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 \(\mathrm{X}\) 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). When a small amount of Phosphorus atoms are replaced with Iodine atoms in the semiconductor, the doping does not correspond to n-type or p-type doping, as both elements have the same number of valence electrons.

Step-by-step solution

Calculate molar mass of the unknown semiconductor

We can calculate the molar mass of the semiconductor using the given mole and mass. The formula to calculate the molar mass is as follows: Molar mass = \( \frac{Mass}{Moles} \) In this case, the mass of the sample is 19.0 g and the moles are 0.132 moles. Plug these values into the formula: Molar mass = \( \frac{19.0 \mathrm{~g}}{0.132 \mathrm{~moles}} \) = 143.94 g/mol

Identify element X from its electron configuration

The electron configuration of element X is given as \([\mathrm{Kr}] 5 s^{2} 4 d^{10}\). We can refer to a periodic table to identify the element with this electron configuration. In this case, the element with this electron configuration is Cadmium (Cd).

Identify element Y using the molar mass calculated in step 1

Since we already know the identity of element X (Cadmium - Cd), we can find the molar mass of element Y using the molar mass calculated in step 1. The formula XY has one atom of each element, so we can write: Molar mass of Y = Molar mass of XY - Molar mass of X The molar mass of Cadmium is 112.41 g/mol. Substitute the values and find the molar mass of Y: Molar mass of Y = 143.94 g/mol - 112.41 g/mol = 31.53 g/mol Refer to the periodic table again to determine the element with a molar mass close to 31.53 g/mol. In this case, the element is Phosphorus (P). Thus, the unknown semiconducting material is Cadmium Phosphide (CdP).

Determine if the doping corresponds to n-type or p-type doping

A small amount of Y (Phosphorus) atoms in the semiconductor is replaced with an equivalent amount of atoms with the electron configuration: \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{5}\). Refer to the periodic table to identify the element with this electron configuration. In this case, the element is Iodine (I). Phosphorus has 5 valence electrons, while Iodine also has 5 valence electrons. Since both elements have the same number of valence electrons, neither an extra electron nor a hole is created when a small amount of Phosphorus atoms are replaced with Iodine atoms. Therefore, the doping does not correspond to n-type or p-type doping.

Key concepts

Electron Configuration

Electron configuration is a method used to describe how electrons are arranged within atoms. It breaks down the way electrons populate an atom's electron shells and subshells, from the lowest to the highest energy levels. Generally, electron configurations are denoted by numbers and letters that represent energy levels and orbitals respectively, like \([\mathrm{Kr}] 5s^{2} 4d^{10}\). This example indicates that the electrons fill up to the noble gas krypton, with additional electrons occupying a series of higher energy orbitals.
Every element has a unique electron configuration which can often be deduced from the periodic table. For example, the electron configuration for Cadmium (Cd) is \([\mathrm{Kr}] 5s^{2} 4d^{10}\). This indicates that after reaching the electron configuration of krypton, the element continues to fill electrons in the 5s and 4d orbitals, totaling 12 additional electrons.

In this exercise, the configuration \([\mathrm{Ar}] 4s^{2} 3d^{10} 4p^{5}\) helps in identifying an element due to its distinct arrangement. Recognizing and understanding these configurations is essential in chemistry for predicting chemical reactivity and bonding.

Molar Mass Calculation

Molar mass is a measure that represents the mass of one mole of a given substance. It is usually expressed in grams per mole (g/mol) and is crucial in stoichiometry for converting between mass and moles in chemical reactions.
To determine the molar mass, you use the formula:
\[ \text{Molar mass} = \frac{\text{Mass}}{\text{Moles}} \]
In this problem, the mass of the semiconducting sample is given as 19.0 grams, and the amount of substance is 0.132 moles. Using these values, the molar mass is calculated as:
\[ \text{Molar mass} = \frac{19.0 \text{ g}}{0.132 \text{ moles}} = 143.94 \text{ g/mol} \]
This calculation allows us to deduce the identities of the elements involved in the unknown material, using the periodic table to match molar masses to known elements.

Further, the identity of element X is given and with its known molar mass, the molar mass of element Y can be isolated, offering insights into the complete formula of the compound, Cadmium Phosphide (CdP). This essential calculation assists in advancing the understanding of material composition and chemical properties.

N-Type and P-Type Doping

Doping in semiconductors involves intentionally introducing impurities to manipulate electrical properties. These can enhance conductivity and create either n-type or p-type semiconductors.
N-type doping adds atoms with extra electrons compared to the host lattice, introducing more negative charge carriers (free electrons). For example, replacing some component atoms with elements that have more valence electrons increases electron density.
P-type doping is the opposite, where dopants have fewer valence electrons. This results in the creation of 'holes', which are positive charge carriers, as electrons from neighboring atoms fill these gaps.

In this scenario, the doping involves replacing Phosphorus (P) atoms, which have 5 valence electrons, with Iodine (I) atoms, which have the same electron number. Since both elements possess equal valence electrons, no additional holes or electrons are created, meaning no significant change in conduction type occurs.

• N-type: more electrons (extra negative carriers)
• P-type: more holes (extra positive carriers)

Understanding these types of doping is key in designing semiconductors for specific electronic components and properties, making it fundamental knowledge in electronics and material science.