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Chapter 10: Problem 141

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

Expert verified
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

01

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#
02

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#
03

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.

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Most popular questions from this chapter

A substance, \(X\), has the following properties: Sketch a heating curve for substance \(\mathrm{X}\) starting at \(-50 .{ }^{\circ} \mathrm{C}\).

Identify the most important types of interparticle forces present in the solids of each of the following substances. a. Ar e. \(\mathrm{CH}_{4}\) b. \(\mathrm{HCl}\) f. \(\mathrm{CO}\) c. HF g. \(\mathrm{NaNO}_{3}\) d. \(\mathrm{CaCl}_{2}\)

Consider the following enthalpy changes: $$ \begin{aligned} \mathrm{F}^{-}+\mathrm{HF} \longrightarrow \mathrm{FHF}^{-} \quad \Delta H=-155 \mathrm{~kJ} / \mathrm{mol} \\ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}+\mathrm{HF} \longrightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}-\mathrm{-HF} \\ \Delta H=-46 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{HOH}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}--\mathrm{HOH}(\text { in ice }) \\ \Delta H=-21 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ How do the strengths of hydrogen bonds vary with the electronegativity of the element to which hydrogen is bonded? Where in the preceding series would you expect hydrogen bonds of the following type to fall?

An ice cube tray contains enough water at \(22.0{ }^{\circ} \mathrm{C}\) to make 18 ice cubes that each have a mass of \(30.0 \mathrm{~g} .\) The tray is placed in a freezer that uses \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) as a refrigerant. The heat of vaporization of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) is \(158 \mathrm{~J} / \mathrm{g} .\) What mass of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) must be vaporized in the refrigeration cycle to convert all the water at \(22.0^{\circ} \mathrm{C}\) to ice at \(-5.0^{\circ} \mathrm{C}\) ? The heat capacities for \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(I)\) are \(2.03 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the enthalpy of fusion for ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\).

What is the formula for the compound that crystallizes with a cubic closest packed array of sulfur ions, and that contains zinc ions in \(\frac{1}{8}\) of the tetrahedral holes and aluminum ions in \(\frac{1}{2}\) of the octahedral holes?
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