Question

A 0.132 -mole sample of an unknown semiconducting material with the formula XY has a mass of \(19.0 \mathrm{g}\). The element 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 molar mass of the semiconducting material XY is calculated as \(19.0 \,g \div 0.132 \,mol = 143.9 \frac{g}{mol}\). Element X is identified as Zn from its electron configuration. The molar mass of Y is calculated as \(143.9 \frac{g}{mol} - 65.38 \frac{g}{mol} (Zn) = 78.5 \frac{g}{mol}\), which corresponds to the element S (sulfur). Thus, the semiconducting material is ZnS. When some Y atoms are replaced by an element with electron configuration [Ar] \(4s^2 3d^{10} 4p^5\), that element is Br (bromine). Br has one more valence electron than S, so this corresponds to n-type doping.

Step-by-step solution

Calculate molar mass of XY

Given that the sample has a mass of 19.0 g and contains 0.132 moles, we can calculate the molar mass of the compound XY. Use the formula: Molar mass = Mass / Moles

Identify element X from its electron configuration

Element X has an electron configuration of [Kr] \(5s^2 4d^{10}\). Referring to the periodic table, we find that this electron configuration corresponds to the element Zn (zinc).

Identify element Y using molar mass

Now that we know the molar mass of XY and have identified X as Zn, we can find the molar mass of Y which will allow us to determine the identity of element Y. Molar mass of XY = Molar mass of X + Molar mass of Y Molar mass of Y = Molar mass of XY - Molar mass_of X Refer to the periodic table to find the identity of element Y using the calculated molar mass of Y.

Determine the type of doping when Y is replaced

When a small amount of Y atoms is replaced by atoms with an electron configuration of [Ar] \(4s^2 3d^{10} 4p^5\), identify this new element by referring to the periodic table. Once identified, compare the valence electrons of the original Y element and the replacement element. If the replacement element has more valence electrons than Y, it corresponds to n-type doping. If it has fewer valence electrons, it corresponds to p-type doping.

Key concepts

Electron Configuration

Electron configuration is a way of representing how electrons are distributed in an atom. It follows specific rules about how electrons fill the atomic orbitals. In our case, the element X has the electron configuration

• [Kr] 5s^2 4d^{10}

This configuration means that the element has filled the orbitals up to the 4d level with electrons.
This specific configuration tells us that X is the element Zinc (Zn) since Zn fills its orbitals up to this specific configuration.
Electron configurations not only help identify elements but also provide insights into the chemical properties of elements. They determine how elements will interact in bonding, and are vital in materials like semiconductors.

N-Type Doping

N-type doping in semiconductors involves adding atoms with more valence electrons than the original atoms in the material. This results in an increase in free electrons.
In the provided exercise, Y atoms, the element being partially replaced, must have fewer valence electrons than the replacing element.

• Replacement: Electrons fill up to [Ar] 4s^2 3d^{10} 4p^5

This shows that the replacing element has more electrons ready to be freed when a voltage is applied, thus introducing more free electrons into the semiconductor.
Hence, in n-type doping, the extra electrons increase the number of negative charge carriers, thereby improving the electrical conductivity of the material.

P-Type Doping

P-type doping is the opposite of n-type doping. It occurs when atoms with fewer valence electrons replace existing atoms in the semiconductor.
This creates 'holes' or absent electrons in the structure. As electrons move to fill these holes, it essentially creates positive charge carriers.
When considering p-type doping in the context of the problem:

• If the replacing element has fewer valence electrons than those being replaced

Holes are formed, and these holes serve as positive charge carriers.
This type of doping is vital because it complements n-type doped regions, allowing for the creation of essential components in semiconductor devices such as diodes and transistors.

Molar Mass Calculation

Calculating the molar mass is essential for identifying unknown elements in compounds and for various analytical processes in chemistry.
Molar mass represents the mass of one mole of a substance, and it is calculated as:

• Molar mass = Mass / Moles

In the exercise, a sample of compound XY has a mass of 19.0 grams and consists of 0.132 moles.
Using the formula, you calculate the molar mass first.
After finding the molar mass, it helps to determine the unknown element Y, once element X, Zinc, is known. Knowing Zinc's molar mass allows one to subtract and solve for Y's molar mass.
Such calculations are not only critical in academic exercises but also are used extensively in quality control and formulation in industries.