Question

to determine the lewis electron dot formula for ammonia, nh₃, the central atom must be determined. the central atom is n because it is the least abundant and because n wants to form 1 bonds. each of the h atoms wants to form 2 bond and so can never be a central atom. the total number of valence electrons is determined by adding up all of the group numbers. the total number of valence electrons in nh₃ is 3. valence electrons are first placed into bonds between atoms, and this accounts for 4 electrons. the final 5 electrons are placed on the n. n now has 6 reds around it, with 7 bonding electron pairs, and 8 nonbonding electron pairs. the x - ray diffractometer sees only 9 atoms around the n, and the shape is 10. a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 45 j. 90 k. 109.5 l. 120 m. 180 n. diatomic o. bent p. planar trigonal q. tetrahedral r. pyramidal s. linear

Explanation:

• 1: Nitrogen (\(N\)) has 5 valence electrons and needs 3 more to complete its octet, so it forms 3 bonds.
• 2: Hydrogen (\(H\)) has 1 valence electron and needs 1 more to complete its duet, so it forms 1 bond.
• 3: Nitrogen (\(N\)) is in group 5 and hydrogen (\(H\)) is in group 1. For \(NH_3\), total valence electrons \(=5+(3\times1)=8\).
• 4: There are 3 \(N - H\) bonds. Each bond has 2 electrons, so \(3\times2 = 6\) electrons in bonds.
• 5: Total valence electrons are 8. After using 6 in bonds, \(8 - 6=2\) electrons are left for \(N\).
• 6: The total number of regions of electron density (REDs) around \(N\) is \(3\) (from \(N - H\) bonds) \(+1\) (lone pair) \(=4\).
• 7: There are 3 \(N - H\) bonds, so 3 bonding electron pairs.
• 8: There is 1 lone pair (non - bonding electron pair) on \(N\).
• 9: The x - ray diffractometer sees the bonded atoms. There are 3 \(H\) atoms bonded to \(N\).
• 10: With 3 bonding pairs and 1 lone pair around \(N\), the molecular shape (based on VSEPR theory) is pyramidal.

Answer:

1. C. 3
2. A. 1
3. H. 8
4. F. 6
5. B. 2
6. D. 4
7. C. 3
8. A. 1
9. C. 3
10. R. pyramidal