A non-feasible reaction among the following is:

1. $2$ $\mathrm{KI}+{\mathrm{Br}}_2\to 2\mathrm{KBr}+{\mathrm{I}}_2$

2. $2$ $\mathrm{KBr}+{\mathrm{I}}_2\to 2\mathrm{KI}+{\mathrm{Br}}_2$

3. $2$ $\mathrm{KBr}+{\mathrm{Cl}}_2\to 2\mathrm{KCl}+{\mathrm{Br}}_2$

4. $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{F}}_2\to 4\mathrm{HF}+{\mathrm{O}}_2$

Standard electrode potentials are:

\(F e^{+ 2} / F e\) , \(E^\circ\) \(=\) \(- 0 .44\)

\(F e^{+ 3} / F e^{+ 2}\), \(E ^\circ\) \(= 0 . 77\)

Choose the correct observation when \(𝐹𝑒^{+2}\) ${\mathrm{}}^{}$, \(𝐹𝑒^{+3}\) and \(Fe_{(solid)}\) are kept together:

1. $F{e}^{+3}$ increases

2. $F{e}^{+3}$ decreases

3. $\frac{F{e}^{+2}}{F{e}^{+3}}$ remains unchanged

4. $F{e}^{+2}$ decreases

The oxidation states(O.S.) of sulphur in the anions ${{\mathrm{SO}}_3}^{2-},$ ${\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}$ $\mathrm{and}$ ${\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}$ follow the order:

1. ${\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{{\mathrm{SO}}_3}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}$

2. ${{\mathrm{SO}}_3}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}$

3. ${\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}<{{\mathrm{SO}}_3}^{2-}$

4. ${\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{{\mathrm{SO}}_3}^{2-}$

Fluorine reacts with ice as per the following reaction

H₂O(s) + F₂(g) → HF(g) + HOF(g)

This reaction is a redox reaction because:

The oxidation number of sulphur and nitrogen in H₂SO₅ and NO₃^- are respectively-

Which of the following correctly represents the formulas of (a) mercury(II) chloride and (b) thallium(I) sulphate?

1. HgCl₂,  Tl₂SO₄

2. Hg₂Cl₂, Tl₂SO₄

3. HgCl₂,  TlSO₄

4. HgCl₂,  Tl₃SO₄

Consider the following reactions:

(a) 6 CO_2(g) + 6H₂O_(l) → C₆H₁₂O_6(aq) + 6O_2(g)

(b) O_3(g) + H₂O_2(l) → H₂O_(l) + 2O_2(g)

These reactions are more appropriately written in the expanded form as:

(a) 6CO_2(g) + 12H₂O_(l) → C₆H₁₂O_6(aq) + 6H₂O_(l) + 6O_2(g)

(b) O_3(g) + H₂O_2(l) → H₂O_(l) + O_2(g) + O_2(g) 

Which of the following is the most appropriate reason for writing them in expanded form?

1. Water molecules are also formed as a product in photosynthesis.

2. Water molecules are not formed as a product in photosynthesis.

3. Oxygen molecules are also formed as a product in photosynthesis.

4. Oxygen molecules are not formed as a product in photosynthesis.

In the manufacture of benzoic acid from toluene, we use alcoholic potassium permanganate as an oxidant because -

1. The cost of adding an acid or a base can be reduced.

2. The reactions can proceed at a faster rate.

3. Both '1' and '2'

4. Neither '1' nor '2'

Consider the following reactions :

2 S₂O₃^2–_(aq) + I_2(s) → S₄O₆^2–_(aq) + 2I^–_(aq)

S₂O₃^2–_(aq) + 2Br_2(l) + 5H₂O_(l) → 2SO₄^2–_(aq) + 4Br^–_(aq) + 10H⁺_(aq)

The same reductant, thiosulphate reacts differently with iodine and bromine because:

1. Br₂ is a stronger oxidizing agent than I₂

2. I₂ is a stronger oxidizing agent than Br₂

3. Both '1' and '2'

4. Neither '1' nor '2'

The correct statement(s) about the given reaction is -

\(\small\mathrm{{X} eO_{6 (aq)}^{4 -} + 2 F^{- 1}_{(aq)} + 6 H^{+}_{(aq )} \rightarrow}~\mathrm{{X} eO_{3 (g)} + {F}_{2 (g)} + 3 H_{2} O_{(l)}}\)${}_{}$

1. ${\mathrm{XeO}}_6^{4-}$ oxidises F^-

2. The oxidation number of F increases from -1  to  zero

3. ${\mathrm{XeO}}_6^{4-}$ is a stronger oxidizing agent that ${\mathrm{F}}^{-}$

4. All of the above