Element Chapter 8 Periodic Properties of the

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© 2014 Pearson Education, Inc.

Sherril Soman

Grand Valley State University

Lecture Presentation

Chapter 8

Periodic

Properties of the

Element

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© 2014 Pearson Education, Inc.

Pg. 373-374

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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)

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Using the Periodic Table to Write Electron Configurations

The electron configuration of Si ends with 3s

2

3p

2

The electron

configuration of Rh

ends with 5s

2

4d

7

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8.2

ns

1

ns

2

ns

2

np

1

ns

2

np

2

ns

2

np

3

ns

2

np

4

ns

2

np

5

ns

2

np

6

d

1

d

5

d

10

4f 5f

Ground State Electron Configurations of the Elements

元素基態的電子組態

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Slide 6 of 35

Periodic Properties of the Elements

元素的週期性質

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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

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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

2

2s

2

2p

6

3s

2

[Ne]3s

2

Br: 1s

2

2s

2

2p

6

3s

2

3p

6

4s

2

3d

10

4p

5

[Ar] 4s

2

3d

10

4p

5

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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

2

2s

2

2p

6

3s

2

3p

4

[Ne]3s

2

3p

4

未成對電子=2

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© 2014 Pearson Education, Inc.

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

2

2s

2

2p

6

3s

2

3p

6

4s

2

3d

10

4p

2

[Ar] 4s

2

3d

10

4p

2

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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

2

2s

2

2p

6

3s

2

3p

1

[Ne] 3s

2

3p

1

Al

3+

: [Ne] or [He] 2s

2

2p

6

diamagnetic(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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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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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

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

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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 )

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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

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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)

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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

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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

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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

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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

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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…

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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。

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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。

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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.

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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.

(57)

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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

(58)

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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.

(59)

© 2014 Pearson Education, Inc.

Electron Configuration and Ion Charge

S atom = 1s

2

2s

2

2p

6

3s

2

3p

4

S atom = 1s

2

2s

2

2p

6

3s

2

3p

4

Mg atom = 1s

2

2s

2

2p

6

3s

2

Mg

2+

cation = 1s

2

2s

2

2p

6

Mg

2+

cation = 1s

2

2s

2

2p

6

Cation 陽離子

Anions 陰離子

(60)

© 2014 Pearson Education, Inc.

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

(61)

© 2014 Pearson Education, Inc.

(62)

© 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

(63)

© 2014 Pearson Education, Inc.

Periodic Trends in Atomic Radius

(64)

© 2014 Pearson Education, Inc.

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

(65)

© 2014 Pearson Education, Inc.

Quantum-Mechanical Explanation for the Group Trend in Atomic Radius

L

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

2

electrons

approximately the same

(66)

© 2014 Pearson Education, Inc.

8.7 Ions: Electron Configurations, Magnetic

Properties, Ionic Radii, and Ionization Energy

(67)

© 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

8.7 Ions: Electron Configurations, Magnetic

Properties, Ionic Radii, and Ionization Energy

(68)

© 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

(69)

© 2014 Pearson Education, Inc.

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

(70)

© 2014 Pearson Education, Inc.

Periodic Trends in Ionic Radius

(71)

© 2014 Pearson Education, Inc.

Periodic Trends in Ionic Radius

(72)

© 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)

(73)

© 2014 Pearson Education, Inc.

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.

(74)

© 2014 Pearson Education, Inc.

(75)

© 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

(76)

© 2014 Pearson Education, Inc.

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.

(77)

© 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.

(78)

© 2014 Pearson Education, Inc.

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.

(79)

© 2014 Pearson Education, Inc.

Trends in Second and Successive

Ionization Energies

(80)

© 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

(81)

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(82)

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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

(83)

© 2014 Pearson Education, Inc.

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

(84)

© 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

(85)

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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

(86)

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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

(87)

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Halogens

(88)

© 2014 Pearson Education, Inc.

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

2

O 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.

(89)

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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

(90)

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Xxpg339 Coulomb’s Law

• Coulomb’s law describes the attractions and repulsions between charged particles.

• For like charges, the potential energy (E) is positive and decreases as the particles get farther apart as r increases.

• For opposite charges, the potential energy is negative and becomes more negative as the particles get

closer together.

• The strength of the interaction increases as the size of the charges increases.

– Electrons are more strongly attracted to a nucleus with a 2+

charge than a nucleus with a 1+ charge.

Figure

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References

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