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Standard XII

Chemistry

Magnetic Moment Examples

[FeF6]3- is p...

Question

$[F e F_{6}]^{3 -}$ is paramagnetic due to the presence of unpaired electrons in the complex. The five electrons remain unpaired because:

fluorine is the most electronegative element

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F^{-}

is a weak field ligand hence does not cause pairing of electrons

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F^{-}

is a strong field ligand hence does not cause pairing of electrons

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pairing does not take place in iron complexes.

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Solution

The correct option is B $F^{-}$ is a weak field ligand hence does not cause pairing of electrons
$F^{-}$ is a weak field ligand hence it does it does not cause pairing of electrons. As $Δ_{o} <$ pairing energy, so electron occupies a higher energy orbital. This is because this requires less energy than occupying a lower energy orbital and pairing with another electron.

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Similar questions

Q. Assertion :Transition metal ion forming octahedral complexes undergo

s p^{3} d^{2}

d^{2} s p^{3}

hybridisation. Reason: Strong field ligands force the unpaired electrons of central metal ion to pair up causing

d^{2} s p^{3}

hybridisation whereas weak field ligands do not affect electronic configuration of the metal ion undergoes in

s p^{3} d^{2}

hybridisation.

Q. Based on VBT,

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

has

C l^{-}

as weak ligands.

F e

s p^{3} d^{2}

hybridized (outer orbital complex) having five unpaired electrons.

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

has

C N^{-}

as strong ligand.

F e

d^{2} s p^{3}

hybridized (inner orbital complex) having only one unpaired electron. Hence,

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

is more paramagnetic than

[F e (C N)_{8}]^{3 -}

.
If true enter 1, else enter 0.

Q. Assertion :

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

s p^{3} d^{2}

hybridised and paramagnetic complex ion. Reason:

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

has four unpaired electrons as

H_{2} O

is a weak field ligand.

Q. In the complex ion

M L_{6}^{n +}, M^{n +}

has five d-electrons and

L

is weak field ligand. According to crystal field theory the magnetic properties of the complex ion corresponds to how many unpaired electrons?

Q. Assertion :Nitrogen does not form

N C I_{5}

N F_{5}

but phosphorus can. Reason: Nitrogen (

1 s^{2} 2 s^{2} 2 p^{3}

) lacks 2d-orbitals in which electrons of 2s can be excited to make five orbitals unpaired. Only three electrons remain unpaired hence only

N X_{3}

is formed. In case of

P (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}), 3 s

electron can be excited into 3d vacant orbitals thus making five unpaired electrons, thus

P X_{5}

is possible.

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[FeF6]3− is paramagnetic due to the presence of unpaired electrons in the complex. The five electrons remain unpaired because:

$[F e F_{6}]^{3 -}$ is paramagnetic due to the presence of unpaired electrons in the complex. The five electrons remain unpaired because: