Question

A 0.132-mol sample of an unknown semiconducting material with the formula XY has a mass of \(19.0 \mathrm{~g} .\) The element \(\mathrm{X}\) has an electron configuration of \([\mathrm{Kr}] 5 s^{2} 4 d^{10} .\) 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 \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{5} .\) Does this correspond to n-type or p-type doping?

Short answer

The unknown semiconducting material is Cadmium Phosphide (CdP), which is formed by elements X (Cadmium, Cd) and Y (Phosphorus, P). When a small amount of Y atoms is replaced with atoms of Iodine, it results in n-type doping.

Step-by-step solution

We are given the electron configuration of element X as [Kr]5s^2 4d^10. Checking the periodic table, the element with this electron configuration is Cd (Cadmium). #Step 2: Calculate Molar Mass of Element X #

Cadmium has an atomic mass of 112.41 g/mol. #Step 3: Determine the Mole Ratio and Mass of Element Y#

We are given that there are 0.132 mol of the unknown semiconductor XY, with a mass of 19.0 g. First, let's find the mass of Cd (X) by multiplying its molar mass by the number of moles: 0.132 mol × 112.41 g/mol = 14.83812 g. Now, we can find the mass of element Y by subtracting the mass of Cd from the total mass: 19.0 g - 14.83812 g = 4.16188 g. #Step 4: Determine the Molar Mass of Element Y#

Since the formula of the semiconducting material is XY, there must be 0.132 mol of element Y also. We can determine the molar mass of element Y by dividing the mass of Y by the number of moles: 4.16188 g / 0.132 mol = 31.53 g/mol. This molar mass corresponds to element P (Phosphorus) in the periodic table. #Step 5: Identify the Unknown Semiconductor Material#

Now that we know the elements, we can identify the semiconducting material as Cadmium Phosphide (CdP). #Step 6: Identify the Doping Type#

An element with electron configuration [Ar]4s^2 3d^10 4p^5 has been introduced to replace a small amount of Y atoms. Referring to the periodic table, we see that this configuration belongs to I (Iodine). To determine whether this results in n-type or p-type doping, we need to compare the number of valence electrons. Element Y (Phosphorus) has 5 valence electrons (from its 4p^3 configuration), while Iodine has 7 valence electrons (4p^5). Therefore, the doping adds excess electrons to the semiconductor, making it an n-type doped semiconductor. In conclusion, the unknown semiconducting material is Cadmium Phosphide (CdP), and the doping corresponds to n-type doping.

Key concepts

Cadmium Phosphide

Cadmium Phosphide, represented as CdP, is a compound formed by combining cadmium (Cd) and phosphorus (P).
It is a member of the family of III-V semiconductors, which are known for their significant electronic properties. Semiconductors like CdP are critical in electronic applications, particularly in devices that convert solar energy into electricity.
In this case, our given semiconducting material is identified as Cadmium Phosphide based on the electron configuration \( [\text{Kr}] \ 5s^{2} \ 4d^{10} \) for Cadmium and a calculated molar mass for Phosphorus. Understanding the composition is essential, as it influences the semiconductor's performance and suitability for various electrical applications.
Considering these traits, Cadmium Phosphide has proven useful across different technological domains. It is also leveraged in the production of laser diodes and other light-emitting devices due to its effective electron transport characteristics.

n-type doping

n-type doping is a process used to enhance the electrical conductivity of semiconductors.
In n-type doping, atoms with extra electrons replace some of the host atoms in the semiconductor lattice. These additional electrons increase the material's negative charge carriers.
For example, in Cadmium Phosphide doping, replacing a portion of phosphorus atoms (element Y) with iodine introduces extra electrons. Iodine, with its electron configuration \([\text{Ar}] \ 4s^{2} \ 3d^{10} \ 4p^{5}\), has more valence electrons than phosphorus. As a result, this substitution enriches the semiconductor with additional free electrons, enhancing its conductivity properties as an n-type material.
Overall, understanding n-type doping allows us to comprehend how such modifications tailor the electrical properties of materials like CdP to optimize them for specific electronic functions.

Electron Configuration

Electron configuration is a notation that represents the distribution of electrons in an atom's atomic orbitals.
It involves arranging electrons in orders of increasing energy, filling the atomic orbitals systematically as outlined by the principles of quantum mechanics.
The electron configuration for cadmium \( [\text{Kr}] \ 5s^{2} \ 4d^{10} \) tells us it has 48 electrons, indicating it fills up to the 5s and 4d orbitals. Similarly, phosphorus' configuration helps determine identities when calculating molar masses by locating elements within the periodic table based on their configurations.
This method reveals details about an element's chemical behavior and interaction capabilities, critical in determining the nature of semiconductors like CdP.

Molar Mass Calculation

Molar mass calculation is essential for determining the mass of individual elements within compounds, crucial for identifying semiconductors.
First, we calculate the molar mass of cadmium by multiplying the number of moles (0.132 mol) by cadmium's atomic weight (112.41 g/mol). This equals approximately 14.83812 g of cadmium in the sample.
After determining the mass in the semiconductor, we subtract cadmium's mass from the total mass (19.0 g) to find the mass of the second element, phosphorus, resulting in 4.16188 g.
Lastly, dividing the mass of phosphorus by the same number of moles finds its molar mass to be approximately 31.53 g/mol, matching phosphorus' characteristic mass.
Understanding this process helps in identifying compounds from their mass and composition, a key task in studying semiconductor materials.