Transcript Document

Chemistry
Classification of elements-II
Session Objectives
Session Opener
Perspective
Understanding of basic properties like atomic size ionisation energy,
electron affinity and electronegativity will help in understanding
general trends in s and p block elements.
Session Objectives
Causes of periodicity
Atomic size,ionic radii,trend in groups and periods
Ionisation energy.
Electron affinity
Electronegativity
Valency and its trend
Anomalous behaviour of first element of group
Diagonal relationship
Causes of periodicity
Repetition of similar valence shell
configuration after regular interval.
Element
Atomic no.
Electronic
configuration
Li
3
1s2,2s1
Na
11
1s2,2s2,2p6,3s1
K
19
1s2,2s2,2p6,3s2,3p6,4s1
Rb
37
1s2,2s2,2p6,3s2,3p6,4s2,
3d10,4p6,5s1
Atomic size
Covalent and van der waal’s radius:
c
a
b
Covalent radius 
Distance between a and b
2
van der Waal's radius=
Distance between b and c
2
Trends of atomic size
Ask your self
Which element has highest covalent radius?
Solution:
Cs
Size of cation
Size of cation
Fe
Protons
26
Fe2+
26
Electrons
26
24
Fe3+
26
23
Size of anion
Protons
Cl
17
17
Electrons
17
18
Cl–
Isoelectronic ions
rC4  rN3  rO2 rF
no.of electrons 10
nuclear charge  6
rNa 
no.of electrons 10
nuclear ch arge  11
10
7
10
8
rMg2 
10
 12
10
9
rAl3
10
 13
• Note for isoelectronic series Na+, Mg2+, Al3+, N3-, O2-, F-,
• N3-> O2-> F-> Na+> Mg2+> Al3+
• Most positive ion the smallest, most negative the largest
Ionisation energy
•Minimum energy required to remove an electron from a groundstate, gaseous atom
•Energy always positive (requires energy)
•Measures how tightly the e- is held in atom (think size also)
•Energy associated with this reaction
IE
-e-
Isolated gaseous atom
Successive ionisation energies
M
–e
IE1
IE3 > IE2 > IE1
M+
–e
IE2
M2+
–e
IE3
M3+
Factors affecting values of
ionisation energy
1. Size
1
Ionisation energy 
Atomic size
Atomic size
Ionisation energy
KJ/mole
Li
1.23
Be
0.89
520
899
Factors affecting values of
ionisation energy
2. Effective nuclear charge
Is net nuclear attraction
towards the valence shell
electrons .
Ionization energy  Effective nuclear charge
Effective nuclear
charge
Ionisation energy
KJ/mole
Li
+3
Be
+4
520
899
Factors affecting values
of ionisation energy
3. Screening effect or
shielding effect
Combined effect of attractive
and repulsive forces between
electron and proton.
Ionisation energy
No. of inner shells
Ionisation energy
KJ/mole

1
Number of inner shells
Li
1
Na
2
520
496
Factors affecting values
of ionisation energy
4. Penetrating power of orbitals
s>p>d>f
5. Complete octet
Elements having ns2,np6 configuration have extremely
high ionisation energy.
Factors affecting values
of ionisation energy
6. Stable Configuration
Ionisation energy
Configuration
Ionisation energy
KJ/mole

1
Stability of configuration
Be
2s2
B
2s22p1
899
801
Trend of ionisation energy
in period and groups
Exceptions
(i) IE
II A
ns2
>
(ii) IE
>
VA
ns2,np3
IE
III A
ns2,np1
IE
VI A
ns2,np4
(iii) Ionisation energy of Al > Ga
Absence of d electrons in Al
Variation of I1 with Z
In a group (column), I1 decreases with increasing Z.
valence e’s with larger n are further from the nucleus, less tightly held
Variation of I1 with Z
Across a period (row), I1 mainly increases with increasing Z.
Because of increasing nuclear charge (Z).
Illustrative example
First ionisation energy of Be is more than Li but the second
ionisation energy of Be is less than Li. Why?
Solution:
IE1
Li – e –
2s1
+
Li – e
1s2
–
IE2
IE1Be > IE1 Li
IE2 Li > IE2 Be
Be – e
2s2
+
Li
2s0
2+
Li
1s1
+
Be – e
2s1
–
–
IE1
IE2
Be +
2s1
Be 2+
2s0
 Be has stable (2s2) configuration.
 Li acquires stable configuration when
it loses one electron.
Electron affinity
•Electron affinity is energy change when an e- adds to a
gas-phase, ground-state atom
•Positive EA means that energy is released, e- addition is
favorable and anion is stable!
•First EA’s mostly positive, a few negative
Successive affinities
1
A(g)  e 
A   g  energy released
EA
2
A  (g)  e  energy supplied 
 A 2   g
EA
e–
EA
Isolated gaseous atom
Factors affecting electron
affinity
Electron affinity
11 nuclear
Effective
Penetrating

charge
power
Stable
configuration
atomic
size
Screening
effect
s>p>d>f
Li
Be
Li
Na
Li
Be
Na
Config.
2s1
2s2
Inner
shells
1 66
2
EA
kJ/mol
-57 1.23
E.N.C
At.
size
1.23
0.89
1.57
EA
-57
-21
EA kJ/mol
kJ/mol -57
-57
-21
66

Trends in electron affinities
•Decrease down a group and increase across a period in general but
there are not clear cut trends as with atomic size and I.E.
•Nonmetals are more likely to accept e-s than metals. VIIA’s like to
accept e-s the most.
Exceptions
1.
2.
3.
EA of Cl > EA of F
Group II A have almost zero
electron affinities due to
stable ns2 configuration of
valence shell.
Group V A have very low
values of electron affinities
due to ns2,np3 configuration
of valence shell.
Do you know?
More the value of electron affinity
greater is the oxidising power.
Electronegativity

It is the relative tendency to attract
shared pair of electrons towards itself.
..
H
H
H
.
xxx
Cl x
xxx
Factors effecting electronegativity
1. Electronegativity 
1
Atomic size
2. Electronegativity is higher for nearly filled
configuration e.g. O(3.5) and F(4.0).
Periodic variation
(i) In period
Li
Valence shell
configuration
Electronegativity
2s1
Be
B
C
2s2 2s2,2p1 2s2,2p2
1.0 1.5
2.0
2.5
(ii) In groups-decreases down the group
N
O
F
2s2,2P3 2s2 ,2P4 2s2,2P5
3.0
3.5
4.0
Pauling scale of electronegativity
Pauling's Electronegativity
EAB = 1/2(EAA + EBB ) +  AB
 AB = 96.49(X A - XB )2 or |X A - XB| =0.102 AB
where X A and XB are constants
characteristic of the atoms A and B.
Mulliken’s scale of electronegativity
Electronegativity represents an average of the binding energy of the
outermost electrons over a range of valence-state ionizations (A+  A
A- in A-B)
In other words, the average of the ionization energy and
the electron affinity.
IE  EA
2
Relation with Pauling's electronegativity
XM 
XP = 0.336(XM- 0.615)
Do you know
1. Smaller atoms have more
electronegativities
2. F is most electronegative element.
3. Decreasing order of electro negativity
F > O > Cl  N > Br > C > I > H
Valency
•
The valency of an element is
decided by number of electrons present in
outermost shell.
•
All the elements of a group have same
valency.
E.g.- All the group I elements show 1
valency.
Li
3
1s2,2s1
Na
11
1s2,2s2,2p6,3s1
K
19
1s2,2s2,2p6,3s2,3p6,4s1
Rb
37
1s2,2s2,2p6,3s2,3p6,4s2,
3d10,4p6,5s1
Valency
•
Valency of s block elements is
same as their group number.
Examples:
Ca is member of group 2
 its valency is 2
•
K is member of group 1
its valency is 1
Valency
•Valency of p block elements is
equal to number of electrons in
valence shell.
e.g.- Al has 3 electrons in valence shell.
Therefore, its valency is 3.
Or
8-number of electrons in valence shell.
e.g- valency of oxygen is 8 – 6 =2
Valency
•
Valency of d and f block elements
variable.
Iron shows the valence 2 and 3
Valency
Valency in period
Element
Li
Be
B
C
N
O
F
Ne
Number of
electrons in
valence shell
1
2
3
4
5
6
7
8
Valency
1
2
3
4
3
2
1
0
Anomalous behaviour of first
element of group
Causes:
•
Smallest size in group.
•
Highest value of ionisation energy in the
group.
•
Absence of vacant d orbitals.
Examples of anomalous behaviour
of first element of group
•
Carbon forms multiple bonds but rest of the members form
only single bonds.
•
Nitrogen does not form NCl5 but phosphorous forms PCl5.
Diagonal relationship
2nd period
Li
Be
B
3rd period
Na
Mg
Al
C
Si
Causes of diagonal relationship
•
Similarity in size.
•
Similarity in ionisation energy.
•
Similarity in electron affinity.
Class Test
Class Exercise - 1
Which has the smallest size?
(a) Na+ (b) Mg2+
(c) Al3+
(d) P5+
Solution:
Size of isoelectronic species decreases with increase in
nuclear charge.
Hence, answer is (d).
Class Exercise - 2
If the first ionization energy of helium is 2.37 kJ/mole,
the first ionization energy of neon in kJ/mole is:
(a) 0.11
(c) 2.68
(b) 2.37
(d) 2.08
Solution:
Ionization energy decreases down the group.
Hence, answer is (d).
Class Exercise - 3
The relative electronegativities of F, O, N, C and H are
(a) F > C > H > N > O
(c) F > N > C > H > O
(b) F > O > N > C > H
(d) F > N > H > C > O
Solution:
The correct order of electronegativities is
F  O  N  C H
4.0
3.5
3.0
Hence, answer is (b).
2.5
2.1
Class Exercise - 4
Which of the following ions has
smallest ionic radius?
(a) Li+
(c) H–
(b) Be2+
(d) All have equal radii
Solution:
More the nuclear charge on cation smaller
will be the size.
Hence, answer is (b).
Class Exercise - 5
Which one of the following is correct order of ionic size?
(a) Ca2+ > K1+ > Cl- > S2(b) S2- > Cl- > K+ > Ca2+
(c) Ca2+ > Cl- > K1+ > S2(d) S2- > Ca2+ > Cl- > K+
Solution:
Size of iso electronic species decreases with increase
in nuclear charge, more interelectronic repulsion in
S and Cl is the reason of their increased size.
Hence, answer is (b).
Class Exercise - 6
The electron affinities of N,O, S and Cl are
(a) N < O < S < Cl
(b) O < N < Cl < S
(c) O = Cl < N = S
(d) O < S < Cl < N
Solution:
The correct order of electron affirmities is
N < O < S < Cl
Hence, answer is (a).
Class Exercise - 7
Which ion has the largest radius?
(a) Ca2+
(c) P3–
(b) F–
(d) Mg2+
Solution:
Anions are larger in size than cation.
Hence, answer is (c).
Class Exercise - 8
In which of the following pairs there
is an exception in the periodic trend
for the ionization energy?
(a) Fe – Ni
(c) Be – B
(b) C – N
(d) O – F
Solution:
Since Be has stable configuration (2s2) as compared to B
(2s2, 2p1).
Hence, answer is (c).
Class Exercise - 9
The first three successive ionisation
energies of an element Z are 520,
7297 and 9810 kJ mol–1 respectively.
The element Z belongs to
(a) group 2
(c) group 15
(b) group 1
(d) group 16
Solution:
Since the difference in first and second ionisation
energies is very high, it belongs to group 1.
Hence, answer is (b).
Class Exercise - 10
Atomic number of element is 108.
This element is placed in ____ block
of periodic table.
(a) s
(b) p
(c) d
(d) f
Solution:
Atomic number — 108
Configuration — 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10,
4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6, 7s2, 5f14, 6d6
Hence, answer is (c).
Class Exercise - 11
Which of the following values in
electron volt per atom represent
the first ionisation energies of
oxygen and nitrogen atom respectively
(a) 14.6, 13.6
(c) 13.6, 13.6
(b) 13.6, 14.6
(d) 14.6, 14.6
Solution:
First ionisation energy of nitrogen is more than oxygen
because of stable (2s2, 2p3) configuration of nitrogen.
Hence, answer is (d).
Class Exercise - 12
The electronic configuration of an
element is (n – 1)d1, ns2 where
n = 4. The element belongs to ____
period of periodic table.
(a) 3
(b) 2
(c) 5
(d) 4
Solution:
The period number is same as maximum value of principal
quantum number.
Hence, answer is (d).
Class Exercise - 13
An element having
atomic number 25 belongs to
(a) s
(b) p
(c) f
(d) d
Solution:
Electronic configuration — 1s2, 2s2, 2p6, 3s2, 3p6,
4s2, 3d5.
Therefore it is d block element.
Hence, answer is (d).
Class Exercise - 14
In its structure an element has 4
shells. Therefore it belongs to
(a) 3rd period (b) 4th period
(c) 2nd period (d) None of these
Solution:
Group 2 is present in s block and for them group
number = number of electrons in valence shell.
Hence, answer is (b).
Class Exercise - 15
A, B, C and D have following electronic
configurations
A : 1s2, 2s2, 2p1
B : 1s2, 2s2, 2p6, 3s2, 3p1
C : 1s2, 2s2, 2p6, 3s2, 3p4
D : 1s2, 12s2, 2p6, 3s2, 3p6, 4s1
Find out the periods of A, B, C and D.
Solution:
Period number is equal to maximum
value of principal quantum number.
Hence, answer is (b).
Element
Element
Element
Element
A — 2nd period
B — 3rd period
C — 3rd period
D — 4th period
Thank you