The correct option(s) regarding the complex

$${\left[ {Co\left( {en} \right){{\left( {N{H_3}} \right)}_3}\left( {{H_2}O} \right)} \right]^{3 + }}\,\,$$ $$\left( {en = {H_2}NC{H_2}C{H_2}N{H_2}} \right)$$ is (are)

To measure the quantity of $$MnC{l_2}$$ dissolved in an aqueous solution, it was completely converted to $$KMn{O_4}$$ using the reaction,

$$MnC{l_2} + {K_2}{S_2}{O_8} + {H_2}O \to KMn{O_4} + {H_2}S{O_4} + HCl$$ (equation not balanced).

Few drops of concentrated $$HCl$$ were added to this solution and gently warmed. Further, oxalic acid ($$225$$ $$mg$$) was added in portions till the colour of the permanganate ion disappeared. The quantity of $$MnC{l_2}$$ (in mg) present in the initial solution is ____________.

(Atomic weights in $$g\,\,mo{l^{ - 1}}:Mn = 55,Cl = 35.5$$ )

Aniline reacts with mixed acid (conc. $$HN{O_3}$$ and conc. $${H_2}S{O_4}$$) at $$288$$ $$K$$ to give P $$\left( {51\% } \right),$$ Q $$\left( {47\% } \right)$$ and R $$\left( {2\% } \right).$$ The major product(s) of the following reaction sequence is (are)

The surface of copper gets tarnished by the formation of copper oxide. $${N_2}$$ gas was passed to prevent the oxide formation during heating of copper at $$1250$$ $$K.$$ However, the $${N_2}$$ gas contains $$1$$ mole % of water vapor as impurity. The water vapor oxidises copper as per the reaction given below : $$2Cu\left( s \right) + {H_2}O\left( g \right) \to C{u_2}O\left( s \right) + {H_2}\left( g \right)$$

$${P_{H2}}$$ is the minimum partial pressure of $${H_2}$$ (in bar) needed to prevent the oxidation at $$1250$$ $$K.$$ The value of $$\ln \left( {{P_{H2}}} \right)$$ is ________.

Given: total pressure $$=1$$ bar, $$R$$ (universal gas constant ) $$=$$ $$8J{K^{ - 1}}\,\,mo{l^{ - 1}},$$ $$\ln \left( {10} \right) = 2.3.\,$$ $$Cu(s)$$ and $$C{u_2}O\left( s \right)$$ are naturally immiscible.

At $$1250$$ $$K:2Cu(s)$$ $$ + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{O_2}\left( g \right) \to C{u_2}O\left( s \right);$$ $$\Delta {G^ \circ } = - 78,000J\,mo{l^{ - 1}}$$

$${H_2}\left( g \right) + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{O_2}\left( g \right) \to {H_2}O\left( g \right);$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\Delta {G^ \circ } = - 1,78,000J\,mo{l^{ - 1}};$$ ($$G$$ is the Gibbs energy)