Question

question 5 (2 points)
the en for h is 1 and the en for n is 2.
the n - h bonds in ammonia have a δen = 3.
according to the δen values in chapter 10, the n - h bonds are 4.
ammonia is pyramidal and is symmetrical/asymmetrical 5.
therefore, ammonia is a polar/nonpolar 6 molecule and behaves like a magnet.
this explains why ammonia dissolves in water.
a. 0 b. 1 c. 0.84 d. 2 e. 2.20 f. 3 g. 3.04 h. 4
i. nonpolar covalent j. polar covalent k. ionic l. asymmetrical
m. symmetrical n. polar o. nonpolar

Explanation:

• For the electronegativity (EN) of H: The standard EN value of H is 2.20.
• For the EN of N: The standard EN value of N is 3.04.
• For \(\Delta EN\) of N - H bond: \(\Delta EN=3.04 - 2.20=0.84\).
• For bond type: When \(0<\Delta EN<1.7\), the bond is polar covalent.
• For symmetry of pyramidal ammonia: Pyramidal shape is asymmetrical.
• For polarity of ammonia: Asymmetrical polar - bonded molecule is polar.

Answer:

1. E. 2.20
2. G. 3.04
3. C. 0.84
4. J. polar covalent
5. L. asymmetrical
6. N. polar