JEE Main 2023 · Chemistry · Coordination Compounds · medium

The sum of the spin-only magnetic moments squared (in BM^2) of [Fe(CN)6]^3-, [CoF6]^3-, [Ni(NH3)6]^2+ and [MnCl4]^2- is

Numerical answer

The sum of the spin-only magnetic moments squared (in

\mathrm{BM^2}

) of

\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: 70

\mu^2=n(n+2)

\mathrm{Fe^{3+}}

(

d^5

) with strong-field

\mathrm{CN^-}

is low-spin:

n=1

\mu^2=3

\mathrm{Co^{3+}}

(

d^6

) with weak-field

\mathrm{F^-}

is high-spin:

n=4

\mu^2=24

\mathrm{Ni^{2+}}

(

d^8

n=2

\mu^2=8

\mathrm{Mn^{2+}}

(

d^5

) with weak-field

\mathrm{Cl^-}

n=5

\mu^2=35

. Sum

=3+24+8+35=70

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