© 2014 Pearson Education, Inc.
Sherril Soman
Grand Valley State University
Lecture Presentation
Chapter 8
Periodic
Properties of the
Element
© 2014 Pearson Education, Inc.
Pg. 373-374
Copyright © 2011 Pearson Canada Inc.
Alkali Metals
Alkaline Earths
Transition Metals
Halogens
Noble Gases
Lanthanides and Actinides
Main Group
Main Group
The Periodic table (pg.346)
Using the Periodic Table to Write Electron Configurations
The electron configuration of Si ends with 3s
23p
2The electron
configuration of Rh
ends with 5s
24d
78.2
ns
1ns
2ns
2np
1ns
2np
2ns
2np
3ns
2np
4ns
2np
5ns
2np
6d
1d
5d
104f 5f
Ground State Electron Configurations of the Elements
元素基態的電子組態
Slide 6 of 35
Periodic Properties of the Elements
元素的週期性質Copyright © 2011 Pearson Canada Inc.
General Chemistry: Chapter 9 Slide 6 of 35
FIGURE 9-12 Atomic properties and the periodic table – a summary
L
L
Large: atomic radius(原子半徑)
Metallic (金屬)
Large: Ionization energy
游離能Electron affinity
電子親和力Nonmetal
lic 非金屬© 2014 Pearson Education, Inc.
Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.1 Electron Configurations
a. Mg
Magnesium has 12 electrons. Distribute 2 of these into the 1 s orbital, 2 into the 2s orbital, 6 into the 2p orbitals, and 2 into the 3s orbital.
Mg 1s2 2s2 2p6 3s2 or [Ne] 3s2
b. P
Phosphorus has 15 electrons. Distribute 2 of these into the 1s orbital, 2 into the 2s orbital, 6 into the 2p orbitals, 2 into the 3s orbital, and 3 into the 3p orbitals.
P 1s2 2s2 2p6 3s2 3p3 or [Ne] 3s2 3p3
c. Br
Bromine has 35 electrons. Distribute 2 of these into the 1s orbital, 2 into the 2s orbital, 6 into the 2p orbitals, 2 into the 3s orbital, 6 into the 3p orbitals, 2 into the 4s orbital, 10 into the 3d orbitals, and 5 into the 4p orbitals.
Br 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 or [Ar] 4s2 3d10 4p5
d. Al
Aluminum has 13 electrons. Distribute 2 of these into the 1s orbital, 2 into the 2s orbital, 6 into the 2p orbitals, 2 into the 3s orbital, and 1 into the 3p orbital.
Al 1s2 2s2 2p6 3s2 3p1 or [Ne] 3s2 3p1 Solution
Write electron configurations for each element.
a. Mg b. P c. Br d. Al
Mg: 1s
22s
22p
63s
2[Ne]3s
2Br: 1s
22s
22p
63s
23p
64s
23d
104p
5[Ar] 4s
23d
104p
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Solution
Since sulfur’s atomic number is 16, it has 16 electrons and the electron configuration 1s2 2s2 2p6 3s2 3p4. Draw a box for each orbital, putting the lowest energy orbital (1s) on the far left and proceeding to orbitals of higher energy to the right.
Distribute the 16 electrons into the boxes representing the orbitals, allowing a maximum of 2 electrons per orbital and remembering Hund’s rule. You can see from the diagram that sulfur has two unpaired electrons.
Two unpaired electrons
For Practice 8.2
Write the orbital diagram for Ar and determine the number of unpaired electrons.
Write the orbital diagram for sulfur and determine the number of unpaired electrons.
Example 8.2 Writing Orbital Diagrams
S: 1s
22s
22p
63s
23p
4[Ne]3s
23p
4未成對電子=2
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.3 Valence Electrons and Core Electrons
Write the electron configuration for Ge. Identify the valence electrons and the core electrons.
Solution
Write the electron configuration for Ge by determining the total number of electrons from germanium’s atomic number (32) and then distributing them into the appropriate orbitals.
Ge 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2
Since germanium is a main-group element, its valence electrons are those in the outermost principal energy level.
For germanium, the n = 1, 2, and 3 principal levels are complete (or full) and the n = 4 principal level is outermost.
Consequently, the n = 4 electrons are valence electrons and the rest are core electrons.
Note: In this book, we write electron configurations with the orbitals in the order of filling. However, electron configurations are sometimes written in order of increasing principal quantum number. The electron configuration of germanium written in order of increasing principal quantum number is Ge 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p2.
For Practice 8.3
Write an electron configuration for phosphorus. Identify the valence electrons and core electrons.
Ge: 1s
22s
22p
63s
23p
64s
23d
104p
2[Ar] 4s
23d
104p
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.4 Writing Electron Configurations from the Periodic Table
Use the periodic table to write the electron configuration for selenium (Se).
Solution
The atomic number of Se is 34. The noble gas that precedes Se in the periodic table is argon, so the inner electron configuration is [Ar]. Obtain the outer electron configuration by tracing the elements between Ar and Se and assigning electrons to the appropriate orbitals. Begin with [Ar]. Because Se is in row 4, add two 4s electrons as you trace across the s block (n = row number). Next, add ten 3d electrons as you trace across the d block
(n = row number − 1). Lastly, add four 4p electrons as you trace across the p block to Se, which is in the fourth column of the p block (n = row number).
Se [Ar] 4s2 3d10 4p4
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.5 Atomic Size
On the basis of periodic trends, choose the larger atom in each pair (if possible). Explain your choices.
a. N or F b. C or Ge c. N or Al d. Al or Ge
Solution
a. N atoms are larger than F atoms because as you trace the path between N and F on the periodic table, you move to the right within the same period. As you move to the right across a period, the effective nuclear charge experienced by the outermost electrons increases, resulting in a smaller radius.
L
Large: atomic radius(原子半徑)
Metallic (金屬)
L
L
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.5 Atomic Size
Continued
b. Ge atoms are larger than C atoms because as you trace the path between C and Ge on the periodic table, you move down a column. Atomic size increases as you move down a column because the outermost electrons occupy orbitals with a higher principal quantum number that are therefore larger, resulting in a larger atom.
L
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.5 Atomic Size
Continued
c. Al atoms are larger than N atoms because as you trace the path between N and Al on the periodic table, you move down a column (atomic size increases) and then to the left across a period (atomic size increases). These effects add together for an overall increase.
L
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.5 Atomic Size
Continued
d. Based on periodic trends alone, you cannot tell which atom is larger, because as you trace the path between Al and Ge you go to the right across a period (atomic size decreases) and then down a column (atomic size
increases). These effects tend to counter each other, and it is not easy to tell which will predominate.
無法判斷
L
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.6 Electron Configurations and Magnetic Properties for Ions
a. Al3+
Begin by writing the electron configuration of the neutral atom. Since this ion has a 3+ charge, remove three electrons to write the electron configuration of the ion. Write the orbital diagram by drawing half-arrows to represent each electron in boxes representing the orbitals. Because there are no unpaired electrons, Al3+ is diamagnetic.
Al [Ne] 3s2 3p1
Al3+ [Ne] or [He] 2s2 2p6
Diamagnetic Solution
Write the electron configuration and orbital diagram for each ion and determine whether each is diamagnetic or paramagnetic.
a. Al3+ b. S2− c. Fe3+
Al: 1s
22s
22p
63s
23p
1[Ne] 3s
23p
1Al
3+: [Ne] or [He] 2s
22p
6diamagnetic(逆磁):電子成對 paramagnetic(順磁):電子未成對.
diamagnetic
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.6 Electron Configurations and Magnetic Properties for Ions
b. S2−
Begin by writing the electron configuration of the neutral atom. Since this ion has a 2− charge, add two electrons to write the electron configuration of the ion. Write the orbital diagram by drawing half-arrows to represent each electron in boxes representing the orbitals. Because there are no unpaired electrons, S2− is diamagnetic.
S [Ne] 3s2 3p4 S2− [Ne] 3s2 3p6
Diamagnetic Continued
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.6 Electron Configurations and Magnetic Properties for Ions
c. Fe3+
Begin by writing the electron configuration of the neutral atom. Since this ion has a 3+ charge, remove three electrons to write the electron configuration of the ion. Since it is a transition metal, remove the electrons from the 4s orbital before removing electrons from the 3d orbitals. Write the orbital diagram by drawing half- arrows to represent each electron in boxes representing the orbitals. There are unpaired electrons, so Fe3+ is paramagnetic.
Fe [Ar] 4s2 3d6 Fe3+ [Ar] 4s0 3d5
Paramagnetic Continued
For Practice 8.6
Write the electron configuration and orbital diagram for each ion and predict whether each will be paramagnetic or diamagnetic.
a. Co2+ b. N3− c. Ca2+
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.7 Ion Size
a. The S2− ion is larger than an S atom because anions are larger than the atoms from which they are formed.
b. A Ca atom is larger than Ca2+ because cations are smaller than the atoms from which they are formed.
c. A Br− ion is larger than a Kr atom because, although they are isoelectronic, Br− has one fewer proton than Kr, resulting in a lesser pull on the electrons and therefore a larger radius.
For Practice 8.7
Choose the larger atom or ion from each pair.
a. K or K+ b. F or F− c. Ca2+ or Cl−
For More Practice 8.7
Arrange the following in order of decreasing radius: Ca2+, Ar, Cl−. Solution
Choose the larger atom or ion from each pair.
a. S or S2− b. Ca or Ca2+ c. Br− or Kr
陰離子>中性原子>陽離子
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.8 Ionization Energy
Solution
On the basis of periodic trends, determine which element in each pair has the higher first ionization energy (if possible).
a. Al or S b. As or Sb c. N or Si d. O or Cl
a. Al or S
S has a higher ionization energy than Al because as you trace the path between Al and S on the periodic table, you move to the right within the same row. Ionization energy increases as you go to the right due to
increasing effective nuclear charge.
L
Ionization energy
游離能L
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.8 Ionization Energy
Continued
b. As or Sb
As has a higher ionization energy than Sb because as you trace the path between As and Sb on the periodic table, you move down a column. Ionization energy decreases as you go down a column as a result of the increasing size of orbitals with increasing n.
L
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.8 Ionization Energy
Continued
c. N or Si
N has a higher ionization energy than Si because as you trace the path between N and Si on the periodic table, you move down a column (ionization energy decreases) and then to the left across a row (ionization energy decreases). These effects sum together for an overall decrease.
L
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.8 Ionization Energy
Continued
d. O or Cl
Based on periodic trends alone, it is impossible to tell which has a higher ionization energy because, as you trace the path between O and Cl, you go to the right across a row (ionization energy increases) and then down a column (ionization energy decreases). These effects tend to counter each other, and it is not obvious which will dominate.
L
無法判斷
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.9 Metallic Character
Solution
On the basis of periodic trends, choose the more metallic element from each pair (if possible).
a. Sn or Te b. P or Sb c. Ge or In d. S or Br
a. Sn or Te
Sn is more metallic than Te because as you trace the path between Sn and Te on the periodic table, you move to the right within the same period. Metallic character decreases as we go to the right.
L
Metallic (金屬)
L
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Example 8.9 Metallic Character
Continued
b. P or Sb
Sb is more metallic than P because as you trace the path between P and Sb on the periodic table, you move down a column. Metallic character increases as we go down a column.
L
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.9 Metallic Character
Continued
c. Ge or In
In is more metallic than Ge because as you trace the path between Ge and In on the periodic table, you move down a column (metallic character increases) and then to the left across a period (metallic character
increases). These effects add together for an overall increase.
L
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.9 Metallic Character
Continued
d. S or Br
Based on periodic trends alone, we cannot tell which is more metallic because as you trace the path between S and Br, you go to the right across a period (metallic character decreases) and then down a column (metallic character increases). These effects tend to counter each other, and it is not easy to tell which will predominate.
L
無法判斷
The Periodic table (pg.346)
Copyright © 2011 Pearson Canada Inc.
Alkali Metals
Alkaline Earths
Transition Metals
Halogens
Noble Gases
Lanthanides and Actinides
Main Group
Main Group
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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro
Example 8.10 Alkali Metal and Halogen Reactions
Solution
Write a balanced chemical equation for each reaction.
a. the reaction between potassium metal and bromine gas b. the reaction between rubidium metal and liquid water c. the reaction between gaseous chlorine and solid iodine
a. Alkali metals react with halogens to form metal halides. Write the formulas for the reactants and the metal halide product (making sure to write the correct ionic chemical formula for the metal halide, as outlined in Section 3.5), and then balance the equation.
2 K(s) + Br2(g) → 2 KBr(s)
b. Alkali metals react with water to form the dissolved metal ion, the hydroxide ion, and hydrogen gas. Write the skeletal equation including each of these and then balance it.
2 Rb(s) + 2 H2O(l) → 2 Rb+(aq) + 2 OH−(aq) + H2(g)
c. Halogens react with each other to form interhalogen compounds. Write the skeletal equation with each of the halogens as the reactants and the interhalogen compound as the product and balance the equation.
Cl2(g) + I2(s) → 2 ICl(g)
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7.1 Nerve Transmission
• Movement of ions across cell membranes is the basis for the transmission of nerve signals.
• Na+ and K+ ions are pumped across membranes in opposite directions through ion channels.
– Na+ out and K+ in
• The ion channels can differentiate Na+ from K+ by their difference in size.
• Ion size and other properties of atoms are periodic properties—properties whose values can be predicted based on the element’s position on the periodic table.
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8.2 The Development of the Periodic Table
週期表Mendeleev (1834–1907)
• Mendeleev’s Periodic Law allows us to predict what the properties of an element will be based on its position on the table.
• It doesn’t explain why the pattern exists.
• Quantum mechanics is a theory that explains why the periodic trends in the properties exist.
– Knowing why allows us to predict what.
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8.3 Electron Configurations: How Electrons Occupy Orbitals
• Quantum-mechanical theory describes the behavior of electrons in atoms.
• The electrons in atoms exist in orbitals.
• A description of the orbitals occupied by electrons is called an electron configuration(電子組態).
Y = fn(n, l, m l ), m s Schrodinger Wave Equation
spin quantum number(自旋量子數) , m
s. -Not in the Schrödinger equation
-Schrödinger equation(n, l, m l )
Quantum Numbers and Electron Orbitals
Principle quantum number, n = 1, 2, 3…
Angular momentum quantum number, l = 0, 1, 2…(n-1)
l = 0, s l = 1, p l = 2, d l = 3, f
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General Chemistry: Chapter 8 Slide 32 of 50
Magnetic quantum number, m l = - l …-2, -1, 0, 1, 2…+l
(x, y, z) = (r, , ) , , ( , , ) = , ( ) , ( ) ( )
l l
l
n l l m m
m l
n r R r
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The Property of Electron Spin
• Spin is a fundamental property of all electrons.
• All electrons have the same amount of spin.
• The orientation of the electron spin is quantized, it can only be in one direction or its opposite.
– Spin up or spin down
• spin quantum number(自旋量子數) , m s .
• m s can have values of +½ or +½ −½.
– Not in the Schrödinger equation – Schrödinger equation(n, l, m
l)
https://www.youtube.com/watch?v=rg4Fnag4V-E
Stern–Gerlach experiment:
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Pauli Exclusion Principle不相容原理
• No two electrons in an atom may have the same set of four quantum numbers.
• Therefore, no orbital may have more than two electrons, and they must have opposite spins.
Allowed Quantum Numbers
Helium has two electrons.
Both electrons are in the first energy level.
Both electrons are in the s orbital of the first energy level.
Because they are in the same orbital, they must
have opposite spins
Orbital energy-level diagram for the first three electronic shells
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General Chemistry: Chapter 8 Slide 35 of 50
We call orbitals with the same energy degenerate.(等能量)
Sublevel Splitting in Multielectron Atoms
s (l = 0) < p (l = 1) < d (l = 2) < f (l = 3)
Multi-electron Atoms (多電子原子)
Schrödinger equation was for only one e - .
Electron-electron repulsion in multi-electron atoms.
Hydrogen-like orbitals (by approximation).
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General Chemistry: Chapter 8 Slide 36 of 50
♦
多電子原子的波函數,只能求得數值上的近似解♦利用量子力學,可準確地計算出多電子原子的能階等性質。
♦多電子原子與氫原子的結構有一主要分別:
氫原子的軌域能階由主量子數 n 決定;
氫原子能階:1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d
多電子原子的軌域能階由主量子數 n 和角動量子數 l 決定。
鉀原子能階:1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p
Orbital energy-level diagram for the first three electronic shells FIGURE 8-36
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General Chemistry: Chapter 8 Slide 37 of 50
38
多電子原子的能階(1)
要考慮核對各電子的吸引力及電子間的 互斥作用。
E=-2.179*10 -18 *(Z-σ) 2 /n 2
E:毎ㄧ個電子的能量(j/個)
Z:核電荷
σ:遮蔽常數
n:主量子數
39
多電子原子的能階(2)
(1)隨主量子數n增大,主階能階變大,K<
L <M <N…,同一主層各副層能階順序 隨l值增大而變大,所以產生能階分裂:
E ns <E np <E nd <E nf
(2)因為屏蔽效應及穿透效應,所以產生能 階交錯:
E ns <E (n-2)f <E (n-1)d <E np
40
屏蔽效應
(1)電子間的互斥而使核對外層電子的吸引 被減弱的作用稱屏蔽效應。
(2)內層電子對外層電子的屏蔽作用較大,
外層電子對內層電子可看做不產生屏蔽。
(3)離核愈近的主層上的電子被其他電子遮
蔽的效應愈小,所以各電子層遮蔽作用
大小為K > L > M > N…
41
穿透效應
(1)外層電子的電子雲具有鑽到內部空間 而更靠近核的現象較電子穿透,電子 穿透降低了其他電子對他的屏蔽作用,
而使軌域能量降低,這種效應稱為穿 透效應。
(2)由電子雲徑向分佈圖發現:ns比np易
穿透,np比nd易穿透。
42
能級交錯舉例說明
E 4s <E 3d :
4s: n=4, s l=0 3d: n=3, d l=2
因為4s主量子數比3d大1,但l角動量量子 數比3d小2,即s orbital 大於 d orbital 穿 透效應增大而使軌域能量降低的作用超
過主量子數增加對軌域能量的升高作用。
Orbital energy-level diagram for the first three electronic shells FIGURE 8-36
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General Chemistry: Chapter 8 Slide 43 of 50
The order of filling of electronic subshells FIGURE 8-37
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General Chemistry: Chapter 8 Slide 43 of 50
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Shielding(遮蔽) and Penetration(穿透)
• Each electron in a multielectron atom experiences both the attraction to the nucleus and repulsion by other electrons in the atom.
• These repulsions cause the electron to have a net reduced attraction to the nucleus; it is shielded from the nucleus.
• The total amount of attraction that an electron feels for the nucleus is called the effective nuclear charge of the electron.
• The closer an electron is to the nucleus, the more attraction it experiences.
• The better an outer electron is at penetrating through the electron cloud of inner electrons, the more attraction it will have for the nucleus.
• The degree of penetration is related to the orbital’s radial distribution function.
– In particular, the distance the maxima of the function are from the nucleus
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Penetration and Shielding
• 2s>2p
• 3s > 3p > 3d
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▲ FIGURE 8.5 General Energy Ordering of Orbitals for
Multielectron Atoms.
Multielectron Electron Configurations
Aufbau process
Electrons occupy orbitals in a way that minimizes the energy of the atom.
Pauli exclusion principle
No two electrons can have all four quantum numbers alike.
Hund’s rule
When orbitals of identical energy (degenerate
orbitals) are available, electrons initially occupy these orbitals singly.
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General Chemistry: Chapter 8 Slide 47 of 50
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Order of Sublevel Filling in Ground State Electron Configurations
Start by drawing a diagram, putting each energy shell on a row and listing the sublevels (s, p, d, f) for that shell in
order of energy (from left to right).
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d
7s Next, draw arrows through
the diagonals, looping back to the next diagonal
each time.
1s <2s< 2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<6d<7s
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Electron Configuration of Atoms in Their
Ground State
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Electron Configurations
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8.4 Electron Configurations, Valence Electrons, and the Periodic Table
• The electrons in all the sublevels with the
highest principal energy shell are called the valence electrons(價電子).
• Electrons in lower energy shells are called
core electrons (內層電子). .
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Insert periodic table at bottom of
page 347
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Irregular Electron Configurations
• Expected
• Cr = [Ar]4s 2 3d 4
• Cu = [Ar]4s 2 3d 9
• Mo = [Kr]5s 2 4d 4
• Ru = [Kr]5s 2 4d 6
• Pd = [Kr]5s 2 4d 8
• Found experimentally
• Cr = [Ar]4s 1 3d 5
• Cu = [Ar]4s 1 3d 10
• Mo = [Kr]5s 1 4d 5
• Ru = [Kr]5s 1 4d 7
• Pd = [Kr]5s 0 4d 10
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• The properties of the elements follow a periodic pattern.
– Elements in the same column have similar properties.
– The elements in a period show a pattern that repeats.
• The quantum-mechanical model explains this because the
number of valence electrons and the types of orbitals they occupy are also periodic.
8.5 The Explanatory Power of the Quantum-Mechanical Model
Properties and Electron
Configuration
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The Noble Gas(鈍氣) Electron Configuration
• The noble gases have eight valence electrons.
– Except for He, which has only two electrons
• They are especially nonreactive.
– He and Ne are practically inert.
• The reason the noble gases are so nonreactive is that the electron
configuration of the noble gases is
especially stable.
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The Alkali Metals(鹼金屬)
• The alkali metals have one more electron than the previous noble gas.
• In their reactions, the alkali metals tend to lose one electron, resulting in the same electron configuration as a noble gas.
– Forming a cation with a 1+ charge
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The Halogens(鹵素)
• Have one fewer electron than the next noble gas
• In their reactions with metals, the halogens tend to gain an electron and attain the
electron configuration of the next noble gas, forming an anion with charge 1−.
• In their reactions with nonmetals, they tend to share electrons with the other nonmetal so that each attains the electron
configuration of a noble gas.
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Electron Configuration and Ion Charge
S atom = 1s
22s
22p
63s
23p
4S atom = 1s
22s
22p
63s
23p
4Mg atom = 1s
22s
22p
63s
2Mg
2+cation = 1s
22s
22p
6Mg
2+cation = 1s
22s
22p
6Cation 陽離子
Anions 陰離子
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Slide 60 of 35 Slide 60 of 35
L
L
Large: atomic radius(原子半徑)
Metallic (金屬)
Large: Ionization energy
游離能Electron affinity
電子親和力
Nonmetal
lic 非金屬8.6 Periodic Trends in the Size of Atoms
and Effective Nuclear Charge
© 2014 Pearson Education, Inc.
© 2014 Pearson Education, Inc.
• Atomic radius increases
down group
Valence shell farther from nucleus
Effective nuclear
charge fairly close
• Atomic radius decreases across period (left to right)
Adding electrons to same valence shell
Effective nuclear charge increases
Valence shell held closer
L
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Periodic Trends in Atomic Radius
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Screening and Effective Nuclear Charge有 效核電荷
Z
effective= Z − S
Z is the nuclear charge
S is the number of electrons in lower energy levels Li 1s 2 2s 1
Z
eff= 3-2 =1
Be 1s 2 2s 2
Z
eff= 4-2 =2
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Quantum-Mechanical Explanation for the Group Trend in Atomic Radius
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Trends in
Atomic Radius:
Transition Metals
• Atomic radii of transition metals are roughly the same size across the d block.
– Much less difference than across main group elements – Valence shell ns 2 , not the (n−1)d electrons
– Effective nuclear charge on the ns
2electrons
approximately the same
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8.7 Ions: Electron Configurations, Magnetic
Properties, Ionic Radii, and Ionization Energy
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L
L
Large: atomic radius(原子半徑)
Metallic (金屬)
Large: Ionization energy
游離能Electron affinity
電子親和力
Nonmetal
lic 非金屬8.7 Ions: Electron Configurations, Magnetic
Properties, Ionic Radii, and Ionization Energy
© 2014 Pearson Education, Inc.
Al atom = 1s 2 2s 2 2p 6 3s 2 3p 1 Al 3+ ion = 1s 2 2s 2 2p 6
• The iron atom has two valence electrons:
Fe atom = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6
• When iron forms a cation, it first loses its valence electrons:
Fe 2+ cation = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6
• It can then lose 3d electrons:
Fe 3+ cation = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5
paramagnetism(順磁).電子都成對
-Will be attracted to a magnetic field
• diamagnetism(逆磁).有不成對電子
– Slightly repelled by a magnetic field
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Trends in Ionic Radius
O 2- > F - > Na + > Mg 2+ > Al 3+
L
L
Large: atomic radius(原子半徑)
Metallic (金屬)
Large: Ionization energy
游離能Electron affinity
電子親和力
Nonmetal
lic 非金屬anions陰離子 > neutral atoms中性原子 > Cations陽離子 越負愈大 越正越小
Isoelectronic = same electron configuration
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Periodic Trends in Ionic Radius
© 2014 Pearson Education, Inc.
Periodic Trends in Ionic Radius
© 2014 Pearson Education, Inc.
L
L
Large: atomic radius(原子半徑)
Metallic (金屬)
Large: Ionization energy
游離能Electron affinity
電子親和力Nonmetal
lic 非金屬Ionization Energy (IE)
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Ionization Energy (IE)
• Minimum energy needed to remove an electron from an atom or ion
Gas state
Endothermic process
Valence electron easiest to remove, lowest IE
M(g) + IE 1 M 1+ (g) + 1 e –
M +1 (g) + IE 2 M 2+ (g) + 1 e –
First ionization energy = energy to remove electron
from neutral atom, second IE = energy to remove
from 1+ ion, etc.
© 2014 Pearson Education, Inc.
© 2014 Pearson Education, Inc.
Exceptions in the First IE Trends
Be
1s 2s 2p B
1s 2s 2p
N
1s 2s 2p
O
1s 2s 2p
• First ionization energy generally increases from left to right across a period
• Except from 2A to 3A, 5A to 6A
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Exceptions in the First Ionization Energy Trends, Be and B
Be
1s 2s 2p
B
1s 2s 2p
Be +
1s 2s 2p To ionize Be, you must break up a full sublevel, which costs extra energy.
B +
1s 2s 2p
When you ionize B, you get a full sublevel, which costs
less energy.
© 2014 Pearson Education, Inc.
Exceptions in the First Ionization Energy Trends, N and O
To ionize N, you must break up a half-full sublevel, which costs extra energy.
N +
1s 2s 2p
O
1s 2s 2p
N
1s 2s 2p
O +
1s 2s 2p
When you ionize O, you get a half-full sublevel, which costs
less energy.
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Trends in Successive Ionization Energies
• Removal of each successive electron costs more energy.
– Shrinkage in size due to having more protons than electrons
– Outer electrons closer to the nucleus; therefore harder to remove
• There’s a regular increase in energy for each successive valence electron.
• There’s a large increase in
energy when core electrons are
removed.
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Trends in Second and Successive
Ionization Energies
© 2014 Pearson Education, Inc.
L
L
Large: atomic radius(原子半徑)
Metallic (金屬)
Large: Ionization energy
游離能Electron affinity
電子親和力Nonmetal
lic 非金屬8.8 Electron Affinities and Metallic Character
M(g) + 1e − M 1− (g) + EA
(EA)電子親和力
© 2014 Pearson Education, Inc.
© 2014 Pearson Education, Inc.
Trends in Electron Affinity
• Alkali metals decrease electron affinity down the column.
– But not all groups do
– Generally irregular increase in EA from second period to third period
• “Generally” increases across period
– Becomes more negative from left to right – Not absolute
– Group 5A generally lower EA than expected because extra electron must pair
– Groups 2A and 8A generally very low EA because
added electron goes into higher energy level or sublevel
• Highest EA in any period = halogen
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Metallic Character
L
L
Large: atomic radius(原子半徑)
Metallic (金屬)
Large: Ionization energy
游離能Electron affinity
電子親和力Nonmetal
lic 非金屬© 2014 Pearson Education, Inc.
Properties of Metals and Nonmetals
• Metals
Malleable and ductile
Shiny, lustrous, reflect light
Conduct heat and electricity
Most oxides basic and ionic
Form cations in solution
Lose electrons in reactions – oxidized
• Nonmetals
Brittle in solid state
Dull, nonreflective solid surface
Electrical and thermal insulators
Most oxides are acidic and molecular
Form anions and polyatomic anions
Gain electrons in reactions – reduced
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8.9 Some Examples of Periodic Chemical Behavior:
The Alkali Metals, the Halogens, and the Noble Gases
Trends in the Alkali Metals,1A
• Very low ionization energies
– Good reducing agents; easy to oxidize – Very reactive; not found uncombined in
nature
– React with nonmetals to form salts
– Compounds generally soluble in water found in seawater
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Trends in the Halogens, 7A
• Very high electron affinities
– Good oxidizing agents; easy to reduce – Very reactive; not found uncombined in
nature
– React with metals to form salts
– Compounds generally soluble in water
found in seawater
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Halogens
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Reactions of Alkali Metals with Halogens 2 Na
(s)+ Cl 2
(g) 2 NaCl
(s)2 M
(s)+ 2 H 2 O
(l
) M +
(aq)+ 2 OH -
(aq)+ H 2
(g)Reactions of Alkali Metals with Water
• Alkali metals are oxidized to the 1+ ion.
• H
2O is split into H
2(g) and OH
−ion.
• The Li, Na, and K are less dense than the water, so they float on top.
• The ions then attach together by ionic bonds.
• The reaction is exothermic, and often the heat released ignites the H
2(g).
• Alkali metals are oxidized to the 1+ ion.
• Halogens are reduced to the 1− ion.
• The ions then attach together by ionic bonds.
• The reaction is exothermic.
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Trends in the Noble Gases, 8A
Liquid helium, a cryogenic liquid, cools substances to temperatures as low as 1.2 K.
• Very unreactive
– Only found uncombined in nature
– Used as “inert” atmosphere when reactions with other gases would be undesirable.
• Melting point and boiling point increase down the column.
– All gases at room temperature – Very low boiling points
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