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Give the formula and describe the structure of a noble gas species which is isostructrural with IBr_2^(-) |
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Answer» Solution : `IBr_2^(-)` : In `IBr_2^(-)` the central I atom has 8 electrons (7 valence electrons + one due to NEGATIVE CHARGE). Two of these form two SINGLE bonds (two bond pairs) with two Br atoms, while the remaining six constitute three lone pairs. Thus, I in `IBr_2^(-)`has two bond pairs and three lone pairs. Therefore, according to VSEPR theory, it is linear. `IBr_2^(-)`has a total of `22 (7 + 2 xx 7 + 1)` valence electrons. A noble gas species having 22 valence electrons is `XeF_2 (8 + 2 xx 7 = 22)`. Like `IBr_2^(-), XeF_2`is also linear 
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