chap 4 periodic tableb

7/26/2019 CHAP 4 Periodic Tableb

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CHAPTER :

PERIODIC TABLE

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The periodic table is a table that arranges all the

known elements in order of increasing proton

number.The elements in the periodic table are arranged in:

group ( vertical column )period ( horizontal row )

The Modern Periodic Table

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The period in the Periodic Table are

numbered 1 to 7

The elements are arranged in order of the

increasing proton number.

The elements in the same period have the

same principle quantum number.

3Li = 1s2 2s1 4Be = 1s

2 2s2 Valence shell = 2Period = 2

P RIO

11Na = 1s2 2s2 2p6 3s1Valence shell = 3

Period = 3

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GROUPS

The groups in the Periodic Table are numberedfrom 1 to 18

The elements in the same group have the same

number of valence electrons.

Group number = number of valence electron, if the

element is in block s and d

Group number = number of valence elecftron + 10,

if the element is in block p

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3Li = 1s2 2s1

11Na = 1s2 2s2 2p6 3s1

Valence e- = 1Group 1

5B = 1s2 2s2 2p1

13Al = 1s2 2s2 2p6 3s2 3p1

Valence e- = 3

Group = 10 + 3

group 13

Example :

22Ti = 1s22s22p63s23p64s23d2

Valence e- = 4

Group 4

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There are 5 main group in periodic table :

Group 1 : alkalimetals

Group 2 : alkali earth metals

Group 3-12 : transition metals

Group 17 : halogens

Group 18 : inert/ noble gas

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1

2

3

45

6

7

group

period

12

3 4 5 6 7 8 9 10 11 12

13 14 15 16 1718

Alkali metals

(except H)

halogens

noble gases

Alkaline earth metals

Transition metals

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BLOCKS

All the elements in the Periodic Table can be

classified into 4 main blocks. These main blocksare block s, p, d and f.

B

o

k

s

Block-d

Block-p

Block-f

1 2 13 14 15 16 17 18

2

3

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s Block :

Groups 1 and 2

Configuration of valence electron : ns1 to ns2

Example:

11Na: 1s2 2s2 2p6 3s1

20Ca: 1s2 2s2 2p6 3s2 3p6 4s2

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p

Block :

Groups 13 to 18

Configuration of valence electrons: ns2 np1 to ns2np6.

Example:13Al: 1s

2 2s2 2p6 3s2 3p1

34Se: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4

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d Block

Also known as a transition elements.

Groups 3 to 12.

Configuration of valence electron: (n-1)d1 ns2 to (n-1)d10 ns2

Example:

23V : 1s2 2s2 2p6 3s2 3p6 3d3 4s2

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f Block

Involve element in the series of lanthanides (4f)

& actinides ( 5f ).

Block-f elements mostly radioactive

Lanthanidesactinides

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Classify the following elements into its appropriate group,

period and block.P 1s2 2s2 2p6 3s2 3p6

Q . 1s2 2s2 2p5

R . 1s2 2s2 2p6 3s2 3p6 4s2

S . 1s2

2s2

2p6

3s2

3p6

3d3

4s2

T .. 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1

EXERCISE !!

Element Group Period block

PQ

R

S

T

18 3 p17 2 p

2 4 s

5 4 d

13 4 p13


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Metal

All the elements on the left side and in themiddle of the periodic table (except for

hydrogen) are metallic elements.

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have properties that fall between those ofmetals and non-metals.

Metalloid

Metalloid

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Further across the period towards the right,

elements gradually lose their metallic characterand gained nonmetallic features.

non - metal

non- metal

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Periodicity

Periodicity is the periodic trend in properties of

elements.

1. Variation in atomic & ionic radii

The size /radius of atom is difficult to be

defined exactly because the electron cloud has

no clear boundary.

To solve this, we measure the distance between

the 2 nuclei in a molecule.

a

Radius, r = a/2

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2 factors that affecting the size of atoms :

1) Effective nuclear charge (Zeff

)

2) Value of the principal quantum number of

the valence electrons

1) Effective nuclear charge (Z

eff

)

is the residual nett charge felt by the valence electron.

Zeff= ZSwhere

Z = no. of proton

S = no. of inner electrons

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EXAMPLE:

12Mg = 1s2 2s2 2p6 3s2

electrons filled at the

inner orbital

Zeff= 1210 = 2

15P = 1s2 2s2 2p6 3s2 3p3

Zeff= 1510 = 5

Zeff , attraction between the nucleus & the valence

electrons become stronger. Thus, atomic size

Size of P is smaller than Mg

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The Screening Effect

The shielding effect is caused by the mutual

repulsion of electrons.The repulsion may occur

between :* electron ofinner orbitals & the electrons

ofvalence orbital

* electron of the same orbital.(less effective )

The shielding effect of the electrons of the inner

orbitals causes the outer electrons to be less

attracted to the nucleus.

2) Value of the principal quantum number of the

valence electrons

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12Mg = 1s2 2s2 2p6 3s2

Inner shell

4Be = 1s22s2

Size of Be is smaller than Mg

The value of n , the shielding effect , the

attraction between the nucleus valence

electrons becomes weaker, the size of atom

EXAMPLE:

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Across the period

3Li1S2 2S14Be1S2 2S2

5B1S2 2S22P16C1S2 2S22P2

7N1S2 2S22P3

Z eff

n / shell

Across the period :

The effective nuclear charge , the attraction

between the nucleus valence electronsbecomes stronger, the size of atoms decrease.

The value of n is same, so it is less important.

1 2 3 4 5

2 2 2 2 2

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3Li

[He] 2S111Na

[Ne]3S119K

[Ar]4S1

Zeff

n / shell

Going down the group : The value of n is , the shielding effects , the

attraction between the nucleus valence

electrons becomes stronger, the size of atoms

increase.

The effective nuclear charge is same, so it is less

important.

Going down the group

1 1 1

2 3 4

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Increasing atomic radius

I

n

e

n

a

o

m

i

c

ra

d

u

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Periodic trends in the ionic radii

a) Size of positive ions

Is formed when an atom loses electron

The mutual repulsions between them decrease &

experience more attraction from the nucleus So, the electron cloud shrinks.

Therefore,

cations is always smaller than

the corresponding neutral atoms.

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b) Size of negative ions

is formed when an atom gains electron.

The mutual repulsions between them increase &

experience less attraction from the nucleus

So, the electron cloud enlarges.

Therefore,

anions are larger than the corresponding neutral atoms

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Isoelectronic Series

Isoelectronic series are groups of atoms or ions

which have the same electronic configuration.

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Isoelectronic series for cations across period 3

11Na+ = 1s2 2s2 2p612Mg2+= 1s2 2s2 2p6

13Al3+ = 1s2 2s2 2p6

14Si4+ = 1s2 2s2 2p6

Isoelectronicconfiguration

Zeff

From Na+ to Si4+ the Zeff , the attraction between

the nucleus valence electrons becomes

stronger, the size of cations decrease.

Na+ > Mg 2+ > Al 3+ > Si 4+

The more positive the charge, the smaller the species29


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17Cl- =

1s2

2s2

2p6

3s2

3p6

16S2- = 1s2 2s2 2p6 3s2 3p615 P3-=

1s2 2s2 2p6 3s2 3p6

P 3- > S 2- > Cl-

Zeff

Isoelectronicconfiguration

The less negative the charge, the smaller the species

From P3+ to Cl- the Zeff , the attraction between

the nucleus valence electrons becomes

stronger, the size of anions decrease.

Isoelectronic series for anion across period 3

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Across the period, sizes of cations or anions

decrease due to the increase of Z

eff

.

P3- > S2- > Cl- > Na+ > Mg2+ > Al3+ > Si4+

P 3- > S 2- > Cl- > Na+ > Mg 2+ > Al 3+ > Si 4+

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Example :

Arrange these ions in order of decreasing ionic radius.

F-, O2-, Na+, Al 3+,Si4+, Mg2+, N3-

Electron configuration: 1s2 2s2 2p6

N3- > O2- > F- > Na+ > Mg2+ > Al3+ > Si4+

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Periodic trends in the ionization energies

The first ionization energy, IE1, is the energy required toremove one mole electron from one mole of neutral,gaseous atom:

X (g) X+ (g) + e- H = +ve

The process is endothermic as energy must be suppliedto remove the electron.

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Second ionization energy (IE2)is the energy required to remove one mole electron from

one mole positive ion (atom that already lost 1e-)

in the gaseous stateX+ (g) X2+ (g) + e- H = +ve

Third ionization energy (IE3)is the energy required to remove one mole electron from

two mole positive ion (atom that already lost 2e-)in the gaseous state

X2+ (g) X3+ (g) + e- H = +ve

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IE , the attraction between thenucleus and the electron

More difficult to remove the electron from the inner shell

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Factors effecting the IE

1) The size of atom (atomic radii)

Atomic radii ,the attraction between the nucleus & the valence e

IE

2) Effective nuclear charge (Zeff)Zeff ,

the attraction between the nucleus & the valence eIE

3) Value of n of the valence en

The attraction between the nucleus & the valence eIE

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going down a group,Ionization energy

WHY ?

Down a group, n , Shielding effect , attractionbetween the nucleus & valence e, atomic size ,

IE

across the period, Ionization energy

WHY ?

Across the period, Zeff , attraction between thenucleus & valence e , atomic size ,IE

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Group 1 2 13 14 15 16 17 18

Elements Li Be B C N O F Ne

IE 520 900 801 1086 1402 1314 1681 2081

Anomalous cases for ionization energy

The irregularity between group 2, 13,15 & 16

It can be explained by the stability of the half-filled & completely filled orbitals

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Explaination :

For Be :The 1st electron is removed from the completelyfilled 2s orbital,which has additional stability.

Whereas,

For B :

The first electron is removed from 2p orbital which ishigher in energy than the 2s.

the 2p electron is more easily removed than the 2selectron.

Be 1s2 2s2 B 1s2 2s2 2p1

IE1 for Be > IE1 for B

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IE1 for N > IE1 for O

Electronic configuration :

N 1s2 2s2 2p3 O 1s2 2s2 2p4

Explaination :

In N,the 1st electron is removed from a half-filled 2porbitals,which has additional stability.

In O, the 4th electron in the 2p orbital is paired.This

electron experience ee repulsion.

So, this e is more easily removed.

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LEARNING OUTCOMES

At the end of the lesson the students should be able

to :

(a) Deduce the electronic configuration of anelement and its position in the periodic

table based on successive ionisation energy

data.

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Ionization energies elements (kJ mol-1)

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1) Slowly increases when e is removed fromthe same shell.

As the positive charge increases, the

remaining electrons are held more

closely by the nucleus & more energy isrequired to remove them.

In general,successive ionization energy

always increases. The trends are :

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2) Drastically increases when e is removed from

another shell.

Has to remove electron from inner orbital,

which has a noble gas configuration.

More difficult to remove electron from astable arrangement.

This gives a clue of number of electron in

valence shell.Thus, we can deduce the

position of element in the periodic table.

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IE 1 2 3 4

(kJmol-1) 899 1757 14845 21000

Example:

Determine

i) electron configuration of the valence electron

for Z

ii) group number of Z in the periodic table

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By determining the IE ratios:

IE2 = 1757 = 1.95

IE1 899IE3 = 14845 = 8.45

IE2 1757

IE4 = 21000 = 1.41

IE3 14845

Highest ratio

Since IE3/ IE2 have the highest ratio, the third

electron is removed from an inner shell.

2 valence electrons are present.Electron configuration: ns2

This element is in group 2

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Example:

Five successive ionization energies (kJmol-1) for atom M

is shown below:

IE1 IE2 IE3 IE4 IE5800 1580 3230 4360 16000

Determine

i) electron configuration of the valence electronfor M

ii) group number of M in the periodic table

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By determining the IE ratios


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By determining the IE ratios:

IE2 = 1580 = 1.98IE1 800

IE3 = 3230 = 2.04IE2 1580

IE4 = 4360 = 1.35IE3 3230

IE5 = 16000 = 3.67IE4 4360

Highest ratio

Since IE5/ IE4 have the highest ratio, the fifth

electron is removed from an inner shell. 4 valence electrons are present.

Electron configuration: ns2np2

This element is in group 1448


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The change in energy of the reaction when an

electron is added to a gaseous atom or ion.

X (g) + e- X- (g)

Electron Affinity

These reactions tend to be exothermic (release

energy) because the first electron to be added

toward a neutral atom experiences an attractionfor the positively charged nucleus.

The higher (more negative) the EA, the more

easily it accepts an electron.

The first electron affinity

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These reactions tend to be endothermic

(absorb energy) because the second electron

is entering a negatively charge ion.

The electron to be added experience a strongrepulsion.Thus,energy is required to overcome

that repulsive force.

The second electron affinity

Br-(g) + e Br2-(g) EA = +ve

However, affinity does not always release

energy. In some cases affinity requires energy.

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Zeff , The attraction towards nucleus , size ,

more energy is released to add the electron,

So, AE or (moreve)

Across a period

1 2 13 14 15 16 17 18

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n , the shielding effect ,The attraction towards nucleus , size ,

more energy is required to add the electron,

so , AE

Down a group

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Indicate which element of each pair has the lower

electron affinity.

a) Na or K

b) Mg or Be

c) Li or Od) Cl or Br

e) Ca or Br

Answers:

a) K

b) Mg

c) Li

d) Br

e) Ca

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Is the relative tendency of an atom to attractelectrons to itself when chemically combined

with another atom.

Atoms with strong attraction for the bondingelectrons have the high electronegativity.

Electronegativity

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9.5

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Zeff , attraction between nucleus & valence

electrons , size ,so, electronegativity

n , the shielding effect ,attraction between

nucleus & valence electrons , size , so,

electronegativity

Across a period

Down a group

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Exercise:

Arrange the elements in order of theirincreasing electronegativity.

a) Na, Li, Cs, K

b) B, F, Li, C

c) Cl, S, Si, Na

Answers:

a) Cs, K, Na, Li

b) Li, B, C, F

c) Na, Si, S, Cl

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LEARNING OUTCOMES

Describe the periodicity of elements across period3 and down groups 1 and 17 for the followingphysical properties:

i. metallic characterii. melting point

iii. boiling point

Describe and explain the acid-base character of

oxides of elements in period 3.

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Physical Properties

Metallic character is closely related to atomic

radius and ionization energy.

The easier it is to remove electrons from an

atom, the more metallic the element.

1. Metallic character

Across the period :

Atomic size

IE

Metallic character

Down a group :

Atomic size

IE

Metallic character 59


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Elements Na Mg Al Si P S Cl Ar

MP

0

C 98 649 661 1410 44 113 -101 -189

BP

0

C 883 1090 2467 2655 280 445 -34 -186

Explain the variation in melting & boiling points of the

elements in terms of the structure & the intermolecularforces.

2. Melting and Boiling point

order of increasing melting/boiling point :

Ar < Cl < P < S < Na < Mg < Al < Si

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V i ti f lti d b ili i t f l t


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Na, Mg, Al have a metallic structure &metallic bonding.

The metallic bonding is formed by theelectrostatic attraction between the positivemetal ions an the sea of electrons in a cloud

a)

Metallic structure (Na to Al)

Variation of melting and boiling point of elements

in period 3 can be discussed in 3 parts:

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Strength of metallic bonding is proportional to

the number of valence electrons.

The more valence electrons,the stronger the

metallic bonding and the higher melting /

boiling point.

From Na to Al The number ofvalence electron, the metallic bonding becomes stronger, the

melting/boiling point

Na+ , Mg2+ , Al3+

The stronger the metallic bonding62

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Silicon has a giant covalent structure.Each atom

is covalently bonded to 4 other silicon atoms intetrahedral arrangement.

These covalent bonds are so strong,thus silicon

having a highest melting and boiling point &

being extremely hard.

b) Gigantic molecular structure (Si)

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)


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P,S,Cl have simple molecular structure (P4, S8,Cl2 )while Ar exist as monoatom.

c) Simple molecular structure ( P to Ar)

covalent bond

Van der Waals forcesS8

Cl2P4

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The covalent bond between the atoms is verystrong but the intermolecular force, that is,

Van der Waals, is very weak.

Van der waals forces increases as size of

molecule increases

Molecular size: Ar < Cl2 < P4 < S8 therefore

melting / boiling point :

Ar < Cl < P < S

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melting and boiling point of :


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Group 1 elements

melting and boiling point of :

Going down the group, the size of atoms

increases, the metallic bond becomes weaker,

the melting and boiling points decrease.66


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Group 17 elements

Going down the group, the molecular size

increases, the Van der Waals forces becomes

stronger, the melting and boiling points increase.67

A id b h t f id f P i d 3


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Acid-base character of oxides of Period 3

element Na Mg Al Si P S Cl

oxide Na2O MgO Al2O3 SiO2 P4O6

P4O10

SO2 Cl2O

Cl2O7

Oxidati

on state

+1 +2 +3 +4 +3

+5

+4

+6

+1

+7

The elements in period 3 burn in oxygen when

heated to form their respective oxides.

Most oxides can be clasified as acidic or basicdepending on whether they ;

Produce acid or base when disslove in water or

act as acids or bases in certain processes. 68


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Adding

H2O

soluble soluble insolub

le

insolu

ble

soluble soluble soluble

oxide Na2O MgO Al2O3 SiO2 P4O6

P4O10

SO2 Cl2O

Cl2O7

Adding

HCl

Has

rxn

Has

rxn

Has

rxn

No

rxn

No

rxn

No

rxn

No

rxn

Adding

NaOHNo

rxn

No

rxn

Has

rxn

Has

rxn

Has

rxn

Has

rxn

Has

rxn

NatureBasic

Oxide

Basic

Oxide

Amph

oteric

Oxide

Acidic

Oxide

Acidic

Oxide

Acidic

Oxide

Acidic

Oxide69


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Oxide Na2O

Add H2O Na2O(s) + H2O(l) 2NaOH(aq)

Add HCl Na2O(s) + HCl(aq) 2NaCl (aq)+ H2O(l)

MgO

MgO(s) + H2O(l) Mg(OH)2(aq)

MgO(s) + 2HCl(ag) MgCl2 (aq)+ H2O(l)

Oxide

Add H2O

Add HCl

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Al2O3

Al2O3(s) + 6HCl(aq) 2AlCl3(aq) + 3H2O(l)

Al2O3(s)+2NaOH(aq) + 3H2O 2Al(OH)4(aq)

Oxide

Add H2O

Add HCl

SiO2

SiO2(s) + 2NaOH(aq) Na2SiO3(aq) + H2O(l)

Oxide

Add H2O

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P4O10 (or P4O6)

P4

O6

+ 6H2

O 4H3

PO3

(Phosphoric(III) acid)

P4O10 + 6H2O 4H3PO4 (Phosphoric(V) acid)

P4O6 + 12NaOH 4Na3PO3 + 6H2O

P4O10 + 12NaOH 4Na3PO4 + 6H2O

SO2

SO2 + H2O H2SO3

SO2 + NaOH Na2SO4 + H2O

Oxide

Add H2O

Add

NaOH

Oxide

Add H2O

Add

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Cl2O, Cl2O7

Cl2O(g) + H2O(l) 2HOCl(aq)

Cl2O7(g)+ H2O(l) HClO4(aq)

Cl2O + NaOH 2NaOCl + H2OCl2O7 + NaOH 2NaClO4 + H2O

Oxide

Add H2O

Add

NaOH

chap 4 periodic tableb

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