Compare the shape and e-dot structure of H₂O and CO₂. They both have 1 central atom with 2 other atoms around it, but they do not look the same. Why not? A) The central atom of H₂O has one single bond and one lone pair, while the central atom of CO₂ two double bonds and no lone pairs.
B) The central atom of H₂O has no lone pairs, while the central atom of CO₂ has two lone pairs that cause repulsion.
C) The central atom of H₂O has two lone pairs that cause repulsion, while the central atom of CO₂ has no lone pairs.
D) The central atom of H₂O has two single bonds and non lone pairs, while the central atom of CO₂ has two double bonds and no lone pairs.

C) The central atom of H₂O has two lone pairs that cause repulsion, while the central atom of CO₂ has no lone pairs.

Explanation:

If we consider H2O and CO2, the first thing that we have to do is to identify the central atom in each compound. In water, the central atom is oxygen. Oxygen has two lone pairs. In CO2, carbon is the central atom and carbon has no lone pairs.

Let us get back to valence shell electron pair repulsion theory. Recall that both CO2 and H2O has four regions of electron density hence they have a tetrahedral electron pair geometry according to VSEPR theory. However, in water, the two lone pairs on oxygen cause repulsion and cause the molecule to be bent with bond angle <<109 degrees.

Triatomic molecules are either linear or bent. CO2 has no lone pairs but displays a linear molecular geometry due to the presence of two double bonds. The bond angle in CO2 is 180 degrees.

the answer for sodium is 0.186

the answer for rubidium is 0.251

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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