please explain in detail, as how to conclude which compound posses which hybridisation???
Hybrid orbitals contain;
(a) all unshared electron pairs (lone pairs)
(b) bonding electron pairs in single bonds (sigma bonds)
(c) one and only one of the electron pairs in a double or triple bond (again sigma bond, pi bonds not included).
In so far as the molecular geometry is concerned, a multiple bond acts as if it were a single bond. In other words, the "extra" electron pairs in a double bond or triple bond have no effect upon the geometry of a molecule.
The geometry of a molecule is fixed by the electron pairs in the hybrid orbitals about the central atom. These electron pairs in the hybrid orbitals may be bonding pairs or lone pairs. These electron pairs are directed to be as far apart as possible according to the VSEPR Theory and determines the shape of the molecules or polyatomic ions.
O = C = O
Carbon has 4 valence electrons and all these electrons are used to make bonds with two O atoms. The central atom C has no lone pair. The type of hybridization of C is "sp" (only two sigma bonds no lone pair).
O - S = O There is a lone pair on S atom (cannot be shown)(O - S bond is a coordinate covalent bond, two bonding electrons are supplied by S atom)
Type of hybridization of S atom: 2 sigma and 1 lone pair, therefore sp2.
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