lecture seventeen chm 151 ©slg topics: 1. electronic configurations, atoms 2. electronic...

LECTURE SeventeenCHM 151 ©slg

TOPICS:

1. Electronic Configurations, Atoms2. Electronic Configurations, Ions

Where the Final Electron Goes:s,f,d,p Blocks of Elements

s

f

dp

Subshells by order of filling,Lowest energy to highest

1s 1s

2s 2s 2p 2p 2p 2p 2p 2p

3s 3s 3p 3p 3p 3p 3p 3p

4s 4s 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 4p 4p 4p 4p 4p 4p


6s 6s 4f 5d 5d 5d 5d 5d 5d 5d 5d 5d 5d 6p 6p 6p 6p 6p 6p

7s 7s 5f

start

finish

Hydrogen, Z=1:

“spectroscopic notation”

1s1Main shell

subshell

Total e’s in subshell

“orbital box diagram”

1s

He Z = 2 1s2

Li Z = 3 1s22s1

Be Z = 4 1s22s2

B Z = 5 1s22s22p1

1s 2s 2p

THE NEXT 4 ELEMENTS

C Z=6 1s22s22p2

N Z=7 1s22s22p3

O Z=8 1s22s22p4

F Z=9 1s22s22p5

Ne Z=10 1s22s22p6

1s 2s 2p 2p 2p

End, Period 2

Note that Helium, 1s2, and Neon, 1s2 2s22p6, found at the end of the first two periods, have completed shell 1 and 2. They are classified as “noble or ‘inert’ gases”, all along with all elements in their periodic table group VIIIA.

All remaining elements in this column, 8A, have completed an outer shell s and p subshell, 8 total outer shell electrons.

This represents a special, very stable arrangement,ns2 np6, which is found to be a goal of elements in forming compounds:

an outer shell of 8 e’s, like the noble gases.

In doing configurations, it is customary to do all electrons from the last noble gas as a summary of the core electrons:

[He2 ] [ Ne10 ] [Ar18] [ Kr 36] [Xe 54] [Rn 86] Nitrogen, Z = 7, becomes [He2] 2s2 2p3

The electrons that are completing these configurations are called “core electrons”: once in place, they do not take place in reactions.

The valence electrons, used for bonding, are those in subshells filling in between two noble gases.

Short form, from last noble gas

Na, Z =11

[Ne10 ] 3s1

Electronic Configurations, the “Long form”

Na, Z=11 1s2 2s2 2p6 3s1

1s2 )2 2s2 2p6 )10 3s1

My personal way of counting e’s using: a summary at end of each period

Core electronsValence electrons

Na Z=11 [Ne10] 3s1

Mg Z=12 [Ne10] 3s2

Al Z=13 [Ne10] 3s2 3p1

Si Z=14 [Ne10] 3s2 3p2

P Z=15 [Ne10] 3s2 3p3

S Z=16 [Ne10]3s2 3p4

Cl Z=17 [Ne10] 3s2 3p5

Ar Z=18 [Ne10] 3s2 3p6

Period 3: Valence e’sCore e’s

Subshells by order of filling,Lowest energy to highest

1s 1s

2s 2s 2p 2p 2p 2p 2p 2p

3s 3s 3p 3p 3p 3p 3p 3p




7s 7s 5f

done

next

At this point, a glance at the period table or a look at our list of (n+ l ) values tells us that the next subshell to befilled is not 3d but 4s: Potassium, Z=19, is in the firstcolumn of the s block, fourth period, and must be utilizinga 4s orbital for its “distinguishing electron.”

The (n+ l ) sums or PT dictate that the filling is as follows:

K Z=19 [Ar18] 4s1

Ca Z=20 [Ar18] 4s2

Sc Z=21 Ar18] 4s23d1

A point of convention:

It is standard practice to arrange the ordering of yourelectronic configurations in order of shells, not inactual order of filling.

Sc, [ Ar18]4s23d1 ORDER OF FILLING

becomes, by convention:

Sc, [Ar18]3d14s2 ORDER OF SHELLS

Order found in texts

Sc Z=21 [Ar18] 3d14s2

Ti Z=22 [Ar18] 3d24s2

V Z=23 [Ar18] 3d34s2

Cr Z=24 [Ar18] 3d44s2 3d54s1

Mn Z=25 [Ar18] 3d54s2

Note exception!

Fe Z=26 [Ar18] 3d64s2

Co Z=27 [Ar18] 3d74s2

Ni Z=28 [Ar18] 3d84s2

Cu Z=29 [Ar18] 3d94s23d104s1

Zn Z=30 [Ar18] 3d104s2

Note exception!

When filling in electrons into the d and f subshells, we run into unexpected “irregularities”: configurationsnot quite what we would predict.

In the “d” block, these occur principally in two locations, column 6B and column 1B:

(6B) “d4s2” “d5s1”

(1B) “d9s2” “d10s1”

The half full and completed d orbital set is lower energy...

In most cases:

Especially stable: half full or filled set of orbitals.

We will meet many such exceptions in the f block!

d5

d10

Predicted: Found:

Cr [Ne10] 3d44s2 [Ne10] 3d54s1

Mo [Kr36] 4d45s2 [Kr36] 4d55s1

W [Xe54] 4f145d46s2 [Xe54] 4f145d46s2

Column 6B:

Predicted: Found:

Cu [Ne10] 3d94s2 [Ne10] 3d104s1

Ag [Kr36] 4d95s2 [Kr36] 4d105s1

Au [Xe54] 4f145d96s2 [Xe54] 4f145d106s1

Column 1B:

After completion of the 3d subshell, we complete the 4th period by filling in distinguishing electrons into the 4p:

Ga Z=31 [Ar18] 3d104s24p1

Ge Z=32 [Ar18] 3d104s24p2

As Z=33 [Ar18] 3d104s24p3

Se Z=34 [Ar18] 3d104s24p4

Br Z=35 [Ar18] 3d104s24p5

Kr Z=36 [Ar18] 3d104s24p6

End, Period 4

Electronic Configuration, Larger Atoms

Suppose we wanted to do Bi, Z = 83:

Note it is in period 6, column 5A, so must finish with “6p3”

Period 5: Repeats pattern, Period 4: 5s2 4d10 5p6

Period 6: Adds first f subshell: 6s2 4f14 5d10 6p6

Bi , Z =83, 6th period, Column 5A

1s 1s

2s 2s 2p 2p 2p 2p 2p 2p

3s 3s 3p 3p 3p 3p 3p 3p




7s 7s 5f

6p3

1A 2A 3A 4A 5A 6A 7A 8A

Period 6

Period 1

Bi, #83: period 6, column 5A “6p3”, Long form, order of filling:

1s2)2 2s2 2p6 )10 3s23p6 )18 4s2 3d104p6 )36 5s2 4d105p6 )54

6s2 4f14 5d106p3 )83

Short form, Order of filling:

[Xe54] 6s2 4f14 5d106p3

Short form, Order of shells:

[Xe54] 4f14 5d10 6s2 6p3 (red e’s: valence electrons)

GROUP WORK

Predict the electronic configuration for:

Te, Z= 52: long, order of filling; short, order of filling short, order of shells

La, Z= 57 short, order of filling short, order of shells Fr, Z= 87 short, order of filling short, order of shells

Circle all valence electrons in short form, order of shells!

KeyTe, Z= 52 (column 6A, period 5, “5p4”)

Long:

1s2 )2 2s2 2p6 )10 3s23p6 )18 4s2 3d104p6 )36

5s2 4d105p4 )52

Short, order of shells:

[Kr36] 4d10 5s25p4

Note: Te is in column 6A, has 6 valence electrons, its “outer shell bonding e’s”

Fr, Z= 87

Short, order of filling or shells:

[Rn86]7s1

Francium, like all column IA elements, has 1 valence electron

La, Z= 57 (Column 3B, period 6, 5d1)

Short, order of filling:

[Xe54]6s2 5d1

Short, order of shells:

[Xe54] 5d1 6s2

Note: La is in column 3B, has 3 valence electrons, its “outer shell bonding e’s” plus its “incomplete subshell electrons in lower shell”

Transition elements use outer s’s and inner d’s for bonding

This configuration for La is an exception: it is expectedto be [Xe54]6s2 4f1 but experimental data (spectral lines)indicate that its correct assignment is the [Xe54]6s2 5d1. La, Z= 57, [Xe54] 5d16s2

Ce, Z= 58, [Xe54] 4f1 5d16s2 Pr, Z= 59, [Xe54] 4f36s2

Nd, Z= 60, [Xe54] 4f46s2

Eu, Z= 63, [Xe54] 4f76s2

Dy, Z= 66, [Xe54] 4f106s2

Yb, Z= 70, [Xe54] 4f146s2

Lu, Z= 71, [Xe54] 4f145d16s2

Hf, Z= 72, [Xe54] 4f14 5d26s2

Check out Table 8.2, p345!

Electronic Configuration of Ions

The electronic configurations we have developedgive the basis for the charges we have alreadyassigned to many elements when they are found asions in an ionic type compound.

The main group elements (columns 1A-8A) lose, gainor share valence electrons in forming compounds so that they can achieve an outer shell configuration ,when possible, of the nearest noble gas.

When electrons are transferred from one element toanother, charged particles called ions are formed.

We have already assigned a positive charge, equal tothe column number, for ions formed from elements in columns 1A, 2A and selected 3A elements.

Note how this correlates with configurations we have done:

Na, Z=11, Column 1A, +1 cation:

1s2 2s2 2p6 3s1-1e

1s 2 2s 2 2p6 Neon configuration Na +

Since all 1A elements share the same outer shellconfiguration, “s1”, they are all expected to form the same charged ion, +1.

All elements in column 2A, with the outer shell configuration of “s2” show a + 2 charge, losing boththese electrons to form the noble gas configuration:

Ba, Z=56, Column 2A, +2 cation:

[Xe54] 6s2 - 2e[Xe54] Xenon configuration

Ba 2+

In the p block, elements filling the p subshell, bothcations and anions are formed:

Aluminum, in group 3A, loses outer s and p electrons to form a + 3 cation:

Al, Z=13, Column 3A, +3 cation:

1s2 2s2 2p6 3s23p1-3e

1s2 2s2 2p6 Neon configuration

Al3+

End, Lecture 17

lecture seventeen chm 151 ©slg topics: 1. electronic configurations, atoms 2. electronic...

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