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Ch.12 - Solids and Solid-State Materials
McMurry - Chemistry 8th Edition
McMurry8th EditionChemistryISBN: 9781292336145당신이 사용하는 게 아니라요?교과서 변경
12장, 문제 7

Examine diagrams for the electron population of the composite s–d band for three metals in question 6. Which metal has the highest melting point? (LO 12.7) (a) Metal 1 (b) Metal 2 (c) Metal 3Energy diagram showing bonding and antibonding states for three metals in a chemistry course.

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1
Examine the energy diagram for the electron population of the composite s–d band for the three metals.
Identify the bonding and antibonding states in the diagram. The bonding states are below the dashed line, and the antibonding states are above it.
Observe the proportion of electrons in the bonding and antibonding states for each metal. Metal I has all electrons in the bonding state, Metal II has a mix of bonding and antibonding electrons, and Metal III has more electrons in the bonding state than in the antibonding state.
Recall that a higher proportion of electrons in the bonding state generally leads to stronger metallic bonding and thus a higher melting point.
Determine which metal has the highest proportion of electrons in the bonding state, as this metal will have the highest melting point.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

s–d Band Theory

The s–d band theory describes the electronic structure of metals, focusing on the interaction between s and d orbitals. In this context, the bonding and antibonding states formed by these orbitals influence the metal's properties, including conductivity and melting point. The distribution of electrons in these bands can indicate the stability and strength of metallic bonds.
추천 영상:
가이드 코스
02:06
d Orbital Orientations

Melting Point and Bonding Strength

The melting point of a metal is closely related to the strength of its metallic bonds, which are influenced by the electron configuration in the bonding and antibonding states. Metals with a higher population of electrons in bonding states typically exhibit stronger interactions between atoms, leading to higher melting points. Thus, analyzing the electron population in the s–d bands can help predict which metal will have the highest melting point.
추천 영상:
가이드 코스
04:29
Boiling Point and Melting Point

Electron Configuration and Stability

Electron configuration refers to the arrangement of electrons in an atom's orbitals. In metals, a stable electron configuration in the bonding states contributes to lower energy and greater stability. The diagrams provided show how the distribution of electrons in bonding versus antibonding states can affect the overall stability of the metal, which is a key factor in determining its melting point.
추천 영상:
가이드 코스
01:23
Electron Orbital Stability
관련 실천
교과서 질문
The following diagrams represent the electron population of the composite s–d band for three metals—Ag, Mo, and Y:

Which diagram corresponds to which metal? (LO 12.7) (a) Ag = 3, Mo = 1, Y = 2 (b) Ag = 2, Mo = 1, Y = 3 (c) Ag = 2, Mo = 3, Y = 1 (d) Ag = 1, Mo = 2, Y = 3
344
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교과서 질문
The molecular orbital diagram of a doped semiconductor is shown below. If the semiconductor is silicon, does the diagram represent n-type or p-type doping and which of the following elements could be dopant? (LO 12.9)

(a) n-type, As (b) n-type, Ga (c) p-type, As (d) p-type, Ga
529
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교과서 질문
Niobium oxide crystallizes in the following cubic unit cell:

What is the formula of niobium oxide, and what is the oxidation state of niobium? (LO 12.5) (a) NbO, Nb = +2 (b) Nb2O, Nb = +2 (c) NbO2, Nb = +4 (d) Nb2O3, Nb = +3
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1
rank
교과서 질문
If the band-gap energy of a gallium phosphide (GaP) semiconductor is 222 kJ/mol, calculate the wavelength of light emitted in a GaP light-emitting diode (LED). (LO 12.11) (a) 186 nm (b) 245 nm (c) 539 nm (d) 854 nm
325
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교과서 질문
Diffraction of X rays with l = 131.5 pm occurred at an angle of 25.5 degrees by a crystal of aluminum. Assuming first-order diffraction, what is the interplanar spacing in aluminum? (LO 12.2) (a) 76.4 pm (b) 183.1 pm (c) 305.5 pm (d) 152.7 pm
283
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교과서 질문
The following diagrams represent the electron population of molecular orbitals for different substances. What diagram corresponds to magnesium oxide, germanium, and tin? (LO 12.8)

(a) Diagram 1 = tin, diagram 2 = magnesium oxide, diagram 3 = germanium (b) Diagram 1 = germanium, diagram 2 = magnesium oxide, diagram 3 = tin (c) Diagram 1 = germanium, diagram 2 = tin, diagram 3 = magnesium oxide (d) Diagram 1 = magnesium oxide, diagram 2 = tin, diagram 3 = germanium
265
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