Question

Complete the table given below for writing the formula of a chemical compound.

Identify the valency	Criss cross the valency	Formula of the compound	Name of the compound
$\mathrm{Cr}$ ${\mathrm{SO}}_4$	$\mathrm{Cr}$ ${\mathrm{SO}}_4$

Answer 1

Part 1: Identify the valency of $\mathrm{Cr}$ and ${\mathrm{SO}}_4$:

• The chromium atom $\mathbf{\left(}\mathbf{Cr}\mathbf{\right)}$ loses three electrons from the outermost shell and forms a chromium ion with three positive charges $\left({\mathrm{Cr}}^{3+}\right)$.
• Thus, the valency of $\mathrm{Cr}$ is $3$.
• ${\mathrm{SO}}_4$ exists as ${\mathrm{SO}}_4^{2-}$ i.e. sulfate ion having two negative charges.
• Thus, the valency of ${\mathrm{SO}}_4^{2-}$ is $2$.

Part 2: Criss-cross the valency:

• The valency of $\mathrm{Cr}$ becomes the number of ${\mathrm{SO}}_4^{2-}$ units.
• The valency of ${\mathrm{SO}}_4^{2-}$ becomes the number of $\mathrm{Cr}$ atoms.

Part 3: Formula of the compound:

• After criss-crossing the valencies, there are two $\mathrm{Cr}$ atoms and three ${\mathrm{SO}}_4^{2-}$ units.
• Thus, the formula of the compound is ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$.

Part 4: Name of the compound:

• While naming the ionic compounds, the name of the cation is written first and then the anion is named.
• Here, the cation is chromium $\left({\mathrm{Cr}}^{3+}\right)$ and the anion is sulfate $\left({\mathrm{SO}}_4^{2-}\right)$.
• Thus, the name of the compound is Chromium(III) sulphate.

Identify the valency	Criss-cross the valency	Formula of the compound	Name of the compound
$\mathrm{Cr}$ ${\mathrm{SO}}_4$ $3$ $2$	$\mathrm{Cr}$ ${\mathrm{SO}}_4$ $2$ $3$	${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$	Chromium(III) sulfate