When sodium is heated in excess of air,which | vedclass

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(D) When sodium is heated in an excess of air,it primarily reacts with oxygen to form sodium peroxide $(Na_2O_2)$ and a small amount of sodium oxide $

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$II, IV$

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(D) When sodium is heated in an excess of air,it primarily reacts with oxygen to form sodium peroxide $(Na_2O_2)$ and a small amount of sodium oxide $(Na_2O)$.The reaction is as follows:$2Na + \frac{1}{2}O_2 \rightarrow Na_2O$$Na_2O + \frac{1}{2}O_2 \rightarrow Na_2O_2$Sodium does not form $NaO_2$ (superoxide) under these conditions,as that is characteristic of larger alkali metals like potassium,rubidium,and cesium. While nitrogen is present in the air,sodium does not react with nitrogen to form $Na_3N$ under normal heating conditions in air.Therefore,the products obtained are $Na_2O$ and $Na_2O_2$,which correspond to $II$ and $IV$.

$I. Na_3N$	$II. Na_2O$
$III. NaO_2$	$IV. Na_2O_2$

List-$I$ (Salt)	List-$II$ (Flame colour wavelength)
$(a)$ $LiCl$	$(i)$ $455.5 \text{ nm}$
$(b)$ $NaCl$	$(ii)$ $670.8 \text{ nm}$
$(c)$ $RbCl$	$(iii)$ $780.0 \text{ nm}$
$(d)$ $CsCl$	$(iv)$ $589.2 \text{ nm}$

When sodium is heated in excess of air,which of the following compounds are obtained?

$I. Na_3N$ $II. Na_2O$

$III. NaO_2$ $IV. Na_2O_2$

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Match List-$I$ with List-$II$:

List-$I$ (Salt) List-$II$ (Flame colour wavelength)

$(a)$ $LiCl$ $(i)$ $455.5 \text{ nm}$

$(b)$ $NaCl$ $(ii)$ $670.8 \text{ nm}$

$(c)$ $RbCl$ $(iii)$ $780.0 \text{ nm}$

$(d)$ $CsCl$ $(iv)$ $589.2 \text{ nm}$

Choose the correct answer from the options given below:

Sodium oxide dissolves in water to give sodium hydroxide,which indicates its:

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When sodium is heated in excess of air,which of the following compounds are obtained? $I. Na_3N$ $II. Na_2O$ $III. NaO_2$ $IV. Na_2O_2$