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Calculate the oxidation number of sulphur in the following molecules ions . (a) H_2S (b) H_2SO_3 (c) SO_4^(2-) (d) Na_2S_2O_3 (e) S_2O_7^(2) (f) H_2SO_4 (g) S_2O_4^(2-). |
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Answer» Solution :(a) `H_2S`. The oxidation number of hydrogen is +1. Let the oxidation number of sulphur be x. `{:(+1,x,),(H_2,S,),((1xx2),"(x)",),(1xx2+x=0,:.,x=2):}` Oxidation number of S in `H_2S `is -2. (B) `H_2SO_3`. Oxidation number of hydrogen is +1 and that of OXYGEN is -2. Let the oxidation number of S be x. `{:(+1,x,-2),(H_2,S,O_3),((+1xx2),(x),(-2xx3)""):}` `(+1xx2)+x+(-2xx3)=0` or `+2+x-6=0` or `x=+4` Oxidation number of S in `H_2SO_3` is +4. (c) `SO_4^(2-)` . The oxidation number of oxygen is -2 and let oxidation number of S be x. `[{:(x,-2),(S,O_4):}]^(2-)` `{:(""x,,(-2xx4)),(x+(-2xx4)=-2,"or",x=-2+8),(,"or",x=+6):}` Oxidation number of S in `SO_4^(2-)` is +6. (d) `Na_2S_2O_3` . The oxidation number of NA is +1 and that of oxygen is -2. Let the oxidation number of S of x. `{:(+1,x,-2),(Na_2,S_2,O_3),((1+xx2),("x"xx2),(-2xx3)""):}` `{:((+1xx2)+("x"xx2)+(-2xx3)=0,,),(""+2+2x-6=0,"or",2x=4),(,,x=+2):}` Oxidation number of S in `Na_2S_2O_3` is +2. (E) `S_2O_7^(2-)` . The oxidation number of oxygen is -2and let the oxidation number of sulphur be x. `[{:(x,-2),(S_2,O_7)]^(2-)` `("x"xx2)(-2xx7)` `"x " xx 2 +(-2xx7) =-2` `2x = -2 +14` `:.""x = 6` Oxidation number of S in `S_2O_7^(2-)` is +6. (f) `H_2SO_4`. The oxidation number of hydrogen is +1 and that of oxygen is -2. Let the oxidation number of S be x. `{:(+1,x,-2),(H_2,S,O_4),((+1xx2),x,(-2xx4)):}` `(+1 xx2)+x + (-2xx4_=0` or `+2+x-8=0` or `x= +6` Oxidation number of S in `H_2SO_4` is +6. (g) `S_2O_4^(2-)` . The oxidation number of oxygen is -2 and let the oxidation number of sulphur be x. `[{:(x,-2),(S_2,O_4):}]^(2-)` ` ("x" xx2) (-2xx4)` `"x" xx 2 + (-2xx4) =-2` `2 x - 8=-2 :. x = +3` `:.` Oxidation number of S in `S_2O_4^(2-)` is +3. |
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