Atom and Periodic Table §ãËÁ iºÒÂ iÁäÁ - Chemistry Kru Boy · 2017-05-11 · Chemistry M.4 Lesson 1 Atom and Periodic Table by Angka Teprattananan angka teprattananan 1

Chemistry M.4 Lesson 1

Atom and Periodic Table

by Angka Teprattananan

1angka teprattananan

Atomic Models¤ o Áo¹ÀÒ¾·Õè¹a¡Ç i·ÂÒÈÒÊµÃ �ÊÃ�Ò§¢ é¹ ¨Ò¡¢�oÁÙÅ·Õèä �¨Ò¡¡ÒÃ

·´Åo§e¾ èoãª �o¸ iºÒÂÅ a¡É³a¢o§oaµoÁ o´Â atomic models ·ÕèÊÃ�Ò§¢ é¹ÊÒÁÒÃ¶»Ã aº»Ãu§ËÃo

e»ÅÕèÂ¹æ»Å§ä � ¶ �ÒÁÕ¼Å¡ÒÃ·´Åo§ãËÁ�æ «è§o¸ iºÒÂo´Âãª � atomic models æººe iÁäÁ�ä �

2องคาร เทพรตนนนท

öË �¹ óÅµ a¹ ( John Dalton )

Atomic�s Theory1. ¸Òµu»Ãa¡oº �ÇÂoaµoÁ«è§e»�¹Ë¹ �ÇÂeÅç¡·ÕèÊu äÁ�ÊÒÁÒÃ¶

æº�§æÂ¡ËÃ o·íÒÅÒÂä´ �2. oaµoÁ¢o§¸Òµue ÕÂÇ¡ a¹äÁÕÊÁºaµieËÁo¹¡ a¹ oaµoÁ¢o§

¸Òµuµ�Ò§ª¹ i´¡ a¹äÁÕÊÁºaµiµ�Ò§¡ a¹3. ÊÒÃ»Ãa¡oºe¡ i´¨Ò¡¡ÒÃÃÇÁ¡ a¹¢o§¸Òµu¤¹Åaª¹ i´

e.g. H2O , CO2(Expect Na ,H2 , Br2 , P4 , S8)


e¨ e¨ ·oÁÊ a¹ (J.J. Thomson)

4

J.J. Thomson È ¡ÉÒ¡ÒÃ¹íÒä¿¿�Ò¢o§æ¡ �Ê·Õèe¡ i´ã¹ cathode ray

tube ÀÒÂãµ�ÊÀÒÇa high voltage and low pressure.

angka teprattananan

Thomson�s Experiment

+‐voltage source

OFF

ON

+

‐eÁ èo¼ �Ò¹Ê¹ÒÁä¿¿ �Ò

¾ºo¹uÀÒ¤·ÕèÁÕ»Ãaüe»�¹Åº æÅaeÃÕÂ¡o¹uÀÒ¤¹ÕéÇ �Ò Electron

æÅaÇ a´¤ �Ò charge to mass ratio of the electron (q/me) = 1.76 x 108 coulombs/gram


Discovery of the Proton– ¤ �¹¾ºo´Â Eugen Goldstein– ÁÕ¡ÒÃ¾º �Canal rays� « è§ÁÕ»ÃaüºÇ¡æÅaeÃÕÂ¡Ç �Ò

Proton.

6

Eugen Goldstein

angka teprattananan

Cathode rays and Canal rays

7

__ cathode anode

+

angka teprattananan

¨Ò¡¡ÒÃ·´Åo§¢o§o¡Å �Êäµ¹ ÊÃu»ä �Ç�Ò - Ãa§ÊÕºÇ¡ËÃoo¹uÀÒ¤ºÇ¡e¡i´¨Ò¡æ¡�Ê « è§eºÕèÂ§eº¹e¢�ÒËÒ¢aéÇÅºä �- Ãa§ÊÕºÇ¡ÁÕ¤�ÒoaµÃÒÊ �Ç¹»Ãaüµ�oÁÇÅäÁ �¤§·Õè ¢é¹oÂÙ�¡aºª¹ i´¢o§æ¡ �Ê- ¶�Òe»ÅÕèÂ¹æ¡ �Êe»�¹äÎo´Ãe¨¹ ä¾ºÇ�Òo¹uÀÒ¤ºÇ¡·Õèe¡i´¢é¹äÁÕ¤�Ò»Ãaüe· �Ò¡ aº

oieÅç¡µÃo¹¾o Õ ¨§eÃÕÂ¡o¹uÀÒ¤ºÇ¡¹ÕéÇ�Ò �o»Ãµo¹�

The oil drop apparatus

ÁiÅiæ¡¹·íÒ¡ÒÃ·´Åo§æÅa¾º»Ãaü¢o§oieÅç¡µÃo¹ : 1.60 x 10-19 C æÅa¤íÒ¹Ç³ËÒÁÇÅ¢o§oieÅç¡µÃo¹ä �e·�Ò¡ aº : 9.1 x 10-28 g


Calculate mass of the Electron1. charge to mass ratio = 1.76x108 coulombs/gram2. charge of electron = 1.60 x 10-19 g


Thomson�s Atomic Model(Plum Pudding Model)

�oaµoÁe»�¹·Ã§¡ÅÁ »Ãa¡oº �ÇÂo»Ãµo¹·ÕèÁÕ»Ãa uºÇ¡æÅaoieÅç¡µÃo¹·ÕèÁÕ»Ãa uÅºoÂÙ�oÂ�Ò§¡Ãa a´¡Ãa¨ÒÂ æÅaÊíÒËÃaºoaµoÁ·Õèe»�¹¡ÅÒ§·Ò§

ä¿¿ �ÒäÁÕ íÒ¹Ç¹o»Ãµo¹e·�Ò¡ aº íÒ¹Ç¹oieÅç¡µÃo¹¾o Õ�10angka teprattananan

Discovery of Nucleus Tested Thomson�s model of atomic

structure with the �gold foil� experiment. ·íÒ¡ÒÃ·´Åo§Âi§Ã a§ÊÕæoÅ¿Òe¢�Òä»Âa§æ¼�¹·o§¤íÒ ¾ºÇ �Ò

Ernest Rutherford


Ã a§ÊÕæoÅ¿ÒÊ �Ç¹ãË�·aÅuoo¡ÁÒe» �¹eÊ �¹µÃ§ Ã a§ÊÕæoÅ¿ÒºÒ§Ê �Ç¹ ÁÕ¡ÒÃeºÕèÂ§eº¹¨Ò¡æ¹Çe´ iÁ Ã a§ÊÕæoÅ¿ÒÊ �Ç¹¹ �oÂÁÒ¡e¡ i´¡ÒÃÊa· �o¹¡Å aº

¨Ò¡¼Å¡ÒÃ·´Åo§ ·íÒãË �ÊÃu»ä´ �Ç �Ò : The atom is mostly empty space All the positive charge, and almost all the mass is in a

small area in the center. He called this a �nucleus� The electrons distributed around the nucleus, and occupy

most of the volume


Rutherford's Atomic Model

�oaµoÁ»Ãa¡oº �ÇÂ¹ iÇe¤ÅÕÂÊ« è§ÁÕ»ÃaüºÇ¡(o»Ãµo¹)oÂÙ�µÃ§¡ÅÒ§ ÁÕ¢¹ÒéÅç¡ÁÒ¡æÅaÁÕÁÇÅÁÒ¡ Ê �Ç¹oieÅç¡µÃo¹·ÕèÁÕ»ÃaüÅºæÅaÁÕÁÇÅ¹ �oÂ e¤Åèo¹·ÕèÃoº¹ iÇe¤ÅÕÂÊ

e»�¹ºÃ ieÇ³¡Ç �Ò§"


The Discovery of the Neutron ¡ÒÃÂ i§o¹uÀÒ¤æoÅ¿Òe¢ �Òä»Â a§æ¼�¹ Be ¾ºo¹uÀÒ¤¹ iÇµÃo¹«è§ÊÒÁÒÃ¶·íÒãË�o»ÃµÃo¹ËÅu´¨Ò¡æ¼�¹

¾ÒÃÒ¿ �¹ä �Neutrons have mass similar to protons. No electrical charge.

14

James Chadwick

angka teprattananan

The Subatomic particlesparticle symbol charge mass(g) mass(amu)electron e -1 9.1x10-27 0.0005proton p +1 1.67x10-24 1.0072neutron n 0 1.67x10-24 1.0086

THE MASS OF THE NEUTRON IS 1839 times greater than an electron.

Composition of the Nucleus:• nuclei are composed of "nucleons": protons and neutrons • atomic mass units: 1 amu = exactly 1/12 the mass of a carbon-

12 nucleus 15angka teprattananan

Atomic Symbol (Nuclear Symbol)

A

ZXMassnumber

Atomicnumber

Element Symbol

• Atomic number = ¨íÒ¹Ç¹o»Ãµo¹(ÁÕ¤ �Òe·�Ò¡ aº¨íÒ¹Ç¹oieÅç¡µÃo¹ÊíÒËÃ aº¸Òµu·Õèe»�¹¡ÅÒ§)

• Mass number = ¼ÅÃÇÁÃaËÇ �Ò§¨íÒ¹Ç¹o»Ãµo¹æÅa¹ iÇµÃo¹


Fill in the blanks for the following nuclear symbols:

Element94Be 14

6C 3517Cl- 74

33As3- 4420Ca2+ 67

31Ga3+

Atomic Number

Mass Number

# of Protons

# of Neutrons

# of Electrons


Isotope , Isotone , Isobar and Isoelectronic

• Isotope are atoms of the same element having different masses, (¨íÒ¹Ç¹¹ iÇµÃo¹äÁ�e·�Ò¡ a¹)

11H p = 1 , e = 1 , n = 021H p = 1 , e = 1 , n = 1

• Isotone are atoms of the different element having equal neutrons.

115B p = 5 , e = 5 , n = 6

126C p = 6 , e = 6 , n = 6


• Isobar are atoms of the different element having equal mass number.

3616S mass no. of 3616S is 36

3618Ar mass no. of 3618Ar is 36

• Isoelectronic are atoms and ion having equal electron.

2010Ne p = 10 , e = 10 , n = 10

2412Mg2+ p = 12 , e = 10 , n = 12


Neclear symbols isotope isotone isobar isoelectronic

146C & 147N

3919K & 4020Ca

3517Cl & 3717Cl

3818Ar & 3216S2-

168O & 188O

4018Ar & 4020Ca

3517Cl- & 3919K+

3115P & 3216S

For each of the following ,check the blank for isotope , isotone , isobar or isoelectronic


Max Plank

È ¡ÉÒ spectrum ¢o§ Electromagnetic Wave Wave. The æÅa¤ÇÒÁÊ aÁ¾ a¹¸ �ÃaËÇ �Ò§ wavelength(λ), the frequency

(ν), and the energy (E)

where c is Speed of light = 3 x 108 m/s

h is Planck's Constant = 6.626 x 10-34 J.s

is Frequency (Hz) λ is wavelength (m)

Max Plank


Energy and frequency of Electromagnetic Wave

Color of spectrum

Wavelength (nm)

Energy (kJ)

Violet Blue Green Yellow Orange

Red

400 - 420420 - 490490 - 580580 - 590590 - 650650 - 700

4.96x10-22 - 4.73x10-22

4.73x10-22 - 4.05x10-22

4.05x10-22 - 3.42x10-22

3.42x10-22 - 3.36x10-22

3.36x10-22 - 3.05x10-22

3.05x10-22 - 2.83x10-22


Calculate about Electromagnetic WaveEx1 Êe»¡µÃ aÁÊÕÁ�Ç§¤ÇÒÁÂÒÇ¤Å è¹ 500 nm äÁÕ¤ÇÒÁ¶Õèe·�Òã´

Ex2 Êe»¡µÃ aÁÊÕæ´§ÁÕ¤ÇÒÁÂÒÇ¤Å è¹ 500 nm äÁÕ¤ÇÒÁ¶ÕèæÅa¾Åa§§Ò¹e·�Òã´


Ex3 The energy of electromagnetic wave is 3x10-22 KJ , Find the color

Ex4 The frequency is 5x1014 Hz , calculate the wavelength , energy and find the color of this spectrum.


Niels Bohr È ¡ÉÒ Hydrogen Spectrum

Niels Bohr


Color of Spectrum

Wavelength (nm)

Energy (kJ) ΔE

RedBluegreen

BlueViolet

656486434410

3.02 x 10-22

4.08 x 10-22

4.57 x 10-22

4.84 x 10-22

10.6 x 10-23

4.9 x 10-23

2.7 x 10-23


Bohr�s Atomic Model

oaµoÁ»Ãa¡oº �ÇÂo»Ãµo¹æÅa¹ iÇµÃo¹ oÂÙ�ÀÒÂã¹¹ iÇe¤ÅÕÂÊ Ê�Ç¹oieÅç¡µÃo¹Ç iè§oÂÙ�Ãoº æ ¹ iÇe¤ÅÕÂÊe»�¹ªaé¹æ ËÃoe»�¹Ãa aº¾Åa§§Ò¹« è§ÁÕ¤ �Ò äÁ�µ�oe¹ èo§¡ a¹


Electron Configuration(o¤Ã§æººoieÅç¡µÃo¹)oieÅç¡µÃo¹äoÂÙ�Ãoºæ ¹ iÇe¤ÅÕÂÊe» �¹ª aé¹æ eÃÕÂ¡Ç �Ò Ãa´ aº¾Å a§§Ò¹ËÅ a¡

�principle energy level�æµ �Åa principle energy level äÁÕoieÅç¡µÃo¹ã¹¨íÒ¹Ç¹·Õè¨íÒ¡ ae» �¹ä»µÒÁ¡®

Rule ; 2n2

n = principle energy levels


Rule; Arrangements of electrons in an atom.1. ¨ aóieÅç¡µÃo¹ã¹Ãa aºªaé¹¾Åa§§Ò¹æÃ¡æãË�eµçÁ¡ �o¹

2. eÁèoÃa aºæÃ¡æ ¨ �éµçÁæÅ�ÇãË� a oieÅç¡µÃo¹Å§ã¹Ãa aºªaé¹¶ a´ä»3. oieÅç¡µÃo¹·ÕèÃa aº¹o¡Êu´ÁÕä � 1-8 æÅaÃa aº¶ aé¢ �ÒÁÒ ÁÕ 8 æÅa 18

µÒÁÅíÒ aº(ÊíÒËÃ aº¸ÒµuËÁÙ�A)

e.g. 11Na: has 11 electrons First energy level 1 (n =1) can fill 2 electrons

Second energy level 2 (n =2) can fill 8 electrons

Third energy level 3 (n =3) can fill 1 electron shot hand 11Na : 2 , 8 , 1


Arrangements of electrons in an atom.1. 3Li : �� 2. 12Mg : ��3. 19K : �� 4. 33As : ��5. 53I : �� 6. 55Cs : ��7. 82Pb : �� 8. 88Ra : �� 9. 22Ti : �� 10. 28Ni : ��


Electron Cloud Model

ã¨¤ÇÒÁÊíÒ¤ a oieÅç¡µÃo¹·ÕèÁÕ¾Åa§§Ò¹µèíÒä¾ººÃ ieÇ³ã¡Å �

¹ iÇe¤ÅÕÂÊ oieÅç¡µÃo¹·ÕèÁÕ¾Åa§§Ò¹ÊÙ§¡Ç �Òä¾ºË�Ò§¨Ò¡

¹ iÇe¤ÅÕÂÊoo¡ÁÒ äÁ�ÊÒÁÒÃ¶ºo¡µíÒæË¹ �§·Õèæ¹ �¹o¹¢o§oieÅç¡µÃo¹

ä � ºo¡ä �e¾ÕÂ§oo¡ÒÊã¹·Õè a¾ºoieÅç¡µÃo¹ o´ÂºÃ ieÇ³·ÕèÁÕ¡Åu�ÁËÁo¡oieÅç¡µÃo¹Ë¹Òæ¹ �¹ÁÕ

oo¡ÒÊ¾ºoieÅç¡µÃo¹ÁÒ¡¡Ç �ÒºÃ ieÇ³·ÕèÁÕ¡Åu�ÁËÁo¡eºÒºÒ§

31Erwin Schrodinger angka teprattananan

Åa¡É³a¢o§ Electron Cloud Model äeÃÕÂ¡Ç �Ò �Orbital� ËÃ o subenergy levels «è§ÁÕ 4 ª¹i´ ä �æ¡ � s , p , d and f

s

p

d

f32angka teprattananan

Electron configurationsoÒÈ aÂËÅa¡¡ÒÃ´ a§¹Õé1. Pauli exclusion principle;

Each orbital can have only 2 electrons and have not the same spin. 2. Aufbau principle;

Electron fill lowest energy levels first.3. Hund�s Rule;

For atoms in ground state, the number of unpaired electrons is the maximum possible and have the same spin.


Arrangements of electrons in Orbital

principle energy levels

(shell)subenergy levels

(subshell)maximum electron

n = 1 s 2

n = 2 s , p 8

n = 3 s , p , d 18

n = 4 s , p , d , f 32

n = 5 s , p , d , f 32

n = 6 s , p , d 18

n = 7 s , p 8

n = 8 s 234angka teprattananan

Lower energy Higher energy1s 2s 2p 3s 3p 4s 3d 4p �.

35angka teprattananan 1S

2S

3S2P

3P

4S

3d

4P5S

4d

Ex. Give the full electron configuration of 27Co

NOTE. principle energy levelsOr 27Co;


»ÃaeÀ·¢o§ electron configuration æº �§e» �¹ 3 »ÃaeÀ·´ a§¹Õé

1) Orbital Diagram.

2) Long notation or spdf configuration.1s22s22p63s2....

3) Shorthand Notation or noble gas core.[Ne]3s23p4

s1

s2


Using Orbital Diagram1. 3Li : �� 2. 12Mg : ��

3. 18Ar : ��

4. 19K : ��

5. 25Mn : ��

6. 26Fe : ��

7. 24Cr : ��


Using Long Notation or spdf configuration 1. 12Mg : �� 2. 18Ar : ��3. 19K : ��4. 33As : ��5. 38Sr : ��6. 55Cs : ��7. 28Ni : ��8. 24Cr : ��9. 29Cu : ��


Using Short hand Notation or Noble Gas core.1. 3Li : �� 2. 12Mg : �� 3. 18Ar : ��4. 19K : �� 5. 33As : �� 6. 28Ni : �� 7. 37Rb : ��8. 53I : ��


(Ç iÇ a²¹Ò¡ÒÃ¢o§µÒÃÒ§¸Òµu)

The History of the Periodic Table


He noticed that chlorine, bromine and iodine had similar properties. And the atomic mass of the middle element was roughly the average of the masses of the others .

He called �Law of Triads�

Cl Chlorine mass = 35.5Br Bromine mass = 79.9I Iodine mass = 126.9

Average mass of chlorine and iodine= (35.5 + 126.9) / 2= 81.9 (close to Br!)

Dobereiner�s other triads included lithium (Li), sodium (Na) and potassium (K), along with calcium (Ca), strontium (Sr) and barium (Ba).

Johaun Dobereiner


He noticed that every 8th element had similar properties, a bit like a musical scale. He listed some of the known elements in rows of 7 as shown below.He called �Law of Octaves� .

His law of octaves work today with the first 20 elements.43

John Newlands

angka teprattananan

Dmitri Mendeleev In 1869 he published a table of the elements

organized by increasing atomic mass. Noticed similar properties appeared at regular

intervals --> �periodic�

Lothar Meyer At the same time, he published his

own table of the elements organized by increasing atomic mass.

44

Mendeleev

Lothar Meyer

angka teprattananan

Mendeleev’s discovery


The table below compares Mendeleev�s prediction with the actual data.


Henry Mosely

¨ aéÃÕÂ§ÅíÒ´ aº¸ÒµuµÒÁeÅ¢oaµoÁ(¨íÒ¹Ç¹o»Ãµo¹)¨Ò¡¹ �oÂä»ÁÒ¡ o´ÂeÃÕÂ§¨Ò¡« �ÒÂä»¢ÇÒã¹µÒÃÒ§¸Òµu¡ÒÃ¨ a´µÒÃÒ§¸Òµuæºº¹ÕéÊÒÁÒÃ¶æ¡ �» �ËÒ¢o§¸Òµu

Tellurium(Te) and Iodine(I) ä´ �¹o¡¨Ò¡¹Õé Mosely Â a§µ aé§e» �¹¡®e¡ÕèÂÇ¡ aºµÒÃÒ§¸Òµu

Periodic Law : ÊÁº aµ i·Ò§¡ÒÂÀÒ¾æÅae¤ÁÕ¢o§¸Òµu ¢é¹oÂÙ�¡ aºeÅ¢oaµoÁæÅa¡ÒÃ¨ aéÃÕÂ§oieÅç¡µÃo¹

47

Henry Mosely

angka teprattananan

Three classes of elements are Metals, Nonmetals and Metalloids

Modern Periodic Table


Metal Elements Good conductors of heat and electricity Have luster, are ductile , malleable , good reflect light All metals are solids at room temperature ,except for

mercury(Hg) Found on left side of periodic table and some on right

side of table

Gold


Nonmetal Elements Have properties that are opposite to those of metals Not good conductors of heat and electricity, poor

reflect light Usually brittle solids or gases ,except for bromine(Br) Found on right side of periodic table � AND hydrogen

SulphurBromine 50angka teprattananan

Metalloids Sometimes called semiconductors Form the �stairstep� between metals and nonmetals Have properties of both metals and nonmetals Examples: B, Si , Ge , As , Sb, Te , Po and At


¡ÒÃe»ÃÕÂºe·ÕÂºÊÁº aµ i¢o§¡ è§oÅËa ¡ aº Al(oÅËa)æÅaI(ooÅËa)

IE1(kJ/mol)

EN Density(g/cm3)

melting�boiling point(oC)

ElecticalConductivity

Type of Compound

Al 584 1.61 2.70 660-2519 √ ionic

B 807 2.04 2.34 2075-4000 √ ionic and covalent

Si 793 1.90 2.33 1414-3265 √ ionic and network covalent

Ge 768 2.01 5.32 938-2833 √ ionic and covalent

As 951 2.18 5.75 358-603 √ ionic and covalent

Sb 840 2.05 6.68 631-1587 √ ionic and covalent

I 1015 2.66 4.93 114-184 X ionic and covalent


Location of Hydrogen in the periodic table

Some properties of Hydrogen , group IA and VIIA

properties Group IA H group VIIA

1. # valence electron 1 1 7

2. Oxidation number in compounds

+1 +1 , -1 +1 , +3 , +5+7 , -1

3. IE1 (kJ/mol) 382-526 1318 1015-1687

4. EN 0.7-1.0 2.1 2.2 -4.0

5. phase solid gas 3 phase

6. Electric conductivity can cannot cannot


¤ÇÒÁÊ aÁ¾ a¹¸ �ã¹æ¹Çµ aé§eÃÕÂ¡Ç �Ò ËÁÙ� æÅaËÁÙ�ã¹µÒÃÒ§¸ÒµuÁÕ·aé§ËÁ´ 18 ËÁÙ� æº �§e»�¹ËÁÙ� A 8 ËÁÙ� æÅaËÁÙ� B 8 ËÁÙ�(ËÁÙ� 8B «éíÒ¡ a¹ 3 ËÁÙ�)

¸Òµu·ÕèoÂÙ�ã¹ËÁÙ�e ÕÂÇ¡ a¹Áa¡äÁÕÊÁºaµi·Ò§¡ÒÂÀÒ¾æÅa·Ò§e¤ÁÕ¤Å�ÒÂ¤Å§¡ a¹

Periodic Table


¤ÇÒÁÊ aÁ¾ a¹¸ �ã¹æ¹Ç¹o¹¢o§¸Òµu eÃÕÂ¡Ç �Ò ¤Òº « è§ÁÕ·aé§ËÁ´ 7 ¤Òºã¹µÒÃÒ§¸Òµu

Periodic Table


The s and p block elements are called �REPRESENTATIVE ELEMENTS (Group A)�

The d and f block elements are called �TRANSITION ELEMENTS (Group B)�

s pd

f

56

Periodic Table

angka teprattananan

REPRESENTATIVE ELEMENTS (Group A ; 8 groups)

57

Alkali MetalsAlkali Earth Metals Halogens

Noble GasesInert Gases

angka teprattananan

Group IA (Alkali Metals)

� e»�¹¢o§æ¢ç§ÊÕe§ i¹æÅaÊa·�o¹æÊ§ä � Õ� o �o¹ÊÒÁÒÃ¶µ a´§ �ÒÂ �ÇÂÁÕ´� ¤ÇÒÁË¹Òæ¹ �¹µèíÒ� e¡ i´» i¡ iÃiÂÒ¡ aºoÒ¡ÒÈä �§ �ÒÂ µ �o§e¡çºäÇ �ã¹¹éíÒÁa¹� ·íÒ» i¡ iÃiÂÒ¡ aº¹éíÒ ä �ÊÒÃÅaÅÒÂ·ÕèÁÕÊÁº aµ ie»�¹eºÊe.g.

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)


� e¡ i´» i¡ iÃiÂÒ¡ aºoo¡« ie¨¹ ä �ÊÒÃ»Ãa¡oºoo¡ä« �ËÅÒÂæºº:4Li(s) + O2(g) 2Li2O(s) (oxide)

2Na(s) + O2(g) Na2O2(s) (peroxide)K(s) + O2(g) KO2(s) (superoxide)

� äÁÕÊÕe©¾ÒaeÁèo¹íÒä»e¼Òã¹e»ÅÇä¿� ÊÒÃ»Ãa¡oº¢o§oÅËaËÁÙ� 1A ·u¡ª¹i´ÅaÅÒÂ¹éíÒä �� ¤ÇÒÁÇ �o§äÇã¹¡ÒÃe¡ i´» i¡ iÃiÂÒe¾ ièÁ¨Ò¡º¹Å§Å�Ò§


Group IIA (Alkali Earth Metals)

� ÁÕ¤ÇÒÁË¹Òæ¹ �¹ÁÒ¡¡Ç �Ò Alkali Metals� ÊÒÃ»Ãa¡oº carbonate , phosphate , sulphid , sulphite

¢o§ Alkali Earth Metal äÁ�ÅaÅÒÂ¹éíÒ� Be äÁ�e¡ i´» i¡ iÃiÂÒ¡ aº¹éíÒ , Mg e¡ i´» i¡ iÃiÂÒª �Òæ¡ aº¹éíÒ æÅa Ca e¡ i´» i¡ iÃiÂÒ¡ aº¹éíÒä � :

Mg(s) + 2H2O(l) Mg(OH)2(aq) + H2(g)Ca(s) + 2H2O(l) Ca(OH)2(aq) + H2(g)


Group VIA (Chalcogen)

� Oxygen , Sulphur and Selenium are nonmetals , Tellurium is Metalloid and Polonium is radioactive element.

� Oxygen ã¹¸ÃÃÁªÒµi¾º 2 ÃÙ» ¤ o O2 and O3. � O3 e¡ i´¨Ò¡ O2 o´Âãª �¡ÃaæÊä¿¿�Ò, eª �¹»ÃÒ¡¯¡ÒÃ³ �¿ �Ò¼�Ò:

3O2(g) 2O3(g) H = +284.6 kJ� O3 e»�¹æ¡ �Ê¾ iÉ.


Group VIIA (Halogens)� Halogen ÊÒÁÒÃ¶Ã aºoieÅç¡µÃo¹æÅa¡ÅÒÂe»�¹äooo¹Åº ä �:

X2 + 2e- 2X-

� Fluorine e»�¹¸Òµu·ÕèÇ �o§äÇã¹¡ÒÃe¡ i´» i¡ iÃiÂÒÁÒ¡·ÕèÊu´:

2F2(g) + 2H2O(l) 4HF(aq) + O2(g) • Halogen 1 oÁeÅ¡uÅä»Ãa¡oº �ÇÂ 2 oaµoÁ ; F2 , Cl2 , Br2

and I2� Fluorine gas is pale-yellow , Chlorine gas is yellow-

green , Bromine liquid is red-brown and solid iodine is black (violet vapor)


• ¤ÇÒÁÇ �o§äÇã¹¡ÒÃe¡ i´» i¡ iÃiÂÒÅ´Å§¨Ò¡º¹Å§Å�Ò§ ¸ÒµuËÁÙ� 7A

2Cl-(aq) + F2(g) 2Br-(aq) + Cl2(g) 2I-(aq) + Br2(g) 2F-(aq) + Cl2(g) 2Cl-(aq) + Br2(g) 2Br-(aq) + I2(g)


Group VIIIA(Noble gases)• Noble gases oÂÙ�e»�¹oaµoÁe ÕèÂÇ(He Ne Ar Kr Xe Rn)• ºÃÃüoieÅç¡µÃo¹eµçÁã¹ s æÅa p orbital• »� 1960 eÃÕÂ¡¸Òµu¡Åu�Á¹ÕéÇ �Ò �inert gases� e¾ÃÒaäÁ�ÊÒÁÒÃ¶e¡ i´» i¡ iÃiÂÒe¤ÁÕ¡ aº¸Òµuo è¹æ

• »�¨ uº a¹ÊÒÁÒÃ¶eµÃÕÂÁÊÒÃ»Ãa¡oº¢o§¸Òµu¡Åu�Á¹Õéä � ; XeF2XeF4 XeF6 KrF2 and HArF


TRANSITION ELEMENTS (Group B ; 8 groups)

65

Transition Metals

InnerTransition MetalsRare-earth elements

angka teprattananan

Electron Configuration and Properties


Phisical Properties of Potassium - Zinc


Oxidation Number of Transition Metals(Stable Oxidation Number in red)


Nomenclature of Elements with Atomic Numbers Greaterer than 100

The Rules for Naming Elements 1. Name directly from the atomic number of the element using the following numerical roots

0 = nil , 1 = un , 2 = bi , 3 = tri , 4 = quad , 5 = pent , 6 = hex , 7 = hept , 8 = oct , 9 = enn

2. The roots are put together in the order of the digits and terminated by �ium� to spell out the name.

Example Atomic Number : 112 Element Name: Ununbium

Element Symbol: Uub 69angka teprattananan

Write the element symbol and name :1. Atomic Number : 114 Element Name : ___________

Element Symbol : ___________2. Atomic Number : 115 Element Name : ___________




Element Symbol : ___________70angka teprattananan

Prediction located of atoms in the periodic table

71

For representative elements (group A)group no. = ¾ i¨ÒÃ³Ò¨Ò¡¨íÒ¹Ç¹oieÅç¡µÃo¹ª aé¹¹o¡Êu´

= ËÃo¼Å¨Ò¡ªaé¹¹o¡Êu´¨Ò¡Ãa aº¾Åa§§Ò¹Â �oÂperiod no. = ¾ i¨ÒÃ³Ò¨Ò¡¨íÒ¹Ç¹Ãa aº¾Åa§§Ò¹

Ex. 17Cl : 1s2 2s2 2p6 3s2 3p5

: 2 , 8 , 7

So, 17Cl is in group 7A and period 3

valence electron

Outer level

Three shells

angka teprattananan

For Transiton metals (group B)group no. = ¾ i¨ÒÃ³Ò¨Ò¡¼Å¢o§oieÅç¡µÃo¹ 2 Ãa aºÊu´·�ÒÂperiod no. = ¾ i¨ÒÃ³Ò¨Ò¡¨íÒ¹Ç¹Ãa aº¾Åa§§Ò¹

e.g. 21Sc : 1s2 2s2 2p6 3s2 3p6 4s2 3d1

So, Sc is in group 3B and period 4.

72

/2+1 = 3

4 principle energy levels

angka teprattananan

Fill in the blank , Determine Group No. and Period No.

1. 7N : �� 6. 25Mn : ��group �� period �� group �� period ��

2. 11Na : �� 7. 26Fe : ��group �� period �� group �� period ��

3. 18Ar : �� 8. 22Ti : ��group �� period �� group �� period ��

4. 20Ca : �� 9. 53I : ��group �� period �� group �� period ��

5. 35Br : �� 10. 29Cu : ��group �� period �� group �� period ��


e»�¹¸Òµu·Õè¹ iÇe¤ÅÕÂÊäÁ�eÊ¶ÕÂÃ äÁÕ¡ÒÃæ¼� alpha, beta or gamma radiation æÅae»ÅÕèÂ¹e»�¹¸Òµu·ÕèeÊ¶ÕÂÃÁÒ¡¢ é¹

Radioactive Elements


Properties of radiation

Type of Radiation Alpha particle Beta particle Gamma ray

Symbol

Mass (amu) 4 1/2000 0

Charge +2 -1 0

Speed slow fast very fast (speed of light)

Ionising ability high medium 0

Penetrating power low medium high

Stopped by: paper aluminium lead

Penetrating power


The behavior of three types of radioactive emissions in an electric field.


Alpha Decaye¡ i´¡ aº¹ iÇe¤ÅÕÂÊ·ÕèÁÕo»Ãµo¹ÁÒ¡e¡ i¹ä» ·íÒãË�e¡ i´æÃ§¼Å a¡ã¹¹ iÇe¤ÅÕÂÊÁÒ¡

¨ §¨íÒe»�¹µ�o§Å´ íÒ¹Ç¹o»Ãµo¹Å§


Beta Decaye¡ i´¡ aº¸Òµu·ÕèÁÕoaµÃÒÊ �Ç¹¹ iÇµÃo¹µ�oo»Ãµo¹ÁÒ¡e¡ i¹ä» ¨ §ÁÕ¡ÒÃe»ÅÕèÂ¹

¹ iÇµÃo¹ãË�e»�¹o»Ãµo¹o´Â¡ÒÃ»Å´»Å�oÂoieÅç¡µÃo¹ËÃoÃ a§ÊÕºÕµÒoo¡ÁÒ


Gamma Decay

e¡ i´¡ aº¹ iÇe¤ÅÕÂÊ¢o§¸Òµu·ÕèÁÕ¾Åa§§Ò¹ÊÙ§e¡ i¹ä» ¨ §ÁÕ¡ÒÃ»Å´»Å�oÂ¾Åa§§Ò¹oo¡ÁÒã¹ÃÙ» high energy photon ËÃ oo¹uÀÒ¤æoÅ¿Ò.

32He* 32He + γ


Partical Symbol Charge mass(amu)*

Alpha α , 42He + 2 4.00276

Beta β , 0-1e - 1 0.000540

Gamma γ 0 0

Positron β+ , 0+1e + 1 0.000540

Neutron n , 10n 0 1.0087

Proton P , 11H + 1 1.0073

Deuteron D , 21H + 1 2.0136

Tritron T , 31H + 1 3.0219

Symbol charge and mass


Nuclear equatione»�¹ÊÁ¡ÒÃ·ÕèæÊ´§¡ÒÃe»ÅÕèÂ¹æ»Å§ÀÒÂã¹¹iÇe¤ÅÕÂÊ¢o§¸Òµu

¡ aÁÁa¹µÃa§ÊÕËÃo¡ÃaºÇ¹¡ÒÃæ¼ �Ãa§ÊÕBalancing Nuclear EquationsEx1:

11H + 94Be ---> 63Li + 42He

Rule: The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.


Balancing Nuclear Equations

A. 2714Si _______ + 0-1e

B. 6629Cu _______ + 0-1e

C. 2713Al + 42He 30

14Si + _______D. 14

6C 136C + ________

E. 22689Ac 226

88Ra + ________

F. 22689Ac 222

87Fr + __________


G. 21383Bi _______ + 42He

H. 20981Tl 209

82Pb + _______I. 23

11Na + 42He 2612Mg + _______

J. 23892U + 16

8O ________ +510n

K. 23892U + 16

8O 23994Pu + ________

L. 23592U + 10n 90

38Sr + 14354Xe + ________


100 g 50 g 25 g14 วน 14 วน

Half life ; t1/2¤ o ÃaÂaeÇÅÒ·Õè radioactive elements e¡ i´¡ÒÃÊÅÒÂµ aÇ

æÅaÁÕ»Ã iÁÒ³Å´Å§¤Ã è§Ë¹ è§¢o§·ÕèÁÕoÂÙ�e´ iÁEx. P-32 has a half life 14 days


Elements Half life Radiation Benefit

U-235 7.1x109 years Alpha Gamma Treatment of Cancers

C-14 5,760 years Beta Archeology

Co-60 5.26 years Gamma Treatment of Cancers

Au-198 2.7 days Beta Gamma Medical Diagnostics

I-125 60 days Gamma Medical Diagnostics

I-131 8.07 days Beta Gamma Medical Diagnostics

P-32 14.3 days Beta Treatment of Cancers

Pu-239 24,000 years Alpha Gamma Generation of Electricity

K-40 1x109 years Beta ArcheologyRa-226 1,600 years Alpha Gamma Treatment of Cancers


µaÇoÂ �Ò§ ¶ �Ò· ié§äoo«o·»¡ aÁÁa¹µÃ a§ÊÕª¹ i´Ë¹è§ 20 ¡Ã aÁ äÇ �¹Ò¹ 28 Ç a¹ »ÃÒ¡¯Ç �ÒÁÕäoo«o·»¹ aé¹eËÅooÂÙ� 1.25 ¡Ã aÁ ¤Ã è§ªÕÇ iµ¢o§äoo«o·»¹ÕéÁÕ¤ �Òe·�Òã´

µaÇoÂ �Ò§ ¨§ËÒ»Ã iÁÒ³ I-131 eÃ ièÁµ�¹ eÁèo¹íÒ I-131 íÒ¹Ç¹Ë¹ è§ÁÒÇÒ§äÇ �e»�¹eÇÅÒ 40.5 Ç a¹ »ÃÒ¡¯Ç �Ò ÁÕÁÇÅeËÅo 0.125 ¡ÃaÁ ¤Ã è§ªÕÇ iµ¢o§ I-131 e·�Ò¡ aº 8.1 Ç a¹


1. After 42 days a 2.0 g sample of phosphorus-32 contains only 0.25 g of the isotope. What is the half-life of phosphorus-32?

2. In 5.49 seconds, 1.20 g of argon-35 decay to leave only 0.15 g. What is the half-life of argon-35?


5. Polonium-214 has a half-life of 164 seconds. How many seconds would it take for 8.0 g of this isotope to decay to 0.25 g?

6. How many days does it take for 16 g of palladium-103 to decay to 1.0 g? The half-life of palladium-103 is 17 days.


Calculations base on half life

Nt = N0 2n

n = T / t1/2

Nt = number remainingN0 = initial numberT = timen = no. time of decayt1/2 = half life

µaÇoÂ �Ò§ ¨§ËÒ»Ã iÁÒ³¢o§ Tc-99 ·ÕèeËÅoeÁèoÇÒ§ Tc-99 ¨íÒ¹Ç¹ 18 ¡Ã aÁäÇ �¹Ò¹ 24 ª aèÇoÁ§ æÅa Tc-99 ÁÕ¤Ã è§ªÕÇ iµ 6 ªaèÇoÁ§


1. After 42 days a 2.0 g sample of phosphorus-32 contains only 0.25 g of the isotope. What is the half-life of phosphorus-32?

2. Polonium-214 has a half-life of 164 seconds. How many seconds would it take for 8.0 g of this isotope to decay to 0.25 g?


Nuclear Fission¡ÃaºÇ¹¡ÒÃ·ÕèoaµoÁ¢o§¸ÒµuË¹ a¡e¡ i ¡ÒÃÊÅÒÂµ aÇãË�oaµoÁ¢o§¸Òµu·ÕèeºÒ

¡Ç �Ò ¾Ã �oÁ»Å´»Å�oÂ¾Åa§§Ò¹oo¡ÁÒ


Nuclear ReactionNuclear Fussion

¡ÃaºÇ¹¡ÒÃ·ÕèoaµoÁ¢o§¸ÒµueºÒ ÃÇÁµ aÇ¡ a¹e»�¹oaµoÁ¢o§¸Òµu·ÕèËÅa¡¢ é¹æÅaÁÕ¡ÒÃ»Å´»Å�oÂ¾Åa§§Ò¹oo¡ÁÒ


Atomic Properties and

Periodic Trends


Atomic size (¢¹ÒóaµoÁ) Ion size (¢¹Ò´äooo¹) Ionization energy (IE) (¾Å a§§Ò¹äoooä¹e«ª a¹) Electron affinity (EA) (Ê aÁ¾ÃÃ¤ÀÒ¾oieÅç¡µÃo¹) Electronegativity (EN) (oieÅç¡o·Ãe¹¡Òµ iÇ iµÕ) Melting point(m.p.) (ü´ËÅoÁeËÅÇ)

and Boiling point(b.p.) (üé´ o´) Oxidation Number(O.N.) (eÅ¢oo¡« ie´ª a¹)

94

Atomic Properties and Periodic Trends

angka teprattananan

95

¡ÒÃ¾ i¨ÒÃ³Ò¢¹ÒóaµoÁ ä¾ i¨ÒÃ³Ò¨Ò¡ atomic radius « è§atomic radius ¤ o ÃaÂa·Ò§¤Ã è§Ë¹è§ÃaËÇ �Ò§ü´¡ è§¡ÅÒ§¢o§oaµoÁ 2 oaµoÁ ·ÕèoÂÙ�µ i´¡ a¹

Atomic Size

angka teprattananan 96

ª¹ i´¢o§Ã aÈÁÕoaµoÁ1. Covalent Radius(Ã aÈÁÕo¤eÇeÅ¹µ �) used for Covalent

compounds. e.g. H2 , F2 , Cl2 , O2

2. Van der Waals Radius(Ã aÈÁÕæÇ¹eóÃ �ÇÒÅÊ �) used for Noble gases. e.g. He , Ne , Ar

3. Metallic Radius(Ã aÈÁÕoÅËa) used for Metal atoms. e.g. Li , Mg , Cu

Cl - Cl

angka teprattananan

Trends of Atomic size

Group trends :The atoms get bigger as we go down a group.

Because the increase in the principal energy levels.

Period Trends :The atoms get bigger as we go from right to left

in a period at same energy level. Because the decrease of nucleus attraction.

97angka teprattananan 98angka teprattananan

1. Which element in each pair has the larger atoms? 1.1 12Mg or 20Ca 1.2 3Li or 8O

1.3 17Cl or 35Br 1.4 11Na or 16S

2. Arrange these atoms in order of increasing size?

11Na , 13Al , 6C , 19K

3. Arrange these atoms in order of increasing size?

33As , 37Rb , 18Ar , 15P


The Octet Rulee»�¹¡ÃaºÇ¹¡ÒÃ·ÕèoaµoÁ¢o§¸Òµu¾ÂÒÂÒÁ¨ aéÃÕÂ§ãË �ÁÕoieÅç¡µÃo¹

ª aé¹¹o¡Êu´ e·�Ò¡ aº 8 eËÁo¹æ¡ �ÊËÁÙ� VIII o´ÂMetals generally give(lose) electrons, Nonmetals take(gain) electrons from other atoms.oaµoÁ·ÕèoÂÙ�ã¹ÃÙ»»Ãaüä¿¿ �Ò eÃÒeÃÕÂ¡Ç �Ò �ion�.


Ions size Metals elements lose valence electrons to form cation.

Cation radius are always smaller than atomic radius.

Non-metal elements gain valence electrons to form anion. Anion radius are always larger than atomic radius.

6.3


Group trends The ions get bigger as we go down a group.

Because the increase in the principal energy levels.Period Trends

The ions get bigger as we go from right to left in a period at same energy level. Because the decrease of nucleus attraction.

Li+

Be2+

B3+

C4+N3- O2- F-


Atoms and Ions size


1. Which atoms or ions in each pair are larger? 1. 12Mg or 12Mg2+ 2. 8O or 8O2-

3. 7N3- or 9F- 4. 11Na+ or 12Mg2+

2. Arrange these atoms and ions in order of increasing size?

12Mg2+ , 13Al3+ , 15P3- , 17Cl-

3. Arrange these atoms and ions in order of increasing size?

3Li+ , 11Na+ , 12Mg , 16S2-


Ionization Energy(IE) ¤ o ¤ �Ò¾Åa§§Ò¹·ÕèoaµoÁ Ùé¢ �Òä»e¾ èoãª�´§oieÅç¡µÃo¹oo¡¨Ò¡oaµoÁã¹Ê¶Ò¹a

æ¡ �Ê «è§æ¹Ço¹�Á¢o§ IE ä¼¡¼a¹¡ aºæ¹Ço¹�Á¢o§¢¹ÒóaµoÁ

e.g.First Ionization Energy(IE1)

Na(g) --> Na+(g) + eSecond Ionization Energy(IE2)

Na+(g) --> Na2+(g) + e


Write IE1 � IE5 of Boron

______ __________________ : IE1 = 807 KJ/mol

______ __________________ : IE2 = 2,433 KJ/mol

______ __________________ : IE3 = 3,666 KJ/mol

______ __________________ : IE4 = 25,033 KJ/mol

______ __________________ : IE5 = 32,834 KJ/mol


IE3 and IE4 is more different , why ?

Find the group number from ionization of following element ?

IE1(MJ/mol)

IE2(MJ/mol)

IE3(MJ/mol)

IE4(MJ/mol)

IE5(MJ/mol)

IE6(MJ/mol)

IE7(MJ/mol)

IE8(MJ/mol)

group

0.744 1.457 7.739 10.547 13.636 18.001 21.710 25.663

1.687 3.381 6.057 8.414 11.029 15.171 17.874 92.047

1.093 2.359 4.627 6.229 37.838 47.285

0.906 1.763 14.855 21.013


ABCD

Trends in IE1 of First 20 Elements

First ionization energy tends to increase from bottom to top within a group. And increase from

left to right across a period.



1. Which element in each pair has the greater ionization energy?

1. 12Mg or 13Al 2. 4Be or 5B

3. 6C or 14Si 4. 2He or 53I

2. Arrange these atoms in order of increasing IE1 ?

33As , 37Rb , 18Ar , 15P , 16S 3. Arrange these atoms in order of increasing IE1 ?

19K , 13Al , 11Na , 12Mg , 2He


Electron Affinity(EA) ¤ o ¤ �Ò¾Åa§§Ò¹·Õè¤ÒÂoo¡ÁÒ eÁèooaµoÁ¢o§¸Òµuã¹Ê¶Ò¹aæ¡ �Ê ÃaºoieÅç¡µÃo¹ 1 o¹uÀÒ¤. ¤ �Ò EA Ê�Ç¹ãË�ÁÕ¤ �Òe»�¹Åº e¹èo§¨Ò¡e»�¹» i¡ iÃiÂÒ ¤ÒÂ¤ÇÒÁÃ �o¹Example ;

O(g) + e O- (g) : EA = -142 KJ/mol

O-(g) + e O2- (g) : EA = 780 KJ/mol

111angka teprattananan 112angka teprattananan

1. Which element in each pair has the greater electron affinity? 1. 12Mg or 13Al 2. 3Li or 8O

3. 6C or 32Ge 4. 18Ar or 53I

5. 11Na or 19K 6. 9F or 53I

2. Arrange these atoms in order of increasing EA ?

19K , 20Ca , 11Na , 15P , 9F


Electronegativity (EN) ¤ o ¤ �ÒæÊ´§¤ÇÒÁÊÒÁÒÃ¶ã¹¡ÒÃæÂ �§ª i§oieÅç¡µÃo¹¢o§oaµoÁã¹

ÊÒÃ»Ãa¡oº ·ÕèÁÕ¡ÒÃÊÃ �Ò§¾ a¹¸ae¤ÁÕ«è§¡ a¹¸Òµu·ÕèÁÕ¤ �Ò EN ÁÒ¡ äÁÕ¤ÇÒÁÊÒÁÒÃ¶ã¹¡ÒÃæÂ �§ª i§

oieÅç¡µÃo¹ä´ � Õ


Trends in Electronegativity of Elements

Electronegativity tends to increase from bottom to top within a group. And increase from left to right across a period.

Because the increase of nucleus attraction. Note; Noble gases are NOT assigned

electronegativities


6.3


1. Which element in each pair has the greater electronegativity?

1. 11Na or 15P 2. 3Li or 8O 3. 6C or 32Ge 4. 9F or 53I

2. Draw arrow to show the bond polarity in each pair elements N---F C---Br O----Cl

Br---Br C---S C----I

3. Arrange these atoms in order of increasing EN ?

12Mg , 20Ca , 17Cl , 9F


Melting Point and Boiling Point

melting point o u³ËÀÙÁi·ÕèÊÒÃe»ÅÕèÂ¹Ê¶Ò¹a¨Ò¡¢o§æ¢ç§e»�¹¢o§eËÅÇ- Helium has the lowest melting point (-272.2oC).- Carbon has the highest melting point (3550oC).

boiling point o u³ËÀÙÁi·ÕèÊÒÃe»ÅÕèÂ¹Ê¶Ò¹a¨Ò¡¢o§eËÅ§e»�¹æ¡ �Ê- Helium has the lowest boiling point (-268.9oC). - Tungsten has the highest boiling point (5927oC).


For metals ;The melting point and boiling point tends to increase from

bottom to top within a group (e¹ èo§ÁÒ¨Ò¡¤ÇÒÁæ¢ç§æÃ§¢o§¾ a¹¸aoÅËa).And increase from left to right across a period

(e¹ èo§ÁÒ¨Ò¡¤ÇÒÁæ¢ç§æÃ§¢o§¾ a¹¸aoÅËaæÅa¨íÒ¹Ç¹oieÅç¡µÃo¹ªaé¹¹o¡·Õèe¾ ièÁ¢é¹).


For nonmetals ;The melting point and boiling point tends to increase from

top to bottom within a group. And increase from right to left across a period.

(e¹ èo§ÁÒ¨Ò¡¤ÇÒÁæ¢ç§æÃ§¢o§æÃ§æÇ¹eóÃ �ÇÒÅ�Ç)

¢ �oÂ¡eÇ �¹ ¤ oËÁÙ� IVA eª�¹ C Si (high melting point , boiling point)e¹ èo§¨Ò¡ÊÒÁÒÃ¶e¡ ié»�¹oÁeÅ¡uÅ¢¹Ò´Â a¡É �·ÕèeÃÕÂ¡Ç �Òo¤Ã§¼Å¡Ã �Ò§µÒ¢ �ÒÂ


Arrange these atoms in order of increasing melting � boiling point ?

1. 19K , 15P , 17Cl

2. 3Be , 10Ne , 13Al

3. 6C , 11Na , 3Li

4. 9F , 53I , 11Na , 13Al

5. 7N , 14Si , 9F , 10Ne


Periodic trendssummary

Ioni

zatio

n en

ergy

Elec

tron

affi

nity

Elec

tron

egat

ivity

m.p

. and

b.p

. of m

etal

Electron affinity

Ionization energy

Electronegativity

m.p. and b.p. of metal

Atom

ic radius

metallic character

m.p. and b.p. of nonm

etal

Atomic radius

metallic character

m.p. and b.p. of nonmetal122angka teprattananan

123

Oxidation numberThe oxidation number of an element indicates the number of electrons lost, gained, or shared as a result of chemical bonding.

angka teprattananan

e¡³± �¡íÒË¹´¤ �ÒeÅ¢oo¡« ie´ªa¹¢o§¸Òµuµ�Ò§æ1. O.N. ¢o§¸ÒµuoiÊÃaÁÕ¤ �Òe»�¹ÈÙ¹Â � eª�¹ S8 O2 Si Br2 Zn 2. O.N. ÁÕ¤ �Òe·�Ò¡ aº»Ãaü¢o§äooo¹ eª�¹ Na+ , Cl- ÁÕ¤ �Ò +1 , -13. O.N. ¢o§¸ÒµuËÁÙ� IA æÅa IIA ã¹ÊÒÃ»Ãa¡oºÁÕ¤ �Ò +1 æÅa +24. O.N. ¢o§äÎo´Ãe¨¹ã¹ÊÒÃ»Ãa¡oºÁÕ¤ �Ò +1 Â¡eÇ �¹ äÎä´Ã´ � e»�¹ -15. O.N. ¢o§oo¡« ie¨¹ã¹ÊÒÃ»Ãa¡oºÁÕ¤ �Ò -2 Â¡eÇ �¹ OF2 e»�¹ +2 ,

e»oÃ �oo¡ä« � (Na2O2 , BaO2 , H2O2) ÁÕ¤ �Ò -1 æÅa«u»e»oÃ �oo¡ä« �( NaO2 , KO2) ÁÕ¤ �Ò -1/2

HCO3-

O = H =

Oxidation numbers of C in HCO3

- ?

124

Find the Oxidation Number of S in SOCl4 ?O.N. of Oxygen = O.N. of Chlorine =

angka teprattananan

Determine the oxidation number of underline element :

1. SO2

2. CaSO4

3. PO43-

4. NH4+

5. Pb(OH)4

6. KMnO4

7. Cu(NO3)

8. K2[Fe(CN)3H2O]


Documents

Atom and Periodic Table §ãËÁ iºÒÂ iÁäÁ - Chemistry Kru Boy · 2017-05-11 · Chemistry M.4 Lesson 1 Atom and Periodic Table by Angka Teprattananan angka teprattananan 1