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The oxide that gives hydrogen peroxide $(H_{2} O_{2})$ on the treatment with a dilute acid $(H_{2} {SO}_{4})$ is –

${MnO}_{2}$

${PbO}_{2}$

${Na}_{2} O_{2}$

${TiO}_{2}$

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Explanation

(C). ${Na}_{2} O_{2} + H_{2} {SO}_{4} \overset{}{\to} {Na}_{2} {SO}_{4} + H_{2} O_{2}$

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Which of the following statement is correct –

Hydrogen has same IP as alkali metals

H has same electronegativity as halogens

H has oxidation no. –1 and +1

It will not be liberated at anode

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Explanation

(C). Because it can lose and gain electrons. It acts as oxidising and reducing agent.

$H + e \to H^{-}, H - e \to H^{+} .$

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The oxidising property of $H_{2} O_{2}$ is best explained by assuming that two oxygen atoms in its molecule are.

Bonded differently

Bonded similarly

Bonded covalently

Bonded by hydrogen bonds

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Explanation

(A). It is believed that out of two hydrogens, one is associated with each oxygen atom at a bond

angle of $94.8 °$ . All the four atoms $H - O - O - H$ do not lie in the same plane.

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$H_{2} O$ has higher boiling point than $H_{2} S$ because.

$H_{2} O$ is a smaller molecule and hence more closely packed

The bond angle in $H_{2} O$ is more than in $H_{2} S$ and hence the former molecules are more tightly packed

Of intermolecular hydrogen bonding in liquid $H_{2} O$

The latent heat of vaporisation is higher for $H_{2} O$ than for $H_{2} S$

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Explanation

(C). The b.p of $H_{2} O$ is higher than that of $H_{2} S$ because of intermolecular hydrogen bonding in

liquid $H_{2} O$ .

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Hydrolysis of one mole of peroxodisulphuric acid produces

two moles of sulphuric acid

two moles of peroxomonosulphuric acid

one mole of sulphuric acid and one mole of peroxomonosulphuric acid

one mole of sulphuric acid, one mole of peroxomonosulphuric acid and one mole of hydrogen peroxide.

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Explanation

(C). $HO - {SO}_{2} - O - O - {SO}_{2} - OH + H_{2} O \overset{}{\to} {HOSO}_{2} {OH}^{+} {HOOSO}_{2} OH Peroxodisulphuric acid sulphuric acid permonosulphuric acid$

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The maximum possible number of hydrogen bonds a water molecule can form is –

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Explanation

(B). Water molecules can form two hydrogen bonds through its two hydrogen atoms and another

two with two lone pairs of electrons on the O atom.

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$H_{2} O_{2}$ acts as antiseptic due to its –

Reducing property

Oxidising property

Bleaching property

Acidic property

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Explanation

(B). $H_{2} O_{2}$ acts as antiseptic because it kills bacteria by oxidation. It gives nascent oxygen which

kills bacteria.

$H_{2} O_{2} \to H_{2} O + O .$

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Which contains both polar and non-polar bonds –

${NH}_{4} Cl$

$HCN$

$H_{2} O_{2}$

${CH}_{4}$

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Explanation

(C). $O - O$ bond is non-polar while $O - H$ bonds are polar.

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The most reactive state of hydrogen is –

Atomic hydrogen

Heavy hydrogen

Molecular hydrogen

Nascent hydrogen

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Explanation

atomic H is very reactive.

One molecule of hydrogen dissociates into two atoms (H₂ → 2H) when an energy equal to or greater than the dissociation energy (i.e., the amount of energy required to break the bond that holds together the atoms in the molecule) is supplied. The dissociation energy of molecular hydrogen is 104,000 calories per mole—written 104 kcal/mole

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In which of the following reactions is $H_{2} O_{2}$ acting as an acid.

$2 {KMnO}_{4} + 3 H_{2} {SO}_{4} + 5 H_{2} O_{2} \to K_{2} {SO}_{4} + {MnO}_{4} + 2 {MnSO}_{4} + 8 H_{2} O + 5 O_{2}$

$2 K_{3} Fe {(CN)}_{6} + 2 KOH + H_{2} O_{2} \to 2 K_{4} Fe {(CN)}_{6} + 2 H_{2} O + O_{2}$

$O_{3} + H_{2} O_{2} \to H_{2} O + 2 O_{2}$

$Ba {(OH)}_{2} + H_{2} O_{2} \to {BaO}_{2} + H_{2} O_{2}$

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Explanation

(D). $Ba {(OH)}_{2} + H_{2} O_{2} \to {BaO}_{2} + 2 H_{2} O$ is similar to acid base (neutralisation reaction). Aqueous

solution of $H_{2} O_{2}$ is weakly acidic.

$H_{2} O_{2} + H_{2} O \to H_{3} O^{+} + {HO}_{2}^{-}$

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