Question

part 1 draw the attraction between a water molecule and a molecule of nf₃ first, draw one molecule and add the partial charges where needed - use the δen to determine the types of bonds. then, draw the second molecule so that the δ+ on one molecule lines up across from the δ- on the other. since we cant draw the molecules here you will answer questions about the drawings that you made on the homework worksheet. nf₃ has 1 bonds with a δen = 2. nf₃ has 4 reds and is symmetrical/asymmetrical (answer is 3) making it a polar/nonpolar (answer is 4) molecule. each f has a 5 charge and the n has a 6 charge. h₂o has 2 7 bonds with a δen = 8. a. nonpolar covalent b. polar covalent c. ionic d. polar e. nonpolar f. 0.97 g. 0 h. 0.01 i. 0.94 j. 0.40 k. 1.24 l. diatomic m. bent n. pyramidal o. planar trigonal p. tetrahedral q. symmetrical r. asymmetrical s. londons t. dipole - dipole u. hydrogen bond v. ion - dipole w. δ+ x. δ- y. no partial charge

Explanation:

• For \(NF_3\), the electronegativity of \(N = 3.04\) and \(F=3.98\). \(\Delta EN=3.98 - 3.04=0.94\). Since \(0.5<\Delta EN<1.7\), the bonds are polar covalent. \(NF_3\) has a trigonal - pyramidal shape (due to 3 bonding pairs and 1 lone pair), so it is asymmetrical and polar. Fluorine is more electronegative than nitrogen, so \(F\) has a \(\delta-\) charge and \(N\) has a \(\delta+\) charge.
• For \(H_2O\), the electronegativity of \(O = 3.44\) and \(H = 2.20\). \(\Delta EN=3.44 - 2.20 = 1.24\). Since \(0.5<\Delta EN<1.7\), the bonds are polar covalent.

Answer:

1. B. polar covalent
2. I. 0.94
3. R. asymmetrical
4. D. polar
5. X. \(\delta-\)
6. W. \(\delta+\)
7. B. polar covalent
8. K. 1.24