Ncert Solutions Chemistry Class 11th

(a) HgCl₂  (b) NiSO₄  (c)SnO₂  (d) Tl₂SO₄  (e) Fe₂ (S0₄)₃  (f) Cr₂O₃.

(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+

Writing the O.N. of each atom above its symbol, we have,          

here, the O.N. of F decreases from 0 in F₂ to -1 in HF and increases from 0 in F₂ to +1 in HOF. Therefore, F₂ is both reduced as well as oxidised. Thus, it is a redox reaction and more specifica

(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 ox

5.19. As the mixture H₂ and O₂ contains 20% by weight of dihydrogen, therefore,

If H₂ = 20g, then O₂ = 80g

No. of moles of H₂ = 20/2 = 10 moles

No. of moles of O₂ = 80/32 = 2.5 moles

Partial pressure of H₂ = [No. of moles of H₂/ (No. of moles of H₂ + No. of moles of O₂)] x

P_tota

(a) In Kl₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K⁺ [I —I <— I]^–. Here, a coordinate bond is formed between I₂ molecule and I^– ion. The oxidatio

Let x be the oxidation number to the underlined elements in the given species:

(a) NaH₂PO₄

(+1) + 2 (+1) + x + 4 (-2) = 0

x + 3 – 8 = 0

x = +5

(b) NaHSO₄

(+1) + (+1) + x + 4 (-2) = 0

x – 6 = 0

x = +6

(c) H₄P₂O₇

4 (+1) + 2x + 7 (-2) = 0

2x -10 =0

x = +5

(d) K₂MnO₄

2 (+1) + x + 4 (-2) = 0

x – 6 = 0

5.18. Given, P₁ = P₂ and V₁ = V₂

We know that P₁V₁ = P₂V₂

Or,                n₁RT₁ = n₂RT₂

i.e.,                  n₁T₁ = n₂T₂

 Substituting n =

5.17. No. of moles of CO₂ = Given mass of CO₂ / Molar mass

                                           = 8.8g / 44g mol^-1 = 0.2 mol

Pressure of CO₂ = 1 bar

RR = 0.083 bar dm³ K^–1 mol^–1

T = 2