Q. No.H2S (g) + Cl2 (g) → 2 HCl (g) + S (s)
The correct statement(s) in the balanced equation is(are):
a. H2S is oxidised.
b. Cl2 is reduced.
c. H2S is reduced.
d. Cl2 is oxidised.
1. a, b
2. b, c
3. c, d
4. a, d
2 Na(s) + H2(g) → 2 NaH (s)
The incorrect statement among the following regarding above mentioned reaction is:
1. Na gets oxidised
2. It is a redox reaction
3. NaH is an ionic hydride
4. None of the above
Match the items in column I with column II.
(O.S = oxidation state)
| Column I |
Column II (Oxidation State) |
| a. O.S of Au in HAuCl4 | i. +2 |
| b. O.S of Tl in Tl2O | ii. +1 |
| c. O.S of Fe in FeO | iii. +3 |
| d. O.S of Cu in CuI |
1. a=i; b=ii; c=iii; d=ii
2. a=iii; b=i; c=ii; d=ii
3. a=iii; b=ii; c=i; d=ii
4. a=ii; b=i; c=iii; d=ii
Consider the given reaction:
\(2 \mathrm{Cu}_2 \mathrm{O}(s)+\mathrm{Cu}_2 \mathrm{~S}(s) \rightarrow 6 \mathrm{Cu}(s)+\mathrm{SO}_2(g)\)
Which of the following statements about the given reaction is correct?
1. Cu is reduced and Cu(I) is an reductant.
2. Sulphur is reduced and S in Cu2S acts as oxidant.
3. Cu(I) is an oxidant and Cu(I) is reduced.
4. Sulphur is reduced and copper is oxidised.
The species among the following that does not show a disproportionation reaction is-
ClO–, ClO2–, \(ClO_{3}^{-}\) and ClO4–
1. ClO–
2. ClO2–
3. ClO4–
4. \(ClO_{3}^{-}\)
Match the items in column I with the items in column II.
| Column I | Column II |
| a. N2 (g) + O2 (g) → 2 NO (g) | i. Disproportionation redox reaction |
| b. 2Pb(NO3)2(s) → 2PbO(s) + 4 NO2 (g) + O2 (g) | ii. Decomposition redox reaction |
| c. NaH(s) + H2O(l) → NaOH(aq) + H2 (g) | iii. Combination redox reactions |
| d. 2NO2(g) + 2OH–(aq) → NO2–(aq) + NO3– (aq) + H2O(l) | iv. Displacement redox reaction |
1. a = iii; b = ii; c = iv; d = i
2. a = iii; b = iv; c = i; d = ii
3. a = ii; b = iii; c = iv; d = i
4. a = iv; b = i; c = iii; d = ii
Consider the following reaction:
Pb3O4 + 8HCl → A + B + 4H2O
A and B are, respectively,:
1. A= PbCl4; B= PbCl2
2. A= PbCl2; B= Cl2
3. A= PbCl4; B=Cl2
4. A = PbCl2; B = O2
Consider the following equation:
Cr2O72–(aq) + 3SO32– (aq)+ 8H+(aq) → A + B +4H2O(l)
The product A and B are, respectively:
1. 2Cr2+; 3SO2
2. 2Cr+; S2O7 2-
3. 2Cr3+; 3SO4 2-
4. \(2 \mathrm{Cr}^{3+} ; 3 \mathrm{HSO}_4{ }^{-}\)
The value of c, d and f in the above mentioned reaction are respectively:
| 1. | c = 1; d = 2; f = 2 | 2. | c = 2; d = 1; f = 2 |
| 3. | c = 2; d = 2; f = 1 | 4. | c = 1; d = 1; f = 1 |
Permanganate(VII) ion, MnO4– in basic solution oxidizes iodide ion, I– to produce molecular iodine (I2) and manganese (IV) oxide (MnO2). The reaction is as follows:
aI–(aq) + bMnO4–(aq) + cH2O(l) → dI2(s) + eMnO2(s) + fOH–(aq)
The value of b, d and f are-
| 1. | b = 2; d = 3; f = 8 | 2. | b = 1; d = 3; f = 8 |
| 3. | b = 3; d = 8; f = 2 | 4. | b = 8; d = 3; f = 2 |
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