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Identify the following: Na_(2)CO_(3)overset(So_(2))rarr A overset(Na_(2)CO_(3))rarr B overset("Elementals")underset(Delta)rarr C overset(I_(2))rarr D Also mention the oxidation state of S in all the compounds. |
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Answer» Solution :`Na_(2)CO_(3)+2SO_(2)+H_(2)O rarr 2NaHSO_(4)+CO_(2)` `2NaHSO_(3)+Na_(2)CO_(3)rarr 2Na_(2)S_(2)O_(3)` `2Na_(2)S_(2)O_(3)+I_(2)overset(Delta)rarr Na_(2)S_(4)O_(6)+2NaI` Oxidation states of sulphur: a. `NaHSO_(3)=1+1+x-6=0, :. X=4` b. `Na_(2)SO_(3)=2+x-6=0, :.x=4` c. `Na_(2)S_(2)O_(3)` (Sodium THIOSULPHATE) Average oxidation numer. `Na_(2)S_(2)O_(3)=2xx1+2x+3(-2)=0` `:. x= +2` But oxidation NUMBER of `S= +2` is wrong because both the S atom cannot be in same oxidation state as is evident form the fact what `Na_(2)S_(2)O_(3)` is treated with dil `H_(2)SO_(4)`, one S atom gets precipitated while the other gets converted into `SO_(2)`. The oxidation numbers of these two S atom can, however, be determined by the chemical bonding method. By chemical bonding method, the structure of `Na_(2)S_(2)O_(3)` is `Na^(o+)O^(Θ)-overset(S)overset(uarr)underset(O)underset(||)S-O^(Θ)Na^(o+)` SINCE there is a coordinate bond between two S atoms. Therefore, the acceptor S atom has an oxidation number of `-2`. The oxidation number of other S atoms can be calculated as follows: `underset(("For Na"))(2xx(-1)) + underset(("For O atoms"))(3xx(-2)) + x + underset(("For coordinate S atom"))(1XX(-2)=0)` OR `+2-6+x-2=0, x= +6` Thus, the two S atoms in `Na_(2)S_(2)O_(3)` have oxidation number of `-2` and `+6`.  d. Structure of tetrahionate ion is `Na^(o+)O^(Θ)-overset(O)overset(||)underset(O)underset(||)(S^(+5))-S^(0)-S^(0)-overset(O)overset(||)underset(O)underset(||)(S^(+5))-O^(Θ)Na^(o+)` Since each of the two terminal S atoms is connected to two oxygen atoms by a double bond and one oxygen atom by a single bond, therefore, the oxidation state of each of these terminal S atoms is `+5`. Since two central S atoms are linked to each other by a single bond, each S is further attached to similar species on either side, the electron pair formaing the `S-S` bond remains in the center and hence each of the two central S-atom has an oxidation state of zero. Therefore, oxidation number of two S-atoms are `(+5 and 0)`. However, the average oxidation state of the four S atoms is 2 S atoms in `+5` oxidation state and 2 S atoms in zero oxidation state. `:. ((2xx5)+(2xx0))/(4)=2.5` |
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