Question

Match the statements of Column A (Atoms) with those of Column B (Molecule Formed).

Column 1	Column 2
$\mathrm{A}$ has 3 protons and $\mathrm{B}$ has 17 protons	${\mathrm{A}}_2\mathrm{B}$
$\mathrm{A}$ has 8 protons and $\mathrm{B}$ has 9 protons	${\mathrm{AB}}_3$
$\mathrm{A}$ has 17 protons and $\mathrm{B}$ has 8 protons	$\mathrm{AB}$
$\mathrm{A}$ has 5 protons and $\mathrm{B}$ has 9 protons	${\mathrm{AB}}_2$

Answer 1

The number protons tells the atomic number of the element because number of protons is equal to the number of electrons.

(A)

• $\mathrm{A}$ has 3 protons and $\mathrm{B}$ has 17 protons: $\mathrm{A}$ has 3 protons means the atomic number of $\mathrm{A}$ is 3. So, $\mathrm{A}$ is Lithium ($\mathrm{Li}$).
• $\mathrm{B}$ has 17 protons means the atomic number of $\mathrm{B}$ is 17 , so $\mathrm{B}$ is Chlorine ($\mathrm{Cl}$).
• So, the compound formed will be Lithium chloride ($\mathrm{LiCl}$) and it is $\mathrm{AB}$-type.

(B)

• $\mathrm{A}$ has 8 protons and B has 9 protons: $\mathrm{A}$ has 8 protons means the atomic number of $\mathrm{A}$ is 8. So, $\mathrm{A}$ is Oxygen ($\mathrm{O}$).
• $\mathrm{B}$ has 9 protons means the atomic number of $\mathrm{B}$ is 9, so $\mathrm{B}$ is Fluorine ($\mathrm{F}$).
• So, the compound formed will be Oxygen fluoride (${\mathrm{OF}}_2$) and it is ${\mathrm{AB}}_2$-type.

(C)

• $\mathrm{A}$ has 17 protons and $\mathrm{B}$ has 8 protons: $\mathrm{A}$ has 17 protons means the atomic number of $\mathrm{A}$ is 17. So $\mathrm{A}$ is Chlorine ($\mathrm{Cl}$).
• $\mathrm{B}$ has 8 protons means the atomic number of $\mathrm{B}$ is 8, so $\mathrm{B}$ is Oxygen ($\mathrm{O}$).
• So, the compound formed will be Chlorine oxide (${\mathrm{Cl}}_2\mathrm{O}$) and it is ${\mathrm{A}}_2\mathrm{B}$-type.

(D)

• $\mathrm{A}$ has 5 protons and $\mathrm{B}$ has 9 protons: $\mathrm{A}$ has 5 protons means the atomic number of $\mathrm{A}$ is 5. So $\mathrm{A}$ is Boron ($\mathrm{B}$).
• $\mathrm{B}$ has 9 protons means the atomic number of $\mathrm{B}$ is 9, so $\mathrm{B}$ is Fluorine ($\mathrm{F}$).
• So, the compound formed will be Boron trifluoride (${\mathrm{BF}}_3$) and it is ${\mathrm{AB}}_3$-type.

Therefore, the answer is $\left(\mathrm{A}\right)\to \left(\mathrm{III}\right),\left(\mathrm{B}\right)\to \left(\mathrm{IV}\right),\left(\mathrm{C}\right)\to \left(\mathrm{I}\right),\left(\mathrm{D}\right)\to \left(\mathrm{II}\right)$.