a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b. BBr3. The B has 3 bonds and no lone pairs, so it is trigonal planar. c. HCN. The C is the central atom and has two bonds (a single to H, triple to N), and no lone pairs, so it is linear. d. The C is the central atom and has three bonds (single to H, double to O), so it is trigonal planar. e. There are two \"central\" N atoms. The first has a single bond to H and double to N, and one lone pair, so it is bent. The second has two bonds (double to each other N), and no lone pairs, so it is linear. H-N=N=N. The first N is bent, the second is linear (so the 3 N atoms all lie in a line, but not the H). Solution a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b. BBr3. The B has 3 bonds and no lone pairs, so it is trigonal planar. c. HCN. The C is the central atom and has two bonds (a single to H, triple to N), and no lone pairs, so it is linear. d. The C is the central atom and has three bonds (single to H, double to O), so it is trigonal planar. e. There are two \"central\" N atoms. The first has a single bond to H and double to N, and one lone pair, so it is bent. The second has two bonds (double to each other N), and no lone pairs, so it is linear. H-N=N=N. The first N is bent, the second is linear (so the 3 N atoms all lie in a line, but not the H)..

1. a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b. BBr3. The B has 3 bonds and
no lone pairs, so it is trigonal planar. c. HCN. The C is the central atom and has two bonds (a
single to H, triple to N), and no lone pairs, so it is linear. d. The C is the central atom and has
three bonds (single to H, double to O), so it is trigonal planar. e. There are two "central" N
atoms. The first has a single bond to H and double to N, and one lone pair, so it is bent. The
second has two bonds (double to each other N), and no lone pairs, so it is linear. H-N=N=N.
The first N is bent, the second is linear (so the 3 N atoms all lie in a line, but not the H).
Solution
a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b. BBr3. The B has 3 bonds and
no lone pairs, so it is trigonal planar. c. HCN. The C is the central atom and has two bonds (a
single to H, triple to N), and no lone pairs, so it is linear. d. The C is the central atom and has
three bonds (single to H, double to O), so it is trigonal planar. e. There are two "central" N
atoms. The first has a single bond to H and double to N, and one lone pair, so it is bent. The
second has two bonds (double to each other N), and no lone pairs, so it is linear. H-N=N=N.
The first N is bent, the second is linear (so the 3 N atoms all lie in a line, but not the H).