$$\underset{\begin{array}{l}\text{Colourless}\\ \text{gases}\left(X\right)\end{array}}{\underbrace{2NO+O_2}}\rightleftharpoons \underset{\begin{array}{l}\text{Brown,}\\ \text{paramagnetic gas}\\ \text{due to presence of}\\ \text{an odd electron}\left(Y\right)\end{array}}{2N{O}_2}\underset{\text{heat}}{\overset{\text{cool}}{\rightleftharpoons }}\underset{\begin{array}{l}\text{Colourless,}\\ \text{diamagnetic}\\ \text{solid}\left(Z\right)\end{array}}{N_2{O}_4}$$

Graphite and boron nitride have similar structure.

$$2C{a^{2 + }} + N{a_2}\left[ {N{a_4}{{\left( {P{O_3}} \right)}_6}} \right] \to 4N{a^ + } + \mathop {N{a_2}\left[ {C{a_2}{{\left( {P{O_3}} \right)}_6}} \right]}\limits_{{\text{water}}\,{\text{soluble}}} $$

$$Xe{F_2}$$  is an oxidising agent.

\[{{P}_{4}}+3{{O}_{2}}\xrightarrow{\text{in presence of }{{N}_{2}}}{{P}_{4}}{{O}_{6}}\]
Here $${N_2}$$ acts as a diluent and thus retards further oxidation. Reaction of $${P_4}$$ under other three conditions.
\[\left( \text{A} \right){{P}_{4}}+3{{O}_{2}}\to {{P}_{4}}{{O}_{6}}\xrightarrow{2{{O}_{2}}}{{P}_{4}}{{O}_{10}}\]
\[\left( \text{C} \right)\]  In moist air, $${P_4}{O_6}$$  is hydrolysed to form $${H_3}P{O_3}$$
$${P_4}{O_6} + 6{H_2}O \to 4\,{H_3}P{O_3}$$
\[\left( \text{D} \right)\]  In presence of $$NaOH$$ ,
$${P_4} + 3O{H^ - } + 3{H_2}O \to P{H_3} + 3{H_2}PO_2^ - $$

$$CO$$ is an example of neutral oxide.

In general, noble gases are not very reactive.
Their chemical inertness is due to the fact that they have completely filled $$n{s^2}n{p^6}$$  electronic configuration of their valence shells. The other reasons are very high ionisation enthalpy and almost zero electron affinity.

$$HBr$$  is not prepared by heating $$NaBr$$  with $${\text{Conc}}{\text{.}}\,{H_2}S{O_4}$$  because $$HBr$$  is a strong reducing agent and reduce $${H_2}S{O_4}\,$$ to $$S{O_2}$$  and is itself oxidised to bromine.
$$\eqalign{ & NaBr + {H_2}S{O_4}\, \to NaHS{O_4} + HBr \cr & {H_2}S{O_4}\, + 2HBr \to S{O_2} + B{r_2} + 2{H_2}O \cr} $$

The structure of $${H_3}P{O_3}$$  is as follows: There are only two $$ - OH$$  groups and hence dibasic. The oxidation number of $$P$$ in this acid is $$ + 3$$ . Whereas $$P$$ can have $$ + 5$$  oxidation state also. Therefore, $${H_3}P{O_3}$$  can be oxidised which means $${H_3}P{O_3}$$  is a reducing agent.

Temporary hardness of water is due to presence of bicarbonates of $$Ca$$  and $$Mg$$  and it is removed by adding $$Ca{\left( {OH} \right)_2}$$  to hard water and precipitating these soluble bicarbonates in the form of insoluble salts.
$$\eqalign{ & Ca{\left( {HC{O_3}} \right)_2}\mathop \to \limits^\vartriangle CaC{O_3} \downarrow + C{O_2} \uparrow + {H_2}O \cr & Mg{\left( {HC{O_3}} \right)_2}\mathop \to \limits^\vartriangle Mg{\left( {OH} \right)_2} \downarrow + 2C{O_2} \uparrow \cr} $$