The element that does not show a disproportionation tendency is/are:
1. | Cl | 2. | Br |
3. | F | 4. | I |
Assertion (A): | The decomposition of hydrogen peroxide to form water and oxygen is an example of a disproportionation reaction. |
Reason (R): | The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in O2 and –2 oxidation state in H2O. |
1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
3. | (A) is True but (R) is False. |
4. | (A) is False but (R) is True. |
The incorrect statement regarding the rule to find the oxidation number among the following is:
1. | The oxidation number of hydrogen is always +1. |
2. | The algebraic sum of all the oxidation numbers carried by elements in a compound is zero. |
3. | An element in its free or uncombined state has an oxidation number of zero. |
4. | Generally, in all its compounds, the oxidation number of fluorine is -1. |
Given below are two statements:
Assertion (A): | In the reaction between potassium permanganate and potassium iodide, permanganate ions act as an oxidising agent. |
Reason (R): | The oxidation state of manganese changes from +2 to +7 during the reaction. |
1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
3. | (A) is True but (R) is False. |
4. | (A) is False but (R) is True. |
Standard reduction potentials of the half-reactions are given below:
The strongest oxidizing and reducing agents, respectively, are:
1. and
2. and
3. and
4. and
(a) | \(E_{k^+/K}^o = - 2.93\ V\); \(E_{Ag^+/Ag}^o = 0.80\ V\) |
(b) | \(E_{Hg^{2+}/Hg}^o = 0.79\ V\); \(E_{Mg^{2+}/Mg}^o = - 2.37\ V\) |
(c) | \(E_{Cr^{3+}/Cr}^o = -0.74\ V\) |
Based on standard electrode potentials given above, the correct arrangement for increasing order of reducing power of
elements is:
1. | \(\mathrm{Ag}<\mathrm{Hg}<\mathrm{Cr}<\mathrm{Mg}<\mathrm{K} \) |
2. | \(\mathrm{Ag}>\mathrm{Cr}>\mathrm{Mg}>\mathrm{Hg}>\mathrm{K}\) |
3. | \(\mathrm{K}>\mathrm{Mg}<\mathrm{Cr}<\mathrm{Hg}>\mathrm{Ag} \) |
4. | \(\mathrm{K}<\mathrm{Mg}<\mathrm{Cr}<\mathrm{Hg}<\mathrm{Ag}\) |
1. | Due to manganese being in its highest oxidation state in MnO₄²⁻. |
2. | Due to manganese being in its highest oxidation state in MnO₄⁻ |
3. | Because the disproportionation reaction of MnO₄²⁻ is endothermic. |
4. | Because the disproportionation reaction of MnO₄²⁻ is exothermic. |
Fluorine reacts with ice as per the following reaction
H2O(s) + F2(g) → HF(g) + HOF(g)
This reaction is a redox reaction because-
1. | F2 is getting oxidized. | 2. | F2 is getting reduced. |
3. | Both (1) and (2)
|
4. | None of the above. |
The formulas for the following compounds are:
(a) Mercury(II) chloride and (b) Thallium(I) sulphate
1. HgCl2, Tl2SO4
2. Hg2Cl2, Tl2SO4
3. HgCl2, TlSO4
4. HgCl2, Tl3SO4
The correct statement(s) about the given reaction is -
\(\mathrm{{X} eO_{6 (aq)}^{4 -} + 2 F^{- 1}_{(aq)} + 6 H^{+}_{(aq )} \rightarrow}~\mathrm{{X} eO_{3 (g)} + {F}_{2 (g)} + 3 H_{2} O_{(l)}}\)
1. oxidises F-
2. The oxidation number of F increases from -1 to zero
3. is a stronger oxidizing agent that
4. All of the above.