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Give the formula and describe the structure of a noble gas species which is isostructrural with ICl_4^(-) |
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Answer» Solution :`ICl_4^(-)` : In `ICl_4^(-)`, the central I atom has in all 8 ELECTRONS (7 valence electrons + one due to negative charge). Four of these form single BONDS with four Cl atoms (four bond pairs) while the REMAINING four constitute two lone pairs. Thus, I in `ICl_4^(-)` has four bond pairs and two lone pairs. Therefore, ACCORDING to VSEPR theory, it is SQUARE planar as shown below. Now, `ICl_4^(-)` has a total of `(7 + 4 xx 7 + 1) = 36` valence electrons. A noble gas species having 36 valence electrons is `XeF_4 (8 + 4 xx 7 = 36)`. Like `ICl_4^(-), XeF_4` is also square planar 
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