The zero magnetic moment of octahedral K 2 NiF 6 is due to-A- low spin d 6 Ni IV complexB- low spin d 8 Ni II complexC- high spin d 8 Ni II complexD- high spin d 6 Ni IV complex

The correct option is A low spin d^6 Ni(IV) complex[NiF_6]^2 ⇒ x 6= 2 x = + 4 In +4 oxidation state, fluoride also behaves as a strong field ligand ⇒ low spin ...

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Standard XII

Physics

Exploiting Ampere's Law - Long Wire

The zero magn...

Question

The zero magnetic moment of octahedral $K_{2} N i F_{6}$ is due to:

low spin

d^{6}

Ni(IV) complex

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low spin

d^{8}

Ni(II) complex

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high spin

d^{8}

Ni(II) complex

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high spin

d^{6}

Ni(IV) complex

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Solution

The correct option is A low spin $d^{6}$ Ni(IV) complex
$[N i F_{6}]^{2 -} \Rightarrow x - 6 = - 2$
x = + 4
In +4 oxidation state, fluoride also behaves as a strong field ligand
$\Rightarrow$ low spin $d^{6}$ Ni(IV) complex
$\Rightarrow$ unpaired electrons = 0 $\Rightarrow$ zero magnetic moment

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Similar questions

The octahedral complex ions

[F e C l_{6}]^{3 -}

and

[F e (C N)_{6}]^{3 -}

are both paramagnetic but the former is high spin and the latter is low spin. In the low spin complex,

[F e (C N)_{6}]^{3 -}

has the d-configuration as :

Q. For which of the following

d^{n}

configuration of octahedral complex(ex), cannot exist in both high spin and low spin forms?

[F e (C N)_{6}]^{- 3}

is low spin complex but

[F e (H_{2} O)_{6}]^{+ 3}

is high spin complex. Explain.

Q. The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two.

The metal ion is :

Q. Carbon donor ligands are strong ligands and usually forms low spin complexes. Among the complexes given below, select the high spin complex.

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The zero magnetic moment of octahedral K2NiF6 is due to:

The correct option is A low spin d6 Ni(IV) complex[NiF6]2−⇒x−6=−2 x = + 4 In +4 oxidation state, fluoride also behaves as a strong field ligand ⇒ low spin d6 Ni(IV) complex ⇒ unpaired electrons = 0 ⇒ zero magnetic moment

The zero magnetic moment of octahedral $K_{2} N i F_{6}$ is due to:

The correct option is A low spin $d^{6}$ Ni(IV) complex
$[N i F_{6}]^{2 -} \Rightarrow x - 6 = - 2$
x = + 4
In +4 oxidation state, fluoride also behaves as a strong field ligand
$\Rightarrow$ low spin $d^{6}$ Ni(IV) complex
$\Rightarrow$ unpaired electrons = 0 $\Rightarrow$ zero magnetic moment