My textbook says that this happens because the number of protons, and thus the total positive charge, increases - a greater attractive force acts on each electron. The book says that this is despite the fact that the number of electrons increases, as each new electron is added to the outermost shell and so the increase in the repulsive force between electrons (electron shielding) is small. But I do not see how this justifies the increase in the force per electron despite the increase in electrons. Help!
2
-
3$\begingroup$ What have you done between getting confused by the textbook and asking for clarification? // Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, objections, question down-voting or even question closure. $\endgroup$– PoutnikMar 28 at 13:13
-
$\begingroup$ The positive charge is concentrated at the center of the atom whereas the negative charges are spread in a much larger volume. $\endgroup$– Buck Thorn ♦Mar 29 at 5:45
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
6
I'll try my best to answer from a conceptual standpoint followed by some well known formulas to help clarify the rationale.
I think your issue stems from the faulty idea that the attractive force of the nucleus is equally divided between the electrons. With this line of thinking I can see why you might think that as the number of protons increase 1:1 with the number of electrons that the overall 'pull' of the nucleus on its outer electron shell remains constant. This is not how the attractive forces between charged particles work. The attractive force between the positively charged nucleus and any given negatively charged electron is only dependent on the magnitude of the two charges and the distance between them. This relationship is described by Coulomb's law:
$F = k\frac{q_1q_2}{r^2}$
Where:
$F = electric force$
$k$ = Coulomb constant
$q_1, q_2 = charges$
$r$ = distance of separation
It's better to think of any given electron as a test charge at a certain distance from the nucleus. Let me illustrate this with an oversimplified example that doesn't take into account shielding from core electrons. Below is a diagram of oxygen showing the charge of its nucleus and its outermost shell of electrons (valence electrons).
We know the nucleus has a positive charge of 8+ and that each electron in its valence shell has a charge of 1-. Also, since valence electrons occupy the same shell, we can assume they are all equidistant from the nucleus. So the attractive force between the nucleus and any given valence electron (test charge) can be described using the above formula:
$q_1, q_2 = 8, -1$
$F_1 = k\frac{(8)(-1)}{r^2}$
Since $k$ is a constant and $r$ doesn't change within an electron shell, I'll leave these values as constants since we'll be comparing two values.
What happens when we move right one group to Fluorine?
Well the charge of the nucleus increases from +8 to +9 and another electron is added to the valence shell. However, since the electron occupies the same shell, its distance from the nucleus is unchanged.
$q_1, q_2 = 9, -1$
$F_2 = k\frac{(9)(-1)}{r^2}$
Comparing the values for Oxygen and Florine we can see:
$F_2 > F_1$, $k\frac{(9)(-1)}{r^2} > k\frac{(8)(-1)}{r^2}$
But what about electron shielding?
We'll both Oxygen and Florine have the same number of core electrons (2). This means that amount of shielding is the same. It follows then, that Florine has a stronger attractive force between its nucleus and valence electrons, but the same amount of core electron shielding. Flourine will exert a larger net 'pull' on its electron cloud, resulting in a smaller atomic radii.
-
$\begingroup$ It is oxygen and fluorine, not Oxygen and Florine. $\endgroup$– PoutnikMar 30 at 3:39
-
$\begingroup$ I disagree.You cannot treat electrons as classical particles when talking about orbitals.... $\endgroup$– VolpinaMar 30 at 12:41
-
$\begingroup$ I'm using these oversimplified equations to express a concept the question poster was struggling with. Of course Coulomb's law can't be used to describe actual electron clouds. It can be used to try and understand general trends and to dispel some of the confusion the original poster had. I'm of the opinion that if this Chemistry 101 concept can't be explained without wave functions, its a failure on the part of the teacher...not the student. $\endgroup$ Mar 30 at 17:42
-
-
$\begingroup$ I would then dedicate this energy to answering the question. $\endgroup$ Mar 31 at 18:24
$\begingroup$
$\endgroup$
Despite believing my answer isnt correct , it is the leading method in calculating the wavefunction of orbitals so I will go mainstream for a while...If you compare the effective nuclear charge of the valence electrons from Wikipedia you will see that the effective charge is increasing as we go down in the periodic table.
This means that if we solve the time independent Schrodinger equation the peak of the wavefunction $\Psi(x)$ of valence electrons will be closer to $x=0$ (the nucleus) as we go down in the periodic table which means that the average distance between the nucleus and the valence electrons will be smaller.
