Exam1-solutions

Exam1-solutions - Midterm Exam I MASC 310 Sept 24, 2009 Dr....

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Midterm Exam I MASC 310 Sept 24, 2009 Dr. Carter SOLUTIONS ______________________________________ Name
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Possibly helpful material Boltzmann’s constant: 8.62x10 -5 eV/atom-K Electronegativities of the elements (listed below the symbol for each element) 2 r cation anion Coord # < 0.155 0.155 - 0.225 0.225 - 0.414 3 4 linear triangular tetrahedral cation anion Coord # 0.414 - 0.732 0.732 - 1.0 6 8 octahedral cubic
Background image of page 2
1. Atomic Bonding (A) Briefly cite the main characteristics and differences between (i) Ionic bonding (ii) Covalent bonding (iii) Metallic bonding and (iv) Van der Waals bonding (secondary bonding) (4 points) The main characteristics and differences between the various forms of primary bonding are: Ionic: there is electrostatic attraction between oppositely charged ions. Covalent: there is electron sharing between two adjacent atoms such that each atom assumes a stable electron configuration. Metallic : the positively charged ion cores are shielded from one another, and also "glued" together by the sea of valence electrons. Van der Waals: Fluctuations in the electron cloud around atoms gives rise to induced dipoles that are weakly attracted to each other. Weaker attraction than other forms of bonding. (B) What type(s) of bonding would be expected for each of the following materials: calcium fluoride (CaF 2 ), cadmium telluride (CdTe), and tungsten(W)? (3 points) The answers to this question relate to the relative positions of the elements in the periodic table CaF 2 – Huge electronegativity (EN) difference between Ca and F => Strongly Ionic CdTe – Small EN difference, centered around group IV => Covalent with slight ionic character W - Metallic (C) Estimate the melting point of Molybdenum given it’s bonding energy of 6.8 eV/atom. Use the data in table 2.3 for substances with a similar bond type. (3 points) The key to this problem is twofold. First, general trends within material categories mean that the ratio of melting point to bond energy should be approximately constant. Second, realize that W is a metal and use the data for metallic bonding (W, Fe, Al, Hg). The simplest answer to the question is to scale the melting point vs bonding energy values for W: Tm=(3410+273)K*(6.8eV/8.8eV)=2845K=2572°C. You can also interpolate between the values for Fe and W. Taking the values for melting point and bond energy in table 2.3, we find 6.8eV*[(3410°C)-(1538°C)]/(8.8eV-4.2eV)=2494°C. You can get more complicated if you want and consider a full interpolation of all values in the metals category (Hg, Al, Fe, and W) which gives Tm=2496°C. The actual melting point of Mo is 2617°C – higher than all of our estimates. Can you think of a physical reason for this discrepancy?
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2. Crystallography and coordination (A) Determine the Miller indices for the planes shown in the following unit cell: (3 points)
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

Exam1-solutions - Midterm Exam I MASC 310 Sept 24, 2009 Dr....

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online