Question

For each of the following molecules draw the Lewis structure on a separate sheet of paper. MAKE SURE TO FOLLOW THE RULES FROM CLASS (ie do not break the octet rule unless necessary to connect all the atoms). Then based on your structure indicate:

the total number of valence electrons.

the electronic and molecular shapes (choose from: linear, trigonal planar, bent, tetrahedral, trigonal pyramidal, trigonal bipyramidal, seesaw, T-shaped, octahedral, square pyramidal, or square planar).

whether or not the molecule is polar (Y/N).

Note: The central atom is the first atom listed, except for HCN, H2CO, and OCN-, where carbon is the central atom (underlined).

Formula Valence electrons Electronic Shape Molecular Shape Polar (Y/N)
HCN
PH3
CHCl3
NH4+
H2CO
SO42-
SeF2
CO2
O2
ClO4-
HBr
PF5
BeH2
PO43-
BH3
Br3-

Answer 1

Kindly check the explanation section.

Step-by-step explanation:

Without mincing words let us dive right into the solution to the question above, taking each compound at a time.

NB: Kindly Check attachment for the Lewis Structure of each of the chemical compounds.

Therefore, the number of valence electrons, electronic shape, molecular shape and whether the molecules are polar(Polarity) is given below for each chemical compound.

(1). Compound: HCN

(a). number of valence electrons = 10.

(b). electronic shape =linear.

(c). molecular shape = linear.

(d). Polarity = Y.

(2). Compound: PH3

(a). number of valence electrons = 8.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Trigonal Pyramidal.

(d). Polarity = Y.

(3). Compound: CHCl3.

(a). number of valence electrons = 26.

(b). electronic shape = tetrahedral.

(c). molecular shape = tetrahedral.

(d). Polarity = Y.

(4). Compound: NH4^+

(a). number of valence electrons = 8

(b). electronic shape = tetrahedral

(c). molecular shape = tetrahedral

(d). Polarity = Y.

(5). Compound: H2CO

(a). number of valence electrons = 12.

(b). electronic shape = Trigonal planar.

(c). molecular shape = Trigonal planar

(d). Polarity = Y.

(6). Compound: SO4^2-

(a). number of valence electrons = 32.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Tetrahedral.

(d). Polarity = N.

(7). Compound: SeF2.

(a). number of valence electrons = 20.

(b). electronic shape = Tetrahedral.

(c). molecular shape = bent.

(d). Polarity = Y.

(8). Compound: CO2.

(a). number of valence electrons = 16.

(b). electronic shape = linear.

(c). molecular shape = linear.

(d). Polarity = N.

(9). Compound: O2

(a). number of valence electrons = 32.

(b). electronic shape = Trigonal planar.

(c). molecular shape = Linear.

(d). Polarity = N.

(10). Compound: ClO4-.

(a). number of valence electrons = 32.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Tetrahedral.

(d). Polarity = N.

(11). Compound: HBr.

(a). number of valence electrons = 8.

(b). electronic shape = Linear.

(c). molecular shape = Linear.

(d). Polarity = Y.

(12). Compound: PF5.

(a). number of valence electrons = 40.

(b). electronic shape = Trigonal Bipyramidal.

(c). molecular shape = Trigonal Bipyramidal.

(d). Polarity = N.

(13). Compound: BeH2.

(a). number of valence electrons = 4.

(b). electronic shape = Linear.

(c). molecular shape = Linear.

(d). Polarity = N.

(14). Compound: PO4^3-.

(a). number of valence electrons = 32.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Tetrahedral.

(d). Polarity = N.

(15). Compound: BH3.

(a). number of valence electrons = 6.

(b). electronic shape = Trigonal planar.

(c). molecular shape = Trigonal planar.

(d). Polarity = N

(16). Compound: Br3-.

(a). number of valence electrons = 32.

(b). electronic shape = Trigonal Bipyramidal.

(c). molecular shape = Linear.

(d). Polarity = N.