JEE Main 2021 · Chemistry · Coordination Compounds · medium

The total number of unpaired electrons present in the three complex ions [Mn(CN)6]^3-, [CoF6]^3- and [Fe(H2O)6]^2+ taken together is

Numerical answer

The total number of unpaired electrons present in the three complex ions

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

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

and

[\mathrm{Fe(H_2O)_6}]^{2+}

taken together is

Show answer and solution

Answer: 10

Decide the spin state from ligand field strength.

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

\mathrm{Mn^{3+}}

d^4

and

\mathrm{CN^-}

is a strong-field ligand, so low spin

t_{2g}^4

with

2

unpaired electrons.

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

\mathrm{Co^{3+}}

d^6

and

\mathrm{F^-}

is weak-field, so high spin

t_{2g}^4e_g^2

with

4

unpaired electrons.

[\mathrm{Fe(H_2O)_6}]^{2+}

\mathrm{Fe^{2+}}

d^6

and water is weak-field, so high spin with

4

unpaired. Total

=2+4+4=10

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