Arrange $$\left( {NNN} \right),\left( {BrCC} \right),\left( {OOO} \right),\left( {CHH} \right),\left( {OOH} \right)$$         in increasing atomic number. The order is ii, iii, v, i, iv.

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In $$ - C{H_3}, - OC{H_3}$$    and $$C{F_3},C{H_3}$$   and $$ - OC{H_3}$$  are electron donating group. Hence, they activate the benzene nucleus. In these, order of activation is $$ - OC{H_3} > - C{H_3}$$    while $$ - C{F_3}$$  group deactivates the benzene nucleus. So, it shows lower rate of electrophilic substitution on benzene ring. Thus, order of electrophilic substitution is

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$${\text{Mass of }}C{O_2}{\text{ formed}} = 0{\text{ }}{\text{.4655}}\,{\text{g}}$$
$${\text{Mass of organic compound taken}}$$       $$ = 0.2115{\text{ }}g$$
$$\eqalign{ & \% \,{\text{of}}\,C = \frac{{12}}{{44}} \times \frac{{{\text{mass}}\,\,{\text{of}}\,\,C{O_2}}}{{{\text{ mass of compound }}}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{12}}{{44}} \times \frac{{0.4655}}{{0.2115}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60.03\% \cr} $$

Mass of compound taken $$ = 0.50\,g$$
Vol. of $${H_2}S{O_4} = 50\,mL$$
Molarity of $${H_2}S{O_4} = 0.5\,M$$
Vol. of $$NaOH$$  required $$ = 60\,mL$$
Molarity of $$NaOH$$  required $$ = 0.5\,M$$
$$\% \,\,{\text{of}}\,\,N = \frac{{1.4 \times M \times 2\left[ {V - \frac{{{V_1}}}{2}} \right]}}{m}$$
By substituting the values in the formula, we get,
$$\eqalign{ & \% \,\,{\text{of}}\,\,N = \frac{{1.4 \times 0.5 \times 2\left( {50 - \frac{{60}}{2}} \right)}}{{0.5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 56 \cr} $$
$$\therefore \% $$  of $$N$$ in the given compound $$ = 56\% $$

$$\eqalign{ & {R_f} = \frac{{{\text{Distance travelled by substance from base line }}\left( x \right){\text{ }}}}{{{\text{Distance travelled by solvent from base line }}\left( z \right)}} \cr & {\text{For}}\,\,x:{R_f} = \frac{x}{z};\,{\text{For}}\,\,y:{R_f} = \frac{y}{z} \cr} $$

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$$\mathop {C{H_3}}\limits_{s{p^3}}^6 - \mathop {CH}\limits_{s{p^2}}^5 = \mathop {CH}\limits_{s{p^2}}^4 - \mathop {C{H_2}}\limits_{s{p^3}}^3 - \mathop C\limits_{sp}^2 \equiv \mathop {CH}\limits_{sp}^1 $$
Thus, the hybridisation of $${C^1},{C^3}$$  and $${C^5}$$ are $$sp,s{p^3},s{p^2}$$   respectively.

Dipole moment of $$p-$$ dichlorobenzene is zero because of symmetrical structure. $$o - $$ and $$m - $$ dichlorobenzene have higher dipole moments than toluene due to high electronegativity of chlorine than $$ - C{H_3}$$  group. Further, the $$o-$$ dichlorobenzene has higher dipole moment due to lower bond angle than the $$m-$$ isomer. Hence, the order of increasing dipole moment is :
$$p-$$ dichlorobenzene (iv) < toluene (i) < $$m-$$ dichlorobenzene (ii) < $$o-$$ dichlorobenzene (iii)