Saved Bookmarks
| 1. |
Calculate the oxidation number of (i) S in H_(2)S, (ii) C in CO_(2) (iii) C in CO_(2) (iii)C in CH_(2)CI_(2), (iv) N in (NH_(4))SO_(4) and (v)P I Na_(3)PO_(4) |
|
Answer» Solution :(i) `S in H_(2)S` Let the oxidation number of S in `H_(2)S` be x writing the oxidation number of each atom above its symbol `overset(+1)H_(2)overset(x)S` sum of oxidation number of various ATOMS in `H_(2)S=2(+1)+x=2+x` But the sum fo the oxidation numbers of various atoms in `H_(2)S` (neutral) is zero (Rule 7) `therefore` 2+x=0 or x=-2 Thus the oxidation numbr of S in `H_(2)S is -2` (II) `C in CO_(2)` Let the oxidaion number of C in `CO_(2)` be x writing the oxidaion number o feach atom above its symbol `therefore`x-4=0or x=+4 Thus the oxidation number of C in `CO_(2)` is +4 (iii)`C in CH_(2)CI_(2)` Let the oxidaition number of C in `CH_(2)CI_(2)` be x writing the oxidatin number of each atom above its sysmbol `overset(x)C overset(+1)H_(2) overset(-1)CI_(2)` `therefore` sum of hte oxidation numbers of various atoms in `CH_(2)CI_(2)` (neutral ) is zero (Rule 7) `therefore`x=0 Thusthe oxidation number of C in `CH_(2)CI_(2)` is zero (iv) N in `(NH_(4))SO_(4)` Let the oxdation number of nitrogen in `(NH_(4))_(2)` be x writing the oxdation number o fhydrogen above its sysmbol and that of `SO_(4)^(2-)` ion above its FROMULA `therefore` Sum of oxidation numbers of all the atoms in `(NH_(4))SO_(4)=2x+2(+1xx4)+(-2)=2x+6` But the sum of oxidation numbers of all the atoms in `(NH_(4))_(2)SO_(4)` (neutral ) is zero (Rule 7) `therefore 2x+6=0 or x =-3` Thus the oxdation number of nitrogen in `(NH_(4)^(2))` is -3 (v)`P in Na_(3)PO_(4)` Let the oxidation number of P in `Na_(3)PO_(4)` be x writing hte oxidation number of each atom above its symbol `overset(+1)Na_(3) overset(x)P overset(-2)O_(4)` Sum of the oxidation numbers of various of various atoms in `Na_(3)PO_(4)=3(+1)+x4(-2)=x-5` |
|