Ncert Solutions Chemistry Class 11th

(i) In SO₂, O.N. of S is +4. In principle, S can have a minimum O.N. of -2 and maximum of +6. Therefore, S in SO₂ can either decrease or increase its O.N. and hence can act both as an oxidising as well as a reducing agent.

(ii) In H₂O₂, the O.N. of O is -1. In principle, O can have a minimum O.N. of -2 and maximum of zero (+1 is possible in O₂F₂ and +2 in OF₂). Therefore, O in H₂O₂ can either decrease its O.N. from -1 to -2 or can increase its O.N. from -1 to zero. Therefore, H₂O₂ acts both as an oxidising as well as a reducing agent.

(iii) In O₃, the O.N. of O is zero. It can only decrease its O.N. from zero to -1 or -2, but cannot increase to +2. Therefore, O₃ acts only as an oxidant.

(iv) In HNO₃, O.N. of N is +5 which is maximum. Therefore, it can only decrease its O.N. and hence it acts as an oxidant only.

(a) H_2? SO₅?  by conventional method.
Let x be the oxidation number of S
2 (+1) + x + 5 (−2) = 0
x = +8
+8 oxidation state of S is not possible as S cannot have an oxidation number more than 6. The fallacy is overcome if we calculate the oxidation number from its structure HO−S (O₂)−O−O−H.
−1+X+2 (−2)+2 (−1)+1=0
x=+6

(b) Dichromate ion
Let x be the oxidation number of Cr in dichromate ion
2x+7 (−2)=−2
x=+6
Hence the oxidation number of Cr in dichromate ion is +6. This is correct and there is no fallacy.

(c) Nitrate ion, by conventional method
Let x be the oxidation number of N in nitrate ion.
x+3 (−2)=−1
From the structure O−−N+ (O)−O−
x+1 (−1)+1 (−2)+1 (−2)=0
x=+5
Thus there is no fallacy.

(a) Here, O is removed from CuO, therefore, it is reduced to Cu, while O is added to H₂ to form H₂O, therefore, it is oxidised. Further, O.N. of Cu decreases from +2 in CuO to 0 in Cu but that of H increases from 0 in H₂ to +1 in H₂0. Therefore, CuO is reduced to Cu but H₂ is oxidised to H₂O. Thus, this is a redox reaction.

(b) Here O.N. of Fe decreases from +3 in Fe₂O₃ to 0 in Fe while that of C increases from +2 in CO to +4 in CO₂. Further, oxygen is removed from Fe₂O₃ and added to CO, therefore, Fe₂O₃ is reduced while CO is oxidised. Thus, this is a redox reaction.

(c) Here, O.N. of B decreases from +3 in BrCl₃to -3 in B₂H₆ while that of H increases from -1 in LiAlH₄to +1 in B₂H₆. Therefore, BCl₃ is reduced while LiAlH₄ is oxidised. Further, H is added to BCl₃ but is removed from LiAlH₄, therefore, BC1₃ is reduced while LiAlH₄ is oxidised. Thus, it is a redox reaction.

(d) Here, each K atom as lost one electron to form K⁺ while F₂ has gained two electrons to form two F^– ions. Therefore, K is oxidised while F₂ is reduced. Thus, it is a redox reaction.

(e) Here, the oxidation number of N increases from –3 in NH₃ to +2 in NO. On the other hand, the oxidation number of O₂ decreases from 0 in O₂ to –2 in NO and H₂O i.e., O₂ is reduced. Hence, the given reaction is a redox reaction.