Question

A 0.132-mole sample of an unknown semiconducting material with the formula XY has a mass of 19.0 g. The element X has an electron configuration of $\left[\mathbf{Kr}\right]\mathbf{5}{s}^{\mathbf{2}}\mathbf{4}{d}^{\mathbf{10}}$. What is this semiconducting material? A small amount of the Y atoms in the semiconductor is replaced with an equivalent amount of atoms with an electron configuration of $\left[\mathbf{Ar}\right]\mathbf{4}{s}^{\mathbf{2}}\mathbf{3}{d}^{\mathbf{10}}\mathbf{4}{p}^{\mathbf{5}}$. Does this correspond to n-type or p-type doping?

Short answer

The semiconductor is CdS.

This corresponds to n-type doping.

Step-by-step solution

Calculating molar mass

$$\begin{gathered} Molar mass = \frac{19}{0.132} \\ = 144 g/mol \end{gathered}$$

The electronic configuration given in the question resembles Cd and

Molar mass of $Cd=112.4g/mol$

Then$Y=144-112.4=31.6g/mol$

So, Y could be Sulfur.

So, XY will be CdS.

The dopant

The electronic configuration in the question resembles Bromine, and Br has one electron more than sulfur. So it will produce n-type doping.