Saved Bookmarks
| 1. |
Calculate the oxidation number of each sulphur atom in the following compounds. (a) Na_(2)S_(2)O_(3) (b) Na_(2)S_(4)O_(6) ( c) Na_(2)SO_(3) (d) Na_(2)SO_(4) |
|
Answer» Solution :(a) `Na_(2)S_(2)O_(3):" "Na^(+)O^(-)-UNDERSET(O)underset(||)overset(S)overset(uarr)(S)-O^(-)Na^(+)` Coordinate covalent bond is present between TWO S atoms so oxidation number of S which accept `e^(-)` is -2 and for another we have to calculate `2(+1)+3(-2)+x+1(-2)=0` `thereforex=+6` (b) `Na_(2)S_(4)O_(6):" "Na^(+)O^(-)-underset(O)underset(||)overset(O)overset(||)(S)-overset(0)(S)-overset(0)(S)-underset(O)underset(||)overset(O)overset(||)(S)-O^(-)-Na^(+)` In this molecules, oxidation number of two S atoms, which is INDICATED in STRUCTURE is zero. So oxidation number of another two S is, `2(+1)+6(-2)+2x+2(0)=0` `therefore2-12+2x=0` `thereforex=+5` ( c) `Na_(2)SO_(3)` : `2(+1)+x+3(-2)=0` `thereforex=+4` (d) `Na_(2)SO_(4)` : `2(+1)+x+4(-2)=0` `therefore2+x-8=0` `thereforex=+6` |
|