Class 11th

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Option (i)

Groups 7,8, 9

Explanation: The metals belonging to group 7, 8 and 9 do not have tendency to form hydrides.

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Option (iii)

Na₆P₆O₁₈

Sodium hexametaphosphate (Na₆P₆O₁₈), commercially called 'calgon', is used to remove hardness from water.

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Option (i)

Ca²⁺

Water becomes 'hard' when it contains calcium and magnesium salts in the form of hydrogen carbonate, chloride, and sulphate.

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Option (iv)

CO (g)+2H₂ (g) → CH₃OH (l)

The reaction shown in equation (iv) shows the synthesis of methanol from water gas.

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Option (ii)

Sulphuric acid

2HSO₄^- (aq) → HO₃SOOSO₃H (aq) → 2HSO₄^- (aq)+2H⁺ (aq) + H₂O₂ (aq)

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Option (iv)

Sodium sulphate and hydrogen peroxide.

Na₂O₂+dil H₂SO₄ → Na₂SO₄ → Na₂SO₄+H₂O₂

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Option (iii)

CO (g)+H₂O (g) → CO₂ (g)+H₂ (g)

The production of dihydrogen can be increased by reacting carbon monoxide of syngas mixtures with steam in the presence of iron chromate as catalyst.

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Option (iii)

Both an oxidizing and a reducing agent

In both acidic and alkaline medium hydrogen peroxide can act as oxidizing and reducing agent.

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Option (ii)

I₂ + H₂O₂ + 2OH^–→ 2I^-+ 2H₂O + O₂

As iodine is reduced from 0 to -1.

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Option (iii)

2I^- + 2H⁺ + H₂O₂ → I₂+ 2H₂O

As the oxidation state of iodine changes from -1 to 0, this shows peroxide is oxidising in nature.