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Bond distance `C-F` in `(CF_(4))` & `Si-F` in `(SiF_(4))` are respective `1.33 Å` & `1.54 Å. C-SI` bond is `1.87 Å`. Calculate the colvalent radius of F atom ignoring the electronegativity differences.A. `0.64 Å`B. `(1.33 +1.54 +1.8)/(3) Å`C. `0.5 Å`D. `(1.54)/(2) Å` |
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Answer» Correct Answer - C `[r_(c) +r_(F) = 1.33 Å` `underset(-----------)(r_(Si)+r_(F) = 1.54 Å)` `r_(c) +r_(Si) +2r_(F) = 2.87 Å` `r_(c) +r_(Si) = 1.87 Å` `1.87 Å + 2r_(F) = 2.87 Å` `2r_(F) = 1.00 Å` `r_(F) = 5.Å]` |
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