Question

Generally, transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?

• A. $Cu{F}_2$
• B. $C{u}_{2}C{l}_2$
• C. $A{g}_{2}S{O}_4$
• D. $Zn{F}_2$

Answer 1

The correct option is A $Cu{F}_2$

• When light passes through the salts of transition elements, some of the wavelengths of visible light is absorbed, using which the unpaired electrons undergo transition, and the salts appear coloured.

• Thus, transition elements form coloured salts due to the presence of unpaired electrons.

Analysing the options

Option $\left(A\right):$

In $A{g}_{2}S{O}_4$ , the oxidation state of $Ag$ is $+1.$

Thus, the electronic configuration of $Ag$ in its $+1$ oxidation state is $\left[Kr\right]4d^{10}.$ Since $A{g}^{+}$ has no unpaired electrons i.e., it will not be coloured.

Option $\left(B\right):$

In $Cu{F}_2,$ the oxidation state of $Cu$ is $+2.$

The electronic configuration of $Cu$ in its $+2$ state is $\left[Ar\right]3d^9.$

Since $C{u}^{2+}$ possesses one unpaired electron, $Cu{F}_2$ will be coloured in the solid state.

Option $\left(C\right):$

In $Zn{F}_2,$ the oxidation state of $Zn$ is $+2.$

The electronic configuration of $Zn$ in its $+2$ oxidation state is $\left[Ar\right]3d^{10}.$

Since $Z{n}^{2+}$ has no unpaired electrons, it will not be coloured.

Option $\left(D\right):$

In $C{u}_{2}C{l}_2,$ the oxidation state of $Cu$ is $+1.$

The electronic configuration of $Cu$ in its $+1$ state is $\left[Ar\right]3d^{10}.$

Since $C{u}^{+}$ has no unpaired electrons, it will not be coloured.

So, option $\left(B\right)$ is the correct answer.