Question

1(a). Based on the solubility properties of the tin(IV) halides, how do they differ from the more commonly encountered metal halides of metals such as sodium or magnesium? 1(b). What is an "adduct"? Identify an example from the chemistry of SnCl4, and one from the chemistry of boron. 1(c). What structures would be predicted for SnCl2, SnCl4, and Cl4Sn(PPh3) using VSEPR theory. (Assume SnCl2 is monomeric).

Answer 1

1,a, Tin halides are highly soluble in organic compounds like benzene,alcohol,acetone ,kerosene etc.

Some halides are sparingly soluble in water or form hydrates with water molecule.

On the other hand,metal halides are readily and highly soluble in water.

b, When two or more molecules are directly added,an adduct is formed. In this addition, all the atoms of all the molecules that are added will be present in the product and this product is called as adduct.

When SnCl₄ is reacted with water, they form an adduct SnCl_4.2H2 O.

All the molecules of SnCl4 and water are present in the adduct.

When tri hydrido boron is added with tetra hydro furan(THF), the formed adduct is the addition of BH₃ with THF molecule.All the atoms of BH₃ and THF are present in the adduct.

c, Monometric SnCl2

Number of atoms present = 3

Since it is monometric,the shape is bent.

SnCl4

Numner of lone pair of electrons = 5

Hence it is tetrahedral

Cl4Sn(PPh₃)

Number of lone pair is 5

Structure is trigonal bipyrimidal.