Transcript Chapter 12 The Periodic Table

The how and why
History
 Russian
scientist Dmitri Mendeleev
taught chemistry in terms of
properties.
 Mid 1800 - molar masses of elements
were known.
 Wrote down the elements in order of
increasing mass (this is wrong!!!!)
 Found a pattern of repeating
properties.
Mendeleev’s Table
 Grouped
elements in columns by similar
properties in order of increasing atomic
mass.
 Found some inconsistencies - felt that
the properties were more important than
the mass, so switched order.
 Found some gaps.
 Must be undiscovered elements.
 Predicted their properties before they
were found.
The modern table
 Elements
are still grouped by
properties.
 Similar properties are in the same
column.
 Order is in increasing atomic number.
 Added a column of elements Mendeleev
didn’t know about.
 The noble gases weren’t found because
they didn’t react with anything.
 Horizontal
rows are called Periods
 There are 7 periods
 Vertical
columns are called
Groups.
 Elements are placed in columns
by similar properties.
 Also called families
 The
elements in the A groups are 8A
called the representative elements 0
1A
2A
3A 4A 5A 6A 7A
The group B are called the
transition elements
 These
are called the inner
transition elements and they
belong here
Periodicity Explained
 Valence
electron cloud
 Outside orbitals
 The orbitals fill up in a regular
pattern
 The outside orbital electron
configuration repeats
 The properties of atoms repeat when
placed in order of Atomic Number
H
Li
1
3
Na
11
K
19
Rb
37
Cs
55
Fr
87
1s1
1s22s1
1s22s22p63s1
1s22s22p63s23p64s1
1s22s22p63s23p64s23d104p65s1
1s22s22p63s23p64s23d104p65s24d10
5p66s1
1s22s22p63s23p64s23d104p65s24d105p
66s24f145d106p67s1
1s2 He 2
Ne
2
2
6
1s 2s 2p
10
1s22s22p63s23p6 Ar18
1s22s22p63s23p64s23d104p6 Kr
36
1s22s22p63s23p64s23d104p65s24d105p6 Xe
54
1s22s22p63s23p64s23d104p65s24d10 Rn
5p66s24f145d106p6 86
s- block
s1
s2
metals all end in s1
 Alkaline earth metals all end in s2
 really have to include He but it fits
better later.
 He has the properties of the noble
gases.
 Alkali
Transition Metals -d block
d1
d2
d3
d5
d5
d6
d7
d8
d9 d10
The p-block
p1 p2
p3
p4
p5
p6
f - block
 inner
transition elements
f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14
Atomic Size
 First
problem where do you start
measuring.
 The electron cloud doesn’t have a
definite edge.
 They get around this by measuring
more than 1 atom at a time.
Atomic Size
}
Radius
Atomic
Radius = half the distance between
two nuclei of a diatomic molecule.
Trends in Atomic Size
Influenced
by two factors.
Energy Level
Higher energy level - electrons
are further away from nucleus.
Charge on nucleus – Zeff
More charge (# of protons) pulls
electrons in closer.
Group trends
 As
we go down a
group
 Each atom has
another energy
level
 So the atoms get
bigger.
H
Li
Na
K
Rb
Periodic Trends
 As
you go across a period the radius
gets smaller.
 Same energy level.
 More nuclear charge = more protons.
 Outermost electrons are closer.
Na
Mg
Al
Si
P
S Cl Ar
Rb
Atomic Radius (nm)
K
Na
Li
Kr
Ar
Ne
H
10
Atomic Number
Ionization Energy
 The
amount of energy required to
completely remove an electron from
a gaseous atom.
 Removing one electron makes a
+1 ion.
 The energy required is called the first
ionization energy.
Ionization Energy
 The
second ionization energy is the
energy required to remove the
second electron.
 Always greater than first IE.
 The third IE is the energy required to
remove a third electron.
 IE1<IE2<IE3
Symbol First
H
He
Li
Be
B
C
N
O
F
Ne
1312
2731
520
900
800
1086
1402
1314
1681
2080
Second
Third
5247
7297
1757
2430
2352
2857
3391
3375
3963
11810
14840
3569
4619
4577
5301
6045
6276
Symbol First
H
He
Li
Be
B
C
N
O
F
Ne
1312
2731
520
900
800
1086
1402
1314
1681
2080
Second
Third
5247
7297
1757
2430
2352
2857
3391
3375
3963
11810
14840
3569
4619
4577
5301
6045
6276
What determines IE

nuclear charge =  IE
  distance from nucleus =  IE
 Filled and half filled orbitals have
lower energy, so achieving them is
easier, lower IE.
 Shielding
Periodic trends: Ionization Energy (IE)
e-
e-
All the atoms in the same period have the
same energy level.
 Same energy level = same shielding, but each
atom gains a proton, therefore….
 Increasing nuclear charge …helps pull e- in
tighter, therefore it is harder to remove.
 So IE generally increases from left to right.
 Exceptions at full and 1/2 fill orbitals.

e-
Periodic trends: Ionization
Energy (IE)
 As
e-
you go down a group first IE
decreases because….
 The electron is further away…and
 There are energy levels between
the nucleus and the e- thus….
 Shielding the e- from + nucleus.
Shielding
 The
electron on the
outside energy level
has to look through
all the other energy
levels to see the
nucleus
Shielding
 The
electron on the
outside energy level
has to look through
all the other energy
levels to see the
nucleus.
 A second electron
has the same
shielding.
Effective Nuclear Charge, Zeff
 Atom
in
 Li
 Be
B
C
N
O
F
Zeff Experienced by Electrons
Valence Orbitals
+1.28
------Increase in
+2.58
Z* across a
+3.22
period
+3.85
+4.49
+5.13
First Ionization energy
He
 He
H
has a greater IE
than H.
 same shielding
 greater nuclear
charge
Atomic number
First Ionization energy
He
Li has lower IE than
H
 more shielding
 further away
 outweighs greater
nuclear charge

H
Li
Atomic number
First Ionization energy
He
Be has higher IE
than Li
 same shielding
 greater nuclear
charge

H
Be
Li
Atomic number
First Ionization energy
He
B has lower IE than
Be
 same shielding
 greater nuclear
charge
 By removing an
electron we make s
orbital half filled

H
Be
B
Li
Atomic number
First Ionization energy
He
H
Be
C
B
Li
Atomic number
First Ionization energy
He
N
H
C
Be
B
Li
Atomic number
First Ionization energy
He
 Breaks
N
H
C O
Be
the
pattern because
removing an
electron gets to
1/2 filled p orbital
B
Li
Atomic number
First Ionization energy
He
N F
H
C O
Be
B
Li
Atomic number
Ne
First Ionization energy
He
 Ne
N F
H
C O
Be
has a lower
IE than He
 Both are full,
 Ne has more
shielding
 Greater distance
B
Li
Atomic number
Ne
First Ionization energy
He

N F
Na has a lower
IE than Li
Both are s1
 Na has more
shielding
 Greater distance

H
C O
Be
B
Li
Na
Atomic number
Atomic number
First Ionization energy
Electron Affinity
 The
energy change associated with
adding an electron to a gaseous
atom.
 Easiest to add to group 7A.
 Gets them to full energy level.
 Increase from left to right atoms
become smaller, with greater nuclear
charge.
 Decrease as we go down a group.
Periodic trends: Electron Affinity(EA)
e-
e-
• From left to right atoms become smaller,
with greater nuclear charge.
e - are attracted by the + charged nucleus.
•therefore EA will increase from left to right
Periodic trends: Electron Affinity(EA)
e-
• Down a group atoms become larger, and
have greater nuclear charge.
• e - are attracted by the increased +charge
but shielding also increases …which has a
greater influence!!
• therefore EA will decrease as you
go down a group or family
e-
Periodic trends: Electron Affinity(EA)
• The energy change associated with adding an
electron to a gaseous atom.
• e - are attracted by a + charge.
– Therefore more protons more attraction
– But remember “shielding”
• Easiest to add to group 7A.
• Gets them to full energy level.
• Increase from left to right atoms become smaller,
with greater nuclear charge.
• Decrease as we go down a group.
Ionization Energy & Electron Affinity
Effective Nuclear Charge
INCREASE
Atomic Radius & Shielding
INCREASES
Size of Isoelectronic ions
 Iso
- same
 Iso electronic ions have the same #
of electrons

Al+3, Mg+2, Na+1, Ne , F-1, O-2 and N-3
 all
have 10 electrons
 all have the configuration 1s12s22p6
Size of Isoelectronic ions
 Positvie
ions have more protons so
they are smaller.
Al+3
Na+1
Mg+2
Ne
F-1
O-2
N-3
Metal Oxides are BASIC
• Na2O(s) + H2O(l)  2 NaOH(aq)
• CaO(s) + H2O(l)  Ca(OH)2 (aq)
• MgO(s) + 2HCl(aq)  MgCl2(aq) + H2O (l)
• NiO(s) + H2SO4 (aq)  NiSO4 (aq) + H2O (l)
Non-Metal Oxides are ACIDIC
• CO2(g) + H2O(l)  H2CO3(aq)
• P4O10(s) + 6 H2O(l)  4 H3PO4 (aq)
• CO2(g) + 2 NaOH (aq)  Na2CO3(aq) + H2O (l)
• SO3(g) + 2 KOH (aq)  K2SO4 (aq) + H2O (l)