Transition metal compounds can show a number of different types of isomerism. Hydration isomerism is where different numbers of water molecules act as ligands. The name chromium (III) chloride is given to several chemical compounds with the formula $$$\mathrm{CrCl}_{3} \cdot \mathrm{XH}_{2} \mathrm{O}$$$, including a number of hydration isomers. Anhydrous chromium(III) chloride, $$$\mathrm{CrCl}_{3}$$$, is a violet solid which can react with water to produce three isomers. $$$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} 3 \mathrm{Cl}^{-}$$$is violet. $$$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}^{2+} 2 \mathrm{Cl}^{-} \cdot \mathrm{H}_{2} \mathrm{O}\right.$$$ is pale green. $$$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]^{+} \mathrm{Cl}^{-} \cdot 2 \mathrm{H}_{2} \mathrm{O}$$$ is dark green. (i) Explain why the three isomers have different colours. (3) ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ (ii) You are provided with three equimolar solutions of the three isomers. Suggest how you would quantitatively determine the relative numbers of free chloride ions in the three isomers using the standard test for a chloride ion. (5) ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................

Chemistry

IGCSE&ALevel

EDEXCEL

Exam No：wch15-01-que-20240113 Year：2024 Question No：17(a)

Answer:

An answer that makes reference to the following points:
- the presence of chloride ligands / change in ligands / different ligands results in a different energy gap between (lower and higher energy) d-orbitals / result in a different splitting in the d-subshell

(1)
- (colour results from the absorption of light by electrons) as they are promoted between d-orbitals / move from lower energy to higher energy (d-orbitals) / move to a higher energy level (d-orbital)

(1)
- so different wavelengths / frequencies (of light) are absorbed / transmitted / reflected (resulting in different colours)
An explanation that makes reference to the following points:
Reagent
- use silver nitrate solution (which reacts with free chloride ions) to give a precipitate

(1)

Practical technique
- use equal volumes of each solution of the three isomers (because they are equimolar solutions)

(1)
- add an excess of silver nitrate solution

(1)
- collect the precipitate / silver chloride by filtration / centrifuge
and
dry the precipitate

(1)
- weigh the silver chloride and calculate the number of moles (of silver chloride / chloride ions / silver ions and so find the ratio)
or
weigh the silver chloride for each isomer and find the ratio

Answer:

Knowledge points:

Solution:

Answer:

Knowledge points:

Solution:

Related Question