(I) ${\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{O}}_3<mspace\; width="1em"></mspace>\stackrel{<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>}{\to }{\mathrm{H}}_2\mathrm{O}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2{\mathrm{O}}_2$

(II) ${\mathrm{H}}_2{\mathrm{O}}_2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{Ag}}_2\mathrm{O}<mspace\; width="1em"></mspace>\stackrel{<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>}{\to }2\mathrm{Ag}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{H}}_2\mathrm{O}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{O}}_2$

Role of hydrogen peroxide in the above reactions is respectively

(a) oxidising in (I) and reducing in (II)

(b) reducing in (I) and oxidising in (II)

(c) reducing in(I) and  (II)

(d) oxidising in(I) and (II)

In acidic medium, H₂O₂ changes ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ to CrO₅ which has two (-O-O-) bonds. Oxidation state of Cr in CrO₅ is

(a) +5

(b) +3

(c) +6

(d) -10

Nascent hydrogen consists of $-$

(A) Hydrogen atoms with excess energy

(B) Hydrogen molecules with excess energy

(C) Hydrogen ions in the excited state

(D) Solvated protons

Maximum concentration of ortho-$H_2$ in ordinary hydrogen is:

(1) 75% ortho-$H_2$ + 25% para-$H_2$

(2) 25% ortho-$H_2$ + 75% para-$H_2$

(3) 50% ortho-$H_2$ + 50% para-$H_2$

(4) 99% para-$H_2$ + 1% ortho-$H_2$

Blackened oil planting can be restored into original form by the action of:

1. Chlorine

2. ${\mathrm{BaO}}_2$

3. ${\mathrm{H}}_2{\mathrm{O}}_2$

4. ${\mathrm{MnO}}_2$

Which of the following statement regarding ${\mathrm{H}}_2{\mathrm{O}}_2$ is wrong.

(1) It is stable in acid medium

(2) It acts as oxidising as well as reducing agent

(3) It has zero dipole moment

(4) Pure ${\mathrm{H}}_2{\mathrm{O}}_2$ is slightly acidic

${\mathrm{Why\; does\; H₂O\; have\; a\; higher\; boiling\; point\; than\; H₂S?}}_{}$

1.  ${\mathrm{H₂O\; molecules\; are\; smaller\; and\; hence\; more\; closely\; packed}}_{}$

2. H₂O has a larger bond angle than H₂S, leading to tighter packing

3.  Due to intermolecular hydrogen bonding in liquid H₂O

4.  H₂O has a higher latent heat of vaporisation than H₂S

${\mathrm{H}}_2{\mathrm{O}}_2$ acts as antiseptic due to its –

(A) Reducing property                           

(B) Oxidising property

(C) Bleaching property                           

(D) Acidic property

Which contains both polar and non-polar bonds –

(A) ${\mathrm{NH}}_4\mathrm{Cl}$                                    

(B) $\mathrm{HCN}$

(C) ${\mathrm{H}}_2{\mathrm{O}}_2$                                     

(D) ${\mathrm{CH}}_4$

Exhausted permutit does not contain  –

1. ${\mathrm{Na}}^{+}$ ion                                     

2. ${\mathrm{Mg}}^{2+}$ ion

3. ${\mathrm{Al}}^{3+}$ ion                                  

4. ${\mathrm{Si}}^{4+}$ ion