Question

Transition elements form colored salts due to the presence of unpaired electrons. Which of the following compounds are expected to be colored in a solid state?

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

Answer 1

The correct option is A

$Cu{F}_2$

The explanation for Correct Answer:

Option(A): $Cu{F}_2$

1. In the complex$Cu{F}_2$, the metal cation is Copper, which is present as $C{u}^{2+}$ and the anion is$F^{-}$.
2. The atomic number of Copper is $29$ with an electronic configuration$[Ar]3d^{10}4s^1$.
3. $C{u}^{2+}$, form, when two electrons are removed from $\mathrm{Cu}$ the resulting electronic configuration, will be$[Ar]3d^9$
4. The$C{u}^{2+}$, have incomplete$\mathrm{d}-\mathrm{orbital}$, and has one unpaired electron, therefore, it will form a colored complex.
5. Hence, it is the correct option.

The explanation for Incorrect Answer:

Option(B): $Zn{F}_2$

1. In the complex$Zn{F}_2$, the metal cation is Zinc, which is present as $Z{n}^{2+}$ and the anion is$F^{-}$.
2. The atomic number of Zinc is $30$ with an electronic configuration of$[Ar]3d^{10}4s^2$.
3. $Z{n}^{2+}$, is formed when $2$ electrons are removed from the outer orbital of$Zn$. The resulting electronic configuration will be$[Ar]3d^{10}$
4. The$Z{n}^{2+}$, have filled$\mathrm{d}-\mathrm{orbital}$, therefore it will not form a colored complex, as no unpaired electron is available.
5. Hence, it is an incorrect option.

Option(C): $C{u}_{2}C{l}_2$

1. In the complex$C{u}_{2}C{l}_2$, the metal cation is Copper which is present as $C{u}^{+1}$ and anion$C{l}^{-}$.
2. The atomic number of copper is $29$ with the electronic configuration$[Ar]3d^{10}4s^1$.
3. From $\mathrm{Cu}$ atom one electron is removed to form $C{u}^{+1}$, and the resulting electronic configuration will be$[Ar]3d^{10}$
4. The$C{u}^{+1}$, have completely filled $\mathrm{d}-\mathrm{orbital}$, therefore it will not form a colored complex.
5. Hence, it is an incorrect option.

Option(D): $A{g}_{2}S{O}_4$

1. The complex$A{g}_{2}S{O}_4$, metal cation is Silver, which is present as $A{g}^{+}$ and the anion is$S{O}_4^{2-}$.
2. The atomic number of Silver is $47$ with electronic configuration$[Kr]4d^{10}5s^1$.
3. From $\mathrm{Ag}$ atom, one electron is removed to form$A{g}^{+}$, and the resulting electronic configuration will be$[Kr]4d^{10}5s^0$
4. $A{g}^{+}$, have completely filled the $\mathrm{d}-\mathrm{orbital}$ therefore, it will not form a colored complex.
5. Hence, it is an incorrect option

Hence, the compound which is colored at solid-state is$Cu{F}_2$.