Each of the following is either linear, angular (bent), planar, tetrahedral, or octahedral. Indicate the correct shape of (a) \(\mathrm{H}_{2} \mathrm{S} ;\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4} ;\) (c) \(\mathrm{HCN} ;\) (d) \(\mathrm{SbCl}_{6}^{-}\); (e) \(\mathrm{BF}_{4}^{-}\)

Short Answer

Expert verified
(a) H2S is angular (bent), (b) N2O4 is planar, (c) HCN is linear, (d) SbCl6- is octahedral, (e) BF4- is tetrahedral.

Step by step solution

01

Analyze the Compound H2S

Sulfur (S) is the central atom. It has two bonds with Hydrogen (H) and two lone pairs of electrons. According to VSEPR theory, such an arrangement leads to a bent (or angular) molecular shape.
02

Analyze the Compound N2O4

The molecule N2O4 can be described as two NO2 molecules bonded together, with Nitrogen (N) being the central atom. Each Nitrogen forms two bonds with Oxygen (O) and has one lone pair, leading to a bent shape for each NO2. The overall shape of N2O4 is therefore described as planar.
03

Analyze the Compound HCN

Carbon (C) is the central atom in HCN, being bonded to both Hydrogen (H) and Nitrogen (N), with no lone pairs on the Carbon. This results in a linear molecular shape.
04

Analyze the Ion SbCl6-

Antimony (Sb) is the central atom in SbCl6-, surrounded by six Chlorine (Cl) atoms. There is no lone pair on the Antimony. According to VSEPR theory, this configuration results in an octahedral molecular shape.
05

Analyze the Ion BF4-

Boron (B) is the central atom in BF4-, surrounded by four Fluorine (F) atoms and no lone pairs. This results in a tetrahedral molecular shape.

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