chemistry
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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Chemistry
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Classification of elements-II
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Session Objectives
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Session Opener
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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.
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Session ObjectivesCauses 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
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Causes of periodicityRepetition of similar valence shell configuration after regular interval.
1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p6,5s1
37Rb
1s2,2s2,2p6,3s2,3p6,4s119K
1s2,2s2,2p6,3s111Na1s2,2s13Li
Electronic configuration
Atomic no.Element
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Atomic sizeCovalent and van der waal’s radius:
a bc
Distance between a and bCovalent radius 2
Distance between b and cvan der Waal's radius= 2
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Trends of atomic size
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Ask your self
Which element has highest covalent radius?
Cs
Solution:
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Size of cation
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Size of cation
Electrons
ProtonsFe
26
26
Fe2+
2624
Fe3+
2326
Cl–
1718
Cl
1717Protons
Electrons
Size of anion
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Isoelectronic ions4 3 2C N O rFr r r
no.of electrons 10 10 10 10nuclear charge 6 7 8 9
2 3Na Mg Alr r rno.of electrons 10 10 10nuclear charge 11 12 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
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Ionisation energy
+ h
Isolated gaseous atom
IE-e-
•Minimum energy required to remove an electron from a ground-state, gaseous atom•Energy always positive (requires energy)•Measures how tightly the e- is held in atom (think size also)•Energy associated with this reaction
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Successive ionisation energies
IE3 > IE2 > IE1
IE1
M – e M+– e M2+ – e M3+
IE2IE3
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Factors affecting values of ionisation energy
1. Size
Ionisation energy 1
Atomic size
Atomic size
Ionisation energy KJ/mole
Li1.23
520
Be0.89
899
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Factors affecting values of ionisation energy
Ionization energy Effective nuclear charge
2. Effective nuclear charge Is net nuclear attraction towards the valence shell electrons .
Effective nuclearcharge
Ionisation energy KJ/mole
Li+3
520
Be+4
899
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Factors affecting values of ionisation energy
3. Screening effect orshielding effect
Combined effect of attractive and repulsive forces between electron and proton.
Ionisation energy 1 Number of inner shells
No. of inner shells
Ionisation energy KJ/mole
Li1
520
Na2
496
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Factors affecting values of ionisation energy
4. Penetrating power of orbitals
s>p>d>f
5. Complete octet
Elements having ns2,np6 configuration have extremelyhigh ionisation energy.
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6. Stable Configuration
Ionisation energy1
Stability of configuration
Factors affecting values of ionisation energy
Configuration
Ionisation energy KJ/mole
Be2s2
899
B2s22p1
801
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Trend of ionisation energy in period and groupsExceptions
(i) IE > IE II A III A ns2 ns2,np1
(ii) IE > IE V A VI A ns2,np3 ns2,np4
(iii) Ionisation energy of Al > Ga
Absence of d electrons in Al
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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
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Across a period (row), I1 mainly increases with increasing Z. Because of increasing nuclear charge (Z).
Variation of I1 with Z
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Illustrative exampleFirst ionisation energy of Be is more than Li but the second ionisation energy of Be is less than Li. Why?
Solution:
Li Li Be Bee –– IE1
2s1 2s0
+ e –– IE1
2s2 2s1
+
Li Li Be Bee –– IE2
1s2 1s1
2++ e –– IE2
2s1 2s0
2++
IE1Be > IE1 Li Be has stable (2s2) configuration.
IE2 Li > IE2 Be Li acquires stable configuration when it loses one electron.
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Electron affinity
Successive affinities
e–
Isolated gaseous atom
EA
•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
1
2
EA
EA 2
A(g) e A g energy releasedA (g) e energy supplied A g
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Effective nuclear
charge Li BeE.N.C 1.23 0.89EA kJ/mol -57 66
Factors affecting electron affinity
Electron affinity
Li NaInner shells 1 2EA kJ/mol -57 -21
1
Screening effect
1atomic size
Li NaAt. size 1.23 1.57EA kJ/mol -57 -21
Penetrating powers>p>d>f
Li BeConfig. 2s1 2s2
EA kJ/mol -57 66
1
Stable configuration
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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. EA of Cl > EA of F2. Group II A have almost zero
electron affinities due to stable ns2 configuration of valence shell.
3. Group V A have very low values of electron affinitiesdue to ns2,np3
configuration of valence shell.
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Do you know?
More the value of electron affinitygreater is the oxidising power.
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ElectronegativityIt is the relative tendency to attract shared pair of electrons towards itself.
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).
H H.. H Cl.x x x
x x xx
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Periodic variation(i) In period
Li Be B C N O FValence shellconfiguration 2s1 2s2 2s2,2p1 2s2,2p2 2s2,2P3 2s2 ,2P4 2s2,2P5
Electronegativity 1.0 1.5 2.0 2.5 3.0 3.5 4.0
(ii) In groups-decreases down the group
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Pauling scale of electronegativity
AB AA BB AB2
AB A B A B AB
A B
Pauling's ElectronegativityE = 1/2(E + E ) +
= 96.49(X - X ) or |X - X | =0.102where X and X are constantscharacteristic of the atoms A and B.
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Mulliken’s scale of electronegativityElectronegativity represents an average of the binding energy of theoutermost 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.
M
P M
IE EAX 2Relation with Pauling's electronegativityX = 0.336(X - 0.615)
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Do you know1. 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
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• 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.
Valency
1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p6,5s1
37Rb
1s2,2s2,2p6,3s2,3p6,4s119K
1s2,2s2,2p6,3s111Na1s2,2s13Li
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• Valency of s block elements issame 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
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Valency
•Valency of p block elements isequal to number of electrons invalence shell.
e.g.- Al has 3 electrons in valence shell.Therefore, its valency is 3. Or8-number of electrons in valence shell.e.g- valency of oxygen is 8 – 6 =2
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Valency
• Valency of d and f block elements variable.
Iron shows the valence 2 and 3
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Valency
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Valency in period
01234321Valency
87654321Number of electrons in valence shell
NeFONCBBeLiElement
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Anomalous behaviour of first element of group
• Smallest size in group.
• Highest value of ionisation energy in the group.
• Absence of vacant d orbitals.
Causes:
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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.
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Diagonal relationship
2nd period
3rd period
CLi
Na Mg
Be
Al Si
B
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Causes of diagonal relationship
• Similarity in size.
• Similarity in ionisation energy.
• Similarity in electron affinity.
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Class Test
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Class Exercise - 1Which has the smallest size?
(a) Na+ (b) Mg2+
(c) Al3+ (d) P5+
Solution:
Size of isoelectronic species decreases with increase innuclear charge.
Hence, answer is (d).
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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 (b) 2.37(c) 2.68 (d) 2.08
Solution:
Hence, answer is (d).
Ionization energy decreases down the group.
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Class Exercise - 3
The relative electronegativities of F, O, N, C and H are
(a) F > C > H > N > O (b) F > O > N > C > H(c) F > N > C > H > O (d) F > N > H > C > O
Solution:
Hence, answer is (b).
The correct order of electronegativities is
4.0 3.0 2.13.5 2.5F O N C H
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Class Exercise - 4
Which of the following ions hassmallest ionic radius?
(a) Li+ (b) Be2+
(c) H– (d) All have equal radii
Solution:
Hence, answer is (b).
More the nuclear charge on cation smallerwill be the size.
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Class Exercise - 5Which 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:
Hence, answer is (b).
Size of iso electronic species decreases with increasein nuclear charge, more interelectronic repulsion inS and Cl is the reason of their increased size.
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Class Exercise - 6The 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:
Hence, answer is (a).
The correct order of electron affirmities isN < O < S < Cl
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Class Exercise - 7
Which ion has the largest radius?
(a) Ca2+ (b) F–
(c) P3– (d) Mg2+
Solution:
Hence, answer is (c).
Anions are larger in size than cation.
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Class Exercise - 8In which of the following pairs thereis an exception in the periodic trendfor the ionization energy?
(a) Fe – Ni (b) C – N(c) Be – B (d) O – F
Solution:
Hence, answer is (c).
Since Be has stable configuration (2s2) as compared to B(2s2, 2p1).
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Class Exercise - 9The first three successive ionisationenergies of an element Z are 520,7297 and 9810 kJ mol–1 respectively.The element Z belongs to
(a) group 2 (b) group 1(c) group 15 (d) group 16
Solution:
Hence, answer is (b).
Since the difference in first and second ionisationenergies is very high, it belongs to group 1.
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Class Exercise - 10
Atomic number of element is 108.This element is placed in ____ blockof periodic table.
(a) s (b) p (c) d (d) f
Solution:
Hence, answer is (c).
Atomic number — 108Configuration — 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10,4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6, 7s2, 5f14, 6d6
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Class Exercise - 11Which of the following values inelectron volt per atom representthe first ionisation energies of oxygen and nitrogen atom respectively
(a) 14.6, 13.6 (b) 13.6, 14.6(c) 13.6, 13.6 (d) 14.6, 14.6
Solution:
Hence, answer is (d).
First ionisation energy of nitrogen is more than oxygenbecause of stable (2s2, 2p3) configuration of nitrogen.
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Class Exercise - 12The electronic configuration of anelement is (n – 1)d1, ns2 wheren = 4. The element belongs to ____period of periodic table.
(a) 3 (b) 2 (c) 5 (d) 4
Solution:
Hence, answer is (d).
The period number is same as maximum value of principalquantum number.
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Class Exercise - 13An element havingatomic number 25 belongs to
(a) s (b) p (c) f (d) d
Solution:
Hence, answer is (d).
Electronic configuration — 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d5.Therefore it is d block element.
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Class Exercise - 14In its structure an element has 4shells. Therefore it belongs to
(a) 3rd period (b) 4th period(c) 2nd period (d) None of these
Solution:
Hence, answer is (b).
Group 2 is present in s block and for them group number = number of electrons in valence shell.
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Class Exercise - 15A, B, C and D have following electronicconfigurationsA : 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:
Hence, answer is (b).
Period number is equal to maximum value of principal quantum number.
Element A — 2nd periodElement B — 3rd periodElement C — 3rd periodElement D — 4th period
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Thank you