X and Y are two metals. When burnt in air,X f | vedclass

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(B) The formation of nitride upon burning in air is a characteristic property of $Li$ and $Mg$ due to their small size and high charge density.$Na$ re

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$Na$ and $Mg$

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(B) The formation of nitride upon burning in air is a characteristic property of $Li$ and $Mg$ due to their small size and high charge density.$Na$ reacts with oxygen to form $Na_2O$ (oxide) and $Na_2O_2$ (peroxide) but does not form a nitride.$Mg$ reacts with both oxygen and nitrogen present in the air to form $MgO$ (oxide) and $Mg_3N_2$ (nitride).Therefore,$X$ could be $Na$ (which forms only oxide) and $Y$ could be $Mg$ (which forms both oxide and nitride).

Column $-I$	Column $-II$
$(a)$. Super oxide	$(p)$. $Na_2O_2$
$(b)$. Peroxide	$(q)$. $CO_2$
$(c)$. Dioxide	$(r)$. $C_3O_2$
$(d)$. Sub oxide	$(s)$. $CsO_2$

$X$ and $Y$ are two metals. When burnt in air,$X$ forms only oxide while $Y$ forms oxide and nitride. The metals $X$ and $Y$ may be

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Heating a metal $X$ in nitrogen gas gives $Y$. The reaction of $Y$ with $H_2O$ produces a colorless gas. When this gas is passed through $CuSO_4$ solution,it produces a blue color. What is $Y$?

Match Column $-I$ with Column $-II$ :

Column $-I$ Column $-II$

$(a)$. Super oxide $(p)$. $Na_2O_2$

$(b)$. Peroxide $(q)$. $CO_2$

$(c)$. Dioxide $(r)$. $C_3O_2$

$(d)$. Sub oxide $(s)$. $CsO_2$

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