JEE Chemistry Redox Reactions MCQs Set 04

Answer: (c) -1

Explanation:
1. H = +1, O = -2, so x + 2(+1) + (-2) + (+1) = 0.
2. x + 1 = 0, so x = -1.

Teacher's Note:
1. Treat OH as a group of charge -1 when counting.
2. Hydroxylamine sits between NH3 (-3) and N2 (0).

Answer: (b) Fluorine

Explanation:
1. Fluorine is the most electronegative element, so it is always -1 in compounds.
2. H varies (+1, -1), C varies from -4 to +4 and O varies (-2, -1, -1/2, +1, +2).

Teacher's Note:
1. Fluorine never shows a positive oxidation state.
2. Oxygen is -2 only in most compounds, not all.

Answer: (c) -4, -2, 0, +2, +4

Explanation:
1. Each H is +1 and each Cl is -1.
2. CH4: x + 4 = 0, x = -4. CH3Cl: x + 3 - 1 = 0, x = -2. CH2Cl2: x + 2 - 2 = 0, x = 0.
3. CHCl3: x + 1 - 3 = 0, x = +2. CCl4: x - 4 = 0, x = +4.

Teacher's Note:
1. Replacing H by Cl raises the oxidation state of carbon by 2 each time.
2. The range -4 to +4 is the full range for carbon.

Answer: (b) Phosphorous acid

Explanation:
1. H3PO3: 3 + x - 6 = 0, x = +3.
2. H3PO4, HPO3 and H4P2O7 each have phosphorus at +5.

Teacher's Note:
1. Add the P-H bonds to the oxidation count: they count as +1 for H.
2. The -ous acid has the lower oxidation state, the -ic acid the higher.

Answer: (c) Oxidation state of S in \( \ce{H2SO5} \) is +8

Explanation:
1. H2SO5 (Caro's acid) contains one peroxide group: 2(+1) + x + 3(-2) + 2(-1) = 0, so x = +6.
2. A sulphur atom cannot be +8 because it has only six valence electrons.

Teacher's Note:
1. Always check for peroxide linkages before assigning -2 to every oxygen.
2. The maximum oxidation state of an element cannot exceed its group number.

Answer: (c) Both Os and Ru

Explanation:
1. OsO4 and RuO4 both contain the metal in the +8 state.
2. +8 is the highest known oxidation state.

Teacher's Note:
1. Group 8 elements reach +8 only with oxygen or fluorine.
2. Os and Ru both form tetroxides.

Answer: (d) 0

Explanation:
1. Sodium amalgam is an alloy of Na and Hg; all components of an alloy are in the zero state.

Teacher's Note:
1. An amalgam is a solution of a metal in mercury, not a compound.

Answer: (a) +2

Explanation:
1. In H-N=C, N is more electronegative than C.
2. 1 + (-3) + x = 0, so x = +2.

Teacher's Note:
1. Use electronegativity, not the written order of atoms.
2. HNC is the isocyanide isomer of HCN.

Answer: (b) \( \frac{200}{94} \)

Explanation:
1. 0.94 x + (-2) = 0.
2. x = 2/0.94 = 200/94.

Teacher's Note:
1. This is a non-stoichiometric (fractional) oxidation state.
2. It means some iron is Fe(II) and some Fe(III).

Answer: (a) +2

Explanation:
1. 2(+1) + 2x + 3(-2) = 0.
2. 2x = 4, so x = +2 (average).

Teacher's Note:
1. The two sulphur atoms are really -1 and +5, the average is +2.
2. Thiosulphate has one S in a sulphate-like and one in a sulphide-like state.

Answer: (a) \( \ce{CaOCl2} \)

Explanation:
1. Bleaching powder is Ca(OCl)Cl.
2. Cl in OCl^- is +1 and Cl in Cl^- is -1.

Teacher's Note:
1. The other compounds have only one kind of chlorine.

Answer: (b) Hydrazoic acid

Explanation:
1. N3H: 3x + 1 = 0, so x = -1/3.

Teacher's Note:
1. Hydrazoic acid is HN3; the average state of N is -1/3.
2. NH3 is -3, NO is +2 and N2O is +1.

Answer: (a) \( \frac{7}{3} \)

Explanation:
1. +3 + 2(+2) + 3x + 7(-2) = 0.
2. 3x = 7, x = 7/3.

Teacher's Note:
1. Fractional values signal a mixture of Cu(II) and Cu(III).

Answer: (a) +7

Explanation:
1. 4(+1) + x + 6(-2) = -1.
2. x = +7.

Teacher's Note:
1. Charge on the ion goes on the right-hand side.

Answer: (c) Hypophosphorous acid

Explanation:
1. H3PO2: 3 + x - 4 = 0, x = +1.
2. H3PO3 is +3, H3PO4 is +5, H4P2O6 is +4.

Teacher's Note:
1. Hypophosphorous acid is a strong reducing agent.

Answer: (c) \( \ce{Fe(CO)5} \)

Explanation:
1. Mohr's salt: +2. K4[Fe(CN)6]: +2. Fe0.94O: about +2.13.
2. Fe(CO)5: 0.

Teacher's Note:
1. Metal carbonyls always have zero-valent metal.

Answer: (d) \( \ce{BaO2} \)

Explanation:
1. BaO2 is a peroxide containing \( \ce{O2^2-} \).
2. H2O and Na2O: -2. O2F2: +1.

Teacher's Note:
1. Peroxides contain an O-O single bond and have -1 on each O.

Answer: (d) +12

Explanation:
1. Peroxydisulphuric acid has one O-O peroxide bond.
2. 2(+1) + 2x + 6(-2) + 2(-1) = 0, x = +6 for each S.
3. Sum = +12.

Teacher's Note:
1. Do not forget the two peroxide oxygens (each -1).

Answer: (c) \( \ce{Cr2O3} \)

Explanation:
1. Cr2O3: 2x - 6 = 0, x = +3.
2. CrO3, CrO2Cl2 and dichromate contain Cr(VI).

Teacher's Note:
1. Cr2O3 is green chromic oxide with Cr(III).

Answer: (c) +3

Explanation:
1. 3(+1) + x + 3(-2) = 0.
2. x = +3.

Teacher's Note:
1. Same pattern as phosphorous acid H3PO3.

Answer: (d) -3, +1, +5, +3

Explanation:
1. PH4^+: x + 4 = +1, x = -3.
2. PO2^3-: x - 4 = -3, x = +1.
3. PO4^3-: x - 8 = -3, x = +5. PO3^3-: x - 6 = -3, x = +3.

Teacher's Note:
1. Write each equation separately and keep the sign of the charge.

Answer: (d) \( \ce{H2S} \), \( \ce{H2S2O3} \), \( \ce{H2SO3} \), \( \ce{H2SO4} \)

Explanation:
1. H2S: -2. H2S2O3: +2 (average). H2SO3: +4. H2SO4: +6.
2. Increasing order: H2S, H2S2O3, H2SO3, H2SO4.

Teacher's Note:
1. List the oxidation numbers first, then sort.

Answer: (b) \( \ce{H2O2} \)

Explanation:
1. Oxygen in H2O2 is -1, an intermediate state between 0 and -2.
2. It can go up to O2 or down to H2O.

Teacher's Note:
1. KMnO4 and K2Cr2O7 have the top state, so they are only oxidants.

Answer: (b) an oxidant in (A) and a reductant in (B)

Explanation:
1. (A): S goes from -2 to +6, H2O2 is reduced to H2O, so H2O2 is the oxidant.
2. (B): Cl2 goes to -1 (reduced), H2O2 is oxidised to O2, so it is the reductant.

Teacher's Note:
1. Look at what happens to the other reactant to decide the role of H2O2.

Answer: (b) \( \ce{H2S} \)

Explanation:
1. HNO2 can oxidise only reductants.
2. H2S is a strong reductant; KMnO4, K2Cr2O7 and Br2 are oxidants and oxidise HNO2 to HNO3.

Teacher's Note:
1. HNO2 (N = +3) is an oxidant towards reductants and a reductant towards oxidants.