Download Patino-CHM2045C-Chapter8

Chapter 8
Atomic Electron Configurations
and Chemical Periodicity
Chapter goals
• Understanding the role magnetism
plays in determining and revealing
atomic structure.
• Understand effective nuclear charge
and its role in determining atomic
properties.
• Write the electron configuration of
neutral atoms and monatomic ions.
• Understand the fundamental physical
properties of the elements and their
periodic trends.
Electron Spin
and the Fourth Quantum Number
• The fourth quantum number is the spin quantum
number which has the symbol ms.
• The spin quantum number only has two possible
values.
ms = +1/2 or −1/2
ms = ± 1/2
• This quantum number tells us the spin and
orientation of the magnetic field of the electrons.
• Wolfgang Pauli discovered the Exclusion
Principle in 1925.
No two electrons in an atom can have the same
set of 4 quantum numbers, n, l, ml, and ms
Electron Spin
• Spin quantum number effects:
– Every orbital can hold up to two
electrons.
• Consequence of the Pauli Exclusion
Principle.
– The two electrons are designated as
having
– one spin up  ms = +1/2
– and one spin down  ms = −1/2
• Spin describes the direction of the
electron’s magnetic field.
Paramagnetism and Diamagnetism
• Unpaired electrons have their spins
aligned   or   (in diff. orbitals)
– This increases the magnetic field of the
atom.
Total spin  0, because they add up.
• Atoms with unpaired electrons are called
paramagnetic .
– Paramagnetic atoms are attracted to a
magnet.
Paramagnetism and Diamagnetism
• Paired electrons have their spins
unaligned . (in the same orbital)
– Paired electrons have no net magnetic
field.
Total spin = 0, because of cancellation,
½ −½=0
• Atoms with no unpaired electrons are
called diamagnetic.
– Diamagnetic atoms are not attracted to a
magnet.
Atomic Orbitals, Spin, and # of Electrons
• Because two electrons in the same orbital
must be paired (due to Pauli’s Exclusion
Principle), it is possible to calculate the
number of orbitals and the number of
electrons in each n shell.
• The number of orbitals per n level is given
by n2 (see table at end of chapter 7.)
• The maximum number of electrons per n
level is 2n2 (two electrons per orbital.)
– The value is 2n2 because of the two
paired electrons per orbital.
#orbitals
ml
n shell l subshell
s
0
1 K 0
s
0
2 L 0
–1,0,1
1 p
0
3 M 0 s
–1,0,1
1 p
2 d
-2,-1,0,1,2
0
4 N 0 s
–1,0,1
1 p
2 d
-2,-1,0,1,2
3 f -3,-2,-1,0,1,2,3
1
1
3
1
3
5
1
3
5
7
Max
n2
#e–
1
2
2 8
4
6
2
6 18
9
10
2
6
16 10 32
14
Atomic Subshell Energies and
Electron Assignments
• The principle that describes how the
periodic chart is a function of electronic
configurations is the Aufbau Principle.
• The electron that distinguishes an element
from the previous element enters the
lowest energy atomic orbital available.
Penetrating and Shielding
• the radial distribution function
shows that the 2s orbital
penetrates more deeply into the
1s orbital than does the 2p
• the weaker penetration of the 2p
sublevel means that electrons in
the 2p sublevel experience more
repulsive force, they are more
shielded from the attractive force
of the nucleus
• the deeper penetration of the 2s
electrons means electrons in the
2s sublevel experience a greater
attractive force to the nucleus and
are not shielded as effectively
• the result is that the electrons in
the 2s sublevel are lower in (more
negative) energy than the
electrons in the 2p
Atomic Subshell Energies and Electron
Assignments
The Aufbau Principle
describes the electron
filling order in atoms. This
is product of the effective
nuclear charge, Z*, Zeff
For the same n, Z* is higher
for s orbital: s > p > d > f
Then, e− in s is the most
attracted by nucleus and has
the lowest energy
Atomic Subshell Energies and Electron
Assignments
One mnemonic
to remember the
correct filling
order for
electrons
in atoms is the
increasing
(n + ) value
Atomic Subshell Energies and Electron
Assignments
or we can use this periodic chart
Atomic Electron Configurations
• Now we will use the Aufbau Principle to
determine the electronic configurations of
the elements on the periodic chart.
• 1st row elements
1s

1
H
2
He 
Configurat ion
1
1s
1s
2
Atomic Electron Configurations
Hund’s rule tells us that the electrons will fill the p and d orbitals
by placing electrons in each orbital singly and with same spin
until half-filled. That is the rule of maximum spin. Then the
electrons will pair to finish the p orbitals.
Electrons in
orbitals of or same
kind, such as p or d
orbitals, in the same
shell (n), have the
same energy; the
are said to be
degenerate.
Atomic Electron Configurations
3rd row elements…
3s
3p
Configurat ion
11 Na Ne 
12
Mg Ne 
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
Ne
Ne
Ne
Ne
Ne
Ne



 

  

  

  

  
Ne 3s1
Ne 3s2
Ne 3s2 3p1
Ne 3s2 3p2
Ne 3s2 3p3
Ne 3s2 3p4
Ne 3s2 3p5
Ne 3s2 3p6
Atomic Electron Configurations
4th row elements…
3d
19 K Ar 
4s

4p
Configurat ion
Ar  4s1
Atomic Electron Configurations
4th row elements…
3d
4s
19 K Ar 

20

Ca Ar 
4p
Configurat ion
Ar  4s1
Ar  4s2
Atomic Electron Configurations
4th row elements…
The five d orbitals are degenerate
3d
4s
19 K Ar 

20
Ca Ar 

Sc Ar  

21
4p
Configurat ion
Ar  4s1
Ar  4s2
Ar  4s2 3d1
Atomic Electron Configurations
4th row elements…
3d
4s
19 K Ar 

20
Ca Ar 

Sc Ar  

Ti Ar   

21
22
4p
Configurat ion
Ar  4s1
Ar  4s2
Ar  4s2 3d1
Ar  4s2 3d 2
Atomic Electron Configurations
4th row elements…
The five d orbitals are degenerate
3d
4s
19 K Ar 

20
Ca Ar 

Sc Ar  

22
Ti Ar   

23
V Ar    

21
4p
Configurat ion
Ar  4s1
Ar  4s2
2
1
Ar  4s 3d
Ar  4s2 3d 2
Ar  4s2 3d 3
Atomic Electron Configurations
4th row elements…
3d
4s
19 K Ar 

20
Ca Ar 

Sc Ar  

22
Ti Ar   

23
V Ar    

21
4p
Configurat ion
Ar  4s1
Ar  4s2
2
1
Ar  4s 3d
Ar  4s2 3d 2
Ar  4s2 3d 3
Atomic Electron Configurations
4th row elements… The [Ar] 4s1 3d5 configuration of
Cr is more stable than [Ar] 4s2 3d4 (expected)
3d
4s
19 K Ar 

20

21
Ca Ar 
Sc Ar  
Ti Ar   

22

23

V Ar    
24
Cr Ar      

4p
Configurat ion
Ar  4s1
Ar  4s 2
Ar  4s 2 3d1
Ar  4s 2 3d 2
Ar  4s 2 3d 3
Ar  4s1 3d 5
There is an extra measure of stability associated
with half - filled and completely filled orbitals.
Atomic Electron Configurations
4th row elements… The [Ar] 4s1 3d10 full d configuration
of Cu is more stable than [Ar] 4s2 3d9 (expected)
3d
25 Mn Ar      
26
27
28
29
4s
4p

Fe Ar      

Co Ar      

Ni Ar      

Cu Ar       
Another exception like Cr and
for essentiall y the same reason.
Configurat ion
Ar  4s2 3d5
Ar  4s2 3d 6
2
7
Ar  4s 3d
Ar  4s2 3d8
Ar  4s1 3d10
Atomic Electron Configurations
4th row elements…
3d
25 Mn Ar      
26
27
28
29
30
4s

Fe Ar      

Co Ar      

Ni Ar      

Cu Ar       
Zn Ar       
4p
Configurat ion
Ar  4s2 3d5
Ar  4s2 3d 6
Ar  4s2 3d 7
Ar  4s2 3d8
Ar  4s1 3d10
Ar  4s2 3d10
Atomic Electron Configurations
4th row elements… (remember Hund’s rule):
  __ is better (lower energy) than  __ __
4p
4p
3d
4s
31 Ga Ar        
4p
Configurat ion
Ar  4s2 3d10 4p1
2
10
2




Ge
Ar








Ar
4s
3d
4p
32
2
10
3




As
Ar









Ar
4s
3d
4p
33
2
10
4
Ar  4s 3d 4p
34 Se Ar          
2
10
5
35 Br Ar           Ar  4s 3d 4p
2
10
6




Kr
Ar









Ar
4s
3d
4p
36
Atomic Electron Configurations
Lanthanides (4f)
2
Ba
[Xe]
6s
56
1 6s2
La
[Xe]
5d
57
1 5d1 6s2
Ce
[Xe]
4f
58
59Pr
[Xe] 4f3 6s2
14 6s2
Yb
[Xe]
4f
70
71Lu
[Xe] 4f14 5d1 6s2
Praseodymium
Ytterbium
Lutetium
Periodic Table
s, p, d, and f-block in the Periodic Table
P 1A
1
1
H 2A
2 3 4
3
4
5
6
7
Li Be
11 12
(P–1)d
Na Mg 3B 4B 5B 6B 7B 8B 8B 8B
19 20 21 22 23 24 25 26 27 28
K Ca Sc Ti V Cr Mn Fe Co Ni
37 38 39 40 41 42 43 44 45 46
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd
55 56 57 72 73 74 75 76 77 78
Cs Ba La Hf Ta W Re Os Ir Pt
87 88 89 104 105 106 107 108 109
Fr Ra Ac Rf Db Sg Bh Hs Mt
(P)s
(P–2)f
58 59
Ce Pr
90 91
Th Pa
1B 2B
29 30
Cu Zn
47 48
Ag Cd
79 80
Au Hg
60 61 62 63 64 65 66
Nd Pm Sm Eu Gd Tb Dy
92 93 94 95 96 97 98
U Np Pu Am Cm Bk Cf
3A
5
B
13
Al
31
Ga
49
In
81
Tl
6A
8
O
16
S
34
Ge As Se
50 51 52
Sn Sb Te
82 83 84
Pb Bi Po
4A
6
C
14
Si
32
5A
7
N
15
P
33
8A
2
He
10
Ne
18
Ar
36
Kr
54
7A
9
F
17
Cl
35
Br
53
I Xe
85 86
At Rn
(P)p
67 68 69 70 71
Ho Er Tm Yb Lu
99 100 101 102 103
Es Fm Md No Lr
Valence Electrons
electrons in shell with highest n, i.e., the outermost
electrons, those beyond the core electrons
1s2 2s2 2p6 3s1
1s2 2s2 2p6 3s2 3p2
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
1s2 2s2
1s2 2s2 2p6 3s2 3p6 4s2 3d7
They determine the chemical properties of an
element. For the representative elements, they
are the ns and np electrons; for transition
elements they are the ns and (n−1)d electrons.
P 1A
1
1
H
2 3
Li
11
3
4
5
6
7
Na
19
K
37
Rb
55
Cs
87
Fr
1s1
2s1
3s1
4s1
5s1
6s1
7s1
# of valence electrons = 1
2A
4
Be
12
Mg
20
Ca
38
Sr
56
Ba
88
Ra
2s2
3s2
4s2
5s2
6s2
7s2
# of valence electrons = 2
3A
5
B
13
Al
31
Ga
49
In
81
Tl
2s2 2p1
3s2 3p1
4s2
4p1
5s2 5p1
6s2 6p1
# of valence electrons = 3
7A
9
F
17
Cl
35
Br
53
I
85
At
# of valence electrons = 7
2s2 2p5
3s2 3p5
4s2 4p5
5s2 5p5
6s2 6p5
For the representative elements, the
# of valence electrons = # of group
The element X has the valence shell
electron configuration, ns2 np4.
X belongs to what group?
1A
8A
chalcogens
2
1
H
3
Li
11
2A
4
Be
12
Na Mg 3B 4B 5B 6B 7B 8B 8B 8B
19 20 21 22 23 24 25 26 27 28
K Ca Sc Ti V Cr Mn Fe Co Ni
37 38 39 40 41 42 43 44 45 46
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd
55 56 57 72 73 74 75 76 77 78
Cs Ba La Hf Ta W Re Os Ir Pt
87 88 89 104 105 106 107 108 109
Fr Ra Ac Unq Unp Unh Uns Uno Une
3A
5
B
13
1B 2B Al
29 30 31
Cu Zn Ga
47 48 49
Ag Cd In
79 80 81
Au Hg Tl
6A
8
O
16
S
34
Ge As Se
50 51 52
Sn Sb Te
82 83 84
Pb Bi Po
4A
6
C
14
Si
32
5A
7
N
15
P
33
7A
9
F
17
Cl
35
Br
53
I
85
At
He
10
Ne
18
Ar
36
Kr
54
Xe
86
Rn
Energy (Orbital) Diagram
4p
3d
4s
3s
3p
E
2p
2s
Be
1s
1s2 2s2
Orbital Box Diagrams
Be
1s
2s
2p
3s
Orbital Box Diagrams
N
1s
2s
2p
Formation of Cations
electrons lost from subshell with highest
n and l first (from valence electrons)
examples
K
K+
1s2 2s2 2p6 3s2 3p6 4s1
[Ar] 4s1
1s2 2s2 2p6 3s2 3p6
[Ar]
Ca
Ca2+
Al
Al3+
In
In3+
1s2 2s2 2p6 3s2 3p6 4s2
[Ar] 4s2
1s2 2s2 2p6 3s2 3p6
[Ar]
1s2 2s2 2p6 3s2 3p1
[Ne]
[Kr] 4d10 5s2 5p1
[Kr] 4d10
Transition Metal Cations
In the process of ionization transition metals
the ns electrons are lost before the (n-1)d
Fe: [Ar] 3d6 4s2  Fe2+: [Ar] 3d6
Fe2+: [Ar] 3d6  Fe3+: [Ar] 3d5
Cu: [Ar] 3d10 4s1  Cu+: [Ar] 3d10
Cu+: [Ar] 3d10
 Cu2+: [Ar] 3d9
Fe, Fe2+, Fe3+, Cu, and Cu2+ are paramagnetic
Two problems of ions, charge, and electron configuration
An anion has a 3− charge and electron configuration
1s2 2s2 2p6 3s2 3p6. What is the symbol of the ion?
The neutral atom has gained 3e- to form the ion, then
the neutral atom had 15 e-. In the neutral atom the # e= # p+ = Atomic number, that is 15. The element is,
then, phosphorus (phosphorus). Symbol of ion is P3−.
A cation has a 2+ charge and its electron
configuration is [Ar] 3d7. What is the symbol of the ion?
Here, the neutral atom has lost 2e-. It is a transition
metal, due to the 3d electrons. Remember they firstly
lose e-s in 4s orbital. Symbol of ion is Co2+.
Neutral atom has 18 + 7 + 2 = 27 e- = 27 p+ = atomic #
[Ar] 3d7 lost
Atomic Properties and Periodic
Trends
Periodic Properties of
the Elements
1. Atomic Radii
2. Ionization Energy
3. Electron Affinity
4. Ionic Radii
Atomic Properties and Periodic Trends
• Establish a classification scheme of the elements
based on their electron configurations.
• Noble Gases
– All of them have completely filled electron
shells. They are not very reactive.
• Since they have similar electronic structures,
their chemical reactions are similar.
– He
1s2
– Ne
[He] 2s2 2p6
– Ar
[Ne] 3s2 3p6
– Kr
[Ar] 4s2 4p6
– Xe
[Kr] 5s2 5p6
– Rn
[Xe] 6s2 6p6
Atomic Properties and Periodic Trends
Representative Elements are
the elements in A groups
on periodic chart.
These elements will have
their “last” electron in an
outer s or p orbital.
These elements have fairly
regular variations in their
properties.
Metallic character, for expl,
increases from right to left
and top to bottom.
Atomic Properties and Periodic Trends
•
•
•
•
d-Transition Elements
Elements on periodic
chart in B groups.
Sometimes called
transition metals.
Each metal has d
electrons.
nsx (n-1)dy configurations
These elements make the
transition from metals to
nonmetals.
Exhibit smaller variations
from row-to-row than the
representative elements.
Atomic Properties and Periodic Trends
•
•
•
•
f - transition metals
Sometimes called inner
transition metals.
Electrons are being
added to f orbitals.
Electrons are being
added two shells below
the valence shell!
Consequently, very
slight variations of
properties from one
element to another.
Atomic Properties and Periodic Trends
Outermost electrons (valence electrons)
have the greatest Influence on the chemical
properties of elements.
Atomic Properties and Periodic Trends
Atomic radii describe the
relative sizes of atoms.
Atomic radii increase within a
column going from the top to
the bottom of the periodic table.
The outermost electrons are
assigned to orbitals with
increasingly higher values of n.
The underlying electrons
require some space, so the
electrons of the outer shells
must be further from the
nucleus.
Atomic Properties and Periodic Trends
Atomic radii decrease
within a row going from
Left to right on the
periodic table.
This last fact seems
contrary to intuition.
How does nature make
the elements smaller
even though the electron
number is increasing?
Atomic Radii
• The reason the atomic radii decrease across a
period is due to shielding or screening effect.
– Effective nuclear charge, Zeff, experienced by
an electron is less than the actual nuclear
charge, Z.
– The inner electrons block the nuclear charge’s
effect on the outer electrons.
• Moving across a period, each element has an
increased nuclear charge and the electrons are
going into the same shell (2s and 2p or 3s and 3p,
etc.).
– Consequently, the outer electrons feel a
stronger effective nuclear charge.
– For Li, Zeff ~ +1
– For Be, Zeff ~ +2
— For B, Zeff ~ +3
Atomic Radii
• Example: Arrange these elements based on
their increasing atomic radii.
– Se, S, O, Te
O < S < Se < Te
In the same group atomic size increases
as n (and Z) increases
─ Br, Ca, Ge, F
F < Br < Ge < Ca
same group
same period
Ionization Energy
• First ionization energy (IE1)
– The minimum amount of energy required to
remove the most loosely bound electron from
an isolated gaseous atom to form a 1+ ion.
• Symbolically:
Atom(g) + energy  ion+(g) + eEndothermic
Mg(g) + 738kJ/mol  Mg+ + e-
IE1= 738kJ/mol
Ionization Energy
• Second ionization energy (IE2)
– The amount of energy required to remove the
second electron from a gaseous 1+ ion.
• Symbolically:
– ion+ + energy  ion2+ + e-
Mg+ + 1451 kJ/mol  Mg2+ + eIE2= 1451 kJ/mol
Atoms can have 3rd (IE3), 4th (IE4), etc. ionization
energies. The values are consecutively getting
larger.
Ionization Energy
Periodic trends for Ionization
Energy:
1) IE2 > IE1
It always takes more energy
to remove a second electron
from an ion than from a
neutral atom.
2) IE1 generally increases
moving from IA elements to
VIIIA elements.
Important exceptions at Be &
B, N & O, etc. due to s and p
and half-filled subshells.
3) IE1 generally decreases
moving down a family.
IE1 for Li > IE1 for Na, etc
First Ionization Energies
of Some Elements
2500
2000
Ionization
Energy
(kJ/mol)
1500
1000
500
0
1 2
3 4
5 6 7
8 9 10 11 12 13 14 15 16 17 18 19 20
Atomic Number
Ionization Energy
• Example: Arrange these elements based on
their (increasing) first ionization energies.
– Sr, Be, Ca, Mg
Sr < Ca < Mg < Be
– Al, Cl, Na, P
Na < Al < P < Cl
– O, Ga, Sr, Se
Sr < Ga < Se < O
Ionization Energy
• The reason Na forms Na+ and not Na2+ is
that the energy difference between IE1 and
IE2 is so large.
– Requires more than 9 times more energy
to remove the second electron than the
first one.
• The same trend is persistent throughout
the series.
– Thus Mg forms Mg2+ and not Mg3+.
– Al forms Al3+ and not Al4+.
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
1312
2371
520
900
800
1086
1402
1314
1681
2080
496
738
577
786
1012
1000
1255
1520
419
Ionization Energies (kJ/mole)
5247
7297
1757
2430
2352
2857
3391
3375
3963
4565
1450
1816
1577
1896
2260
2297
2665
3069
11810
14840
3659
4619
4577
5301
6045
6276
6912
7732
2744
3229
2910
3380
3850
3947
4600
21000
25020
6221
7473
7468
8418
9376
9540
10550
11580
4356
4954
4565
5146
5770
5879
32810
37800
9443
10980
11020
12190
13360
13620
15030
16080
6272
6996
6544
7240
7971
47300
53250
13320
15160
15230
16610
18000
18370
19790
21270
8490
9330
8810
9619
64340
71300 84850
17860 92000
20110
21700
23290
23780
25410
28080
11020
11970
11380
25490
25660
27460
29250
29840
31720
33600
13840
14950
Electron Affinity (EA)
• Electron affinity is the amount of energy
absorbed or emitted when an electron is
added to an isolated gaseous atom to form
an ion with a 1- charge.
• Sign conventions for electron affinity.
– If EA > 0 energy is absorbed (difficult)
– If EA < 0 energy is released (easy)
• Electron affinity is a measure of an atom’s
ability to form negative ions.
• Symbolically:
atom(g) + e-  ion-(g)
EA (kJ/mol)
Electron Affinity
• General periodic trend for electron affinity is
– the values become more negative from left
to right across a period on the periodic
chart (affinity for electron increases).
– the values become more negative from
bottom to top at a group on the periodic
chart.
−Noble gases have EA > 0 (full electron
confg)
• An element with a high ionization energy
generally has a high affinity for an electron,
i.e., EA is largely negative. That is the case for
halogens (F, Cl, Br, I), O, and S.
Electron Affinity
F (Z= 9) and Cl (Z = 17) have the most negative EA
Noble gases, He (2), Ne (10), and Ar (18), EA > 0; also Be, Mg, N
They are all first Electron Affinity.
A(g)- + e-  A2-(g) EA2(kJ/mol) is the 2nd
Electron Affinity
Two examples of electron affinity values:
Mg(g) + e- + 231 kJ/mol  Mg-(g) EA = 231kJ/mol
Br(g) + e-  Br-(g) + 323 kJ/mol
EA = -323 kJ/mol
Br has a larger affinity for e− than Mg. The greater the
affinity an atom has for an e− , the more negative EA is,
the smaller it is.
Ionic Radii
Cations (positive ions) are always smaller than
their respective neutral atoms. When one or more
electrons are removed, the attractive force of the
protons is now exerted on less electrons.
Element
Atomic
Radius (Å)
Ion
Ionic
Radius (Å)
Na
11 p+, 11e-
Mg
12p+, 12 e-
Al
13 p+, 13e-
1.86
1.60
1.43
Na+
Mg2+
11 p+, 10e- 12 p+, 10 e1.16
0.85
Al3+
13 p+, 10e0.68
Ionic Radii
Anions (negative ions) are always larger than
their neutral atoms.
F 1s2 2s2 2p5 + e−  F− 1s2 2s2 2p6 same Z
nine electrons
Element
Atomic
Radius(Å)
Ion
Ionic
Radius(Å)
ten electrons
N
7 p+, 7e0.75
N37 p+, 10e1.71
The three
O
F
0.73
0.72
O2F−
8 p+, 10e- 9 p+, 10e1.26
1.19
ions are isoelectronic
Ionic Radii
Cation (positive ions) radii decrease from left to
right across a period.
Increasing nuclear charge attracts the
electrons and decreases the radius.
Rb+ and Sr2+ are isoelectronic, same # of e-s
Ion
Ionic
Radii(Å)
Rb+
Sr2+
In3+
Z = 37 p+
Z = 38 p+
Z = 49 p+
1.66
1.32
0.94
Ionic Radii
Anion (negative ions) radii decrease from left
to right across a period.
Increasing electron numbers in highly
charged ions cause the electrons to repel
and increase the ionic radius.
For these isoelectronic anions…
10 e− and 7 p+
8 p+
9 p+
Ion
N3-
O2-
F−
Ionic
Radii(Å)
1.71
1.26
1.19
Ionic Radii
Example: Arrange these ions in order of
decreasing radius.
Ga3+, K+, Ca2+
K+ > Ca2+ > Ga3+
Cl−, Se2−, Br−, S2−
Se2− > Br−
isoelectronic
>
S2− > Cl−
isoelectronic, same # of electrons
Se2−(34 p+) > Br−(35 p+); they have 36 e− each.
S2−(16 p+) > Cl−(17 p+); they have 18 e− each.
Br− > S2− because Br− is in the 4th period, S2− is in the 3rd.
Ionic Radii of isoelectronic species
Isoelectronic species have the same number of
electrons. Here are some examples with the
number of (protons) and + or − charges
N3−(Z=7) > O2−(Z=8) > F−(Z=9) > Ne(Z=10) neutral >
Na+(Z=11) > Mg2+(Z=12) > Al3+ (Z=13) all have 10e−
The nuclear charge (+) increases from left to right,
so does attraction force to electrons: r decreases.
S2−(Z=16) > Cl− (Z=17) > Ar0 (Z=18) > K+ (Z=19) >
Ca2+ (Z=20) > Sc3+ (Z=21) all of them have 18e−