶3s²3p⁶

Electronic configuration of Ni⁺² = [Ar]3d⁸ Outer electronic configuration of Ni⁺² = 3d⁸

Since there are 4 CN ions so they can either form tetrahedral or square planar geometry. And CN^- is a strong field ligand (according to experimental data of spectro-chemical series) it causes pairing of the 2 unpaired electrons.

It undergoes dsp² (one d orbital, one s and two p orbitals used by the ligands) hybridization and forms square planar structure.

Since all the electrons are paired so it is diamagnetic.

Paramagnetic compounds- those compounds which have one or more no. of unpaired electrons in their atomic orbitals.

Diamagnetic compounds-those compounds in which all the electrons in their atomic orbitals are paired.

Co-ordination number	Type of hybridisation	Distribution of hybrid orbitals in space
4	sp3	Tetrahedral
4	dsp3	Square planar
5	sp3d	Trigonalbipyramidal

6	sp3d2	Octahedral
6	d2sp3	octahedral

In case of [Ni (Cl)₄]^2- ion, Cl^- is a weak field ligand so it will not pair the unpaired electrons of Ni⁺² ion. Therefore it undergoes sp³ hybridization.

Overall charge balance: X + 4 (-1) = -2

X = + 2.

Since there are 2 unpaired electrons in the d orbital so it is a paramagnetic compound.

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<p>According to the valence band theory, the central metal atom or ion under the influence of ligands can use its (n-1)d, ns, np (inner orbital complex) or ns, np, and (outer orbital complex)orbitals for hybridisation to form equivalent set of orbitals of definite geometry.</p><p>In [Ni (CN)₄]^2-, oxidation state of Ni can be calculated as :</p><p>Using overall charge balance as the whole ion has overall -2 charge:</p><p>x + 4 (-1) = -2 (? CN^- has -1 negative charge) x = + 2</p><p>Ni is in + 2 oxidation state.</p><p>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</p><p>Since there are 4 CN ions so they can either form tetrahedral or square planar geometry. And CN^- is a strong field ligand (according to experimental data of spectro-chemical series) it causes pairing of the 2 unpaired electrons.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1686712807phpP3IMIR_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1686712807phpP3IMIR.jpeg" alt="" width="409" height="230"></picture></div></div><p>It undergoes dsp² (one d orbital, one s and two p orbitals used by the ligands) hybridization and forms square planar structure.</p><p>Since all the electrons are paired so it is diamagnetic.</p><p>Paramagnetic compounds- those compounds which have one or more no. of unpaired electrons in their atomic orbitals.</p><p>Diamagnetic compounds-those compounds in which all the electrons in their atomic orbitals are paired.</p><table><tbody><tr><td><p><em>Co-ordination number</em></p></td><td><p><em>Type of hybridisation</em></p></td><td><p><em>Distribution of hybrid orbitals in space</em></p></td></tr><tr><td><p>4</p></td><td><p>sp³</p></td><td><p>Tetrahedral</p></td></tr><tr><td><p>4</p></td><td><p>dsp³</p></td><td><p>Square planar</p></td></tr><tr><td><p>5</p></td><td><p>sp³d</p></td><td><p>Trigonalbipyramidal</p></td></tr></tbody></table><table><tbody><tr><td width="202"><p>6</p></td><td width="202"><p>sp³d²</p></td><td width="202"><p>Octahedral</p></td></tr><tr><td width="202"><p>6</p></td><td width="202"><p>d²sp³</p></td><td width="202"><p>octahedral</p></td></tr></tbody></table><p>In case of [Ni (Cl)₄]^2- ion, Cl^- is a weak field ligand so it will not pair the unpaired electrons of Ni⁺² ion. Therefore it undergoes sp³ hybridization.</p><p>Overall charge balance: X + 4 (-1) = -2</p><p>X = + 2.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1686712948phpLDjZfP_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1686712948phpLDjZfP.jpeg" alt="" width="332" height="137"></picture></div></div><p>Since there are 2 unpaired electrons in the d orbital so it is a paramagnetic compound.</p>

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