Homework Question 3 (20 points): Bond Type Predict the most likely bond type for
ID: 1082283 • Letter: H
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Homework Question 3 (20 points): Bond Type Predict the most likely bond type for the following. Also, calculate the percentage ionic character (a) Cu (Copper) (b) KCl (Potassium Chloride) (c) Si (Silicon) (d) CdTe (Cadmium Telluride) (e) ZnTe (Zinc Telluride) Atomic Number and Electronegativity are follows Element Atomic Number Electronegativity 29 19 17 Copper (Cu) Potassium (K Chlorine (CI) Silicon (Si) Cadmium (Cd) Zinc (Zn) Tellurium (Te 0.9 48 30 52 1.5 2.0 90 IC = {1-exp[-(0.25)(XA-XB)2]} × 100Explanation / Answer
a) Copper is a metal so it has metallic bond between the copper atoms.Such a metallic bonding is the electrostatic attraction between delocalized valence electrons or the conduction electron and the metallic cation .
b)KCl has K with an electronegativity of 0.9 and Cl with an electronegativity of 2.9,which is a significant difference.
%IC={1-exp[-(0.25)(2.9-0.9)^2]}*100={1-0.368}*100=63.21%
Its has ionic bonding between K+ and Cl- ions
c) Silicon is not a metal but a metalloid,so it shares its 4 valence electrons with neighboring Silicon atoms in its crystal lattice forming covalent bonds.
d)CdTe
Electronegativity (Cd)=1.5
Electronegativity (Te)=2.0
% ionic character={1-exp[-(0.25)(2.0-1.5)^2]}*100=6.05% (very low) ,so covalently bonded atoms in CdTe
e)ZnTe
Electronegativity (Zn)=1.7
Electronegativity (Te)=2.0
% ionic character={1-exp[-(0.25)(2.0-1.7)^2]}*100=2.22% (very low) ,so covalently bonded atoms in ZnTe
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