Polarity of bond depends on difference in electronegativity of the two concerned atoms. $${H_2}O$$  is more polar than $${H_2}S$$  because oxygen $$\left( {{\text{in}}\,O - H} \right)$$   is more electronegative than sulphur $$\left( {{\text{in}}\,S - H} \right).$$

Enthalpy of bond dissociation of deuterium is higher $$\left( {443.35\,kJ\,mo{l^{ - 1}}} \right)$$    than hydrogen $$\left( {435.88\,kJ\,mo{l^{ - 1}}} \right).$$

$$\mathop {2{H_2}{O_2}}\limits_{68\,g} \to 2{H_2}O + \mathop {{O_2}}\limits_{22.4\,L\,\,{\text{at}}\,\,STP} $$
$$22.4\,L\,{O_2}\,{\text{is given by}}\,68\,g\,\,{\text{of}}\,\,{H_2}{O_2}$$
$$10\,L\,{O_2}\,{\text{is given by}}\,\frac{{68}}{{22.4}} \times 10$$       $$ = 30.35\,g\,{L^{ - 1}}$$

$$\eqalign{ & N = \frac{{{W_2} \times 1000}}{{E.{w_2} \times VmL}} \cr & = \frac{{34 \times 1000}}{{17 \times 1120}} \cr & = \frac{{200}}{{12}} \cr & 1{\text{ }}N{\text{ of }}{H_2}{O_2} = 5.6{\text{ }}vol{\text{ strength}} \cr} $$
$$\frac{{200}}{{12}}N\,{\text{of}}\,{H_2}{O_2} = 5.6 \times \frac{{200}}{{12}} = $$       $$10\,vol{\text{ strength}}$$

Option (i), (ii) and (iii) are correct.

Ionic character increases with the size of the cation as we move down the group.

Heavy water is $${D_2}O$$  deuterium oxide.

The reducing agent loses electron during redox reaction i.e. oxidises itself.
$$\eqalign{ & \left( {\text{A}} \right){H_2}\mathop {{O_2}}\limits^{ - 1} + 2{H^ + } + 2{e^ - } \to 2{H_2}\mathop O\limits^{ - 2} \left( {{\text{Red}}{\text{.}}} \right) \cr & \left( {\text{B}} \right){H_2}\mathop {{O_2}}\limits^{ - 1} \to \mathop {{O_2}}\limits^0 + 2{H^ + } + 2{e^ - }\left( {{\text{Ox}}{\text{.}}} \right) \cr & \left( {\text{C}} \right){H_2}\mathop {{O_2}}\limits^{ - 1} + 2{e^ - } \to 2\mathop O\limits^{ - 2} {H^ - }\left( {{\text{Red}}{\text{.}}} \right) \cr & \left( {\text{D}} \right){H_2}O_2^{ - 1} + 2O{H^ - } \to \mathop {{O_2}}\limits^0 + {H_2}O + 2{e^ - }\left( {{\text{Ox}}{\text{.}}} \right) \cr} $$

$$100\,g$$  of $${H_2}{O_2}$$  sample solution contains $$ = x\,g$$  of $${H_2}{O_2}$$
$$3.4\,g$$  of $${H_2}{O_2}$$  sample solution contains $$ = \frac{x}{{100}} \times 3.4$$
$$\eqalign{ & {\text{Weight of}}\,{H_2}{O_2} = \frac{{3.4\,x}}{{100}} \cr & Eq.\,{\text{of}}\,{H_2}{O_2} = \frac{{3.4\,x}}{{100}} \times \frac{1}{{17}} \cr & mEq\,\,{\text{of}}\,{H_2}{O_2} \cr & = \frac{{3.4x}}{{100 \times 17}} \times 1000 \cr & = \frac{{34\,x}}{{17}} = 2x \cr & mEq\,\,{\text{of}}\,\,KMn{O_4} = x \times N \cr & x \times N = 2x \cr & N = 2 \cr} $$

On moving from left to right in a period. the acidic character of hydrides increases. $$N{H_3}$$  is basic, $${H_2}O$$  is neutral and $$HF$$  is acidic.