JEE Main 2025 · Chemistry · Coordination Compounds · medium

The sum of the number of unpaired electrons in the complexes [Fe(CN)6]^3-, [CoF6]^3-, [Ni(NH3)6]^2+ and [MnCl4]^2- is

Numerical answer

The sum of the number of unpaired electrons in the complexes

[\mathrm{Fe(CN)_6}]^{3-}

[\mathrm{CoF_6}]^{3-}

[\mathrm{Ni(NH_3)_6}]^{2+}

and

[\mathrm{MnCl_4}]^{2-}

Show answer and solution

Answer: 12

[\mathrm{Fe(CN)_6}]^{3-}

\mathrm{Fe^{3+}}

d^5

\mathrm{CN^-}

strong field, low spin

t_{2g}^5

, so

1

unpaired.

[\mathrm{CoF_6}]^{3-}

\mathrm{Co^{3+}}

d^6

\mathrm{F^-}

weak field,

t_{2g}^4e_g^2

, so

4

[\mathrm{Ni(NH_3)_6}]^{2+}

d^8

octahedral,

t_{2g}^6e_g^2

, so

2

[\mathrm{MnCl_4}]^{2-}

\mathrm{Mn^{2+}}

d^5

in a weak-field tetrahedral field, high spin, so

5

. Total

1+4+2+5=12

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