1.

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`

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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