IB Topic 3 Periodicity

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Transcript IB Topic 3 Periodicity

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Topic 3
Periodicity
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4.1 The Periodic Table: Learning
Goals
Understand
how the elements in the
periodic table are arranged
Understand
the terns ‘group’ and
‘period’
Understand
how the electronic
configuration of an element relates to
its position in the periodic table
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The Periodic Table
 Arranged
in order of increasing atomic
number
 Vertical
 Most
columns and horizontal periods
of the elements are metals
 The
elements bordering between metals and
non-metals are sometimes referred to as semimetals or metalloids (Si, Ge, Sb)
 Some
of the groups have specific names
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History of Chemistry
In
1869 Russian scientist Dmitri
Mendeleev published the original
periodic table
 He
arranged the elements based on
increasing atomic weight
 He also left gaps for elements that had not
yet been discovered and predicted their
properties
The
modern periodic table is arranged
by increasing atomic number
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The Periodic Table and Electron
Configurations
The
group number of an element indicates the
number of electrons in the element’s highest
main energy level (outer shell)
 Group
1= 1 outer electron
 Group 2= 2 outer electrons, etc.
The
period number indicates the number of
main energy levels (shells) in an atom
 Period
1= 1 main energy level
 Period 2= 2 main energy levels, etc.
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4.2 Physical Properties: Learning
Goals
 Define
first ionization energy and electronegativity of
an element
 Understand
trends in atomic radius, first ionization
energy and electronegativity across a period
 Understand
trends in atomic radius, first ionization
energy and electronegativity down group 1 and
group 7
 Explain
the variation of melting points for elements
across period 3 and down group 1 and group 7
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Electronegativity
 In
covalent bonds between 2 different types of
atoms, the atoms do not attract the electron pair
equally
 How strongly electrons are attracted depends on
the size of individual atoms and their nuclear
charge
 Electronegativity
decreases down a group
 Because, the size of atoms increase down a group
 The larger an atom is, the less pull from the
nucleus thus, lower electronegativity
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Example: HF vs. HCL
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Electronegativity Across a Period
Increases
across a period
 Because, increases
in nuclear charge with no
significant change in shielding
 Shielding remains approximately constant
because atoms in the same period have the
same number of inner shells
 Also, the number of protons attracting the
electrons increases
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First Ionization Energy
 The
energy required to remove the outermost
electron from a gaseous atom
 Energy
 Down
for: M(g) M+(g) + e-
a group:
 I.E. decreases
 Because,
size of the atom increases so the outer
electron is further from the attraction of the nucleus
 Across
a period:
 I.E. increases
 Because,
the nuclear charge increases
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Atomic Radius
 Size
of an atom; mostly due to electron orbits
 Trends:


Increases down a group
 Because, there are more electrons when going down a group
Decreases across a period
 Because increase in nuclear charge with no increase in shielding
(similar as electronegativity and ionization energy)
 Across

period 3, Na is the larges and Cl is the smallest
Ar is not counted because it does not bond, thus the radius cannot be
measured
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Ionic Radius
Measure
Positive
of the size of an ion
ions are smaller than their atomic
radii
Negative
radii
ions are greater than their atomic
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Ionic Radii Trends
 Not
as simple as neutral atomic radii
 Due
to the type of ions changing from side to side
 Ex. Na+ and Mg2+
 They have the same number of electrons, but Mg2+ has
the highest nuclear charge (pull of nucleus on
electrons) and is smallest
 Ex. P3- and S2 Equal electrons, but S2- has the stronger nuclear charge,
thus is smaller
 Trend
is generally similar to atomic radii
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Melting Point
 Depends
 Does
 The
on the type of bonding in an element
not have a consistent trend
alkali metals are metallic m.p. decrease
 Because, the
delocalized electron sea is further from the
pull of the nucleus
 The
halogens increases down the family
 Because
the relative atomic masses of X2 increases and
van der Waal’s forces increases
 van der Waal’s Intermolecular forces resulting from
temporary dipole-dipole interactions
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Variation in M.P. across period 3

Must look at bonding to understand changes

Increases from Na to Mg to Al all metallically bonded

Silicon increases sharply has a giant covalent structure which are
very difficult to break

There is a large decrease from Si to P (covalent) van der Waal’s
forces are weak, thus not a lot of energy is required to break them

Slight increase from P to S covalent compound, but has a higher
molecular mass and requires more E to break the bonds

Decrease from S to Cl to Ar also covalently bound, but much
smaller masses thus, easier to break
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Graph the boiling points
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Na 98

Mg 649
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Al 660
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Si 1410
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P 44
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S 119

Cl -101

Ar -189
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4.3 Chemical properties of
elements in group 1 and group 7
Chemical
properties of elements in the
same group
Chemical
reactions depend on the amount
of outer electrons an atom contains
Thus, families have basically the same
properties
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Reactions of elements in group 1
 Very
reactive metals that react readily with oxygen,
water, halogens, and other things
 Nearly
all of the reactions involve the single outer
electron being removed
 Reactions
are more vigorous going down the
column due to lesser ionization energy
 Reaction
 React
with oxygen:
vigorously with oxygen
 4M(s) + O2(g) 2M2O(s)
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Con’t
Reaction
Rapid
with water
reactions with water
2M(s) + 2H2O(l) 2MOH(aq) + H2(g)
The alkali metal hydroxides are
strong bases
Reactions get more vigorous
descending down the column
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Reactions of Group 7 Elements
 Reactions
generally are due to the elements
gaining an electron to have an octet
 Reactivity
 Fluorine
decreases down the family
is the most reactive element known
 Basically
reacts with every element on the periodic table
 Variations
in reactivity are not as easily explained
as group 1
 Several
factors must be considered when explaining
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Halogen Reactions
React
with alkali metals to form salts
2M(s)
+ X2(g) 2MX(s)
Salts formed are white or colorless
Displacement
Reactions
Reactions
between a solution of a halogen
and a solution containing halide ions
Ex: Cl2(aq) + 2KBr(aq)  2KCl(aq) + Br2(aq)
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4.4 Properties of the Oxides of
Period 3 Elements
Look
at table 4.3 p.156
Sodium
oxide, magnesium oxide,
aluminum oxide
Have
giant ionic structures
Ions held by strong electrostatic forces
causing high melting points and boiling
points
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Covalent Oxides
 Look
at 4.4 p. 158
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Reactions of Period 3 Oxides with
Water
In
general, metallic oxides are
basic and non-metallic oxides are
acidic
Ex. Na2O(s)
2NaOH(aq)
P4O10(s)+
+ H2O(l) 
6H2O(l)  4H3PO4(aq)
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HL
first years
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4.5 Properties of the Chlorides of
Period 3 Elements
 Chlorides


of period 3
Table 4.6 p. 161
Bonding of aluminum chloride is complex and undergoes
changes of structure and bonding as it changes states

We assume bonding is covalent molecular in all states
 In
the Lewis structure Al only has 6 electrons and will
bond with another AlCl3 to form a dimer
 Overall
 The
structure is non-planar
bonds are weak and easily broken, thus low
melting/ boiling points
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Con’t
 SiCl4
and PCl3 covalent molecular liquids
 SiCl4
is tetrahedral and PCl3 is trigonal pyramidal
 van der Waals’ forces are weak, so they have low melting/
boiling points
 Neither conduct electricity
 PCl5
 Solid
complex bonding
at room temperature
 Forms a covalent molecular liquid when melted
 van der Waals’ are stronger than PCl3, has a slightly higher
melting/ boiling point
 Does not conduct electricity
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Con’t
Cl2
covalent molecular gas
 Diatomic
molecule
 Weak van der Waals’ forces in liquid
chlorine
 Low molecular mass
 Bonds can be broken at temperatures
below room temperature
 Does not conduct electricity
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Reactions of Period 3 Chlorides
with Water

Table 4.7 p. 162

NaCl(s) Na+(aq) + Cl-(aq)

[Mg(H2O)6]2+(aq)  [Mg(H2O)5(OH)]+(aq) + H+(aq)

AlCl3(s) + 3H2O(l)  Al(OH)3(s) +3HCl(g)

AlCl3(s) + 6H2O(l)  [Al(H2O)6]3+(aq) + 3Cl-(aq)

SiCl4(l) + 4H2O(l)  Si(OH)4(s) + 4HCl(aq)

PCl3(l) + 3H2O(l)  H3PO3(aq) +3HCl(aq)

PCl5(l) +4H2O(l)  H3PO4(aq) + 5HCl(aq)

Cl2(aq) + H2O(l) HCl(a) + HOCL(aq)
(with excess water)
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4.6 The Transition Elements:
Learning Goals
 Describe
the characteristic properties of transition metals
 Explain
why transition metals have variable oxidation states
 Explain
why formation and describe with shape of complex
ions
 Explain
why transition metal complex ions are colored
 Describe
some uses of transition metals and their
compounds and catalysts
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The Transition Metals
The ‘d-block’ elements
3d
to 7d elements
Defined
as: an element that forms at least one
stable oxidation state (other than 0) with a
partially filled d subshell
Scandium
metals
and zinc excluded from transition
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Properties of Transition Metals
 All
typical metals; high m.p., b.p. and densities
 Ionization
 Radii
energies increase from Ti to Cu, but only slightly
decrease from Ti to Cu, but only slightly
 Can
exhibit more than one oxidation number in compounds/
complexes
 Form
complex ions
 Usually
 In
form colored complexes/ compounds
compound/ complex form can act as catalysts in many
reactions
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Ionization of Transition Elements
 The
transition elements form positive ions
Common transition metal ions
Element
Ion(s)
Cr
Cr2+ and Cr3+
Mn
Mn2+
Fe
Fe2+ and Fe3+
Co
Co2+
Cu
Cu+ and Cu2+
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Variable Oxidation Numbers
All
transition metals show an oxidation
state of 2+
Because
they have a full 4s subshell and
removal of these electrons would result in
a 2+ charge
Table
4.25 Pg. 167
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Cause of Multiple O.N.
4s
and 3d subshells are close in energy
Removing
electrons from either require
nearly the same amount of energy
Thus, lost electrons depend on several
factors
Lattice enthalpy
Ionization energy
Hydration of enthalpy
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Complex Ions
 Consist
of a central transition metal ion surrounded by
a ligand
 Ligand
negative ions or neutral molecules that use
lone pairs of electrons to bond to a transition metal ion
to form a complex ion


Dative covalent bonds are formed between the ligand and the
transition metal ion
Look and Fe complex ion
 Except
for Ti, al transition metals form an octahedral
complex ion to form [M(H2O)6]2+ in solution
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Oxidation number of a Transition
Metal in a Complex Ion
 May
be determined based on charge of ligand
 Ligands
are either negative or neutral
Neutral Ligands
1- Ligands
H2O
Cl-
NH3
CN-
CO
Br-
 Deduce
the charge of Ni in the complex ion
[Ni(CN)4]2-
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Shapes of Complex Ions
Transition
metal complexes do not
obey the VSEPR theory rules
Six
coordinate (dative covalent bonds)
complexes are nearly always octahedral
Four coordinate complexes are either
tetrahedral or square planar
Table 4.9 P. 169
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Formation of Complex Ions
 May
undergo substitution reactions
 Ex. Water
replaced with another ligand
 [Cu(H2O)6]2+(aq) + 4Cl-(aq)  [CuCl4]2-(aq) + 6H2O(l)
 The
ligands are considered Lewis bases and
according to Le Chatelier’ principle, a strongly
acidic solution means that the concentration of H+
ions is high and the position of equilibrium is
shifted to the left-hand side
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Formation of Colored Complexes
 Complex
ions form colored solutions as a result of their 3d
orbitals being split into 2 groups by ligands
 The
electrons are promoted from the group of lesser energy
to the group with higher energy, causing an emission of
colored light
 Ions
with no electrons in the 3d subshell are colorless
 Similarly, ions
with 10 electrons in the 3d subshell are
colorless because their electrons will not move between the
orbitals
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Catalytic Ablility
 The
elements and their compounds/ complexes are
able to act as catalysts
 Ex.
 Iron
is the catalyst in the Haber reaction to produce
ammonia
 Vanadium (V) oxide can be used in the Contact process of
convert sulfur(IV) oxide to sulfuric(VI) acid
 Manganese(IV) oxide is used in the decomposition of
hydrogen peroxide
 Nickel is used in the hydrogenation of alkenes to form
alkanes
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Catalysts of Haber and Contact
Processes
 Catalysts
are used to speed up a reaction without
themselves being used up
 With
a catalyst, less energy and time are required to
gain product
 The
use of the catalysts in both the Haber and
Contact processes is to run the reaction at a lower
temperature and increase the yield of product
 Improves
the overall cost effectiveness of these
processes in industry