Import Question JSON

$\text{Hint: Higher the difference of electronegativity, the higher will be the ionic nature.}$ $1.\ \text{Na and Ca} \rightarrow \text{S block = No reaction.}$ $2.\ \text{K and O}_2\ \text{form normal oxide, peroxide, superoxide which are all ionic.}$ $3.\ \text{O}_2\ \text{and Cl}_2 \rightarrow \text{P - block = Covalent.}$ $4.\ \text{Al and I}_2 \rightarrow \text{Covalent.}$ $\text{Ionic compounds are generally formed between a metal (which tends to lose electrons) and a non-metal (which tends to gain electrons).}$ $\text{1. Na (Group 1 metal) and Ca (Group 2 metal): Both are metals, so they will not form an ionic bond with each other. They would typically form alloys or no reaction.}$ $2.\ \text{K (Potassium) is a Group 1 alkali metal, and O}_2\ \text{(Oxygen) is a non-metal. Potassium will readily lose electrons to form K}^+\ \text{ions, and oxygen will gain electrons to form O}^{2-}\ \text{ions. They can form ionic compounds like potassium oxide (K}_2\text{O), potassium peroxide (K}_2\text{O}_2\text{), or potassium superoxide (KO}_2\text{), all of which are ionic.}$ $3.\ \text{O}_2\ \text{and Cl}_2\text{: Both are non-metals. Non-metals typically form covalent bonds with each other by sharing electrons.}$ $4.\ \text{Al (Aluminum) is a metal, and I}_2\ \text{(Iodine) is a non-metal. While Al and I}_2\ \text{can react, the bond formed between Al and I}$ $\text{is primarily covalent in nature, especially for anhydrous aluminum iodide.}$ $\text{Therefore, the pair of elements that can form an ionic compound is K and O}_2\text{.}$