VIII: Periodic Table Trends

Transcript VIII: Periodic Table Trends

Topic IX: Electronic Configuration, Ions and Atoms

LECTURE SLIDES

• •

Electronic Configurations: Atoms Ions Magnetism Kotz & Treichel, Chapter 8, Sections 8.1-8.5

ELECTRONIC CONFIGURATIONS OF THE ELEMENTS We can now describe the arrangement of all the electrons around the nucleus of any given atom in terms of shells and subshells .

Recall that the periodic table can be a valuable tool in leading us along the “ aufbau ” or “ building up ” of electrons in discreet “probable locations” around the nucleus...

Matching: Subshells and PT Positions

s f d p

Subshells by order of filling, Lowest energy to highest

t star 1s 2s 2s 7s 7s 6d 5f 6d 6d 6d 6d 6d 6d 1s 2p 2p 2p 2p 2p 2p 3s 3s 4s 4s 3d 5s 5s 4d 4d 4d 4d 4d 4d 4d 4d 4d 4d 3p 3p 3p 3p 3p 3p 3d 3d 3d 3d 3d 3d 3d 3d 3d 4p 4p 4p 4p 4p 4p 5p 5p 5p 5p 5p 5p 6s 6s 5d 4f 5d 5d 5d 5d 5d 5d 5d 5d 5d 6p 6p 6p 6p 6p 6p finish

Electronic Configurations

We can describe the electron arrangement in an atom by a listing of each filled subshell in turn, from lowest energy to highest ...

We will call this listing of arrangements “ electronic configurations ” and it can be done in two modes: “ spectroscopic notation ” or “ orbital box diagram ”

Hydrogen, Z=1: “ spectroscopic notation ” Main shell

1s

1 Total e’s in subshell subshell “ orbital box diagram ” 1s

THE NEXT 4 ELEMENTS: Spectroscopic notation:

Z = 2

1s 2 Li

Z = 3 1s

2 2s 1 Be B

Z = 4 Z = 5 1s

2 2s 2

1s

2 2s 2 2p 1

Orbital box diagram 1s 2s 2p

HUND’S RULE

Following the placement of the first electron into the p subshell with Boron, the question then becomes, “does the next electron into the p subshell go to the second p orbital or does it fill up the first p orbital?” Hund’s Rule answers that question: in filling multi-orbital subshells, always put one electron into each , same spin , then begin filling each “half full” orbital.

Filling “p” orbitals: 1st, 2nd 3rd electron in place: 4th electron in place

C N O F

Z=6 Z=7 Z=8 Z=9

Z=10 1s

2 2s 2 2p 2

1s

2 2s 2 2p 3

1s

2 2s 2 2p 4

1s

2 2s 2 2p 5

1s

2 2s 2 2p 6

1s 2s 2p 2p 2p End, Period 2

Note that Helium , 1s 2 , and Neon , 1s 2 2s 2 2p 6 , 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, ns 2 np 6 , which is found to be a goal of elements in forming compounds:

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

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

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

In doing configurations, it is customary to do all electrons from the last noble gas as a summary of the core electrons: [He 2 ] [ Ne 10 ] [Ar 18 ] [ Kr 36 ] [Xe 54 ] [Rn 86 ] Nitrogen, Z = 7, becomes [He 2 ] 2s 2 2p 3

Electronic Configurations, the “Long form” Na, Z=11 1s 2 2s 2 2p 6 3s 1 1s 2 ) 2 2s 2 2p 6 ) 10 3s 1 My personal way of counting e’s using: a summary at end of each period Core electrons Short form, from last noble gas Na, Z =11 [Ne 10 ] 3s 1 Valence electrons

Period 3:

Na Mg Al Si P S Cl Ar Z=11

Core e’s Valence e’s [Ne 10 ] 3s 1

Z=12

[Ne 10 ] 3s 2

Z=13

[Ne 10 ] 3s 2 3p 1

Z=14

[Ne 10 ] 3s 2 3p 2

Z=15

[Ne 10 ] 3s 2 3p 3

Z=16

[Ne 10 ] 3s 2 3p 4

Z=17

[Ne 10 ] 3s 2 3p 5

Z=18

[Ne 10 ] 3s 2 3p 6

1s 2s 2s

Subshells by order of filling, Lowest energy to highest

1s 2p 2p 2p 2p 2p 2p 3s 3s 4s 4s 3d 3p 3p 3p 3p 3p 3p 3d 3d 3d 3d 3d 3d 3d 3d 3d 4p 4p 4p 4p 4p 4p 5s 5s 4d 6s 6s 5d 4f 4d 4d 4d 4d 4d 4d 4d 4d 4d 5d 5d 5d 5d 5d 5d 5d 5d 5d 5p 5p 5p 5p 5p 5p 6p 6p 6p 6p 6p 6p don e 7s 7s 6d 5f next

Electronic Configurations, Period 4 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 be filled is not 3d but 4s : Potassium, Z=19, is in the first column of the s a 4s block, fourth period, and must be utilizing orbital for its “distinguishing electron.” The ( n+ l ) sums or PT dictate that the filling is as follows:

K Ca Sc Z=19

[Ar 18 ] 4s 1

Z=20

[Ar 18 ] 4s 2

Z=21

Ar 18 ] 4s 2 3d 1

A point of convention: It is standard practice to arrange the ordering of your electronic configurations in order of shells, not in actual order of filling .

Sc, [ Ar

]

4s 2 3d 1 FILLING ORDER OF becomes, by convention

Sc, [Ar

]

3d 1 4s 2 SHELLS ORDER OF Order found in texts

Sc Ti V Cr

Core e’s Valence e’s (d, s)

Z=21 [Ar

]

3d 1 4s 2

Z=22 [Ar

]

3d 2 4s 2

Z=23 [Ar

]

3d 3 4s 2

Z=24 [Ar

]

3d 4 4s 2

3d

4s

1 Mn

Z=25 [Ar

]

3d 5 4s 2

Note exception!

Fe Co Ni Cu

valence core

Z=26 [Ar

]

3d 6 4s 2

Z=27 [Ar

]

3d 7 4s 2

Z=28 [Ar

]

3d 8 4s 2

Z=29 [Ar

]

3d 9 4s 2

3d

4s

1 Zn

Z=30 [Ar

]

3d 10 4s 2

Filled subshell: core Note exception!

When filling in electrons into the d and f subshells, we run into unexpected “irregularities” : configurations not quite what we would predict.

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

“d

s

” “d

s

”

(1B)

“d

s

” “d

s

”

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

Especially stable: half full or filled set of orbitals.

d 5 d 10 We will meet many such exceptions in the d , f blocks!

Cr

Column 6B:

Predicted: [Ne

]

3d 4 4s 2

Mo [Kr

]

4d 4 5s 2

Found: [Ne

]

[Kr

]

W [Xe

] 4f

14 5d 4 6s 2

[Xe

] 4f

14 5d

Column 1B:

Predicted: Cu [Ne

]

3d 9 4s 2

Ag [Kr

]

4d 9 5s 2

Found: [Ne

]

[Kr

]

Au [Xe

] 4f

14 5d 9 6s 2

[Xe

] 4f

14 5d

After completion of the 3d subshell, we complete the 4th period by filling in distinguishing electrons into the 4p : GROUP WORK 9.1

For Bromine, Br, Z=35:

•

Do the complete (“long”) form of the electronic configuration, order of filling

•

Do the short form, starting with [Ar 18 ], order of filling

•

Do the short form, starting with [Ar 18 ], order of shells

•

Circle the valence e’s in short form, order of shells

The complete 4p subshell:

Ga Ge As Se Br Kr

Core e’s Valence e’s s , p

Z=31 [Ar

]

3d 10 4s 2 4p 1

Z=32 [Ar

]

3d 10 4s 2 4p 2

Z=33 [Ar

]

3d 10 4s 2 4p 3

Z=34 [Ar

]

3d 10 4s 2 4p 4

Z=35 [Ar

]

3d 10 4s 2 4p 5

Z=36 [Ar

]

3d 10 4s 2 4p 6

Electronic Configurations, Period 5

We saw that Period 4 configurations completed with: [Ar 18 ] 4s 2 3d 10 4p 6 (order of filing, PT ) [Ar 18 ] 3d 10 4s 2 4p 6 ( order of shells )

Period 5: Repeats pattern, Period 4:

[Kr 36 ] 5s 2 4d 10 5p 6 (order of filling, PT) [Kr 36 ] 4d 10 5s 2 5p 6 (order of shells)

GROUP WORK 9.2

Do a short electronic configuration, order of filling , then order of shells for: Ag, Z= 47 Te, Z= 52 Circle valence electrons!

Electronic Configuration, Period 6

Period 6: Adds first f subshell: [Xe 54 ] 6s 2 4f 14 5d 10 6p 6 (order of filling, PT) [Xe 54 ] 4f 14 5d 10 6s 2 6p 6 (order of shells) An aggravating exception to the order of filling occurs with La, Z= 57 ... This exception repeats itself with Ac, Z= 89 . Both of these elements predictably would be “f” block fillers, but instead both are in column 3B , elements starting a “d” set of subshell fillers...

La, Z= 57 (Column 3B, period 6, 5d 1 ) Short, order of filling: [Xe 54 ] 6s 2 5d 1 Short, order of shells: [Xe 54 ] 5d 1 6s 2

This configuration for La to be [Xe 54 ] 6s 2 4f 1 is an exception: it is expected but experimental data (spectral lines) indicate that its correct assignment is the [Xe 54 ] 6s 2 5d 1.

La, Z= 57, [Xe 54 ] 5d 1 6s 2 Ce, Z= 58, [Xe 54 ] 4f 1 5d 1 6s 2 Pr, Z= 59, [Xe 54 ] 4f 3 6s 2 Nd, Z= 60, [Xe 54 ] 4f 4 6s 2 Eu, Z= 63, [Xe 54 ] 4f 7 6s 2 Dy, Z= 66, [Xe 54 ] 4f 10 6s 2 Yb, Z= 70, [Xe 54 ] 4f 14 6s 2 Lu, Z= 71, [Xe 54 ] 4f 14 5d 1 6s 2 Hf, Z= 72 , [Xe 54 ] 4f 14 5d 2 6s 2 Check out Table 8.2, p345!

Overview, Electronic Configuration of Atoms

Let’s start by doing a complete set of configurations for lead, Pb, Z= 82 The first step: find element in periodic table noting: a) which block (s, p, d, f) b) which period c) which column

Pb , Z =82, 6th period, Column 4A 1A 2A 3A 4A 5A 6A 7A 8A 1s 2s 2s 1s Period 1 2p 2p 2p 2p 2p 2p 3s 3s 4s 4s 3d 3p 3p 3p 3p 3p 3p 3d 3d 3d 3d 3d 3d 3d 3d 3d 4p 4p 4p 4p 4p 4p 4d 4d 4d 4d 4d 4d 4d 4d 4d 5p 5p 5p 5p 5p 5p 5s 5s 4d 6s 6s 5d 4f 5d 5d 5d 5d 5d 5d 5d 5d 5d 6p 6p 6p 6p 6p 6p Period 6 7s 7s 6d 5f

6d 6d 6d 6d 6d 6d

6p 2

Places last in subshell e Pb, Z= 82 p block 6th period Locates shell (n# for s,p = period #) Column 4A “6p 2 ” Locates last e in place in orbital 6p 6p 6p

Long form, order of filling, from PT: 1s 2 ) 2 2s 2 2p 6 ) 10 3s 2 3p 6 ) 18 4s 2 3d 10 4p 6 ) 36 5s 2 4d 10 5p 6 ) 54 6s 2 5d 1 4f 14 5d 9 6p 2 ) 82 4f 14

5d

10 summarize!

Short, order of filling:

Xe

54 6s 2

4f

5d

10 6p 2 Short, order of shells:

Xe

4f

5d

10 6s 2 6p 2 Four outer shell, valence electrons; group 4A; once the inner shell d’s and f’s are full: core electrons

Period 7, incomplete and all elements radioactive, follows the Period 6 pattern: [Rn 86 ] 7s 2 5f 14 6d 10 (order of filling, PT) [Rn 86 ] 5f 14 6d 10 7s 2 (order shells...)

Group Work 9.3

Do Short form order of filling , short form order of shells , circle valence e’s for Rf , Rutherfordium, Z=104

Electronic Configuration of Ions

The electronic configurations we have developed give the basis for the charges we have already assigned to many elements when they are found as ions in an ionic type compound .

The main group elements ( columns 1A-8A ) lose, gain or 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 to another, charged particles called ions are formed.

We have already assigned a positive charge, equal to the column number , for ions formed from elements in columns 1A (1+), 2A (2+) and selected 3A (3+) elements. Note how this correlates with configurations we have done: Na, Z=11 , Column 1A, +1 cation: -1e 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 Neon configuration Na +

Column

: All Ions,

+1 H Z=1 Li Z=3 Na Z=11 K Z=19 Rb Z= 37 Cs Z= 55 Fr Z= 87 1s 1 He 2 2s 1 Ne 10 3s 1 Ar 18 4s 1 Kr 36 5s 1 Xe 54 6s 1 Rn 86 7s 1 H 1+ Li 1+ Na 1+ K 1+ Rb 1+ Cs 1+ Fr 1+

Nearest noble gas

He Ne Ar 2 10 18 Kr 36 Xe 54 Rn 86

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

All elements in column 2A , with the outer shell configuration of “s 2 ” show a + 2 charge, losing both these electrons to form the noble gas configuration: Ba, Z=56 , Column 2A, +2 cation: [Xe 54 ] 6s 2 - 2e Ba 2+ [Xe 54 ] Xenon configuration

Column

: All Ions,

+2 Be Z=4 Mg Z=12 Ca Z=20 Sr Z= 38 Ba Z= 56 Ra Z= 88 He 2 2s 2 Ne 10 3s 2 Ar 18 4s 2 Kr 36 5s 2 Xe 54 6s 2 Rn 86 7s 2 Be 2+ Mg 2+ Ca 2+ Sr 2+ Ba 2+ Ra 2+ He 2 Ne 10 Ar 18 Kr 36 Xe 54 Rn 86

In the p block , elements filling the p subshell, both cations and anions are formed: let’s consider the cations first...

Aluminum, in group 3A, loses outer s and p electrons to form a + 3 cation : Al, Z=13 , Column 3A, +3 cation: -3e 1s 2 2s 2 2p 6 3s 2 3p 1 Al 3+ 1s 2 2s 2 2p 6 Neon configuration Other metals in the p block show variable charges, losing either the p e’s only or the p’s and s’s

P Block metals : Variable Charges

Tl Z=81 Xe Tl Z=81 Xe 54 54 4f 4f 14 14 5d 5d 10 10 6s 6s 2 2 6p 6p 1 1 Tl Tl 1+ 3+ Xe 54 4f 14 5d 10 6s 2 Xe 54 4f 14 5d 10 Pb Z=82 Xe 54 4f 14 5d 10 6s 2 6p 2 Pb Z=82 Xe 54 4f 14 5d 10 6s 2 6p 2 Bi Z=83 Xe 54 4f 14 5d 10 6s 2 6p 3 Pb 2+ Xe 54 4f 14 5d 10 6s 2 Pb 4+ Xe 54 4f 14 5d 10 Bi 3+ Xe 54 4f 14 5d 10 6s 2 Bi Z=83 Xe 54 4f 14 5d 10 6s 2 6p 3 Bi 5+ Xe 54 4f 14 5d 10 Also: Ga, In: 1 + , 3 + ; Sn, 2 + , 4 + ; lower charges more common

• •

NOTE: In arranging our atoms by the “short form, order of shells ” we have put our configuration into a format which allows immediate information on ion formation: always add e’s to the outermost subshell to form anions always remove e’s from the outermost shell first to make cations...

Anions of the P Block

The non-metals in columns 5, 6 and 7 are most likely to form monoatomic anions; the metalloids of these groups are less likely to be found as these anions.

In all cases, these elements gain sufficient e’s to become “ isoelectronic ” with following noble gas.

“ ISOELECTRONIC ”: same number of electrons

Anions Isoelectronic with Neon: N, Z=7 , Column 5A, -3 anion : +3e 1s 2 2s 2 2p 3 1s 2 2s 2 2p 6 Neon configuration O, Z=8 , Column 6A, -2 anion : +2e 1s 2 2s 2 2p 4 1s 2 2s 2 2p 6 Neon configuration F, Z=9 , Column 7A, -1 anion : +1e 1s 2 2s 2 2p 5 1s 2 2s 2 2p 6 Neon configuration N 3 O 2 F 1-

Anions Isoelectronic with Argon: P, Z=15 , Column 5A, -3 anion : +3e 1s 2 2s 2 2p 6 3s 2 3p 3 1s 2 2s 2 2p 6 3s 2 3p 6 Argon configuration S, Z=16 , Column 6A, -2 anion : +2e 1s 2 2s 2 2p 6 3s 2 3p 4 1s 2 2s 2 2p 6 3s 2 3p 6 Argon configuration Cl, Z=17 , Column 7A, -1 anion : +1e 1s 2 2s 2 2p 6 3s 2 3p 5 1s 2 2s 2 2p 6 3s 2 3p 6 Argon configuration P 3 S 2 Cl 1-

Let us consider the “variable charge” transition elements : these metals can utilize both their outer s and their inner d electrons for ion formation, and they are not as likely to revert to a noble gas in the process: In forming ions, the outermost electrons, the “s” e’s , are lost first to form a +2 ion which most transition elements exhibit. The other charges arise from subsequent loss of 1 or more d electrons .

’s lost first, then

’s

Fe, Z=26 Column 8A, +2 cation : [Ar 18 ] 3d 6 4s 2 -2e [Ar 18 ] 3d 6 Fe 2+ Fe, Z=26 Column 8A, +3 cation : [Ar 18 ] 3d 6 4s 2 -3e [Ar 18 ] 3d 5 Fe 3+ Note that in the second cation, Fe 3+ , The 3d subshell consists of 5 unpaired, same spin electrons, leading us next to a consideration of the topic of magnetism ...

But first: Group Work...

GROUP WORK 9.4

Do: Short form, order of filling , then order of shells for atom, then form ion: Ag, Ag + I, I 1 Zn, Zn 2+ Se, Se 2-

Magnetism: Predictable by Electronic Configurations...

Substances may be classified under this heading three ways: a) diamagnetic: slightly repelled by a strong magnet b) paramagnetic: attracted to a magnetic field, c) ferromagnetic: strongly attracted to magnetic field

Most substances fall into the category of “diamagnetic”, meaning that they appear to be un attracted to ordinary “kitchen” magnets, are are actually slightly repelled by strong magnetic fields generated in the laboratory.

A significant number of metals and compounds are attracted to strong magnetic fields in the lab although they are not attracted to weak magnets of the “refrigerator” variety: they are “paramagnetic.”

On the other hand, the compounds or metals exhibiting “ferromagnetism” are used to make up ordinary household magnets and are attracted to weak and strong magnetic fields.

Examples of this type are the salt magnetite, Fe 3 O 4 , and “Alnico” an alloy of Al, Ni and Co .

The characteristic which separates compounds and elements into these categories turns out to be one that is predicted by electronic configurations: the presence or absence ion or the atom...

of unpaired electrons in the

Diamagnetism: no unpaired electrons in atom or either ion of compound; Paramagnetism: one or more unpaired electron in atom or either ion of compound; Ferromagnetism: many unpaired electrons in atom or either ion of compound.

We find unpaired electrons in the metallic elements when subshells are unfinished: All metals are predictably paramagnetic, except for those in columns 2A (s 2 ) and 2B ( d 10 s 2 ) Nonmetallic elements, except oxygen , form polyatomic molecules with no unpaired e’s, all diamagnetic .

Most compounds have no unpaired electrons and are diamagnetic as well; they have been lost, gained or shared in the bonding process. The notable exceptions are compounds containing cations of the d block metals.

Transition elements have many unpaired electrons in both atoms and ions, and offer best structures for ferromagnetism : Fe 2 O 3 : Fe [Ar 18 ] 3d 6 4s 2 - 3 e [Ar 18 ] 3d 5 Fe 3+ d subshell

Alnico: Al , 2s 2 2 p 1 ; p's shown Co , Z=27, 4s 2 3d 6 Ni , Z=28, 4s 2 3d 7

Group Work 9.5

Do orbital box diagram for last subshell to be filled for the following elements, then decide which are:

• •

“diamagnetic” ( not attracted to a magnet, no unpaired e’s) “paramagnetic” (unpaired e’s, attracted) Cs, Ca, Cu, C, Kr, Co