
The complex showing a spin-only magnetic moment of 2.83BM is:
(A) $Ni{{(CO)}_{4}}$
(B) ${{[Ni{{(Cl)}_{4}}]}^{2-}}$
(C) $Ni{{(PP{{h}_{3}})}_{4}}$
(D) ${{[Ni{{(CN)}_{4}}]}^{2-}}$
Answer
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Hint: Synthesis of early understandings of chemical bond formation by valence bond theory. To explain the nature of bonding in coordination compounds is the Valence Bond theory (VB), which theory explains the structure and magnetic properties of coordination compounds. The structure of and bond linkages of coordination compounds are easily explained with the help of the Valence Bond theory.
Complete step by step answer:
Postulates of valence bond theory of coordination compounds:
(i) The central metal ion of a suitable number of atomic orbitals are hybridized to provide empty hybrid orbitals.
(ii) The ligands which are donated electron pairs to these empty hybrid orbitals and are directed towards the positions of ligands according to the geometry of the complex.
(iii) (n-1) d orbitals which are known as inner d-orbitals used in hybridization, then the complex is called inner orbital or spin complex.
(iv) When the coordination compounds with one or more unpaired electrons d-orbital, is said to be paramagnetic.
Spin only magnetic moment of complex compounds will be calculated by the expression,
$\mu =\sqrt{n(n+2)}$
Where n= number of unpaired electrons and the value of spin-only magnetic moment units in BM
Hence, the complex showing a spin-only magnetic moment of 2.83BM is ${{[Ni{{(Cl)}_{4}}]}^{2-}}$
So, the correct answer is “Option B”.
Note: According to valence band theory, there are two types of ligands- strong field ligands and weak field ligands. A ligand that is capable of forcing the electrons of a central metal ion or atom or complex to pair up is a strong field ligand and weak field ligand is not capable of making electrons of metal ions or atoms to pair up.
Complete step by step answer:
Postulates of valence bond theory of coordination compounds:
(i) The central metal ion of a suitable number of atomic orbitals are hybridized to provide empty hybrid orbitals.
(ii) The ligands which are donated electron pairs to these empty hybrid orbitals and are directed towards the positions of ligands according to the geometry of the complex.
(iii) (n-1) d orbitals which are known as inner d-orbitals used in hybridization, then the complex is called inner orbital or spin complex.
(iv) When the coordination compounds with one or more unpaired electrons d-orbital, is said to be paramagnetic.
Spin only magnetic moment of complex compounds will be calculated by the expression,
$\mu =\sqrt{n(n+2)}$
Where n= number of unpaired electrons and the value of spin-only magnetic moment units in BM
| Ion/complex | Configuration | The oxidation state of metal | No of unpaired electrons (n) | Spin-only magnetic moment$\mu =\sqrt{n(n+2)}$ |
| $Ni{{(CO)}_{4}}$ | $4{{s}^{2}}3{{d}^{8}}$ (strong field ligand) | 0 | 0 | 0 BM |
| ${{[Ni{{(Cl)}_{4}}]}^{2-}}$ | $4{{s}^{2}}3{{d}^{8}}$ (weak field ligand) | +2 | 2 | $\mu =\sqrt{n(n+2)}=\sqrt{2(2+2)}=2.83BM$ |
| $Ni{{(PP{{h}_{3}})}_{4}}$ | $4{{s}^{2}}3{{d}^{8}}$ (strong field ligand) | 0 | 0 | 0 BM |
| ${{[Ni{{(CN)}_{4}}]}^{2-}}$ | $4{{s}^{2}}3{{d}^{8}}$ (strong field ligand) | 0 | 0 | 0 BM |
Hence, the complex showing a spin-only magnetic moment of 2.83BM is ${{[Ni{{(Cl)}_{4}}]}^{2-}}$
So, the correct answer is “Option B”.
Note: According to valence band theory, there are two types of ligands- strong field ligands and weak field ligands. A ligand that is capable of forcing the electrons of a central metal ion or atom or complex to pair up is a strong field ligand and weak field ligand is not capable of making electrons of metal ions or atoms to pair up.
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