Electronic Configurations

Electronic Configurations
5 Electronic Configurations and the Periodic Table 5.1 Relative Energies of Orbitals 5.2 Electronic Configurations of Elements 5.3 The Periodic Table 5.4 Ionization Enthalpies of Elements 5.5 Variation of Successive Ionization Ethalpies with Atomic Numbers 5.4 Atomic Size of Elements

Relative Energies of Orbitals
5.1 Relative Energies of Orbitals

In one-electron systems (e. g
In one-electron systems (e.g. H, He+), there are no interactions(no shielding effects) between electrons. All subshells(s, p, d, f,…) of the same principal quantum shell have the same energy. The subshells are said to be degenerate.

2s and 2p subshells are degenerate
Evidence In the Lyman series, only one spectral line is observed for the transition from n = 2 to n = 1. 2s and 2p subshells are degenerate 2s 2p 1s

In multi-electron systems, there are interactions(shielding effects) between electrons.
Different subshells of the same principal quantum shell occupy different energy levels. The energies of subshells or orbitals follow the order : s < p < d < f

Relative energies of orbitals
5.1 Relative energies of orbitals (SB p.106) Relative energies of orbitals

Relative energies of orbitals
5.1 Relative energies of orbitals (SB p.106) Relative energies of orbitals Electrons enter 4s subshell before filling up 3d subshell.

Both 4s and 3d electrons are shielded from the nuclear attraction by the inner core (2,8,8)

4s electron is more penetrating than 3d electron, spending more time closer to the nucleus.
4s electron experiences stronger nuclear attraction 4s electron is more stable.

Three rules to build up electronic configurations
1. Aufbau (building up) Principle 2. Hund’s Rule 3. Pauli’s Exclusion Principle

Aufbau (building up) Principle
Electrons enter the orbitals in order of ascending energy.

s p d f g h i Numbers read downwards Letters read across 1 2 3 2 4 5
6 7 2

1s, s p d f g h i 1 2 3 4 5 6 7 2

1s, 2s, s p d f g h i 1 2 3 4 5 6 7 2

1s, 2s, 2p, 3s, s p d f g h i 1 2 3 4 5 6 7 2

1s, 2s, 2p, 3s, 3p, 4s, s p d f g h i 1 2 3 4 5 6 7 2

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s s p d f g h i 1 2 3 4 5 6 7 2

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s s p d f g h i 1 2 3 4 5 6 7 2

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s s p d f g h i 1 2 3 4 5 6 7 2

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s
s p d f g h i 1 2 3 4 5 6 7 8 2

Building up of electronic configurations
5.1 Relative energies of orbitals (SB p.106) Building up of electronic configurations

Electrons-in boxes diagram
2. Hund’s rule : - Orbitals of the same energy must be occupied singly and with the same spin before pairing up of electrons occurs. Carbon 1s 2s 2p Electrons-in boxes diagram

3. Pauli’s exclusion principle : -
Electrons occupying the same orbital must have opposite spins. W. Pauli , Nobel prize laureate in Physics, 1945 Check Point 5-1

Electronic Configurations of Elements
5.2 Electronic Configurations of Elements

Ways to Express Electronic Configurations
5.2 Electronic configurations of elements (SB p.108) Ways to Express Electronic Configurations 1. The s, p, d, f notation Na 1s2, 2s2, 2p6, 3s1 1s2, 2s2, 2px2, 2py2, 2pz2, 3s1

Q.17(a) K 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 Q.17(b) Fe 1s2, 2s2, 2p6, 3s2, 3p6, 3d6, 4s2 orbitals of the same quantum shell are placed together

2. Using a noble gas ‘core’
Na [Ne] 3s1 Ca [Ar] 4s2

Q.18(a) Si [Ne] 3s2, 3p2 outermost shell Q.18(b) V [Ar] 3d3, 4s2

 3. Electrons – in – Boxes representation N
All boxes should be labelled Boxes of the same energies are put together. 2px 2py 2pz 

Q.19 Hund’s rule is violated Hund’s rule is violated Pauli’s exclusion principle is violated

Q.20(a) Phosphorus 1s 2s 2p 3p 3s

Q.20(b) Chromium 1s 2s 3s 2p 3p 3d 4s The half-filled 3d subshell has extra stability due to the more symmetrical distribution of charge. The energy needed to promote an electron from 4s to 3d is more than compensated by the energy released from the formation of half-filled 3d subshells.

Q.20(b) Chromium + energy [Ar] 3d4, 4s2  [Ar] 3d5, 4s1 + energy 1s 2s
2p 3p 3d 4s

Q.20(c) Copper 1s 2s 3s 2p 3p 3d 4s The full-filled 3d subshell has extra stability due to the more symmetrical distribution of charge. The energy needed to promote an electron from 4s to 3d is more than compensated by the energy released from the formation of full-filled 3d subshells.

Q.20(c) Copper + energy [Ar] 3d9, 4s2  [Ar] 3d10, 4s1 + energy 1s 2s

Empty orbital(s) in a partially filled subshell should be shown
Silicon 3p 3s [Ne] Empty orbital(s) in a partially filled subshell should be shown

Silicon 3p 3s [Ne] + energy  Energy difference : 3p – 3s > 3d – 4s

21(a) 21(b) Ar (i) S2 1s2, 2s2, 2p6, 3s2, 3p6 (ii) Cl 1s2, 2s2, 2p6, 3s2, 3p6 (iii) K+ 1s2, 2s2, 2p6, 3s2, 3p6 (iv) Ca2+ 1s2, 2s2, 2p6, 3s2, 3p6

Same electronic configurations
S2 , Cl , Ar , K+ , Ca2+ Same electronic configurations Isoelectronic Q.22

Represented by ‘electrons-in-boxes’ diagrams
5.2 Electronic configurations of elements (SB p.110) Represented by ‘electrons-in-boxes’ diagrams

Check Point 5-2 5.2 Electronic configurations of elements (SB p.110)

Building up of electronic configurations

A brief history of the Periodic Table
Ancient Greece, Aristotle : - Four elements Fire, Water, Air, Earth,

A brief history of the Periodic Table
Ancient Greece, Aristotle : - Four elements Air, Fire, Earth, Water Buddha : 地、水、火、風、空 Quintessence (The fifth element)

Seven Planetary Elements of Alchemists

Moon  Silver Mars  Iron Sun  Gold Venus  Copper Jupiter  Tin Mercury  Mercury Saturn  Lead

Other Alchemical Elements
As Sb Bi Pt S P

Law of Triads (Dobereiner, 1829)
Element Molar mass (g/mol) Density (g/cm³) chlorine 35.453 0.0032 bromine 79.904 3.1028 iodine 4.933 calcium 40.078 1.55 strontium 87.62 2.54 barium 3.594 The molar mass and density of the middle one  average of the other two.

Law of Octaves (Newlands, 1865)
Elements of similar physical and chemical properties recurred at intervals of eight Group 1A Group 2A Group 3A Group 4A Group 5A Group 6A Group 7A Li Be B C N O F Na Mg Al Si P S Cl

First Periodic Table (Mendeleev, 1869)
Periodicity : Chemical properties of elements are periodic functions of their atomic masses. Elements arranged in terms of their properties (not exactly follow the order of atomic mass) Elements with similar properties are put together in vertical groups Gaps were left in the table for ‘missing elements’

‘missing elements’ predicted by Mendeleev Ekaboron (atomic mass = 44) Scandium (44.96) Ekaaluminium (68) Gallium (69.3)

‘missing elements’ predicted by Mendeleev 3. Ekamanganese (100) Technetium (98) Ekasilicon (72) Germanium (72.59)

7 groups or 8 groups ?

Discovery of the Noble Gases
Lord Rayleigh William Ramsay Nobel Laureate in Physics, 1904 Nobel Laureate in Chemistry, 1904

??? 1894 % error  0.5% Density ( g / dm3) Air N2 1.2572 NH3 N2
- (O2, CO2, H2O) N2 1.2572 NH3 N2 decompose 1.2508 % error  0.5% ???

Argon is present in air Confirmed by spectroscopy RAM : Ar(39.95) > K(39.10) Unlike group 2 elements, Ar shows no reactivity. Placed before K and after Cl A new group in the Periodic Table Group 0

Rn discovered in 1900 by F.E. Dorn
Group 1A Group 2A Group 3A Group 4A Group 5A Group 6A Group 7A Group 0 Li Be B C N O F Ne Na Mg Al Si P S Cl Ar Helium discovered in 1895 Ne, Kr, Xe discovered in 1898 All by Ramsay Rn discovered in 1900 by F.E. Dorn Po, Ra discovered in 1898 by Pierre & Marie Curie

Congratulations ! Nobel Laureate in Chemistry, 2010

Modern Periodic Table Elements arranged in order of increasing atomic number 91 elements discovered up to 1940 Most are naturally occurring except Po(84), At(85), Rn(86), Fr(87), Ra(88), Ac(89), Pa(91) – from radioactive decay Pm(61) discovered in 1945 as a product in nuclear fission - not found in nature

Transuranium Elements
92 U 99 98 96 94 95 Pu 93 Np Discovered by McMillan and Seaborg

92 U 93 Np 94 Pu Discovered by McMillan and Seaborg Nobel Laureates in Chemistry, 1951 From University of California, Berkeley, United States of America

was considered the heaviest elements
92 U 93 94 95 96 98 99 Uranium, discovered in 1789, was considered the heaviest elements

Named after Uranus (天王星) Discovered in 1781
92 U Named after Uranus (天王星) Discovered in 1781 Was Considered the Farthest Planet from The Earth in the Solar System

92 U 93 Np 94 Pu Neptunium : Discovered in 1940 by McMillan Neptune(海王星) : The Next Planet out from Uranus

92 U 93 Np 94 Pu Plutonium : Discovered in 1941 by McMillan & Seaborg Pluto(冥王星) : Was considered the next ‘Planet’ out from Neptune

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Americium (1944) Nobel Laureates in Chemistry, 1951 University of California, Berkeley, United States of America

It was named americium because it is just below europium in the Periodic Table.

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Curium Marie Curie Nobel Laureate in Physics, 1903 Nobel Laureate in Chemistry, 1911

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Berkelium (1949) Nobel Laureates in Chemistry, 1951 University of California, Berkeley, United States of America

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Californium (1950) Nobel Laureates in Chemistry, 1951 University of California, Berkeley, United States of America

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr McMillan and Seaborg Nobel Laureates in Chemistry, 1951

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Einsteinium (1952 by Albert Ghiorso) Albert Einstein Nobel Laureate in Physics, 1921

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Fermium (1952 by Albert Ghiorso) Enrico Fermi Nobel Laureate in Physics, 1938 Developer of the first nuclear reactor, 1942

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr General Consultant of the Manhattan Project Hiroshima – little boy Nagasaki – fat man

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Mendelevium (1955 by Albert Ghiorso) Mendeléev Discovery of Periodicity 1869

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Nobelium (1958 by Albert Ghiorso) Alfred Nobel Inventor of Dynamite, 1867 The Man Behind the Nobel Prize

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Lawrencium (1961 by Albert Ghiorso) @ Lawrence Radiation Laboratory University of California, Berkeley,

95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr Ernest Orlando Lawrence Developer of cyclotron Nobel Laureate in Physics, 1939 University of California, Berkeley,

1964 – 1996 AD Teams from Russia(USSR), USA & Germany Synthesis of Rf(104), Db(105), Sg(105), Bh(106), Hs(108), Mt(109), Ds(110), Rg(111) & Uub(112). Rg(111) = Roentgenium

1999 – 2003 AD Russia(USSR) & USA Synthesis of Uut(113), Uuq(114), Uup(115), Uuh(116)

Naming of Elements – IUPAC System
0 = nil 1 = un 2 = bi 3 = tri 4 = quad 5 = pent 6 = hex 7 = sept 8 = oct 9 = enn ium ium 111 = unununium (Uuu) = Roentgenium (Rg)

111 = unununium (Uuu) d-block 112 = ununbium (Uub) 113 = ununtrium (Uut) 114 = ununquadium (Uuq) p-block 115 = ununpentium (Uup) 116 = ununhexium (Uuh) 6B

5.3 The Periodic Table (SB p.112)

s-block & p-block elements are called representative elements
5.3 The Periodic Table (SB p.112) s-block & p-block elements are called representative elements s-block p-block d-block f-block

f-block elements are called inner-transition elements
Rare Earth Metals(稀土金屬) f-block elements are called inner-transition elements

中東有石油中國有稀土鄧小平

17 Rare Earth Metals(稀土金屬)
鑭(La)、鈰(Ce)、鐠(Pr)、釹(Nd)、鉕(Pm)、釤(Sm)、銪(Eu)、釓(Gd)、鋱(Tb)、鏑(Dy)、鈥(Ho)、鉺(Er)、銩(Tm)、鐿(Yb)、鑥(Lu)、鈧(Sc)、釔(Y) 17 Rare Earth Metals(稀土金屬)

Q.23(a) They are named after the outermost orbitals to be filled

No. of the last subshell to be filled
Q.23(b) No d- block f-block Period no. n No. of the last subshell to be filled n – 1 n  4 n – 2 n  6

d-block starts in Period 4 (n  4)
Transition metals f-block starts in Period 6 (n  6) Lanthanides : Period (rare earth metals) Actinides : Period 7

Q.23(c) True only for IB to VII B

Q.23(c) IIIB Sc [Ar] 3d1, 4s2 IVB Ti [Ar] 3d2, 4s2 VB V [Ar] 3d3, 4s2 VIB Cr [Ar] 3d5, 4s1 VIIB Mn [Ar] 3d5, 4s2 3 4 5 6 7

Q.23(c) IB Cu [Ar] 3d10, 4s1 IIB Zn [Ar] 3d10, 4s2 Electrons in fully-filled 3d subshells cannot be removed easily.  They are not treated as outermost shell electrons

Q.23(c) Not true for VIIIB elements VIIIB Fe [Ar] 3d6, 4s2 Co [Ar] 3d7, 4s2 Ni [Ar] 3d8, 4s2

Let's Think 1 Check Point 5-3

Visual Elements Periodic Table
The Song of Elements by Tom Lehrer The Song of Elements – on YouTube Visual Elements Periodic Table

Periodicity as illustrated by
Variation in atomic radius with atomic number Variation in ionization enthalpy with atomic number

Atomic Size of Elements
5.6 Atomic Size of Elements

Atomic radius is defined as half the distance between two nuclei of the atoms joined by a single covalent bond or a metallic bond

Atomic radii of noble gases were obtained by calculation

Atomic radii  across both Periods 2 and 3

Q: Explain why the atomic radius decreases across a period.
5.6 Atomic size of elements (p. 122) Q: Explain why the atomic radius decreases across a period. Moving across a period, there is an increase in the nuclear attraction due to the addition of proton in the nucleus.( in nuclear charge) The added electron is placed in the same quantum shell. It is only poorly repelled/shielded/screened by other electrons in that shell. The nuclear attraction outweighs the increase in the shielding effect between the electrons. This leads to an increase in the effective nuclear charge.

Effective nuclear charge, Zeff, is the nuclear charge experienced by an electron in an atom.
In the present discussion, only the outermost electrons are considered.

+3 Li The outer 2s electron sees the nucleus through a screen of two inner 1s electrons.

Two electrons in the inner shell
nucleus 2s electron of Li outside the inner shell

+3 Li The outer 2s electron is repelled/shielded/screened by the inner 1s electrons from the nucleus

The nuclear charge experienced by the 2s electron is  +1
+3 Li Li  +1 The nuclear charge experienced by the 2s electron is  +1

The inner 1s electrons shield the outer electrons almost completely
Be +2 +4 Be The inner 1s electrons shield the outer electrons almost completely

Be +1.5 +2 Be The two electrons in the same shell (2s) shield each other less poorly. Zeff  1.5

Atomic radii  down a group

Q: Explain why the atomic radius increases down a group.
5.6 Atomic size of elements (p. 122) Q: Explain why the atomic radius increases down a group. Moving down a group, an atom would have more electron shells occupied. The outermost shell becomes further away from the nucleus. Moving down a group, although there is an increase in the nuclear charge, it is offset very effectively by the screening effect of the inner shell electrons.

Sharp  in atomic radius when a new Period begins

5.4 Ionization enthalpies of elements (SB p.117)
Q: Explain why there is sharp  in atomic radius when a new Period begins The element at the end of a period has the smallest atomic radius among the elements in the same period because its outermost electrons are experiencing the strongest nuclear attraction.

Q: Explain why there is sharp  in atomic radius when a new Period begins The element at the beginning of the next period has one extra electron in an outer shell which is far away from the nucleus. Although there is also an increase in the nuclear charge, it is very effectively screened by the inner shell electrons. Check Point 5-6

Ionization Enthalpies of Elements
5.4 Ionization Enthalpies of Elements

Across a Period, there is a general  in I. E
Across a Period, there is a general  in I.E. leading to a maximum with a noble gas.

Effective nuclear charge  from left to right across the Period

First I.E.  down a group

The outermost electrons are further away from the nucleus and are more effectively shielded from it by the inner electrons

The first ionization enthalpies generally decrease down a group and increase across a period

Fully-filled subshell
5.4 Ionization enthalpies of elements (SB p.117) Q: Explain why there is sharp  in IE when a new Period begins The element at the end of a period has a stable duplet or octet structure. Much energy is required to remove an electron from it as this will disturb the stable structure. He  1s2 (duplet) Ne  2s2, 2p6 (octet) Fully-filled shells Ar  3s2, 3p6 (octet) Fully-filled subshell

Q: Explain why there is sharp  in I.E. when a new Period begins The element at the beginning of the next period has one extra electron in an outer quantum shell which is far away from the nucleus. Although there is also an increase in the nuclear charge, it is very effectively shielded by the inner shell electrons. Thus the outermost electron experiences a much less nuclear attraction.

Irregularities : -

Q: Explain why there is a trough at Boron(B) in Period 2. Be : 1s2, 2s2 B : 1s2, 2s2, 2p1

1s More penetrating More diffused

In multi-electron systems,
penetrating power : - s > p > d > f

3d electrons are more diffused (less penetrating)
3d electrons are more shielded by 1s electrons

Q: Explain why there is a trough at Boron(B) in Period 2. It is easier to remove the less penetrating 2p electron from B than to remove a more penetrating 2s electron from a stable fully-filled 2s subshell in Be.

Irregularities : -

Q: Explain why there is a trough at Oxygen(O) in Period 2.
5.4 Ionization enthalpies of elements (SB p.117) Q: Explain why there is a trough at Oxygen(O) in Period 2. e.c. of N : 1s2, 2s2, 2px1, 2py1, 2pz1 e.c. of O : 1s2, 2s2, 2px2, 2py1, 2pz1 The three 3p electrons in N occupy three different orbitals, thus minimizing the repulsion between the electrons(shielding effect). It is more difficult to remove an electron from the half-filled 2p subshell of N.

Alternately, the removal of a 2p electron
5.4 Ionization enthalpies of elements (SB p.117) Q: Explain why there is a trough at Oxygen(O) in Period 2. e.c. of N : 1s2, 2s2, 2px1, 2py1, 2pz1 e.c. of O : 1s2, 2s2, 2px2, 2py1, 2pz1 Alternately, the removal of a 2p electron from O results in a stable half-filled 2p subshell.

Variation of Successive Ionization Enthalpies with Atomic Numbers
5.5 Variation of Successive Ionization Enthalpies with Atomic Numbers

Successive I.Es. Show similar variation patterns with atomic number.
5.5 Variation of successive ionization enthalpies with atomic numbers (p. 120) Successive I.Es. Show similar variation patterns with atomic number. 3rd I.E. > 2nd I.E. > 1st I.E.

Each represents a pair of isoelectronic species
5.5 Variation of successive ionization enthalpies with atomic numbers (p. 120) Plots of successive I.E. are shifted by one unit in atomic number to the right respectively. e.g. Be+ = Li (2, 1) B+ = Be (2, 2) C+ = B (2, 3) Each represents a pair of isoelectronic species

Invert relationship between atomic radius and first I.E.
Why is the atomic radius of helium greater than that of hydrogen, despite of the fact that the first I.E. of helium is higher than that of hydrogen ?

Q.24 (a) A would have the largest atomic number. It is because A has the lowest first ionization enthalpy. Group I It is because 1st I.E. << 2nd I.E.

Q.25 B is most likely to form B3+ It is because 3rd I.E. << 4th I.E. A and D are in Group I It is because 1st I.E. << 2nd I.E.

Q.26 D is a noble gas. It is because D has a higher I.E. than those of A, B and C and has a much higher I.E. than E. A B C D E F N O F Ne Na Mg P S Cl Ar K Ca

The END

Check Point 5-1 Answer Back
5.1 Relative energies of orbitals (SB p.108) Back Check Point 5-1 Write the electronic configurations and draw “electrons-in –boxes” diagrams for (a) nitrogen; and (b) sodium. Answer Nitrogen: 1s22s22p3 (b) Sodium: 1s22s22p63s1

5.2 Electronic configurations of elements (SB p.110) Back Check Point 5-2 Give the electronic configuration by notations and “electrons-in-boxes” diagrams in the abbreviated form for the following elements. silicon; and copper. Answer Silicon: [Ne]3s23p3 (b) Copper: [Ar]3d104s1

Back Let's Think 1 If you look at the Periodic Table in Fig. 5-5 closely, you will find that hydrogen is separated from the rest of the elements. Even though it has only one electron in its outermost shell, it cannot be called an alkali metal, why? Answer Hydrogen has one electron shell only, with n =1. This shell can hold a maximum of two electrons. Hydrogen is the only element with core electrons. This gives it some unusual properties. Hydrogen can lose one electron to form H+, or gain an electron to become H-. Therefore, it does not belong to the alkali metals and halogens. Hydrogen is usually assigned in the space above the rest of the elements in the Periodic Table – the element without a family.

Check Point 5-3 Outline the modern Periodic Table and label the table with the following terms: representative elements, d-transition elements, f-transition elements, lanthanide series, actinide series, alkali metals, alkaline earth metals, halogens and noble gases. Answer

Back Check Point 5-3

Check Point 5-4 Give four main factors that affect the magnitude of ionization enthalpy of an atom. Answer The four main factors that affect the magnitude of the ionization enthalpy of an atom are: (1) the electronic configuration of the atom; (2) the nuclear charge; (3) the screening effect; and (4) the atomic radius.

Check Point 5-4 Explain why Group 0 elements have extra high first ionization enthalpies and their decreasing trend down the group. The first ionization enthalpies of Group 0 elements are extra high. It is because Group 0 elements have very stable electronic configurations since their orbitals are completely filled. That means, a large amount of energy is required to remove an electron from a completely filled electron shell of [ ]ns2np6 configuration. Going down the group, the first ionization enthalpies of Group 0 elements decreases. It is because there is an increase in atomic radius down the group, the outermost shell electrons experience less attraction from the nucleus. Further, as there is an increase in the number of inner electron shells, the outermost shell electrons of the atoms are better shielded from the attraction of the nucleus (greater screening effect). Consequently, though the nuclear charge increases down the group, the outermost shell electrons would experience less attraction from the positively charged nucleus. That is why the first ionization enthalpies decrease down the group. Answer

5.4 Ionization enthalpies of elements (SB p.118) Back Check Point 5-4 Predict the trend of the first ionization enthalpies of the transition elements. Answer (c) The first ionization enthalpies of the transition elements do not show much variation. The reason is that the first electron of these atoms to be removed is in the 4s orbital. As the energy levels of the 4s orbitals of these atoms are more or less the same, the amount of energy required to remove these electrons are similar.

Example 5-5 Answer For the element 126C,
5.5 Variation of successive ionization enthalpies with atomic numbers (p. 121) Example 5-5 For the element 126C, (i) write its electronic configuration by notation. (ii) write its electronic configuration by “electrons-in- boxes” diagram. Answer (i) 1s22s22p2 (ii)

5.5 Variation of successive ionization enthalpies with atomic numbers (p. 121)
Example 5-5 The table below gives the successive ionization enthalpies of carbon. (i) Plot a graph of log [ionization enthalpy] against number of electrons removed. (ii) Explain the graph obtained. 1st 2nd 3rd 4th 5th 6th I.E. (kJ mol-1) 1090 2350 4610 6220 37800 47000 Answer

Example 5-5 (i)

Back Example 5-5 (ii) The ionization enthalpy increases with increasing number of electrons removed. It is because the effective nuclear charge increases after an electron is removed, and more energy is required to remove an electron from a positively charged ion. Besides, there is a sudden rise from the fourth to the fifth ionization enthalpy. This is because the fifth ionization enthalpy involves the removal of an electron from a completely filled 1s orbital which is very stable.

Check Point 5-5 Answer Give the “electrons-in-boxes” diagram of 26Fe.
5.5 Variation of successive ionization enthalpies with atomic numbers (p. 122) Check Point 5-5 Give the “electrons-in-boxes” diagram of 26Fe. Fe2+ and Fe3+ have 2 and 3 electrons less than Fe respectively. If the electrons are removed from the 4s orbital and then 3d orbitals, give the electronic configurations of Fe2+ and Fe3+. Fe : Fe2+ : Fe3+ : Answer

Check Point 5-5 (c) Which ion is more stable, Fe2+ or Fe3+? Explain briefly. (c) Fe3+ ion is more stable because the 3d orbital is exactly half-filled which gives the electronic configuration extra stability. Answer

Check Point 5-5 Given the successive ionization enthalpies of Fe: (i) plot a graph of successive ionization enthalpies in logarithm scale against the number of electrons removed; (ii) state the difference of the plot from that of carbon as shown in P. 121. 1st 2nd 3rd 4th 5th 6th I.E. (kJ mol-1) 762 1560 2960 5400 7620 10100 Answer

Number of electrons removed
5.5 Variation of successive ionization enthalpies with atomic numbers (p. 122) Check Point 5-5 (i) Number of electrons removed 1 2 3 4 5 6 log (I.E.) 2.88 3.19 3.47 3.73 3.88 4.00

Check Point 5-5 (ii) The ionization enthalpy increases with increasing number of electrons removed. This is because it requires more energy to remove an electron from a higher positively charged ion. In other words, higher successive ionization enthalpies will have higher magnitudes. However, the sudden increase from the fourth to the fifth ionization enthalpies occurs in carbon but not in iron. This indicates that when electrons are removed from the 4s and 4d orbitals, there is no disruption of a completely filled electron shell. Hence, there are no irregularities for the first six successive ionization enthalpies of iron. Back

Check Point 5-6 Answer Explain the following:
5.6 Atomic size of elements (p. 123) Check Point 5-6 Explain the following: (a) The atomic radius decreases across the period from Li to Ne. Answer (a) When moving across the period from Li to Ne, the atomic sizes progressively decrease with increasing atomic numbers. This is because an increase in atomic number by one means one more electron and one more proton in atoms. The additional electron would cause an increase in repulsion between the electrons in the outermost shell. However, since each additional electron goes to the same quantum shell and is at approximately the same distance from the nucleus, the repulsion between electrons is relatively ineffective to cause an increase in the atomic radius. On the other hand, as there is an additional proton added to the nucleus, the electrons will experience a greater attractive force from the nucleus (increased effective nuclear charge). Hence, the atomic radii of atoms decrease across the period from Li to Ne.

Check Point 5-6 Answer Back Explain the following:
5.6 Atomic size of elements (p. 123) Back Check Point 5-6 Explain the following: (b) The atomic radius increases down Group I metals. Answer (b) Moving down Group I metals, the atoms have more electron shells occupied. The outermost electron shells become further away from the nucleus. Besides, the inner shell electrons will shield the outer shell electrons more effectively from the nuclear charge. This results in a decrease in the attractive force between the nucleus and the outer shell electrons. Therefore, the atomic radii of Group I atoms increase down the group.

Electronic Configurations

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Electronic Configurations

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