91Ó°ÊÓ

Bromine is a non-metal and rubidium is a metal. So,they form an ionic bond

and the compound formed is called as rubidium bromide.

Electronic configuration of Rbis $\left[\text{Kr}\right]{\text{5s}}^{\text{1}}$. Thus it donates one electron to form${\text{Rb}}^{\text{+}}$ .

Electronic configuration of Bromine is$\left[\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{5}}$ . It accepts one electron to form${\text{Br}}^{\text{-}}$ .

Thus one atom of Rb combines with one atom of Br to form a neutral molecule rubidium bromide $\left(\text{RbBr}\right)$.

It can be shown by criss-cross method that interchanging the oxidation numbers of the cation and anion would give the formula of the neutral molecule.${\text{¸é²ú }}^{\text{+}}{\text{ B°ù}}^{−}\text{ }→\text{ R²úµþ°ù}$

Oxygen is a non-metal and aluminum is a metal they form an ionic bond

and the name of the compound formed isaluminum oxide.

Electronic configuration of Alis $\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{1}}$. Thus it donates three electron to form${\text{Al}}^{\text{+3}}$ .

Electronic configuration of oxygenis $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^4$. It accepts two electron to form${\text{O}}^{\text{-2}}$ .

Thus two atoms of Al combines with three atoms of O to form a neutral molecule aluminum oxide$\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)$.

It can be shown by criss-cross method that interchanging the oxidation numbers of the cation and anion would give the formula of the neutral molecule.

${\text{´¡±ô }}^{\text{+3}}{\text{ O}}^{\text{-2}}\text{ }→{\text{‼î±ô}}_{\text{2}}{\text{O}}_{\text{3}}$

The formula is $A{l}_2{O}_3$

Iodine is a non-metal and calcium is a metal they form an ionic bond

and the name of the compound formed is calcium iodide.

Electronic configuration of Cais $\left[\text{Ar}\right]{\text{4s}}^{\text{2}}$. Thus it donates two electron to form${\text{Ca}}^{\text{+2}}$ .

Electronic configuration of iodineis $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^4$. It accepts one electron to form ${\text{I}}^{\text{-}}$ .

Thus one atom ofCa combines with two atoms of I to form a neutral molecule calcium iodide$\left({\text{CaI}}_{\text{2}}\right)$.

It can be shown by criss-cross method that interchanging the oxidation numbers of the cation and anion would give the formula of the neutral molecule${\text{°ä²¹â€‰}}^{\text{+2}}{\text{ I}}^{\text{-}}\text{ }→{\text{ C²¹±õ}}_{\text{2}}$.

The formula is$Ca{I}_2$ .