Answer
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Hint: Alkalide ions are anions of alkali metals. Simply, we can say that these are the compounds in which alkali metal atoms are negative ions (that are anions). Alkalides are found to be reactive in nature.
Step by step solution:
- Alkalides are found to be thermally labile due to the high reactivity of the alkalide anion.
- It is also found that these are able to break the covalent bonds like C-O bonds in cryptand (which is a ligand).
- There are several alkalides known for alkali metals like- Potassium (${{K}^{-}}$), sodide ($N{{a}^{-}}$), Caeside ($C{{s}^{-}}$),Rubidide ($R{{b}^{-}}$).
- We can also see here that alkalides are usually diamagnetic in nature. Let’s discuss about that:
- As we know that the electronic configuration of Na is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$, there is one unpaired electron present due to which it is paramagnetic.
- Whereas, in case of Sodadide that is $N{{a}^{-}}$, we can write the electronic configuration as: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$, here we can see that there are no unpaired electrons present, hence we can say that it is diamagnetic in nature.
Hence, we can conclude that the correct option is (A) that the alkalide ion is Diamagnetic.
Additional information:
- There were also several alkalides synthesized like: inverse sodium hydride, Sodium-crypt nitride etc.
Note:
- The alkalides of the other alkali metals like Lithides and Francides are not yet discovered.
- We can say that these are having the oxidation state of -1.
Step by step solution:
- Alkalides are found to be thermally labile due to the high reactivity of the alkalide anion.
- It is also found that these are able to break the covalent bonds like C-O bonds in cryptand (which is a ligand).
- There are several alkalides known for alkali metals like- Potassium (${{K}^{-}}$), sodide ($N{{a}^{-}}$), Caeside ($C{{s}^{-}}$),Rubidide ($R{{b}^{-}}$).
- We can also see here that alkalides are usually diamagnetic in nature. Let’s discuss about that:
- As we know that the electronic configuration of Na is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$, there is one unpaired electron present due to which it is paramagnetic.
- Whereas, in case of Sodadide that is $N{{a}^{-}}$, we can write the electronic configuration as: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$, here we can see that there are no unpaired electrons present, hence we can say that it is diamagnetic in nature.
Hence, we can conclude that the correct option is (A) that the alkalide ion is Diamagnetic.
Additional information:
- There were also several alkalides synthesized like: inverse sodium hydride, Sodium-crypt nitride etc.
Note:
- The alkalides of the other alkali metals like Lithides and Francides are not yet discovered.
- We can say that these are having the oxidation state of -1.
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