JEE Main 2014 — Chemistry Inorganic Chemistry

2014mcqeasy

The correct statement about the magnetic properties of [Fe(CN) $_{6}$ ] $^{3-}$ and [FeF $_{6}$ ] $^{3-}$ is : (Z = 26).

Official previous-year question

Held on 9 Apr 2014 · Verified 6 Jul 2026.

Options

A
Both are paramagnetic.
B
Both are diamagnetic.
C
[Fe(CN) $_{6}$ ] $^{3-  }$ is paramagnetic,

[FeF $_{6}$ ] $^{3- }$ is diamagnetic.
D
[Fe(CN) $_{6}$ ] $^{3-  }$ is diamagnetic,

[FeF $_{6}$ ] $^{3- }$ is paramagnetic.

Did you get this right?

Solution

In both complexes Fe is in +3 oxidation state.

F $^{-}$ ion is weak field ligand so no pairing of electrons in inner 3d-orbitals & outer orbital complex. sp $^{3}$ d $^{2}$ hybridisation is formed.

JEE Main 2014 Chemistry, Coordination Compounds — solution figure 1

have 5 unpaired electrons, so paramagnetic

${\mu }_{{[{\mathrm{FeF}}_{6}]}^{-3}}=\sqrt{5(5+2)}=\sqrt{35}\text{B}.\text{M}$

JEE Main 2014 Chemistry, Coordination Compounds — solution figure 2

One unpaired Electron, CN $^{-}$ ion strong ligand field. So will pair the d-orbital electrons & form inner orbital hybridization d $^{2}$ sp $^{3}$ .

${\text{μ}}_{{ [ \text{Fe} { ( \text{CN} ) }_{6} ] }^{ 3 - }} = \sqrt{ 1 ( 1 + 2 )}$ (1 unpaired e $^{-}$ so para magnetic)

$= \sqrt{ 3 } B.M$ .

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JEE Main Inorganic Chemistry in other years

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JEE Main 2014 — Chemistry Inorganic Chemistry

2014mcqeasy

The correct statement about the magnetic properties of [Fe(CN) $_{6}$ ] $^{3-}$ and [FeF $_{6}$ ] $^{3-}$ is : (Z = 26).

Official previous-year question

Held on 9 Apr 2014 · Verified 6 Jul 2026.

Options

A
Both are paramagnetic.
B
Both are diamagnetic.
C
[Fe(CN) $_{6}$ ] $^{3-  }$ is paramagnetic,

[FeF $_{6}$ ] $^{3- }$ is diamagnetic.
D
[Fe(CN) $_{6}$ ] $^{3-  }$ is diamagnetic,

[FeF $_{6}$ ] $^{3- }$ is paramagnetic.

Did you get this right?

Solution

In both complexes Fe is in +3 oxidation state.

F $^{-}$ ion is weak field ligand so no pairing of electrons in inner 3d-orbitals & outer orbital complex. sp $^{3}$ d $^{2}$ hybridisation is formed.

JEE Main 2014 Chemistry, Coordination Compounds — solution figure 1

have 5 unpaired electrons, so paramagnetic

${\mu }_{{[{\mathrm{FeF}}_{6}]}^{-3}}=\sqrt{5(5+2)}=\sqrt{35}\text{B}.\text{M}$

JEE Main 2014 Chemistry, Coordination Compounds — solution figure 2

One unpaired Electron, CN $^{-}$ ion strong ligand field. So will pair the d-orbital electrons & form inner orbital hybridization d $^{2}$ sp $^{3}$ .

${\text{μ}}_{{ [ \text{Fe} { ( \text{CN} ) }_{6} ] }^{ 3 - }} = \sqrt{ 1 ( 1 + 2 )}$ (1 unpaired e $^{-}$ so para magnetic)

$= \sqrt{ 3 } B.M$ .

Your note

JEE Main Inorganic Chemistry in other years

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Work through every JEE Main Inorganic Chemistry PYQ, year by year.

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JEE Main 2014 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main Inorganic Chemistry in other years

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JEE Main 2014 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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