Questions: How many d electrons are in the valence shell of the Cd2+ cation? How many unpaired electron spins? Number of d electrons: Number of unpaired spins:

How many d electrons are in the valence shell of the Cd2+ cation? How many unpaired electron spins?
Number of d electrons:
Number of unpaired spins:

Transcript text: How many $d$ electrons are in the valence shell of the $\mathrm{Cd}^{2+}$ cation? How many unpaired electron spins? \begin{tabular}{|ll|} \hline Number of $d$ electrons: & $\square$ \\ Number of unpaired spins: & $\square$ \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Determine the Electron Configuration of Neutral Cadmium (Cd)

Cadmium (Cd) has an atomic number of 48. The electron configuration for a neutral cadmium atom is: \[ \text{Cd: } [\text{Kr}] 4d^{10} 5s^2 \]

Step 2: Determine the Electron Configuration of the $\mathrm{Cd}^{2+}$ Cation

When cadmium forms a $\mathrm{Cd}^{2+}$ cation, it loses two electrons. These electrons are removed from the outermost shell first, which is the 5s orbital: \[ \mathrm{Cd}^{2+}: [\text{Kr}] 4d^{10} \]

Step 3: Count the Number of $d$ Electrons

In the $\mathrm{Cd}^{2+}$ cation, the electron configuration is $[\text{Kr}] 4d^{10}$. This indicates that there are 10 electrons in the 4d subshell.

Step 4: Determine the Number of Unpaired Electron Spins

In the 4d subshell, when it is fully filled with 10 electrons, all the electrons are paired. Therefore, there are no unpaired electron spins.