(a)
Interpretation:
The elements needed to combine with the given set of elements in order to form good semiconductor should be determined.
Concept Introduction:
Semiconductors are substances that conduct electricity either by addition of an impurity or by the effects of temperature on it. Semiconductors have small energy gap between valence and conduction band hence its electrical conductivity lies between conductor and insulator.
Addition of impurity to a semiconductor is termed as doping. Doping alters the conductivity of a semiconductor. The addition of an element having either more or less number of valence electrons than the natural semiconductor decides the combination as the following two types of semiconductor.
- n- type semiconductor: (conduction is due to movement of extra electrons)
The element added will have more valence electron than the natural semiconductor. Therefore, the extra electron from the added element resides in conduction band and increases the conductivity.
Example: Silicon (natural semiconductor) and Phosphorus
- p-type semiconductor: (conduction is due to movement of holes)
The element added will have less valence electron than the natural semiconductor. Here, instead of extra electron, there will be “holes” at the places, where a semiconductor is replaced by added element. A p-type semiconductor increases conductivity because the holes (effective positive charge; lies at valence band) move through the natural semiconductor rather than electrons.
Example: Silicon (natural semiconductor) and Gallium
(b)
Interpretation:
The elements needed to combine with the given set of elements in order to form good semiconductor should be determined.
Concept Introduction:
Semiconductors are substances that conduct electricity either by addition of an impurity or by the effects of temperature on it. Semiconductors have small energy gap between valence and conduction band hence its electrical conductivity lies between conductor and insulator.
Addition of impurity to a semiconductor is termed as doping. Doping alters the conductivity of a semiconductor. The addition of an element having either more or less number of valence electrons than the natural semiconductor decides the combination as the following two types of semiconductor.
- n- type semiconductor: (conduction is due to movement of extra electrons)
The element added will have more valence electron than the natural semiconductor. Therefore, the extra electron from the added element resides in conduction band and increases the conductivity.
Example: Silicon (natural semiconductor) and Phosphorus
- p-type semiconductor: (conduction is due to movement of holes)
The element added will have less valence electron than the natural semiconductor. Here, instead of extra electron, there will be “holes” at the places, where a semiconductor is replaced by added element. A p-type semiconductor increases conductivity because the holes (effective positive charge; lies at valence band) move through the natural semiconductor rather than electrons.
Example: Silicon (natural semiconductor) and Gallium
(c)
Interpretation:
The elements needed to combine with the given set of elements in order to form good semiconductor should be determined.
Concept Introduction:
Semiconductors are substances that conduct electricity either by addition of an impurity or by the effects of temperature on it. Semiconductors have small energy gap between valence and conduction band hence its electrical conductivity lies between conductor and insulator.
Addition of impurity to a semiconductor is termed as doping. Doping alters the conductivity of a semiconductor. The addition of an element having either more or less number of valence electrons than the natural semiconductor decides the combination as the following two types of semiconductor.
- n- type semiconductor: (conduction is due to movement of extra electrons)
The element added will have more valence electron than the natural semiconductor. Therefore, the extra electron from the added element resides in conduction band and increases the conductivity.
Example: Silicon (natural semiconductor) and Phosphorus
- p-type semiconductor: (conduction is due to movement of holes)
The element added will have less valence electron than the natural semiconductor. Here, instead of extra electron, there will be “holes” at the places, where a semiconductor is replaced by added element. A p-type semiconductor increases conductivity because the holes (effective positive charge; lies at valence band) move through the natural semiconductor rather than electrons.
Example: Silicon (natural semiconductor) and Gallium

Want to see the full answer?
Check out a sample textbook solution
Chapter 24 Solutions
CHEMISTRY:ATOMS FIRST-2 YEAR CONNECT
- 10. The most important reason why Br- is a better nucleophile than Cl-is ___. A. polarizability; B. size; C. solvation; D. basicity; E. polarity. Please include all steps. Thanks!arrow_forwardPredicting the qualitative acid-base properties of salts Consider the following data on some weak acids and weak bases: base acid Ка K₁₁ name formula name formula nitrous acid HNO2 4.5×10 4 pyridine CHEN 1.7 × 10 9 4 hydrofluoric acid HF 6.8 × 10 methylamine CH3NH2 | 4.4 × 10¯ Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on. solution 0.1 M NaNO2 0.1 M KF pH choose one v choose one v 0.1 M C5H5NHBr 0.1 M CH3NH3CI choose one v ✓ choose one 1 (lowest) 2 ☑ 3 4 (highest) 000 18 Ararrow_forward4. The major product from treatment of 2-propanol with the Jonesreagent is ___.A. acetone; B. none of the other answers is correct C. propene; D.propanoic acid; E carbon dioxide. Please include all steps! Thank you!arrow_forward
- 7. All of the following compounds that are at the same oxidation levelare ___.u. methyl epoxide, v. propyne, w. propanal, x. propene,y. 2,2-dihydroxypropane, z. isopropanol?A. u,v,w,y; B. u,v,w; C. v,w,y,z; D. v, z; E. x,y,z Please include all steps. Thank you!arrow_forward9. Which one of the following substituents is the worst leaving group inan SN2 reaction? A. -NH2; B. -OH; C. –F; D. NH3; E. H2O Please include all steps. Thanks!arrow_forwardUsing the general properties of equilibrium constants At a certain temperature, the equilibrium constant K for the following reaction is 2.5 × 105: CO(g) + H2O(g) CO2(g) + H2(g) Use this information to complete the following table. Suppose a 7.0 L reaction vessel is filled with 1.7 mol of CO and 1.7 mol of H2O. What can you say about the composition of the mixture in the vessel at equilibrium? What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. CO2(9)+H2(g) CO(g)+H₂O(g) What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 3 CO(g)+3H2O(g) = 3 CO2(g)+3H2(g) There will be very little CO and H2O. x10 There will be very little CO2 and H2. 000 Neither of the above is true. K = ☐ K = ☐ 18 Ararrow_forward
- 8. When ethane thiol is treated with hydrogen peroxide the product is___.A. ethane disulfide; B. diethyl sulfide; C. ethane sulfoxide; D. ethanesulfate; E. ethyl mercaptan. Please include all steps. Thanks!arrow_forward5. The major product of the three step reaction that takes place when 1-propanol is treated with strong acid is?A. dipropyl ether; B. propene; C. propanal; D. isopropyl propyl ether;E. 1-hexanol Please include all steps. Thank you!arrow_forward6. The formula of the product of the addition of HCN to benzaldehydeis ___.A. C8H7NO; B. C8H6NO; C. C14H11NO; D. C9H9NO; E. C9H8NO Please include all steps. Thank you!arrow_forward
- Predicting the qualitative acid-base properties of salts Consider the following data on some weak acids and weak bases: base acid K K a name formula name formula nitrous acid HNO2 4.5×10 hydroxylamine HONH2 1.1 × 10 8 hypochlorous acid HCIO 8 3.0 × 10 methylamine CH3NH2 | 4.4 × 10¯ 4 Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on. 0.1 M KCIO solution PH choose one 0.1 M NaNO2 0.1 M CH3NH3Br 0.1 M NaBr choose one ✓ choose one v ✓ choose one 1 (lowest) ☑ 2 3 4 (highest)arrow_forwardFor this Orgo problem, don't worry about question 3 below it. Please explain your thought process, all your steps, and also include how you would tackle a similar problem. Thank you!arrow_forwardUsing the general properties of equilibrium constants At a certain temperature, the equilibrium constant K for the following reaction is 0.84: H2(g) + 2(g) 2 HI(g) = Use this information to complete the following table. Suppose a 34. L reaction vessel is filled with 0.79 mol of HI. What can you say about the composition of the mixture in the vessel at equilibrium? There will be very little H2 and 12. ☐ x10 There will be very little HI. Neither of the above is true. What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 2 HI(g) H₂(9)+12(9) K = What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 2 H2(g)+212(9) 4 HI(g) K = ☐ ☑arrow_forward
- Physical ChemistryChemistryISBN:9781133958437Author:Ball, David W. (david Warren), BAER, TomasPublisher:Wadsworth Cengage Learning,Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning





