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Give the formula and describe the structure of a noble gas species which is isostructrural with BrO_3^(-) |
Answer» Solution :`BrO_3^(-)`: The central Br atom has SEVEN electrons. Four of these electrons form two double bonds or coordinate bonds with two oxygen atoms while the fifth ELECTRON forms a single bond with `O^(-)` ion. The remaining two electrons form one lone pair. Thus, in all there are THREE bond pairs and one lone pair around Br atom in `BrO_3^(-)`. Therefore, according to VSEPR theory, `BrO_3^(-)` is pyramidal. Now, BRO, has a total of `26 (7 + 3 xx 6 + 1 = 26)` valence electrons. A noble gas species having 26 valence electrons is `XeO_3 (8 + 3 xx 6 = 26)`. Like `BrO_3^(-), XeO_3` is also pyramidal. 
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