JEE Main 2025 Apr 7 Shift 2, Chemistry Q24: Colour, Magnetism And Stability Of Co-ordination Compounds

JEE Main2025Apr 7, Shift 2Chemistry

The number of paramagnetic metal complex species among [Co(NH $_{3}$ ) $_{6}$ ] $^{3 +}$ , [Co(C $_{2}$ O $_{4}$ ) $_{3}$ ] $^{3 -}$ , [MnCl $_{6}$ ] $^{3 -}$ , [Mn(CN) $_{6}$ ] $^{3 -}$ , [CoF $_{6}$ ] $^{3 -}$ , [Fe(CN) $_{6}$ ] $^{3 -}$ and [FeF $_{6}$ ] $^{3 -}$ with same number of unpaired electrons is ____.

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Solution

Working out the electron configuration for each species using crystal field theory:

[Co(NH $_{3}$ ) $_{6}$ ] $^{3 +}$ : Co $^{3 +}$ ( $d^{6}$ ) with strong field NH $_{3}$ gives $t_{2 g}^{6} e_{g}^{0}$ , 0 unpaired electrons (diamagnetic).

[Co(C $_{2}$ O $_{4}$ ) $_{3}$ ] $^{3 -}$ : Co $^{3 +}$ ( $d^{6}$ ) with the strong field chelating oxalate gives $t_{2 g}^{6} e_{g}^{0}$ , 0 unpaired electrons (diamagnetic).

[MnCl $_{6}$ ] $^{3 -}$ : Mn $^{3 +}$ ( $d^{4}$ ) with weak field Cl $^{-}$ gives high-spin $t_{2 g}^{3} e_{g}^{1}$ , 4 unpaired electrons.

[Mn(CN) $_{6}$ ] $^{3 -}$ : Mn $^{3 +}$ ( $d^{4}$ ) with strong field CN $^{-}$ gives low-spin $t_{2 g}^{4} e_{g}^{0}$ , 2 unpaired electrons.

[CoF $_{6}$ ] $^{3 -}$ : Co $^{3 +}$ ( $d^{6}$ ) with weak field F $^{-}$ gives high-spin $t_{2 g}^{4} e_{g}^{2}$ , 4 unpaired electrons.

[Fe(CN) $_{6}$ ] $^{3 -}$ : Fe $^{3 +}$ ( $d^{5}$ ) with strong field CN $^{-}$ gives low-spin $t_{2 g}^{5} e_{g}^{0}$ , 1 unpaired electron.

[FeF $_{6}$ ] $^{3 -}$ : Fe $^{3 +}$ ( $d^{5}$ ) with weak field F $^{-}$ gives high-spin $t_{2 g}^{3} e_{g}^{2}$ , 5 unpaired electrons.

Comparing the paramagnetic species, [MnCl $_{6}$ ] $^{3 -}$ and [CoF $_{6}$ ] $^{3 -}$ both have exactly 4 unpaired electrons, the only matching pair among all the paramagnetic complexes.

$∴$ The number of such species is 2.

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JEE Main 2025 Apr 7 Shift 2, Chemistry Q24: Colour, Magnetism And Stability Of Co-ordination Compounds

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