Question

Select correct statement(s):
A. Oxides $$\left({\text{M}}_2\text{O}\right)$$ and peroxides $$\left({\text{M}}_2{\text{O}}_2\right)$$ of alkali metals are diamagnetic and colourless.
B. Superoxides $$\left(\text{M}{\text{O}}_2\right)$$ of alkali metals are paramagnetic.
C. $$\text{Li}$$ and $$\text{Na}$$ do not form superoxides.
D. All the above

Answer 1

Hint:Different products are formed when alkali metals react with oxygen. These products include alkali metal oxides $$\left({\text{M}}_2\text{O}\right)$$, alkali metal peroxides $$\left({\text{M}}_2{\text{O}}_2\right)$$ and alkali metal superoxides $$\left(\text{M}{\text{O}}_2\right)$$.

Complete step by step solution:
A. Oxides $$\left({\text{M}}_2\text{O}\right)$$ and peroxides $$\left({\text{M}}_2{\text{O}}_2\right)$$ of alkali metals are diamagnetic and colourless.
In diamagnetic compounds, all the electrons are paired. This is possible for compounds having even numbers of electrons. Also a compound is coloured when it contains one or more unpaired electrons. If all the electrons are paired, then the compound is colorless. Thus, compounds with even numbers of electrons are colorless.
Consider sodium oxide $$\left(\text{N}{\text{a}}_2\text{O}\right)$$ as an example of alkali metal oxide.
Each sodium atom accounts for 11 electrons and oxygen atom accounts for 8 electrons.
Calculate the total number of electrons in sodium oxide.
$$2\left(11\right)+8=30$$
Thus, one molecule of sodium oxide contains 30 electrons. This is an even number of electrons. Hence, sodium oxide is diamagnetic and colorless.
Consider sodium peroxide $$\left(\text{N}{\text{a}}_2{\text{O}}_2\right)$$ as an example of alkali metal peroxide.
Each sodium atom accounts for 11 electrons and oxygen atom accounts for 8 electrons.
Calculate the total number of electrons in sodium peroxide.
$$2\left(11\right)+2\left(8\right)=38$$
Thus, one molecule of sodium peroxide contains 38 electrons. This is an even number of electrons. Hence, sodium peroxide is diamagnetic and colorless.
Hence, option A represents a correct statement.

B. Superoxides $$\left(\text{M}{\text{O}}_2\right)$$ of alkali metals are paramagnetic.
In paramagnetic compounds, one or more unpaired electrons are present. This is possible for compounds having.
Consider sodium superoxide $$\left(\text{Na}{\text{O}}_2\right)$$ as an example of alkali metal superoxide.
Sodium atoms account for 11 electrons and each oxygen atom accounts for 8 electrons.
Calculate the total number of electrons in sodium superoxide.
$$11+2\left(8\right)=27$$
Thus, one molecule of sodium superoxide contains 27 electrons. This is an odd number of electrons. Hence, sodium superoxide is paramagnetic.
Hence, option B represents a correct statement.

C. $$\text{Li}$$ and $$\text{Na}$$ do not form superoxides.
$$\text{Li}$$ and $$\text{Na}$$ have small atomic sizes.
Superoxides are formed by metals having large atomic size.
Hence, option C represents a correct statement.

Hence, the correct option is the option D All the above

Note:
You can use alkali metal peroxides in the preparation of other peroxides. You can also use alkali metal peroxides for bleaching and air purification.