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Assign oxidation number to the underlined elements in each of the following species: a.`NaH_(2)PO_(4)` b. `NaHul(S)O_(4)` c. `H_(4)ul(P_(2))O_(7)` d. `K_(2)ul(Mn)O_(4)` e. `ul(Ca)O_(2)` f. `Naul(B)H_(4)` g. `H_(2)ul(S_(2))O_(7)` h. `KAl(ul(S)O_(4))_(2).12H_(2)O` |
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Answer» a. Let the oxidation number of `P` be `x`. Writing the oxidation number of each atom above its symbol, we have `overset(+1)(Na)overset(+1)(H_(2))overset(x)(P)overset(-2)(O_(4))` Sum of oxidation numbers of varios atoms in `NaH_(2)PO_(4) =1(+1)+2(+1)+1(x)+4(-2)` `=x-5` But the sum of oxidation number of various atoms in `NaH_(2)PO_(4)` (netural) is zero `:. x-5=0` or `x=+5` Thus, the oxidation number of `P` in `NaHPO=+5`. b. `overset(+1)(Na)overset(+1)(H)overset(x)(S)overset(-2)(O_(4))` `:. 1(+1)+1(+1)+x+4(-2)=0` `or x=+6` Thus, the oxidation number of S in `NaHSO_(4)=+6`. c. `overset(+1)(H_(4))overset(x)(P_(2))overset(-2)(O_(7))` `:. 4(+1)+2(x)+7(-2)=0` or `x=+5` Thus, the oxidation number of `P` in `H_(4)P_(2)O_(7)=+5`. d. `overset(+1)(K_(2))overset(x)(Mn)overset(-2)(O_(4))` `:. 2(+1)+1(x)+4(-2)=0` or `x=+7` Thus, the oxidation number of `Mn` in `K_(2)MnO_(4)=+7`. e. Let the oxidation number of `Ca` be `x`. Since `O` in peroxides has an oxidation of `-1`. Thus, `overset(x)(Ca)overset(-1)(O_(2))` `:. x+2(-1)=0` or `x=+2` Thus, the oxidation number of calcium in `CaO_(2)=+2`. f. In `NaBH_(4), H` is present as hydride ion. Therefore, its oxidation number is `-1`. Thus, `overset(+1)(Na)overset(x)(B)overset(-1)(H_(4))` `:. 1(+1)+x+4(-1)=0` or `x=+3` Thus, the oxidation number of `B` in `NaBH_(4)=+3`. g. `overset(+1)(Na_(2))overset(x)(S_(2))overset(-2)(O_(7))` `:. 2(+1)+2(x)+7(-2)=0` or `x=+6` Thus, the oxidation number of `S` in `Na_(2)S_(2)O_(7)=+6`. h. `overset(+1)(K)overset(+3)(Al)(overset(x)(S)overset(-2)(O_(4)))_(2).12(overset(+1)(H_(2))overset(-2)(O))` or `+1+3+2x+8(-2)+12(2xx1-2)` or `x=+6` Alternatively, since `H_(2)O` is a neutral molecule, therefore, sum of oxidation numbers of all the atoms in `H_(2)O` may be taken as zero. As such water molecules may be ignored while computing the oxidation number of `S`. `:. +1+3+2x-16=0` or `x=+6` Thus, the oxidation number of `S` in `KAl(SO_(4))2.12 H_(2)O=+6`. |
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