
(a)
Interpretation: The band that is the valence band should be determined.
Concept Introduction:The band that forms from bonding molecular orbital in a semiconductor is called the valence band
(b)
Interpretation: The band that is the conduction band should be determined.
Concept Introduction: The band that forms from anti-bonding molecular orbital in a semiconductor is called the conduction band.
(c)
Interpretation: The region of the diagram that represents the band gap should be determined.
Concept Introduction: The band gap is the gap between the valence band and conduction band.
(d)
Interpretation: The band that consists of bonding molecular orbital should be determined.
Concept Introduction:The band that forms from bonding molecular orbital in a semiconductor is called the valence band that contains bonding molecular orbital
(e)
Interpretation: The type of semiconductor represented by the diagram should be explained.
Concept Introduction:An n-type semiconductor is one in which the dopant element has more number of valence electrons than the host element and this leads to an increase in number of negatively charged electrons in the conduction band that can move easily. So, in an n-type semiconductor conduction band is not completely empty.
A p-type semiconductor is one in which the dopant element has less number of valence electrons than the host element and this leads to electron vacancies, known as holes, in the valence band. Thus in a p-type semiconductor the valence band has electron vacancies and is not completely filled with electrons.
(f)
Interpretation: The elements that could be dopant from the given elements
Concept Introduction:The doped semiconductor given in the diagram is a p-type semiconductor. It means the dopant element must have less valence electron than germanium.

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Chapter 20 Solutions
Chemistry: The Central Science, Books a la Carte Plus MasteringChemistry with eText -- Access Card Package (13th Edition)
- -AG|F=2E|V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: Acidic solution -0.93 +0.38 -0.51 -0.06 H3PO4 →H4P206 H3PO3 H3PO2 → P→ PH3 -0.28 -0.50 → -0.50 Basic solution 3-1.12 -1.57 -2.05 -0.89 PO HPO →→H2PO2 P PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P2O6 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH, 0.0 -0.5- 2 3 9 3 -1.5 -2.0 Pa H,PO H,PO H,PO -3 -1 0 2 4 Oxidation state, N 2 b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) c) Elemental phosphorus tends to disproportionate under basic conditions. Use data in…arrow_forwardThese two reactions appear to start with the same starting materials but result in different products. How do the chemicals know which product to form? Are both products formed, or is there some information missing that will direct them a particular way?arrow_forwardWhat would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 3 1 2 2. n-BuLi • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure. 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Priva ×arrow_forward
- Predict the products of this organic reaction: Explanation Check IN NaBH3CN H+ ? Click and drag to start drawing a structure. D 5 C +arrow_forwardPredict the products of this organic reaction: H3O+ + ? • Draw all the reasonable products in the drawing area below. If there are no products, because no reaction will occur, check the box under the drawing area. • Include both major and minor products, if some of the products will be more common than others. • Be sure to use wedge and dash bonds if you need to distinguish between enantiomers. No reaction. Click and drag to start drawing a structure. dmarrow_forwardIarrow_forward
- Draw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option commandarrow_forwardAG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, Narrow_forward4. For the following complexes, draw the structures and give a d-electron count of the metal: a) Tris(acetylacetonato)iron(III) b) Hexabromoplatinate(2-) c) Potassium diamminetetrabromocobaltate(III) (6 points)arrow_forward
- 2. Calculate the overall formation constant for [Fe(CN)6]³, given that the overall formation constant for [Fe(CN)6] 4 is ~1032, and that: Fe3+ (aq) + e = Fe²+ (aq) E° = +0.77 V [Fe(CN)6]³ (aq) + e¯ = [Fe(CN)6] (aq) E° = +0.36 V (4 points)arrow_forward5. Consider the compounds shown below as ligands in coordination chemistry and identify their denticity; comment on their ability to form chelate complexes. (6 points) N N A B N N N IN N Carrow_forward1. Use standard reduction potentials to rationalize quantitatively why: (6 points) (a) Al liberates H2 from dilute HCl, but Ag does not; (b) Cl2 liberates Br2 from aqueous KBr solution, but does not liberate C12 from aqueous KCl solution; c) a method of growing Ag crystals is to immerse a zinc foil in an aqueous solution of AgNO3.arrow_forward
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