HW 2 Solutions

HW 2 Solutions - Chapter 2, Solution .11. {a} Ionic bonding...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 2, Solution .11. {a} Ionic bonding arises from the elecu'ostatic attraction between oppositely charged ions. In the process of ion formation an electron or a number of electrons may be transferred from a highly electroposititre element to a highly electronegative one. The ionic bond in solids is nondirectional. {a} Covalent bonding is a primary type of bonding which arises floor the reduction in energy associated with the overlapping of half-filled orbitals of two atoms. In this bond: there is an electron exchange interaction. The covalent bond is a directional type of bond. {'b] Metallic bonding is a primary type of bonding involving the interaction of die valence electron or elecu'ons of one atom 1a-itlo many stu'roturding atoms. 'This interaction leads to a reduction in energy of the system considered. The valence bonding electrons of these bonds are sometimes regarded as an “electron gas" bonding the positive ion cores l{atoms less their valence electrons} of atoms. The metallic bond is nondirectional. Chapter 1, Solution 36. Afier ionization: the Cl" ion is larger because the electron-to-proton ratio of the chlorine atom is decreased the ionization process. Chapter 2, Solution .17. The attractive force between the ion pair is found by applying Coulomb‘s law: 'I _ _.Z'_z:e‘ Factrmirn _ 1 danger: Where Er=-1 for K: E:=-1 for BI: and no = r3- + rm. =o.133nm+o.raamn = D.3]9nm=3.19:-l{|"m m Substituting: _ _ _ -l? I Fanzm.;=—l: 11": mm’m C] =2.l.t><1tl"’ N oregano“t3.-N.m1}[3.:axto"“ m]: Chapter 2, Solution 1.1. l. The structure ofiouic solids is partly determined by the radii of the cations and anions of the ionic solid. Only certain ranges of cation-to-anion radius ratios are allowed by packing considerations. 2. Electrical neutrality innsth maintained. Thus: if the cations and anions have different valences. the number of anions sturoturding a particular cation will he restricted. Chapter 2, Solution 52. A possible correlation between the melting points and the elecu'on configurations ofthe elements from scandium {Z = 21] to copper {Z = 39} is that unpaired 3d electrons cause covalent hybridized bonds. and hence give higher melting points to these transition metals. Chapter 2, Solution I51. Applying Pauling's equation to C'tiTe and lnP compounds, For CdTe [I — IS} '31; Ionic character = {l—e'c'H'J'H '3']: :IIIIIZIIII'l-LJ = 5.1% For in? [3 — j}. as Ionic character = {t—fi-“fl-H '--“ ){1 uses) = sass ‘Jfliile a 2 — 5 compound typically has a hith ionic character than a 3 — i the relativele high electronegatii'itjt of phosphorous causes In? to be more ionic in nature. Chapter 3, Solution Ii. The three most common crystal structures found in metals are: body-centered cubic {EEC}, face-centered cubic {FCC}, and hexagonal close-packed (HEP). Examples of metals having these structures include the following. BCC: cr—ironvanadium. timgsten, niobitun, and chromium. FCC: copper. alumintun, lead, nickel, and silver. HCP: magnesitun, tat—titanium, zinc, beryllium, and cadmium. Chapter 3, Solution T". A ECC crystal stmcture has two atoms in each unit cell. Chapter 3, Solution 3. 35 EU: crystal structure has a coordination number of eight. Chapter 3, Solution 10. Letting :1 represent the edge length of the BCC unit cell and R the moly'btientun atomic radius, 1 as Jim: =43 or u=—R 4 J? (illicit?l urn] = [L323 nm Chapter 3, Solution l-l. Each tunt cell ofthe FCC crystal structure contains four atoms. Chapter 3, Solution 15. The FCC crystal structure has a coordination number of make. Chapter 3, Solution 15". For an FCC unit cell has-"log an edge length a an containing strontitun atoms, \I'Eo =4R or J: §:R=:?_l—{CI.215 um} = [LEEDS nm Chapter 3, Solution 20. By definition the atomic packing factor is given as: volume of atoms in FCC unit cell A ' k' f = mm: Pa: mg 5mm s-‘oltune ofthe FCC unit cell These voltuues, associated 1with the four-atom FCC unit ce]l, are 1"an =4|::HR5:|=$ TR3 and I'imitcoll =fl5 3 3 . . . 4R where it represents the lattice constant. Suhaututmg :1 = . 5 MRS I-MiICDill =fl = NE The atomic packing factor then becomes. “law-15 “1:” 1J5 ‘: HE hPFIIT-‘CC unitceflj= q WE, I: 6 d . J- . = [LT-l Chapter 3, Solution 22. The coordination number associated with the HEP crystal structure it t'trelt'e. ...
View Full Document

This note was uploaded on 04/16/2008 for the course MSE 150 taught by Professor Wittig during the Spring '07 term at Vanderbilt.

Page1 / 3

HW 2 Solutions - Chapter 2, Solution .11. {a} Ionic bonding...

This preview shows document pages 1 - 3. Sign up to view the full document.

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