Formation of SnCl2 is possible only theoretically. This reaction direction is mainly limited by: (1) hydrolysis of SnCl2 under moistured air to Sn(OH)Cl (so you will obtaine a mixture of tin salts of various composition and (2) oxidation of highly active SnCl2 reductant by O2/H2O to tin (IV) (for example to SnOCl2 and similar compounds). 

Hello Alex,

Thanks for your kind answer to my question. As I understand we can only obtain oxidized tinchloride products. It makes sense to me. If possible, can you reference it with a article or book ? 

Another question I wanted to ask is ;

-How can I be sure a SnCl2 is with two crystal water molecules ? Can I distinguish via IR ?

Thanks a lot

Hi Ihsan

I think you can review the electrons configuration of Sn that is [Kr] 4d10 5s2 5p2 6s0 6p0. Based on this configuration, Sn can bind two Cl forming SnCl2 leaving two vacant orbitals of 5p and 6s in the ground state. The capability of the two vacant orbitals to accept two water molecules are very good so that we possible to obtain SnCl2. 2H2O through simple method (without any special treatment). However in the excited state Sn can bind four Cl forming SnCl4 leaving four vacant orbitals of 6s and 6p so that we possible to obtain SnCl4. 4H2O with special treatment. In addition, you can confirm the hydrate molecules using FTIR spectroscopy. The existance of hydrate molecules will be presented by hydrogen bond fibration that appear in around 3426-3430 cm-1 broad wavelength range.

As Dr. Tolstov mentioned above, there is only way by air exposure: hydrolysis + oxidation by O2. By this way SnCl2 * 2H2O cannot be obtained. By the way, do you know that its structure looks as [Sn(H2O)Cl2] * H2O? Its formation is possible to determine by means of XRD analysis.

Its crystal structure is presented in my post (jpg-file). Moreover, its crystallographic data are described in pdf-file. And I'm kindly attaching a standard XRD profile (at Ka Cu) of this compound (dat-file). Surely this info will serve you well.

Sincerely,

MN

I agree with Dr Alexander Tolstov.