chemistry: trends in the periodic tale

8/10/2019 Chemistry: Trends in the Periodic Tale

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Chemistry for Engineering

TRENDS IN THE PERIODIC TABLE


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LEARNING OUTCOMES

By the end of this session, the student should be able to:

1. Identify the trends of properties of elements in theperiodic table.

2. Arrange different elements according to the size of theiratomic and ionic radii.

3. Arrange elements according to their ionization energies.

4. Arrange elements according to their electronegativity.

5. Determine the strength and type of a bond.

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Trends in

periodictable

Atomic radius

Ionic radius

Ionization energyelectronegativity

Type of bond

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THE ATOMIC RADIUS

The electron cloud surrounding the nucleus

does not have a sharp boundary.

The atomic size is considered as the volume

containing about 90 % of the total electrondensity around the nucleus.

The atomic radius is one-half the distance

between the two nuclei in two adjacent metal

atoms or in a diatomic molecule.

Examples: For Cuoin copper metal, r=0.128 nm

For Cl2, r= 0.099 nm.

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ATOMIC RADII (in pm) OF REPRESENTATIVE ELEMENTS

(same trend for ionic radii).

1. In the same period, the effective nuclear charge increases from left to right the

added valence electron at each step is more strongly attracted by the nucleus than the

one before, so the radii decrease from left to right.

2. In the same group, the size increases with increasing number of electron orbtits

radius increases from top to bottom.


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ATOMION

cation

Removing 1 or more e-from anatom reduces electron-electron

repulsion butthe nuclear charge(p+) remains the same, so the e-

cloud shrinks

the cation is smaller than its

corresponding atom.

Na+(0.095 nm) vs. Nao (0.186 nm)

anion

Gaining 1 or more e-

increases the radius, because

the nuclear charge (p+)

remains the same but the

repulsion resulting from the

additional e

-

enlarges thedomain of the electron cloud.

The anion is larger than its

corresponding atom

Cl-(0.181 nm) vs , Clo(0.099nm)


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Arrange the following sets of atoms and ions in order ofincreasing size using the periodic table

1. Mg, Al, Ca.

2. S, Cl, S2-

3. Fe, Fe2+

, Fe3+

1. Comparison with reference to Mg:Al is to the right to Mg, thus smaller

than Mg. Ca is below Mg, thus larger. The order is Al < Mg < Ca.

2. Comparison with reference to S atom: Cl is to the right to S, thus is

smaller. The S2-

anion is larger than the S atom.

The order is Cl


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IONIZATION ENERGY (IE)

Ionization energy is the minimum

energy (in kJ/mol) required to removean electron from a gaseous atom in its

ground state.

X + energy X++ 1 e-

The 1st ionization energy (endothermic

process = requires E) is the E change for

the removal of the outermost e- from a

gaseous atom to form a +1 ion. The

more difficult it is to remove e-, the

larger the ionization energy.

2ndand 3rdIE (to remove a 2ndand a 3rd

e-) are certainly more difficult due to

the increasing attraction exerted by the

nucleus at every loss of 1 e- after the

other. 819-10-2014

Why IE decreases as we go

down a group?

A greater separation

between the e- and the

nucleus = weaker attraction.


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ELECTRON DISTRIBUTION IN AN ATOM

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nucleus


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THE ELECTRONIC CONFIGURATION OF

ELEMENTS

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(only outer shell level of electrons is represented)

THE CHEMICAL PROPERTIES OF ANY ATOM

ARE DETERMINED BY THE CONFIGURATION

OF THE ATOMS VALENCE ELECTRONS.


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EXCEPTIONS TO TREND OF IE (1stexception)

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between Group 2A and 3A elements in the same period

Ex: Be vs. B and Mg vs. Al.

IE of 3A < 2A in the same

period.

because they all have a single e-

in the outermost p subshell ( ns2

np1 ), which is well shielded by

the inner e-and the ns2e-.

less energy is needed to

remove a single p e- than to

remove an s e- from the sameenergy level.


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EXCEPTIONS TO TREND OF IE (2ndexception)

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between Groups 5A and 6A elements in the same period

ex: N vs. O and P vs. S

IE of 6A < 5A in the same period.

In 5A ( ns2 np3 ), the p e- are in 3

separate orbitals while in 6A ( ns2 np4),

the additional e- is paired with 1 of the 3p e- The proximity of two e- in the

same orbital leads to high electrostatic

repulsion, which makes it easier to

ionize an atom of the Group 6A element,

even though the nuclear charge hasincreased by one unit.


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Complete:

1. Ionization energy (IE) a.across the periodic tablefrom left to right and b.. moving down the periodic

table.

2. Comparing the trends of ionic radii and ionization energy it is

clear that there is an

c

. correlation between them.3. The d.. the atom, the more tightly its electrons are

held to the positively charged nucleus and the more difficult

they are to remove thus the e.. the ionization energy.

4. In a

f

atom (ex: group I, period 5), the electron isrelatively far from the nucleus, so the lower its g..

to remove it from the atom.

Answers:

a) increases, b) decreases, c) inverse, d) smaller, e) larger, f) large, g) ionization.

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Test yourself


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Which atom with following electron configurations has the

largest first ionization energy, explain your choice.

1s22s22p6 1s22s22p63s1 1s22s22p63s2

The atom with the largest value of IE is 1s22s22p6 (this is a Ne),because it is found at the right end of period 2, and the IEincreases from left to right in the same period of the periodictable.

As for the other 2 configurations including 3s electrons they willbe of lower IE because it is easier to loose e- from an incompleteshell.

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Test yourself


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ELECTRONEGATIVITY (ELECTRON AFFINITY)

Electron affinity is the energy change that occurs when an e- is accepted by

an atom in the gaseous state to form an anion.X(g)+ e

- X-(g)Electronegativity increases from left to right in the periodic table and vary

little within a given group. The halogens (Group 7A) have the highest

electronegativity.

In a covalent bond, the more electronegative the atom, the more it attracts

the shared e- towards it resulting in unfair sharing polarizationof the bond

with the most electronegative acquiring apartial negative charge -- vs. a +

on the least electronegative.

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IONIC BOND by full e- transfer

from the least to the most

electronegative (i.e. large

difference in electronegativity)

COVALENT BOND between

identical atoms, i.e. no

difference in electronegativity

(NON POLARIZED)COVALENT BOND (POLARIZED,

i.e. some difference inelectronegativity)


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Order the following bonds according to bond polarity

H-H, S-H, Cl-H, O-H, and F-H

(2.1) (2.1) (2.5)(2.1) (3.0)(2.1) (3.5)(2.1) (4.0)(2.1)

0 0.4 0.9 1.4 1.9

Thus the order is

H-H < S-H < Cl-H < O-H < F-HNon polar covalent bondPolar covalent bond

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Test yourself


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REFERENCES

1. Chemistry, 10thed., Raymond Chang, ISBN 978-0-

07-017264-7, McGraw Hill. Chapter 8.

2. Lecture 2 by Prof. Rasha Elnashar, GUC, WS 2013.

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chemistry: trends in the periodic tale

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