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The zero magnetic moment of octahedral K2NiF6 is due to
  • a)
    low spin d6 Ni(IV) complex
  • b)
    Low spin d8 Ni(II) complex
  • c)
    High spin d8 Ni(II) complex
  • d)
    High spin d6 Ni(IV) complex
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The zero magnetic moment of octahedral K2NiF6 is due toa)low spin d6 N...
Ni(IV) complex

[NiF6]2−
x−6=−2
x = + 4
In +4 oxidation state, fluoride also behaves as a strong field ligand. So low spin is possible.


Ni(IV) complex with d6 configuration


unpaired electrons = 0

Zero magnetic moment.
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The zero magnetic moment of octahedral K2NiF6 is due toa)low spin d6 N...
Introduction:
The zero magnetic moment of octahedral K2NiF6 is due to the low spin d6 Ni(IV) complex. This means that the nickel ion in the compound has a d6 electron configuration and is in the +4 oxidation state. To understand why this leads to a zero magnetic moment, we need to consider the factors that influence the magnetic behavior of transition metal complexes.

Magnetic behavior of transition metal complexes:
The magnetic behavior of transition metal complexes is determined by the number of unpaired electrons present in the d orbitals of the metal ion. These unpaired electrons give rise to magnetic moments and can interact with an external magnetic field.

Spin and orbital contributions:
In a transition metal complex, the magnetic moment arises from two sources: spin and orbital contributions. The spin contribution arises from the alignment of the spins of the unpaired electrons, while the orbital contribution arises from the alignment of the orbital angular momentum of the electrons.

High spin and low spin complexes:
The number of unpaired electrons in a transition metal complex is determined by the ligand field splitting of the d orbitals. In octahedral complexes, the d orbitals are split into two sets of three: a lower energy set (eg) and a higher energy set (t2g).

In a high spin complex, the electrons populate the t2g orbitals before pairing up in the eg orbitals. This leads to a greater number of unpaired electrons and a larger magnetic moment. On the other hand, in a low spin complex, the electrons fill up the eg orbitals before populating the t2g orbitals. This results in a smaller number of unpaired electrons and a smaller magnetic moment.

Explanation of the answer:
In the case of octahedral K2NiF6, the nickel ion is in the +4 oxidation state and has a d6 electron configuration. This means that there are six electrons in the d orbitals of the nickel ion.

In an octahedral ligand field, the d orbitals are split into t2g and eg sets. In the case of a low spin d6 complex, the electrons fill up the eg orbitals before populating the t2g orbitals. This results in the pairing of four electrons in the t2g orbitals and two unpaired electrons in the eg orbitals.

The presence of two unpaired electrons gives rise to a magnetic moment. However, in the case of K2NiF6, the fluoride ligands are strong field ligands that cause a large ligand field splitting. This splitting is so large that the energy required to pair up the two unpaired electrons in the eg orbitals is less than the energy gained by filling up the t2g orbitals.

As a result, the two unpaired electrons in the eg orbitals pair up, leading to a d6 low spin configuration. In this configuration, there are no unpaired electrons and hence no magnetic moment. Therefore, the zero magnetic moment of octahedral K2NiF6 is due to the low spin d6 Ni(IV) complex.
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The zero magnetic moment of octahedral K2NiF6 is due toa)low spin d6 Ni(IV) complexb)Low spin d8 Ni(II) complexc)High spin d8 Ni(II) complexd)High spin d6 Ni(IV) complexCorrect answer is option 'A'. Can you explain this answer?
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