Question #fb8a1

1 Answer
Oct 6, 2016

Answer:

You can do it like this:

Explanation:

Your question does not use Slater's Rules. Since you want to understand these in your sub - note I will use these rules in my answer.

Consider an atom like sodium in the gaseous state:

ths.talawanda.org

Sodium has 11 protons in its nucleus i.e #sf(Z=11)#. The outer electron in green is attracted to the nucleus by electrostatic forces which depend on the magnitude of the charges and the distance between them.

The presence of the other 10 electrons in red has the effect of "shielding" or "screening" the outer electron from the attractive force of the nucleus.

This means that it experiences an attraction for the nucleus which is less than you would expect from the 11 protons.

This is referred to as the "effective nuclear charge" or #sf(Z_(eff)#.

So we can write:

#sf(Z_(eff)=Z-s)#

Where #sf(s)# is the screening constant and is the total screening effect of all the other electrons.

In 1930 John Slater worked out a method of calculating this screening effect which are known as "Slater's Rules".

Electrons are grouped together in increasing order of #sf(n)# and #sf(l)# values. #sf(n)# tells you the number of the energy level and #sf(l)# denotes the sub - shell such as #sf(s, p, d)#. Because #sf(s)# and #sf(p)# electrons are close in energy they are grouped together:

#sf([1s][2s,2p][3s,3p][3d][4s,4p][4d][4f][5s,5p])# and so on.

Each group has its own shielding constant. This depends on the sum of these contributions:

1.

The number of electrons in the same group.

You might think that the 2 electrons in helium are not able to shield each other since they are the same distance from the nucleus using the Bohr model:

montessorimuddle.org

However if you look at the probability of finding an #sf(s)# electron at a given distance from the nucleus you get:

intro.chem.okstate.edu

Point C is referred to as "The Bohr Radius" but you can see that there is some probability of the electron being at points closer to the nucleus such as A and B.

This means that the #sf(s)# electrons in the same orbital can effectively "screen" each other. The same applies to the #sf(s)# and #sf(p)# electrons in the same groups.

2.

The number and type of electrons in the preceding groups.

The total shielding constant is, therefore, the overall shielding effect from #sf(1)# and #sf(2)#.

For #sf(1)# the shielding constant is the sum of :

0.35 for each other electron within the same group, except for #sf[1s]# when this is 0.3

For #sf(2)# if the group is of the #sf(["ns,np"])# type you add an amount of 0.85 for each electron with a principal quantum number of #sf((n-1))# and an amount of 1.00 for each electron with a quantum number of #sf((n-2))# or less.

For #sf([d])# and #sf([f])# electrons just add 1.00 for each electron which is closer to the nucleus.

Lets see how this applies to the examples in your comment:

#sf(""_9F^(-))# is #sf([1s^(2)][2s^(2)2p^(6)])# in which the electrons have been grouped according to the rules.

So #sf(Z=9)#

Within the [2s and 2p] group a single 2p electron will be shielded by 7 other electrons each contributing 0.35 and below the group there are 2 x 1s electrons each contributing 0.85.

#:.##sf(s=(7xx0.35)+(2xx0.85)=4.15)#

#:.##sf(Z_(eff)=9-4.15=4.85)#

#sf(""_11Na^(+))# is also #sf([1s^(2)][2s^(2)2p^(6)])#

But this time #sf(Z=11)#

So

#sf(Z_(eff)=11-4.15=6.85)#

This shows that a 2p electron in #sf(Na^+)# experiences a greater effective nuclear charge than a 2p electron in #sf(F^-)#.