The idea of electron den sity, or charge density, was introduced in Chapter 8.
The behavior of electrons in atoms can be described by mathematical functions.
The square of an atomic orbital function is related to the probability of finding an electron in the threedimensional region.
We usually refer to the region that has the highest chance of finding the electron as the shape of the orbital.
We can map the total electron density throughout a molecule, not just the density of a single orbital.
The map is obtained by probing an electron density surface with a positive point charge.
The change in energy that Move probe over molecule occurs when a unit positive to measure potential charge is brought to this point, starting from another point that is infinitely far removed from the molecule.
The ammonia molecule has the same surface area as the atomic orbitals discussed in Chapter 8.
The map shows the distribution of electron charge.
Information about the distribution of electron charge in a molecule can be found in an electrostatic potential map.
If the potential at a point is positive, it is likely that an atom carries a net positive charge.
An arbitrary color scheme is used in the display of a map.
Blue is used to color the unit of positive charge at regions of the most positive electrostatic potential, while red is used to move a regions of the most negative electrostatic potential.
Intermediate colors have the same speed from one value of the potential.
For the molecule for which the map is calculated, the HCl is negative to positive.
All of the molecules were compared.
The range is -157 to 157 kJ mol-1.
The nitrogen atom has a negative charge.
Let's take a look at the computed electrostatic potential maps.
The uniform distribution of electron charge density is depicted by the uniform color distribution in the map.
This is typical for a nonpolar bond and occurs in all diatomic molecules with the same atoms.
The distribution of electron charge density of the sodium chloride molecule is not uniform.
The chlorine in the red extreme of negative charge is almost exclusively in the blue extreme of positive charge.
The map shows that the transfer of electron density from the sodium atom to the chlorine atom is not complete.
The NaCl bond is not completely ionic.
The bond is 80% ionic.
The electron charge density of the molecule is shown by the color of the map.
The hydrogen atom has a partial positive charge.
The yellow-green color shows that the chlorine atom has a partial negative charge.
The bond in HCl is clearly depicted in the map.
The H atom has less affinity for electrons than the Cl atom.
The ability of atoms to lose or gain electrons when they are part of a molecule is one of the more meaningful predictions about bond polarities.
A+B is favored.
The equation tells us that a nonpolar bond will result when the electron affinity is the same for both atoms involved in the bond.
The quantity 1Ei - Eea2 is a measure of the ability of an atom to attract electrons.
The electronegativity of the atom is related to it.
There are many ways to convert qualitative comparisons to actual numerical values of the elements.
The EN values range from 0.7 to 4.0 by Pauling.
The more metallic the element is, the less energy the A-B bond has.
When we interpret electronegativity B(g), these are the expected trends.
The age of the A-A and B-B bond between electron affinity and electronegativity is clearly seen when we consider energies.
B to F 11681 kJ mol-12.
With the help of a periodic table, you can decide which is the most negatively charged atom of each of the elements.
The ionic character is 100.
The bond between the difference atoms is very small.
The bond Although Figure 10-7 is described as polar covalent.
21g2, duce bonds that are essentially ionic, are expected to pro it in Li.
If you don't have a collection of EN values, you should be able to predict the essential character of a bond between two atoms.
The ENH is 2.1, the ENCl is 3.0, and the ENO is 3.5.
C/EN is 3.0 - 2.1.
O bond, C/EN is 3.5 The more polar bond is O bond.
Determine the percent ionic character from the figure.
In this example, we used EN values to make a decision.
It is possible to determine which end of a bond will be slightly negative.
The H end of the bond will be slightly positive.
If you want to see the variation of bond polarity with electronegativity, consider the electrostatic potential maps.
The color on the H atom ranges from HCl has the greatest ionic dark blue in the electrostatic potential map.
The halogen atom becomes less red as the negative charge decreases.
The variation of polarity within a group of related molecules can be shown on eoStatic potential maps.
When charge separation within a molecule contributes to understanding the topic at hand, we will use computed electrostatic potential maps.
There are two maps shown.
One corresponds to NaF and the other to NaH.
The bond with the greatest electronegativity difference will be more polar and will show a greater range of colors.
If you can find the number of ENH, it's 2.1; if you can find the number of ENNa, it's 0.9; and if you can find the number of ENF, it's 4.0.
Na bond, C/EN, is 2.1 Na bond, C/ EN, is 4.0.
The more polar bond is Na bond.
We conclude that the map on the left is a representation of NaF.
The charge density surface around the H "atom" in NaH appears larger than the F "atom" in NaF.
It is more appropriate to think in terms of H F ion in comparison to the bonds in the two molecules.
Various studies suggest that the NaH and NaF bonds are between 50% and 80% ionic.
Studies on solid NaH suggest that the ion's radius is between that of Cl F H and 133 pm.
In this section we combine the ideas introduced in the preceding three sections with a few new concepts to write a variety of Lewis structures.
Lewis structures have some essential features that we have already encountered.
There are only two outer-shell electrons for hydrogen.
Most of the bonds are formed by C, N, O, P, and S atoms.
The atoms are arranged in the order in which they are bonds to one another.
In a structure with more than two atoms, we need to distinguish between the central and terminal atoms.
Consider CH3 CH2OH as an example.
The following structural formula is the same as its skeletal structure.
The O atom is printed in red.
All six H atoms are printed in blue.
There are more facts about central atoms, terminal atoms, and skeletal structures.
Terminal atoms are hydrogen atoms.
An H atom can only hold two electrons in its valence shell, so it can only form one bond to another atom.
Those with the lowest electronegativity are the central atoms.
H atoms can be only terminal atoms in the skeletal structure.
The central atoms are the lowest in electronegativity.
The O atom is a central atom and has the highest electronegativity.
It is not possible for O to be a terminal atom in structure because it would have to exchange places with an H atom.
It's a good fact to keep in mind when writing Lewis structures.
The more compact structure on the right is the one observed for phosphoric acid.
At this point, we should incorporate a number of the ideas that we have considered so far into a specific approach to writing Lewis structures.
This strategy is designed to give you a place to begin, as well as consecutive steps to follow to achieve a plausible Lewis structure.
Determine the number of electrons in the structure.
There are 4 valence electrons for each C atom, 8 for the two C atoms, 1 for each H atom, 6 for the six H atoms, and 6 for the lone O atom.
An additional 3 valence electrons must be brought into the structure.
There are 5 valence electrons for the N atom and 1 for each H atom, or 4 for all four H atoms.
The central and terminal atoms are identified.
Write a structure that is plausible.
The structure at this point is a satisfactory Lewis structure if there are just enough valence electrons to complete octets for all the atoms.
It is necessary to give all atoms complete octets in order to create a plausible Lewis structure.
The procedure for writing Lewis structures is summarized in Figure 10-8.
It takes a lot of practice to become proficient at electrons.
Write structures of molecule that only have a skeleton.
The terminal atoms can be identified.
The total number of valence electrons should be subtracted from the number of electrons used.
Lewis structure drawing is complete.
The scheme for constructing Lewis structures is applied here.
Determine the number of electrons.
The total number of electrons is 18.
The terminal and central atoms are identified.
The C atoms have a lower electronegativity than the N atoms, so they are central atoms.
There are 12 valence electrons to be assigned.
Only 12 valence electrons are enough to complete the octets of the N atoms.
To form bonds with the central C atoms, move lone pairs of electrons from the terminal N atoms.
Each C atom has four electrons in it's valence shell and needs four more to complete an octet.
If we move two lone pairs from each N atom into its bond with a C atom, then each C atom requires two additional pairs of electrons.
The construction of correct Lewis structures is an important skill that all chemists have to master.
Determine the number of electrons.
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