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
47
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
60
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
b)
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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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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
NAOH72
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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
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