Chemistry M.C.Q (1 Marks)

Reduction Half reaction : $Cr _2 O _7^{2-}+6 e ^{-} \longrightarrow 2 Cr ^{3+}$

Oxidation Half reaction: $SO _3^{2-} \longrightarrow SO _4^{2-}+2 \overline{ e } \times 3$

Overall reaction : $Cr _2 O _7^{2-}+3 SO _3^{2-} \longrightarrow 2 Cr ^{3+}+3 SO _4^{2-}$

- To balance 'O' atoms, adding $H _2 O$ on LHS

$Cr _2 O _7^{2-}+3 SO _3^{2-} \longrightarrow 2 Cr ^{3+}+3 SO _4^{2-}+4 H _2 O$

- To balance 'H' atoms, adding $H ^{+}$on $RHS$

$Cr _2 O _7^{2-}+3 SO _3^{2-}+8 H ^{+} \longrightarrow 2 Cr ^{3+}+3 SO _4^{2-}+4 H _2 O$

$\therefore \quad a =1$

$b =3$

$c =8$

$\underline CH _{4}(-4)$

$\underline CCl _{4}(+4)$

$-4$ to $+4$

$C_2O_4^{-2}\rightarrow CO_2\; 2e^-\; loss\dots(2)$

multiplying $( 1 )$ by $2$ and $( 2 )$ by $5$ to balance $e^-$

$2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{-2} \longrightarrow 2 \mathrm{Mn}^{+2}+10 \mathrm{CO}_{2}$

on balancing charge;

$2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{-2}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{+2}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}$

Soluble complex 

$\mathrm{I}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}$ water soluble

Here, the oxidation state of every atom remains the same so, it is not a redox reaction.

$\stackrel{0}{\mathrm{Fe}}+\mathrm{CuSO}_{4} \rightarrow \stackrel{+2}{\mathrm{FeSO}_{4}}+\mathrm{Cu}$

$\stackrel{0}{\mathrm{Fe}}+5 \mathrm{CO} \rightarrow \stackrel{0}{\mathrm{Fe}}(\mathrm{CO})_{5}$

$\stackrel{0}{Fe}+\mathrm{H}_{2} \mathrm{O} \rightarrow \stackrel{+3}{\mathrm{Fe}_{2} \mathrm{O}_{3}}+\mathrm{H}_{2}$

above reactions shows the oxdiation state of iron for all given condition.

$\mathrm{KMnO}_{4}$ is $5$

$(a)$ For $\mathrm{FeSO}_{3}$

$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}$      (No. of $e^-$  $s$ involved $=1$ )

$\mathrm{SO}_{3}^{2-} \longrightarrow \mathrm{SO}_{4}^{2-}$       (No. of $e^-$  $s$ involved $=2$ )

Total number of $e^{-}$ $s$ involved $=1+2=3$

$(b)$ For  $\mathrm{FeC}_{2} \mathrm{O}_{4}$

$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}$       (No. of  $e^-$  $s$ involved $=1$ )

$\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow 2 \mathrm{CO}_{2}$ (No. of  $e^-$  $s$ involved $=2$)

Total number of $e^{-}$ $s$ involved $=1+2=3$

$(c)$ For $\mathrm{Fe}\left(\mathrm{NO}_{2}\right)_{2}$

$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}$      (No. of  $e^-$  $s$ involved $=1$ )

$2 \mathrm{NO}_{2} \rightarrow 2 \mathrm{NO}_{3}^{-}$    (No. of  $e^{-}$  $s$ involved $=4$)

Total number of $e^-$  $s$ involved $=1+4=5$

(d) For $FeSO_4$,

$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}$      (No. of  $e^-$  $s$ involved $=1$)

Total number of $e^-$  $s$ involved $= 1$

As $FeSO_4$, requires least number of electrons thus, it will require least amount of $\mathrm{KMnO}_{4}$.

$\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{O}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{O}+2 \mathrm{O}_{2}$

$\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{Ag}_{2} \mathrm{O} \longrightarrow 2 \mathrm{Ag}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$

$\mathrm{H}_{2} \mathrm{O}_{2}$ acts as reducing agent in all those reactions in which $\mathrm{O}_{2}$ is evolved.

This deep blue coloured complex has the following structure

Oxidation state of $Cr$ is $+6$ due to the presence of two peroxide linkages which can be calculated as $Cr$ peroxide normal $x+(-1) 4+(-2)=0, x-6=0$ and $x=+6$

$5 \mathrm{Na}_{2} \mathrm{SO}_{3}+3 \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{KMnO}_{4} \rightarrow 5 \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+3 \mathrm{H}_{2} \mathrm{O}$

So, number of moles of sodium sulphite that reacts with one mole of $\mathrm{KM} \mathrm{nO}_{4}=\frac{5}{2}$

The oxidation state of $Cr$ in $K _2 Cr _2 O _7$ is calculated as :

$2+2 x +(-2) \times 7=0$

$\Rightarrow 2+2 x -14=0$

$\Rightarrow 2 x -12=0$

$\Rightarrow 2 x =+12$

$\Rightarrow x =+6$

$\therefore$ Oxidations state of $Cr =+3$

Potassium dichromate reacts with hydrogen fluoride to produce potassium fluorotrioxochromate $(VI)$ and water. The reaction involved is :

$K _2 Cr _2 O _7+2 HF \rightarrow 2 K \left[ CrO _3 F \right]+ H _2 O$

The oxidation state of $Cr$ in $K \left[ CrO _3 F \right]$ can be calculated as follows:

$+1+ x +(-2) \times 3+(-1)=0$

$\Rightarrow+1+ x -6-1=0$

$\Rightarrow x -6=0$

$\Rightarrow x =+6$

$\therefore$ Oxidations state of $Cr =+3$

Since, there is no change in the oxidation state of $Cr$ in the reactant and in the product, the color of the solution doesn't change. Therefore, the color of acidified $K _2 Cr _2 O _7$ is not changed by $HF$.

$NH_3$ being basic in solution, does not change yellow colour of $CrO_4\,^{2-}(aq.)$ .

$(b)\,\,CuS{O_4} + {\text{KCN}} \to {{\text{K}}_2}{\text{S}}{{\text{O}}_4} + {\text{Cu}}{({\text{CN}})_2} \downarrow \xrightarrow[\begin{subarray}{l} 
  Intramolecular \\ 
  \operatorname{Re} dox 
\end{subarray} ]{\begin{subarray}{l} 
  Spon\tan eous \\ 
  (R.T.) 
\end{subarray} }{\text{Cu}}({\text{CN}}) \downarrow  + {({\text{CN}})_2} \uparrow $

$(c)\,{\text{Pb}}{\left( {{\text{C}}{{\text{H}}_3}{\text{COO}}} \right)_2} + {\text{KI}}\xrightarrow[{{\text{Non - redox}}\,reaction}]{{Ion\,Exchange\,reaction}}{\text{Pb}}{{\text{I}}_2} \downarrow  + {\text{C}}{{\text{H}}_3}{\text{COOK}}$

$(d)\,{\text{A}}{{\text{g}}_2}{\text{O}} + {\text{S}}{{\text{O}}_2}\xrightarrow{{\operatorname{Re} dox}}{\text{Ag}} + {\text{S}}{{\text{O}}_3} \uparrow $

$\mathrm{Fe}^{3+}(a q .)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S} \longrightarrow \mathrm{Fes} \downarrow$

$\mathrm{Mn}^{2 +}(a q .)+\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{NH}_{3} \longrightarrow \mathrm{MnO}(\mathrm{OH})_{2} \downarrow$ or $\mathrm{MnO}_{2} \downarrow .2 \mathrm{H}_{2} \mathrm{O}$

$\mathrm{Fe}^{2 +}(a q .)+\mathrm{Na}_{2} \mathrm{O}_{2} \mathrm{sol} \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3} \downarrow$

$A{g_2}{C_2}{O_4}\xrightarrow{\Delta }2Ag + 2C{O_2}$

$pb{(N{O_3})_2}\xrightarrow{\Delta }pbO + 2N{O_2} + \frac{1}{2}{O_2}$

$A{g_2}C{O_3}\xrightarrow{\Delta }2Ag + C{O_2} + \frac{1}{2}{O_2}$

$C_2H_5OH + 4I_2 + 6OH^-\to  CHI_3 + HCO_2\,^-+ 5I^-+ 5H_2O$

The balanced chemical reaction is as given below.

$\mathrm{P}_{4}+3 \mathrm{NaOH}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{PH}_{3}+3 \mathrm{NaH}_{2} \mathrm{PO}_{2}$

$5 \mathrm{BiO}_{3}^{-}+14 \mathrm{H}^{+}+2 \mathrm{Mn}^{2+} \rightarrow 5 \mathrm{Bi}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{MnO}_{4}^{-}$

In this equation, the number of atoms of each element on the left side are equal to the number of atoms on the right side.

The charges are also balanced on the two sides of equation.

Hence, option B is the correct answer.

$2 \mathrm{MnO}_{4}^{-}+5 \mathrm{CO}_{4}^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{SO}_{4}^{2-}+3 \mathrm{H}_{2} \mathrm{O}$

From the balanced chemical equation we can say that:

$\because 5$ moles $\mathrm{SO}_{4}^{2-}$ reacts with 2 moles of $\mathrm{KMnO}_{4}$.

$\therefore 1$ mole $\mathrm{SO}_{4}^{2-}$ reacts with $2 / 5$ mole of $\mathrm{KMnO}_{4}$

Hence, the correct option is $\mathrm{A}$

$Cr _2 O _7^{2-}+ Sn ^{+2}+ H ^{+} \rightarrow Sn ^{-4}+ Cr ^{+3}+ H _2 O$

Step $1:$ Assign the oxidation state using oxidation number of $O =-2$.

we get:

$Cr _2 O _7^{2-}+ Sn ^{+2+}+ H ^{+} \rightarrow Sn ^{+4}+ Cr ^{+3}+ H _2 O$

Reduction half-reaction:

$Cr _2 O _7^{2-} \rightarrow 2 Cr ^{-3}:$ Gain of 6 electrons

Oxidation half reaction:

$Sn ^{+2} \rightarrow Sn ^{+4}:$ Loss of $2$ electrons

Step $2:$ Equalise the number of electrons as:

Oxidation half reaction:

$3 Sn ^{-2} \rightarrow 3 Sn ^{+4}$

Step $3:$ balance $O$ atoms by adding $H _2 O$ and then $H$ by $H ^{+}$

$Cr _2^{+6} O _7^{2-}+14 H ^{+} \rightarrow 2 Cr ^{-3+}+7 H _2 O$

Step $4:$ overall reaction:

$Cr _2^{+6} O _7^{2-}+14 H ^{+}+3 Sn ^{+2} \rightarrow 2 Cr ^{3+}+3 Sn ^{+4}+7 H _2 O$

Thus $3$ mole of $Sn ^{2+}$ will reduce 1 moles of $K _2 Cr _2 O _7$.

Therefore,$1$ mole of $Sn ^{2+}$ will reduce $\frac{1}{3}$ moles of $K _2 Cr _2 O _7$.

$( C$ and $H )$

$2C{u^ + }(aq.)\xrightarrow{{R.T.}}Cu + C{u^{2 + }}(aq.)$

$3\mathop {Mn}\limits^{ + 6} O_4^{2 - }(aq.)\xrightarrow[{dispn.Rn.}]{{R.T.}}2\mathop {Mn}\limits^{ + 7} O_4^ - (aq.) + \mathop {Mn}\limits^{ + 4} {O_2} + 4O{H^ - }$

$Ca(OCl)Cl\xrightarrow{{Water}}C{a^{2 + }}(aq.) + C{l^ - }(aq.) + OC{l^ - }(aq.)$

$\therefore \,$ New oxidation state of $N$ is $-2+4  \Rightarrow 2$

$n$ factor for $1\, mol HCl =\frac{2}{4}=\frac{1}{2}$

Eq. wt. of $HCl =\frac{ M }{ n \cdot \text { factor }}=\frac{ M }{1 / 2}=2\, M$

Lead $(IV)$ oxide and the lead atom shares its $4$ electrons $2$ each with oxygen atoms. In Peroxide there a bond sharing of two oxygen atoms as in:

Hydrogen Peroxide $H - O - O - H$

Water is an oxide $H - O - H$

Likewise $PbO _2$ is an oxide $P$ b has a valency $4$

$O = Pb = O$

which can be written as $\mathrm{NH}_{4}^{+} \mathrm{NO}_{3}^{-}$ For $\mathrm{NH}_{4}^{+}$

Let oxidation state of $\mathrm{N}=\mathrm{x}$

we know that oxidation state of $\mathrm{H}=+1$

Sum of total oxidation state of all atoms $=$ Overall charge on the compound. $x+4 \times(1)=1$

$\mathrm{x}=-3$

Let oxidation state of $\mathrm{N}$ in $\mathrm{N} \mathrm{O}_{3}^{-}=\mathrm{y}$, we know that oxidation state of $\mathrm{O}=-2$

Applying above formula $y+3 \times(-2)=1$

$y=+5$

Oxidation state of $\mathrm{N}$ in $\mathrm{NH}_{4} \mathrm{N} \mathrm{O}_{3}=-3,+5$

$\mathrm{EW}=\frac{32}{4}=8$

Equivalent mass of $\mathrm{KMnO}_{4}=\frac{\text { Molecular mass }}{\text { Change in oxidation number }}$ $=\frac{158}{5}=31.6$

$\begin{array}{ll}\mathrm{CrO}_{2} \mathrm{Cl}_{2} & +6 \\ \mathrm{MnO}_{4}^{-} & +7 \quad \text { (Maximum) } \\ \mathrm{MnO}_{2} & +4 \\ {\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}} & +3 \\ {\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}} & +3 \\ \mathrm{MnO} & +2\end{array}$

$\Rightarrow \quad x=-2$

$\mathrm{NH}_{3} \Rightarrow x+3(+1)=0 \Rightarrow x=-3$

$\mathrm{N}_{3} \mathrm{H} \Rightarrow 3 x+1(+1)=0$

$\Rightarrow 3 x+1=0 \Rightarrow x=-1 / 3$

$\mathrm{NH}_{2} \mathrm{OH} \Rightarrow x+2+1(-2)+1=0$

$\Rightarrow x+1=0$

$\Rightarrow x=-1$

Change in oxidation no. of

reductant $=$ moles of electrons transferred by

$1$ mol reductant $=3$

$\mathrm{Fe}: \mathrm{x}+6+(4 \times-2)=0$

$\Rightarrow \mathrm{x}=8-6$

$\Rightarrow \mathrm{x}=+2$

$\mathrm{Cr}: 2+2 \mathrm{x}+(4 \times-2)=0$

$\Rightarrow 2 \mathrm{x}=8-2$

$\Rightarrow \mathrm{x}=6 / 2$

$\Rightarrow \mathrm{x}=+3$

$\mathop {Cl{O^ - }}\limits^{ + 1} ,\, + \mathop {HCl{O_3}}\limits^{ + 5} ,\,\mathop {NaCl{O_3}}\limits^{ + 5} $

$+1 \; \stackrel{-1}{-2}\,+2$

direction of co-ordinate bond is from more $EN$ atom to less $EN$ atom. So there is no development of charge due to this co-ordinate bond.

So in $\mathrm{HNC} \Rightarrow$

$\mathrm{O} \cdot \mathrm{N} \text { of } \mathrm{H}=+1$

$\mathrm{O} \cdot \mathrm{N}$ of $\mathrm{N}=-3$

$\mathrm{O} \cdot \mathrm{N}$ of $\mathrm{C}=+2$

$\therefore x+4(-2)=-3=>x=+5$

Let oxidation number of $\sin \mathrm{SO}_{4}^{2-}$ be y.

$\therefore y+4(-2)=-2=>y=+6$

Let oxidation number of Crin $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ be z.

$\therefore 2 z+7(-2)=-2=>z=+6$

$\Rightarrow \Delta \mathrm{EN}=0$

$O. {N}$ of $\mathrm{S}=0$

$\mathrm{S}_{2} \mathrm{F}_{2} \Rightarrow(\mathrm{O} . \mathrm{N} \text { of } \mathrm{F} \text { is always }-1)$

$2 x+(-1) \times 2=0$

${2 x=2} $

${x=+1}$

So $O.{N}$ of $\mathrm{S}$ in $\mathrm{S}_{2} \mathrm{F}_{2}=+1$

$\mathrm{H}_{2} \mathrm{S}$

$ 1 \times 2+x =0 $

$ x=-2$

So $O. {N}$ of $\mathrm{S}$ in $\mathrm{H}_{2} \mathrm{S}=-2$

${\text{Equivalent}}\,\,{\text{weight}}\,\,{\text{of }}{K_2}C{r_2}{O_7}$

$ = \frac{{{\text{Molecular}}\,\,{\text{Mass}}}}{6}$$ = \frac{{294.2}}{6} = \frac{M}{6}$.

$2 + x - 2 \times 4 = 0$

$x = + 6$.

$2 \times 3{e^ - } = + 6$.

     (b) $H\mathop N\limits^ * O$;$1 + x - 2 = 0$;$x = + 1$

     (c) $\mathop {N{H_3}}\limits^{ * \,\,\,\,\,\,} $; $x + 3 = 0$; $x = - 3$

     (d) ${\mathop {N_2}\limits^ *}{O_5}$; $2x - 10 = 0$;$2x = 10$; $x = \frac{{10}}{2}$; $x = 5.$

Suppose molecular weight is $M$ Oxidation number of ${I_2}$ in $IO_4^ - $ in Acidic medium i.e., $I \times ( - 8) + 1{e^ - } = + 7$

So eq. wt. $ = M/7$.

as we know oxygen in peroxide is hare $-1$ oxidation no.

So, $BaO _2$ and $H _2 O_2$ are peroxide have $-1$ oxidation in oxygon.

Equivalent wt. = $\frac{{{\rm{Mol}}{\rm{. wt}}{\rm{.}}}}{{\rm{2}}}$=$\frac{{34}}{2}$ $=17.$

$x + ( - 3) = 0 \Rightarrow x = + 3$

$( + 1) \times 2 + x = 0$

$2 + x = 0$; $x = - 2$

$F{{e}^{++}}\xrightarrow{{}}\,[Ar]\,3{{d}^{6}}\,4{{s}^{0}}$

$F{{e}^{+++}}\xrightarrow{{}}\,[Ar]\,3{{d}^{5}}4{{s}^{0}}$

In $ +2 $ state $Fe$ is called Ferrous & in $+3$  state as ferric.

$x + 3( + 1) = 0$, $x = - 3$.

$x + 2 - 2 = 0$; $x = 0\,\,in\,\,C{H_2}C{l_2}$.

$1 + x - 2 \times 4 = 0$; $x = 8 - 1 = + 7$.

$2 \times ( + 1) + x + 4 \times ( - 2) = 0$

$ + 2 + x - 8 = 0$; $x = 8 - 2 = + 6$

Electronic configuration of sulphur in ${H_2}S{O_4}$ is

$1{s^2},\,\,2{s^2},\,\,2{p^6}$.

$\mathop {{H_2}S{O_4}}\limits^{\,\, * } = + 6$; $\mathop {{H_2}S}\limits^{\,\,\,\,\,\,\,\,\, * } = - 2$.

$3x + 1 = 0$

$3x = - 1,\, \Rightarrow x = - \frac{1}{3}\,\,{\rm{in}}\,\,{N_3}H$

$x + 2\,( + 1) + 1\,( - 2) + 1(1) = 0$

$x = - 1\,\,{\rm{in}}\,\,N{H_2}OH$

$x \times 2 + 4\,(1) = 0$$x = - \frac{4}{2} = - 2\,{\rm{in}}\,\,{N_2}{H_4}$

$x + 3\,(1) = 0$$x = - 3\,\,in\,\,N{H_3}$

Hence, highest in ${N_3}H$.

$HCHO = 0$

$CHC{l_3} = + 2$

$C{H_3}OH = - 2$

${C_{12}}{H_{22}}{O_{11}} = 0$

$2x + 5 = + 1$; $2x = 1 - 5$

$2x = - 4$; $x = - 2$.

$1 + x - 2 \times 4 = 0$; $x = 8 - 1 = + 7$.

$x - 4 = 0$                                 $4 + 2x = 0$
$x = + 4$                                     $2x = - 4$
                                                    $x = \frac{{ - 4}}{2} = - 2$

$\mathop {{H_2}S{O_4}}\limits^ * = + 6$

$\mathop {N{a_2}{S_2}{O_3}}\limits^{\,\, * } = + 2$

$\mathop {N{a_2}{S_4}{O_6}}\limits^{\,\, * } = + \frac{5}{2}$.

$\mathop {{N_3}H}\limits^{ * \,\,\,\,\,\,\,\,} $

$3x + 1 = 0$, $3x = - 1$, $x = - \frac{1}{3}$.

$H - O - \mathop {\mathop S\limits_{||}^{||} }\limits_O^O  - O - O - \mathop {\mathop S\limits_{||}^{||} }\limits_O^O  - O - H$

:-So the oxidation number of S should be :$\mathop {2 \times ( + 1)}\limits_{({\rm{for}}\,\,H)} + \mathop {2 \times X}\limits_{({\rm{for}}\,\,S)} + \mathop {6 \times ( - 2)}\limits_{({\rm{for}}\,\,O)} + \mathop {2 \times ( - 1)}\limits_{({\rm{for}}\,\,O - O)} = 0$ or $X = + 6$.

$1 + x - 2 \times 4 = 0;\,\,1 + x - 8 = 0$
$x = 8 - 1 = + 7$ oxidation state.

$x + [( - 2) \times 4] = - 2$$2x + ( - 2) \times 7 = - 2$$x = 8 - 2 = + 6$$2x = 14 - 2 = 12$,$x = \frac{{12}}{2} = + 6$

In this reaction oxidation and reduction are not involved because there is no change in oxidation number.

$NO _3{ }^{-} \rightarrow+5$

$NH _3 \rightarrow-3$ [ Number of moles of electrons taken $=5-(-3)=8$ ]

$NH _2 OH \rightarrow-1$ [ Number of moles of electrons taken $=5-(-1)=6$ ]

$NO \rightarrow+2$ [ Number of moles of electrons taken $=5-(+2)=3]$

$NO _2 \rightarrow+4$ [ Number of moles of electrons taken $\left.=5-(4)=1\right]$

$(a)$ $2{\text{HCl}} + {\text{Pb}}{{\text{O}}_2} \to {\text{PbO}} + {\text{C}}{{\text{l}}_2} \uparrow  + {{\text{H}}_2}{\text{O}}$

$(b)$ $HCl + conc.{H_2}S{O_4} \to No\,reaction$

$(c)$ $4{\text{HCl}} + {\text{Mn}}{{\text{O}}_2} \to {\text{MnC}}{{\text{l}}_2} + 2C{l_2} \uparrow  + 2{{\text{H}}_2}{\text{O}}$

$(d)$ $2{\text{HCl + }}{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7} + 12{{\text{H}}^ + } \to 2{\text{C}}{{\text{r}}^{3 + }}(aq.) + 2{{\text{K}}^ + }(aq.) + {\text{C}}{{\text{l}}_2} \uparrow  + 7{{\text{H}}_2}{\text{O}}$

$SO_2$ is colourless gas.

No change in oxidation number

$I _2$ undergoes reduction hence acts as oxidising agents.

$S _2^{+2} O _6^{2-} \rightarrow S _4^{+2.5} O _6^{2-}$

It undergoes oxidation hence acts as a reducing agent.

$\mathrm{SO}_{2}:$ Sulphur is present in +4 oxidation state so it can act as both oxidizing as well as reducing agent.

$\mathrm{H}_{2} \mathrm{O}_{2}:$ Oxygen is present in -1 oxidation state so it can act as both oxidizing as well as reducing agent.

$2NaN{O_2} + 2KI + 2{H_2}S{O_4}\xrightarrow{{}}\,N{a_2}S{O_4} + {K_2}S{O_4}$

$ + 2NO + 2{H_2}O + {I_2}$

Reducing property.

${H_2}{O_2} + NaN{O_2}\xrightarrow{{}}NaN{O_3} + {H_2}O$

$\mathop {HAs{O_2}}\limits^{ + 3}  + \mathop {Sn}\limits^{ + 2} \xrightarrow{{}}\,\mathop {As}\limits^0  + \mathop {Sn}\limits^{2 + }  + {H_2}O$

Hence, here $HAs{O_2}$ is acting as oxidizing agent.