1 Electronic Configurations and the Periodic Table 5.1Relative Energies of Orbitals 5.2Electronic...

Electronic Configurations Electronic Configurations

and the Periodic Tableand the Periodic Table

5.15.1 Relative Energies of Orbitals Relative Energies of Orbitals

5.25.2 Electronic Configurations of Elements Electronic Configurations of Elements

5.35.3 The Periodic TableThe Periodic Table

5.45.4 Ionization Enthalpies of ElementsIonization Enthalpies of Elements5.55.5 Variation of Successive Ionization EthalpiesVariation of Successive Ionization Ethalpies with Atomic Numbers with Atomic Numbers

5.45.4 Atomic Size of ElementsAtomic Size of Elements

5.5.11Relative EnergieRelative Energie

s of Orbitalss of Orbitals

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.

In the Lyman series,

only one spectral line is observed for the transition from n = 2 to n = 1.

Evidence

2s and 2p subshells are degenerate

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 orbitalsRelative energies of orbitals5.1 Relative energies of orbitals (SB p.106)

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

1.Aufbau (building up) PrincipleElectrons enter the orbitals in order of ascending energy.

s p d f g h i

Letters read across

Numbers read downwards

5 5 5 5

6 6 6 6 6

7 7 7 7 7 7

s p d f g h i1

5 5 5 5

6 6 6 6 6

7 7 7 7 7 7

s p d f g h i1

5 5 5 5

6 6 6 6 6

7 7 7 7 7 7

1s, 2s,

s p d f g h i1

5 5 5 5

6 6 6 6 6

7 7 7 7 7 7

1s, 2s, 2p, 3s,

s p d f g h i1

5 5 5 5

6 6 6 6 6

7 7 7 7 7 7

1s, 2s, 2p, 3s, 3p, 4s,

s p d f g h i1

5 5 5 5

6 6 6 6 6

7 7 7 7 7 7

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s

s p d f g h i1

5 5 5 5

6 6 6 6 6

7 7 7 7 7 7

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s

s p d f g h i1

5 5 5 5

6 6 6 6 6

7 7 7 7 7 7

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

s p d f g h i1

5 5 5 5

6 6 6 6 6

7 7 7 7 7 7

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

Building up of electronic Building up of electronic configurationsconfigurations

5.1 Relative energies of orbitals (SB p.106)

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 5-1Check Point 5-1

5.5.22 Electronic Electronic ConfigurationConfigurations of Elementss of Elements

Ways to Express Electronic ConfigurationsWays to Express Electronic Configurations

5.2 Electronic configurations of elements (SB p.108)

1. The s, p, d, f notation

Na 1s2, 2s2, 2p6, 3s1

1s2, 2s2, 2px2, 2py

2, 2pz2, 3s1

K 1s2, 2s2, 2p6, 3s2, 3p6, 4s1

Fe 1s2, 2s2, 2p6, 3s2, 3p6, 3d6, 4s2

Q.17(a)

Q.17(b)

orbitals of the same quantum shell are placed together

2. Using a noble gas ‘core’

Na [Ne] 3s1

Ca [Ar] 4s2

Si [Ne] 3s2, 3p2

V [Ar] 3d3, 4s2

Q.18(a)

Q.18(b)

outermost shell

3. Electrons – in – Boxes representation

N1s 2s 2p

(1) All boxes should be labelled

(2) Boxes of the same energies are put together.

2px 2py 2pz

Hund’s rule is violatedHund’s rule is violated

Pauli’s exclusion principle is violated

Q.20(a)

1s 2s 2p 3p3s

Phosphorus

Q.20(b)

Chromium1s 2s 3s2p 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

[Ar] 3d4, 4s2 [Ar] 3d5, 4s1 + energy

+ energy

1s 2s 3s2p 3p 3d 4s

Q.20(c)

Copper

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.

1s 2s 3s2p 3p 3d 4s

[Ar] 3d9, 4s2 [Ar] 3d10, 4s1 + energyQ.20(c)

Copper

+ energy

1s 2s 3s2p 3p 4s

Silicon3p3s

Empty orbital(s) in a partially filled subshell should be shown

Silicon3p3s

+ energy

Energy difference : 3p – 3s > 3d – 4s

21(a)(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

21(b) Ar

S2 , Cl , Ar , K+ , Ca2+

Same electronic configurations

IsoelectronicQ.22

Represented by ‘electrons-in-boxes’ Represented by ‘electrons-in-boxes’ diagramsdiagrams

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

http://www.chemcollective.org/applets/pertable.php

Building up of electronic Building up of electronic configurationsconfigurations

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Quintessence (The fifth element)

、空

Seven Planetary Elements of Alchemists

Sun Gold

Moon Silver Venus

Copper

Mars Iron

Jupiter TinMercury Mercury Saturn

Other Alchemical Elements

Law of Triads (Dobereiner, 1829)

Element Molar mass(g/mol)

Density(g/cm³)

chlorine 35.453 0.0032

bromine 79.904 3.1028

iodine 126.90447 4.933

calcium 40.078 1.55

strontium

87.62 2.54

barium 137.327 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

Li Be B C N O F

Na Mg Al Si P S Cl

Group 1A

Group 2A

Group 3A

Group 4A

Group 5A

Group 6A

Group 7A

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

1. Ekaboron (atomic mass = 44)

Scandium (44.96)

2. Ekaaluminium (68)

Gallium (69.3)

First Periodic Table (Mendeleev, 1869)

‘missing elements’ predicted by Mendeleev

First Periodic Table (Mendeleev, 1869)

3. Ekamanganese (100)

Technetium (98)

4. Ekasilicon (72)

Germanium (72.59)

7 groups or 8 groups ?

Discovery of the Noble Gases

Lord Rayleigh William RamsayNobel Laureate

in Physics, 1904Nobel Laureate in Chemistry, 1904

Air - (O2, CO2, H2O)

Density ( g / dm3)

1.2572

NH3 N2 1.2508

% error 0.5% ???

decompose

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

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

Rn discovered in 1900 by F.E. Dorn

Helium discovered in 1895

Ne, Kr, Xe discovered in 1898All by Ramsay

Po, Ra discovered in 1898 by Pierre & Marie Curie

Congratulations !

Nobel Laureate in Chemistry,

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

Modern Periodic Table

Transuranium Elements

Discovered by McMillan and Seaborg

99989694969594

Discovered by McMillan and Seaborg

Nobel Laureates in Chemistry, 1951

From University of California, Berkeley,

United States of America

U93 94 95 96 94 96 98 99

Uranium, discovered in 1789,

was considered the heaviest elements

Named after Uranus ( 天王星 )

Discovered in 1781

Was Considered the Farthest Planet from The Earth in the Solar System

Neptune( 海王星 ) : The Next Planet out from Uranus

Neptunium : Discovered in 1940 by McMillan

Pluto( 冥王星 ) : Was considered the next ‘Planet’ out from Neptune

Plutonium : Discovered in 1941 by McMillan & Seaborg

University of California, Berkeley,

Americium (1944)

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

Nobel Laureate in Physics, 1903

Nobel Laureate in Chemistry, 1911

Curium Marie Curie

Berkelium (1949)

Californium (1950)

McMillan and Seaborg

Einsteinium (1952 by Albert Ghiorso)

Albert Einstein

Developer of the first nuclear reactor, 1942

Fermium (1952 by Albert Ghiorso)Enrico Fermi

General Consultant of the Manhattan Project

Hiroshima – little boy

Nagasaki – fat man

Discovery of Periodicity 1869

Mendelevium (1955 by Albert Ghiorso)

Mendeléev

Inventor of Dynamite, 1867

The Man Behind the Nobel Prize

Nobelium (1958 by Albert Ghiorso)

Alfred Nobel

Lawrencium (1961 by Albert Ghiorso)

@ Lawrence Radiation Laboratory University of California, Berkeley,

Ernest Orlando Lawrence

Developer of cyclotron

1964 – 1996 ADTeams 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 System0 = nil

1 = un

2 = bi

3 = tri

4 = quad

5 = pent

6 = hex

7 = sept8 = oct9 = enn

ium ium

111 = unununium (Uuu) = Roentgenium (Rg)

111 = unununium (Uuu)112 = ununbium (Uub)113 = ununtrium (Uut)

114 = ununquadium (Uuq)115 = ununpentium (Uup)

116 = ununhexium (Uuh)

d-block

p-block

The Periodic TableThe Periodic Table

5.3 The Periodic Table (SB p.112)

d-block

p-block

f-block

s-block

s-block & p-block elements are called representative elements

f-block elements are called inner-transition elements

Rare Earth Metals( 稀土金屬 )

中東有石油中國有稀土

鄧小平

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

Q.23(b) No

d- block f-block

Period no. n n

No. of the last subshell to be filled

n – 1n 4

n – 2n 6

d-block starts in Period 4 (n 4)

Transition metals

f-block starts in Period 6 (n 6)

Lanthanides : Period 6 (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

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)

VIIIB Fe [Ar] 3d6, 4s2

Co [Ar] 3d7, 4s2

Ni [Ar] 3d8, 4s2

Not true for VIIIB elements

The Song of Elements by Tom Lehrer

The Song of Elements – on YouTube

Visual Elements Periodic Table

Periodicity as illustrated by

(i) Variation in atomic radius with atomic number

(ii) Variation in ionization enthalpy with atomic number

5.5.66 Atomic Size Atomic Size

of Elementsof 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.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.

5.6 Atomic size of elements (p. 122)

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.

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

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

The inner 1s electrons shield the outer electrons almost completely

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.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.

• 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

Q: Explain why there is sharp in atomic radius when a new Period beginsQ: 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.

5.4 Ionization enthalpies of elements (SB p.117)

Q: Explain why there is sharp in atomic radius when a new Period beginsQ: 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.

5.5.44 Ionization Ionization Enthalpies of Enthalpies of

ElementsElements

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

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 beginsQ: 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.Q: Explain why there is a trough at Boron(B) in Period 2.

• Be : 1s2, 2s2

B : 1s2, 2s2, 2p1• Be : 1s2, 2s2

B : 1s2, 2s2, 2p1

More diffused

More penetrating

In multi-electron systems,

penetrating power : -

s > p > d > f

22 4 r

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.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.Q: Explain why there is a trough at Oxygen(O) in Period 2.

• e.c. of N : 1s2, 2s2, 2px1, 2py

1, 2pz1

e.c. of O : 1s2, 2s2, 2px2, 2py

1, 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.

• e.c. of N : 1s2, 2s2, 2px1, 2py

1, 2pz1

e.c. of O : 1s2, 2s2, 2px2, 2py

1, 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.

Q: Explain why there is a trough at Oxygen(O) in Period 2.Q: Explain why there is a trough at Oxygen(O) in Period 2.

e.c. of N : 1s2, 2s2, 2px1, 2py

1, 2pz1

e.c. of O : 1s2, 2s2, 2px2, 2py

1, 2pz1

Alternately, the removal of a 2p electronfrom O results in a stable half-filled 2psubshell.

e.c. of N : 1s2, 2s2, 2px1, 2py

1, 2pz1

e.c. of O : 1s2, 2s2, 2px2, 2py

1, 2pz1

Alternately, the removal of a 2p electronfrom O results in a stable half-filled 2psubshell.

5.5.55Variation of Variation of

Successive Successive Ionization Ionization

Enthalpies with Enthalpies with Atomic NumbersAtomic Numbers

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.

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

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 ?

Invert relationship between atomic radius and first I.E.

(a) A would have the largest atomic number.

It is because A has the lowest first ionization enthalpy.

(b) Group I

It is because 1st I.E. << 2nd I.E.

(a) B is most likely to form B3+

It is because 3rd I.E. << 4th I.E.

(b) A and D are in Group I

It is because 1st I.E. << 2nd I.E.

(a) 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.

(b) A B C D E F

N O F Ne Na Mg

P S Cl Ar K Ca

The END

Write the electronic configurations and draw “electrons-in –boxes” diagrams for

(a) nitrogen; and

(b) sodium.

Answer

5.1 Relative energies of orbitals (SB p.108)

(a) Nitrogen: 1s22s22p3

(b) Sodium: 1s22s22p63s1

Give the electronic configuration by notations and “electrons-in-boxes” diagrams in the abbreviated form for the following elements.

(a) silicon; and

(b) copper.

Answer

(a) Silicon: [Ne]3s23p3

(b) Copper: [Ar]3d104s1

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?

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.

Answer

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

(a)Give four main factors that affect the magnitude of ionization enthalpy of an atom.

Answer

(a) 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.

(b)Explain why Group 0 elements have extra high first ionization enthalpies and their decreasing trend down the group.

Answer

(b) 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.

(c)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.

For the element 126C,

(a)(i) write its electronic configuration by notation.

(ii) write its electronic configuration by “electrons-in- boxes” diagram. Answer(a) (i) 1s22s22p2

(b)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. Answer

1st 2nd 3rd 4th 5th 6th

I.E. (kJ

mol-1)

1090 2350 4610 6220 37800

(b) (i)

(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.

(a) Give the “electrons-in-boxes” diagram of 26Fe.

(b)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+.

Answer

(a) Fe :

(b) Fe2+ :

Fe3+ :

(c) Which ion is more stable, Fe2+ or Fe3+? Explain briefly.

Answer

(c) Fe3+ ion is more stable because the 3d orbital is exactly half-filled which gives the electronic configuration extra stability.

(d)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.

Answer

1st 2nd 3rd 4th 5th 6th

I.E. (kJ

mol-1)

762 1560 2960 5400 7620 10100

(d) (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

(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 su

ccessive ionization enthalpies of iron.

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.

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.

1 Electronic Configurations and the Periodic Table 5.1Relative Energies of Orbitals 5.2Electronic...

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