Saved Bookmarks
| 1. |
Calculate the oxidation number of sulphur, chromium and nitrogen in H_(2)SO_(5),Cr_(2)O_(7)^(2-)andNO_(3)^(-). Suggest structure of these compounds. Count for the fallacy. |
|
Answer» SOLUTION :(i) `H_(2)underlineSO_(5)` : As per rules : `H_(2)SO_(5)to2(+1)+S+5(-2)=0` `thereforeS=+8` Which is impossible as suphur.s oxidation number should not greater than +6. So, structure should is as following : `H-O-underset(O)underset(||)overset(O)overset(||)(S)-overset(-1)(O)-overset(-1)(O)-H` `therefore2(+1)+S+2("Peroxide O")+3(O)=0` `therefore2(+1)+S+2(-1)+3(-2)=0` `therefore+2+S-2-6=0` `thereforeS=+6` (ii) `Cunderliner_(2)underlineO_(7)^(-2)` : As per rules : `Cr_(2)O_(7)^(-2)=2Cr+7(-2)=-2` = `2Cr-14=-2` Cr = +6 Chemical structure : `overset(-2)(O)=underset(O^(-1))underset(|)overset(O^(-2))overset(||)(Cr)-overset(-2)(O)-underset(O^(-1))underset(|)overset(O^(-2))overset(||)(Cr)=overset(-2)(O)` (iii) `NunderlineO_(3)^(-)` : As per rules : `NunderlineO_(3)^(-)=N+3(-2)=-1` `thereforeN=+5` Chemical structure : 
|
|