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Q. Explain about coordination compound?
These compounds are usually formed by the donation of pair of electrons to the central metal ion by the group of ions or molecules. The bond so formed is called a coordinate bond and hence the name coordination compound. However, such a simple donor-acceptor bond approach does not seem to apply to all types of coordination compounds. In the following discussion we shall not only restrict ourselves to those complexes where the central atom or ion is a metal or more specifically a transition metal, but also exclude a class of compounds known as double salts, addition products and organometallic compounds. It will be useful to briefly define these compounds to avoid any future confusion. Double salts, as you know, are compounds such as alums where two salts crystallise together as one solid but when dissolved in water they dissociate into their constituent ions. For example, a compound like potash alum, K2Al2 (SO4)42.4H20 or KAI(S04)2.12H20, when dissolved in water dissociates to give K+, Al3+. SO2-4 ions. A complex shows different properties from the original constituents and generally does not dissociate into component ions. In case of addition products, two apparently stable compounds such as NH3 and BF3 when brought together result in the formation of a compound called addition product or simply adduct with the formulation NH3. BF3. The formation of such a compound is no doubt due to donation of a pair of electrons from nitrogen atom of NH3 molecule to the boron atom 6f BF3; but we cannot assign the role of a central atom to any particular atom on the resultant molecule. Organometallic compounds are those compounds where the central metal atom or ion is directly attached to at least one carbon atom of a molecule. A complex compound could be neutral, cationic or anionic depending upon the total charge carried by the complex species.
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Molecular Formula Molecular formula of a substance provides the actual number of atoms that present in one molecule of the substance. Molecular formula = n * Empirical fo
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