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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

low spin

d^{8}

Ni(II) complex

No worries! We‘ve got your back. Try BYJU‘S free classes today!

high spin

d^{8}

Ni(II) complex

No worries! We‘ve got your back. Try BYJU‘S free classes today!

high spin

d^{6}

Ni(IV) complex

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

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.

Join BYJU'S Learning Program

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