JEE Advanced 2022 Paper 2 · Chemistry · Coordination Compounds · medium

Three isomeric hydrates of chromium(III) chloride have the formula CrCl3·6H2O, with Cr always octahedral: [Cr(H2O)6]Cl3, [CrCl(H2O)5]Cl2· H2O and…

Numerical answer

Three isomeric hydrates of chromium(III) chloride have the formula

\mathrm{CrCl_3\cdot6H_2O}

, with

\mathrm{Cr}

always octahedral:

[\mathrm{Cr(H_2O)_6}]\mathrm{Cl_3}

[\mathrm{CrCl(H_2O)_5}]\mathrm{Cl_2\cdot H_2O}

and

[\mathrm{CrCl_2(H_2O)_4}]\mathrm{Cl\cdot2H_2O}

. A

0.0100\ \mathrm{m}

aqueous solution of one isomer (assume complete ionisation of the outer-sphere chloride, no further dissociation of the complex ion, ideal behaviour) freezes at

-0.0558\,^\circ\mathrm{C}

. Take

K_f(\mathrm{H_2O})=1.86\ \mathrm{K\,kg\,mol^{-1}}

. The number of water molecules per formula unit of this isomer that are NOT bonded to the chromium ion is _____.

Show answer and solution

Answer: 1

The depression in freezing point is

\Delta T_f=i\,K_f\,m

. So

i=\dfrac{0.0558}{1.86\times0.0100}=3.0

.

Count the particles each isomer yields on ionisation.

[\mathrm{Cr(H_2O)_6}]\mathrm{Cl_3}

gives

1+3=4

ions,

[\mathrm{CrCl(H_2O)_5}]\mathrm{Cl_2\cdot H_2O}

gives the complex cation plus

2\ \mathrm{Cl^-}

, so

i=3

, and

[\mathrm{CrCl_2(H_2O)_4}]\mathrm{Cl\cdot2H_2O}

gives

1+1=2

ions. Only the middle isomer matches

i=3

. The hydrated water in the formula is not part of the coordination sphere (the coordination number of six is already filled by

\mathrm{Cl}

and five

\mathrm{H_2O}

), so exactly one water molecule per formula unit is a water of crystallisation, not bonded to Cr. The answer is

1

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