Z - Wits Structural Chemistry

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Z - Wits Structural Chemistry

Chapter 7

Periodic Properties of the

Elements


Development of the Periodic Table

• Currently 115 elements known.

• The majority of the elements were discovered

between 1735 and 1843.

• Developments in technology and advances in

chemistry allowed for further discovery of

elements.

• Challenge to chemists: How to organize 115

different elements in a meaningful way


The Periodic Table

• Dmitri Ivanovich Mendeleev 1834 - 1907

• Arranging elements in ascending

order of atomic number produces a

periodic pattern.

• Arrange elements to reflect the

trends in chemical and physical

properties.

• Lothar Meyer

First attempt (Mendeleev and Meyer) arranged

the elements in order of increasing atomic

weight.

Certain elements were

missing from this scheme.


Gallium (Ga) and Germanium (Ge) were not known!!!

In 1871, Mendeleev noted that As properly belonged underneath

P and not Si, which left a missing element underneath Si.

He predicted a number of properties for this element. In 1886

Ge was discovered. The properties of Ge match Mendeleev’s

predictions well.

Property

Atomic weight

Density (g/cm 3 )

Specific heat (J/gK)

Melting point (°C)

Color

Formula of oxide

Density of oxide (g/cm3)

Formula of chloride

Boiling point of chloride (°C)

Mendeleev’s prediction (1871)

72

72.59

5.5

5.35

0.305

0.309

Properties of Ge

(discovered 1886)

High

947

Dark Gray Grayish white

XO 2

GeO 2

4.7

4.70

XCl 4

GeCl 4

A little under 100 84


Modern periodic table: elements arranged in

order of increasing atomic number.


Effective Nuclear Charge

Recall from Chapter 6:

Effective nuclear charge is the charge experienced

by an electron in a many-electron atom.

The effective nuclear charge is not the same as the

charge on the nucleus because of the shielding

effect of the inner electrons.

Electrons are attracted to the nucleus, but repelled

by the electrons that screen it from the nuclear

charge.


The nuclear charge experienced by an electron depends

on its distance from the nucleus and the number of core

electrons.

Electron considered to be in a net electric field consisting

of the effect of the nucleus and the other electrons.

As the average number of screening electrons (S)

increases, the effective nuclear charge (Z eff ) decreases.

As the distance from the nucleus increases, S increases

and Z eff decreases.

Z eff < Z

Z eff accounts for

electron-electron

repulsions

- screening constant


Li

e -

Effective Nuclear Charge

e - For 3 Li: 1s 2 2s

Core Electrons : 1s 2

Z eff = Z – S

= 3 – 2

= +1

e -

e -

Be

e -

e -

e - For 4 Be: 1s 2 2s 2

Core Electrons : 1s 2

Z eff = Z – S

= 4 – 2

= +2

Electrons in the same

shell hardly screen

each other from the

nucleus.


For Mg: [Ne] 3s 2

Electrons in the same shell

hardly screen each other from

the nucleus. i.e., hardly affect S.

The S value is usually close to the number of core electrons.

The chemistry of the elements is dominated by the outer

(valence) electrons


For Na: [Ne] 3s

Calculated Z eff = +2.5

The 3s electron has a definite

probability of being close to the

nucleus. This decreases S and

consequently Z eff is larger.

“Slater’s Rules”


3p

-

For 15 P: [Ne] 3s 2 3p 3

Core Electrons : 1s 2 2s 2 2p 6

3p

-

-

3s

10-

15+

3s

-

-

3p

Z eff = Z – S

= 15 – 10

= +5

Calculated Z eff

(3s) = +5.6

Calculated Z eff (3p) = +4.9

The 3s electrons can penetrate the

Ne electron core.

S(3s) decreases and Z eff is larger.

Notice the 3s electrons are

attracted more strongly by the

nucleus than the 3p electrons.


Give it some thought P 265

Which would you expect to experience a greater effective

nuclear charge, a 2p electron of a Ne atom or a 3s electron

of a Na atom

Na: [Ne] 3s

Z eff = Z – S

= 11 – 10 = +1

2p

-

2p

-

-

2s

-

2p

2-

10+

-

2p

2s

-

-

2p

-

2p

Ne : 1s 2 2s 2 2p 6

Ne : [He] 2s 2 2p 6

Core Electrons : 1s 2

Z eff = Z – S

= 10 – 2 = +8

NB: Easier to remove the 3s electron of

Na than to remove a 2p electron of Ne.


Trends in Effective Nuclear Charge

Changes in effective nuclear charge very significant across a

period.

Z eff increases with atomic number since the number of core

electrons remains the same but the actual nuclear charge

increases.

Z eff (Na) = +1 Z eff (Mg) = +2 Z eff (Al) = +3 Z eff (Si) = +4 Z eff (P) = +5

Changes in effective nuclear charge are less significant

going down a group.

Smaller increases occur since larger electron cores are

less effective at screening outer (Valence electrons) from

the nuclear charge.

Z eff (Li) = +1 Z eff (Na) = +1 Z eff (K) = +1

Z eff (Rb) = +1


Additional Example

Suppose inner electrons completely effective at screening

valence electrons in sulfur.

(a) What would be values of S and Z eff

(b) From calculations find that S for a 3p e – in S is 10.52.

Why is this different to the answer in (a)


Consider a simple diatomic molecule.

• The atomic radius or van

der Waals radius is the

non-bonding radius of the

atom.

• The distance between the

two nuclei d is called the

bond distance.

• If the two atoms which

make up the molecule are

the same, then half the

bond distance is called the

bonding or covalent

radius of the atom.

Sizes of Atoms and Ions


Periodic Trends in Atomic Radii

Atomic size varies consistently through the periodic table.

As we move down a group, the atoms become larger.

As we move across a period, atoms become smaller.

Two factors to consider:

principal quantum number, n

effective nuclear charge, Z eff


As the principal quantum number n increases (we move

down a group and more shells are being filled), the

distance of the outermost electron from the nucleus

becomes larger.

The atomic radius increases.

As we move across the periodic table, the number of core

electrons remains constant but the nuclear charge increases.

There is an increased attraction between the nucleus and

the outermost electrons.

The atomic radius decreases


Trends in the Sizes of Ions

Ionic size (ionic radius) is the distance between ions in an

ionic compound.

Ionic radius size depends on nuclear charge, number of

electrons, and orbitals that contain the valence electrons.

Cations are smaller than the parent ion because there is a

decrease in electron-electron repulsion as an electron is lost

from the atom.

Anions gain electrons; electron-electron repulsion increases

and they are larger than the parent ion.


+ r + r + r +


• For ions of the same charge, ion size increases

down a group.

• All the members of an isoelectronic series have

the same number of electrons. As nuclear charge

increases in an isoelectronic series the ions

become smaller:

O 2- > F - > Na + > Mg 2+ > Al 3+

same number of electrons but increasing nuclear charge


-

2s

-

2p

2 -

2p

-

NB: Ionic radii decrease as the

effective nuclear charge increases

for isoelectronic series: Consider

the 2p electrons for the isoelectronic

series: O 2- > F - > Na + > Mg 2+ > Al 3+

2p

-

2p

-

12 11 89 13 +

-

2s

2p

-

All ions above have 10 electrons.

Electron configuration:

[He] 2s 2 2p 6

Core Electrons : 1s 2

-

2p

NB: Increase in Z eff causes ion to become

more compact. i.e., smaller ionic radius

Z eff (O 2- ) = Z – S

= 8 – 2 = +6

Z eff (F - ) = Z – S

= 9 – 2 = +7

Z eff (Na + ) = Z – S

= 11 – 2 = +9

Z eff (Mg 2+ ) = Z – S

= 12 – 2 = +10

Z eff (Al 3+ ) = Z – S

= 13 – 2 = +11


Example

1. True or False (and why)

a) Cl - is larger than Cl

b) S 2- is larger than O 2-

c) Ca 2+ is larger than K +

2. In this representation of

a chemical reaction

between a metal and a

non-metal, which sphere

represents the metal and

which the non-metal

What type of reaction is

this


Ionisation Energy

Ionisation energy – the minimum energy required to remove an electron

from the ground state of a gaseous atom or ion.

The first ionisation energy, I 1 , is the amount of energy required to remove an

electron from a gaseous atom:

Na (g) → Na + (g) + e-

The second ionisation energy, I 2 , is the energy required to remove an electron

from a +1 gaseous ion:

Na + (g) → Na 2+ (g) + e -

etc for I 3 , I 4 , …

The larger the ionisation energy, the more difficult it is to remove the

electron from the atom.

I 1 < I 2 < I n as additional electrons are removed from cations of increasing

positive charge.


Variations in Successive Ionisation Energies

There is a sharp increase in ionisation energy when

a core electron is removed.

11

Na: [Ne] 3s 1 loses 1e - readily

Core electrons closer to nucleus.

12

Mg: [Ne] 3s 2 loses 2e - readily Greater effective nuclear charge.

13

Al: [Ne] 3s 2 3p 1 loses 3e - readily

15

P: [Ne] 3s 2 3p 3


Example

Which will have the greater 3 rd ionisation energy, Ca or S

For 20 Ca: [Ar] 3s 2 For 16 S: [Ne] 3s 2 3p 4

Can lose 2 valence e -

3 rd e - is a core electron.

Can lose 6 valence e -

3 rd e - is still a valence electron.

∴ 3 rd Ionisation energy of Ca is higher than the 3 rd

ionisation energy of S

NB: All elements show a large increase in their ionisation

energy when electrons are removed from their noble gas

core.


Periodic Trends in Ionization Energies

• Ionisation energy decreases down a group. This

means that the outermost electron is more readily

removed as we go down a group.

As the atom gets bigger, it becomes easier to remove an

electron from the highest energy orbital.

(Nuclear charge remains the same but atomic radius increases)

• Ionisation energy generally increases across a

period.

As we move across a period, Z eff increases. Therefore, it

becomes more difficult to remove an electron.

(Nuclear charge increases and atomic radius decreases)


Figure 7.10, p. 272

Irregularities

N

Be

O

Mg

P

S

B

Al

Representative elements (s- and p-block) show greater

variation in I 1 than do the transition metals


Irregularities

Decrease in ionisation energy from Be to B and Mg to Al.

Be: 1s 2 2s 2

B: 1s 2 2s 2 2p

Mg: 1s 2 2s 2 2p 6 3s 2

Al: 1s 2 2s 2 2p 6 3s 2 3p

Additional electron occupies a previously vacant

p-subshell

that is of higher energy.

The s electrons are more effective at shielding the p

electrons than each other. Therefore, configuration s 2 p 0

becomes more favorable.


N: [Ne] 2s 2 2p 3

Irregularities

Decrease in ionisation energy from N to O and P to S.

O: [Ne] 2s 2 2p 4 P: [Ne] 3s 2 3p 3

S: [Ne] 3s 2 3p 4

Additional electron results in electron pairing in

the p 4 configuration.

When a second electron is placed in a p orbital, as

occurs in s 2 p 4 , the electron-electron repulsion increases.

When this electron is removed, resulting s 2 p 3 is more

stable than starting s 2 p 4 configuration.


Electron Affinities

Electron affinity is the energy change when a gaseous

atom gains an electron to form a gaseous ion:

Cl (g) + e - → Cl - (g) ∆E = -349kJ/mol

Electron affinity can either be exothermic (as the above

example) or endothermic. The more negative the electron

affinity the greater the attraction between the added electron

and the nucleus.


Noble Gases: Highly Endothermic. +ve electron affinities

Ne (g) + e - → Ne - (g)

∆E >0

Ne : 1s 2 2s 2 2p 6

Ne - : 1s 2 2s 2 2p 6 3s

In Ne - electron occupies the

higher energy 3s subshell.

Unstable and does not form…

Z eff = Z – S

= 10 – 10

= 0


Halogens: Strongly exothermic. -ve electron affinities

F (g) + e - → F - (g) ∆E < 0

F : 1s 2 2s 2 2p 5

F - : 1s 2 2s 2 2p 6

Z eff = Z – S

= 9 – 2

= +7

Incoming electron experiences high Z eff as it completes the p

subshell; reflects high tendency of halogens to form –1 ions.

Halogens all acquire the stable noble gas configuration..


GROUP 2: Endothermic or only slightly exothermic

Incoming electron going into higher energy p

subshell; experiences relatively low Z eff


GROUP 5: Endothermic (N) or only slightly exothermic

Incoming electron going into an half filled p 3 subshell

resulting electron-electron repulsions that lower the

electron affinity for this electron.


End of Chapter 7:

Periodic

Properties of the

Elements

BLB Sections

7.6 - 7.8 to be

read outside

of class.

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