Complex A has a composition of \(\mathrm{H_{12}O_6Cl_3Cr}\). If the complex on treatment with conc.\(\mathrm{H_2SO_4}\) loses 13.5% of its original mass, the correct molecular formula of A is :
[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]
[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]
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- \(\mathrm{[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O}\)
- \(\mathrm{[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O}\)
- \(\mathrm{[Cr(H_2O)_3Cl_3] \cdot 3H_2O}\)
- \(\mathrm{[Cr(H_2O)_6]Cl_3}\)
The Correct Option is B
Solution and Explanation
Mass lost = 13.5% of 265 = 35.8 g, which is about 36 g = 2 x 18 g.
So the complex contains 2 molecules of water of crystallisation (removed by conc. H2SO4), and the other 4 H2O are coordinated.
With 3 Cl in total and 6 coordination sites: 4 H2O + 2 Cl as ligands + 1 Cl as counter ion.
Formula: \(\mathrm{[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O}\).
Answer: (2) [Cr(H2O)4Cl2]Cl.2H2O
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