Tin(IV) chloride, also known as stannic chloride, is a compound of tin and chlorine with the formula SnCl 4.Despite what the name implies, it is not an ionic salt that contains Sn 4+ ions, but a covalent compound containing covalent Sn-Cl bonds. With the above information, the rest is pretty much just math. Ionic compounds need a neutral, balanced charge. Selenide usually forms a 2- ion. Tin(IV) has a positive four charge. Both the stannous and the stannic chlorides are primarily covalent bonding, neither consists of a lattice of discrete tin cations and chloride anions.
in the case of SnCl4, the el- of Sn is 1.96. this causes bond polarisation. Tin(II) chloride isn't an ionic compound. instead of a completely ionic bond, where the electrons from the metal are completely transferred to the non-metal, the bond is said to be ionic with covalent character (in this case it is closer to covalent, so we regard it as that). The electronegativity difference between them is [Cl (3.16), Sn (1.6)] equal to 1.2. Tin (Sn) is a metal but it is a representative metal and chlorine is a non-metal that are not too far apart in the periodic table.
with such a high el-, tin atoms attract bonding electrons back toward itself. In higher-level chemistry, this could easily be considered polar covalent, since the electronegativity difference is about 1 and there’s a long scale about it. A covalent molecular compound SnCl4, tin(IV) chloride, stannic chloride, tin tetrachloride. The electronegativity difference is only 1.2 so it would be expected to be covalent rather than ionic. Source(s): Huheey, "Inorganic Chemistry" 4/e for electronegativity values, Greenwood and Earnshaw "Chemistry of the Elements" for the structures of SnCl2. Because of this, it has a much lower melting point than typical metal chlorides, and is a colorless liquid at room temperature when anhydrous. Tin (II) bromide would be ionic, but Tin (IV) bromide would probably have polar bonds (which are midway between pure ionic and pure covalent bonds - think of them as covalent bonds where the electrons spend most time near the bromine atoms).
This is because tin is a poor metal (being in the p-block of the periodic table along with the non-metals).
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