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Total number of valence electrons present in `6.4 gm` of peroxide ion `O_(2)^(-2)` is -A. `2.8 N_(A)`B. `3.2 N_(A)`C. `0.2 N_(A)`D. `3.6 N_(A)` |
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Answer» Correct Answer - A Mole `= (6.4)/(32) = 0.2` Number of valence electrons in `O_(2)^(2) = 14` Total number of valence electron `= 0.2 xx 14 xx N_(A) = 2.8 N_(A)` |
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