Which statement explains why NaBr is classified as a compound?
1 answer:
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Answer:
Pi bonds (π bonds) are covalent chemical bonds where two lobes of an orbital involved in the bond overlap with two lobes of the other orbital involved. These orbitals share a nodal plane that passes through the nuclei involved. Are generally weaker than sigma links, because their negatively charged electronic density is further from the positive charge of the atomic nucleus, which requires more energy.
They are frequent components of multiple bonds, as is the molecule indicated in our exercise.
The characteristics that distinguish pi bonds from other kinds of interactions between atomic species are described below, beginning with the fact that this union does not allow the free rotation movement of atoms, such as carbon. For this reason, if there is rotation of the atoms, the bond is broken.
Explanation:
In order to describe the formation of the pi bond, first we must talk about the hybridization process, as this is involved in some important links.
Hybridization is a process where hybrid electronic orbitals are formed; that is, where orbitals of atomic sub-levels s and p can get mixed. This causes the formation of sp, sp2 and sp3 orbitals, which are called hybrids.
In this sense, the formation of pi bonds occurs thanks to the overlapping of a pair of lobes belonging to an atomic orbital over another pair of lobes that are in an orbital that is part of another atom.
This orbital overlap occurs laterally, so the electronic distribution is mostly concentrated above and below the plane formed by the linked atomic nuclei, and causes the pi bonds to be weaker than the sigma bonds.
When talking about the orbital symmetry of this type of junction, it should be mentioned that it is equal to that of the p-type orbitals as long as it is observed through the axis formed by the bond. In addition, these junctions are mostly made up of p orbitals.
Since pi bonds are always accompanied by one or two more links (one sigma or another pi and one sigma), it is relevant to know that the double bond that is formed between two carbon atoms has less bond energy than that corresponding to two Sometimes the sigma link between them.
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Explanation:
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The attached file have the solved problem.
<u>Answer:</u> The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.
<u>Explanation:</u>
We are given:
Initial concentration of sulfur dioxide = 0.500 M
Initial concentration of nitrogen dioxide = 0.500 M
Initial concentration of sulfur trioxide = 0.00500 M
Initial concentration of nitrogen monoxide = 0.00500 M
The chemical reaction follows:
<u>Initial:</u> 0.500 0.500 0.005 0.005
<u>At eqllm:</u> 0.500-x 0.500-x 0.005+x 0.005+x
The expression of equilibrium constant for the above reaction follows:
We are given:
Putting values in above equation, we get:
Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration
So, equilibrium concentration of sulfur dioxide =
Equilibrium concentration of nitrogen dioxide =
Equilibrium concentration of sulfur trioxide =
Equilibrium concentration of nitrogen monoxide =
Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.