Question

5. Determine the steric numbers and the molecular geometries of the following molecules and ions (a) CHCI (b) NO, (c) NH (d) SO 6. Identify hybridization of the central atoms in the following molecules: (a) CO. (b) NO2 (e) 0 (d) CIO

Answer 1

Solution 5

Sic number = No.of 'r'bonds + No. of Lone pairs (a) CHely 5 I No.of. Valence electrons on Carbon = 4. No. of bonded atoms to the Carbon =4 No. of Lone pair = 4-4 = 0 Hence it will be Abu type Steric Number - uobonds to Lone pair = = 4 o Lone pair Hence Geometry → Tetrahedral.

b) No5 Number of electrons on central Nitrogen = 5+1=6 Each brygen alom bave teclency to form two bands Hlince No.of lone pairs of e?s = 6-3/2) = 0 Type of molecule = Abg Hence Stesíc number = 3. Geometry = Toyonal planer 10: (1 C) NH3 No. of Valence electrons on Nitrogen = 5 No of atoms bonded to nitoogen = 3 No. of Lone peior electrons = 5+3 = 2 e's No. of Lone pairs = 2 = 1 pair. molecule type → AB3L Stesíc NO-(3 o bonds + 1 come pair) = 4 Geometry Tetrahedral H H 20 H

2 d) SO2 No. of Valence electrons on Sulphus = 6 eis cuch Oxygen atom is capable of fooming two bonds. So number of cone bauro electrons = 6-2(2) . = 6-4 2e's . Hence No.of come parts = L = 1 lone pair molecular type → ABL .. Stevic number → (2 5 bonds + I lone part) 7_3 molecular Geometry → Tõigonal planer