(s) about coordination compounds.

One or more options correct

Select the correct statement(s) about coordination compounds.

A. The spin-only magnetic moments of $\mathrm{[NiCl_4]^{2-}}$ , $\mathrm{[Ni(CN)_4]^{2-}}$ and $\mathrm{Ni(CO)_4}$ are $2.83$ , $0$ and $0\ \mathrm{BM}$ respectively
B. $\mathrm{[Mn(CN)_6]^{3-}}$ and $\mathrm{[MnF_6]^{3-}}$ have the same spin-only magnetic moment
C. Ignoring pairing energy, the crystal field stabilisation energies of $\mathrm{[Co(NH_3)_6]^{3+}}$ and $\mathrm{[CoF_6]^{3-}}$ are $-2.4\Delta_o$ and $-0.4\Delta_o$ respectively
D. $\mathrm{[Co(en)_2Cl_2]^+}$ ( $\mathrm{en}$ = ethane-1,2-diamine) has three stereoisomers in total

Show answer and solution

Answer: A, C, D

(A) is correct.

\mathrm{[NiCl_4]^{2-}}

is tetrahedral

\mathrm{Ni^{2+}}

(

d^8

) with two unpaired electrons, so

\mu=\sqrt{8}=2.83\ \mathrm{BM}

;

\mathrm{[Ni(CN)_4]^{2-}}

is square planar (diamagnetic) and

\mathrm{Ni(CO)_4}

contains

\mathrm{Ni(0)}

d^{10}

, so both have zero moment. (B) is wrong: strong-field cyanide makes

\mathrm{Mn^{3+}}

(

d^4

) low spin with

t_{2g}^4

, two unpaired electrons,

2.83\ \mathrm{BM}

, while fluoride gives high spin

t_{2g}^3e_g^1

with four unpaired electrons,

4.90\ \mathrm{BM}

. (C) is correct: low-spin

d^6

t_{2g}^6

, CFSE

=-2.4\Delta_o

; high-spin

d^6

t_{2g}^4e_g^2

, CFSE

=4(-0.4)+2(0.6)=-0.4\Delta_o

. (D) is correct: trans gives one achiral isomer and the cis isomer exists as an enantiomeric pair, which gives three stereoisomers.

Original question written in the JEE Advanced 2021 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake