When chlorine gas reacts with hot and concentrated $$NaOH$$   solution, it disproportionates into chloride $$\left( {C{l^ - }} \right)$$  and chlorate $$\left( {ClO_3^ - } \right)$$  ions.

In this process, oxidation number of chlorine changes from 0 to - 1 and 0 to + 5.
NOTE
In disproportionation reactions, the same element undergoes oxidation as well as reduction.

\[2Al+6HC{{l}_{\left( g \right)}}\xrightarrow{\text{Heat}}2AlC{{l}_{3}}+3{{H}_{2}}\]

$$CO_3^{2 - }$$  and $$NO_3^ - $$  are isoelectronic as both contain 32 electrons and isostructural as both have trigonal planar structures with $$s{p^2}$$  hybridisation of the central atom.

Chlorine reacts with excess of ammonia to produce ammonium chloride and nitrogen.

$$\mathop {{P_4}}\limits^0 + 3NaOH + 3{H_2}O \to $$     $$P\mathop {{H_3}}\limits^{ + 3} + 3Na{H_2}P\mathop {{O_2}}\limits^{ - 1} $$
It is redox reaction.
(i) Oxidation number increases during oxidation reaction and decreases during reduction reaction.
(ii) In a neutralisation reaction acids and bases reacts together to form salt and water.

No explanation is given for this question. Let's discuss the answer together.

$${}_7N = 1{s^2}2{s^2}2{p^3};\,\,\,\,\,{}_{15}P = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}$$
NOTE: In phosphorous the $$3d$$- orbitals are available. Hence phosphorous can from pentahalides also but nitrogen can not form pentahalide due to absence of $$d\, - $$orbitals

$$N{O_2}$$  is reddish brown coloured gas. Rest of the oxides are colourless.

Diamond does not show electrical conductivity due to the absence of free electrons. Sodium and potassium are metallic conductors while graphite is a non-metallic conductor.

$$3HC{l_{\left( {conc.} \right)}} + HN{O_{3\left( {conc.} \right)}}$$     \[\to \underset{\begin{smallmatrix} \left( X \right) \\ \text{Aqua regia} \end{smallmatrix}}{\mathop{NH{{O}_{3}}\cdot 3HCl}}\,\]
Aqua regia is used for dissolving noble metals like gold, platinum. $$HCl$$  decomposes salts of weaker acids e.g., carbonates, hydrogen carbonates, sulphites, etc.