Q2.88P - Chemistry: Molecular Nature Of Matter And Change

Question

Give the name and formula of the compound formed from the following elements

a. ${}_{\mathbf{37}}\mathit{Q}\mathit{a}\mathit{n}\mathit{d}{}_{\mathbf{35}}\mathit{R}$

b. ${}_{\mathbf{8}}\mathit{Q}\mathit{a}\mathit{n}\mathit{d}{}_{\mathbf{13}}\mathit{R}$

c.${}_{\mathbf{8}}\mathit{Q}\mathit{a}\mathit{n}\mathit{d}{}_{\mathbf{13}}\mathit{R}$

Step-by-Step Solution

Answer

a. The name of the compound is rubidium bromide and the formula is$RbBr$  

b. The name of the compound is aluminum oxide and the formula is$A{l}_2{O}_3$  

c. The name of the compound is calcium iodide and the formula is$Ca{I}_2$ 

Step 1: Compounds formed between Q 37   a n d   R 35

a. The element with atomic number 37 is rubidium and with atomic number 35 is bromine.

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.${\text{Rb\hspace{0.17em}}}^{\text{+}}{\text{\hspace{0.17em}Br}}^{-}\to \text{\hspace{0.17em}RbBr}$

 

Step 2: Compounds formed between Q 8   a n d   R 13

b. The element with atomic number 8 in is oxygen and the element with atomic number 13 is aluminum.

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.

 ${\text{Al\hspace{0.17em}}}^{\text{+3}}{\text{\hspace{0.17em}O}}^{\text{-2}}\to {\text{\hspace{0.17em}Al}}_{\text{2}}{\text{O}}_{\text{3}}$

 The formula is $A{l}_2{O}_3$ 

Step 3: Compounds formed between Q 20   a n d   R 53

C. The element with atomic number 20 is calcium and theelement with atomic number 53 is iodine.

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${\text{Ca\hspace{0.17em}}}^{\text{+2}}{\text{\hspace{0.17em}I}}^{\text{-}}\to {\text{\hspace{0.17em}CaI}}_{\text{2}}$.

 

The formula is$Ca{I}_2$ .