Electron Configurations:

Problem 5.64:  Why does the number of elements in successive periods of the periodic table increase by the progression 2, 8, 18, 32?

This problem is a little bit misleading.  For rows 1-7, the lengths of the periods are actually, 2, 8, 8, 18, 18, 32, and 32, respectively.  Be that as it may, the number of orbitals in each period is n² and each of these can take two electrons to give 2n² elements.  Since n² goes as 1, 4, 9, 16, the lengths of the corresponding rows (periods) would be 2, 8, 18, and 32 as advertised.

Note also that the square of numbers correspond to the number of orbitals in each subshell.  That is

1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7.

Nothing mystical about this; that's just the way numbers operate!

For this problem, it is best to remember that various columns in the periodic table belong to what we shall call "blocks."  For instance, the groups 1A and 2A belong to the "s-block" and groups 3A-8A to the "p-block."  The various blocks correspond to the orbitals being filled and are best described using the following diagram:

This view of the ordering of orbitals by energy is quite simply followed.  You proceed left to right, top to bottom (just following the ordering of the atomic numbers).  At least, I think that this is simple...

We simply follow the "block version" of the periodic table just shown.  Armed with this weapon this problem should be exceedingly easy!

Problem 5.70:  Give the expected ground state electron configurations for the following elements:
 

(a)	Ti
(b)	Ru
(c)	Sn
(d)	Sr
(e)	Se

Note that it is easier to put the core electrons with the nearest previous inert gas.  [Ar] is shown explicitly above.  For the record,

There is really no point in rewriting this long set of symbols over and over and over...

This is much like the previous problem.  We shall do in the same way and it is left as an exercise to know the appropriate inert gas core's electronic configuration.

(a)	Z = 55
(b)	Z = 40
(c)	Z = 80
(d)	Z = 62

Problem 5.73:  Draw orbital-filling diagrams for atoms with the following atomic numbers.  Show each electron as an up or down arrow and use the abbreviation of the preceding noble gas to represent inner-shell electrons.

(These are neither easy to type or to draw.  I'll endeavor to do my best!)

(a)	Z = 25
(b)	Z = 56
(c)	Z = 28
(d)	Z = 47

Note that silver, Ag, follows a corollary of Hund's rule:  "Half-filled and fully-filled subshells are particularly stable."  In this case, a 5s electron migrates to the 4d subshell to give it its requisite 10 electrons.

Element #116 would be expected to belong to group 6A (which you can see if you look at the periodic table).  The nearest previous inert gas is radon (Rn).  That is element #86 and we need to fill 30 more orbitals!  Here is how that would be done.

This is easier than it looks.  The 7s, 5f, and 6d subshells are all filled and, from what we know of 6A elements, these all have their last p-shell as p⁴.

Fr is element #87.  The length of the period following it is 32.  Thus, we would expect its atomic number to be 87 + 32 = 119.  Its electronic configuration would be

At the right, I had to "fake" a symbol for the nearest inert gas, the as yet undiscovered element #118.

The answer given here is a good guess but could be incorrect since we are reasoning from analogy.  We note that two periods have to pass before d-orbitals begin to fill.  This is also true of the spot where f-orbitals commence filling.

The second row where f-orbitals fill happens to be row #7 (the one starting with Fr).  The first element where 5f-orbitals enter in is actinium (Ac, Z = 89).  Adding 32 to this would give element #121.  This is as good a guess as any for when the filling of g-orbitals should start!

There would be 9 g-orbitals.  These would hold 18 electrons.  Thus, if things begin filling at Z = 121, we would expect the first element with a fully filled set of g-orbitals to be element #139 (in analogy with Lu and Lr with f-orbitals).