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2014/08/01
7.1 Pauli Exclusion Principle
No two electrons in an atom can have the same
set of 4 quantum numbers (n, ℓ, mℓ, ms) –
governs arrangement of electrons in atoms.
Chapter 7: Structure of
Atoms and periodic trends.
End of chapter exercises: 1, 5, 11,
15, 19, 21, 27, 29, 31, 35, 37, 43,
58, 69
Electron Configuration = the way electrons are
distributed among the various orbitals of an atom.
• Ground state – most stable state – electrons in
lowest possible energy states.
• Orbitals are filled in order of increasing energy
• Electrons assigned to shells (defined by n) of
increasing energy
• Within a shell, electrons assigned to subshells
(defined by ℓ) of successively higher energy.
• Electrons assigned to ensure total energy of
atom is always as low as possible
Atomic Electron Configurations.
• Hydrogen (H): Z = 1 (1 p+ & 1 e-) 1 e- in 1s
orbital – can use orbital notation ↑ or spdf
notation 1s1
• Lithium (Li) and Group 1 A elements ….
• Li (1st element period 2) Z = 3 (p+ = 3, e- = 3)
• 1st 2 e-s in 1s orbital, 3rd e- in 2s:
• Abbreviated form: [He]2s1 i.e. previous noble
gas (which includes core e-s) with valence
electrons
For a given orbital, i.e. 1s or 2pz … n, ℓ and mℓ are
fixed, therefore ms must have different values.
There are only 2 ms values
→ orbital can only hold 2 electrons
→e-s must have opposite spin
e.g. 3 p-orbitals → each hold 2 e-s → 6 e-s in psubshell
5 d-orbitals → 10 electrons
n-subshells in the nth electron shell → n2 orbitals in a
shell → maximum 2n2 electrons in any shell.
Bohr model – the energy of H-atom depends on n
….(E = - Rhc/n2)
• Atoms with more than 1 electron, sub-shell
energies depend on both n and ℓ.
• Subshell energy order – 2 general rules
• Electrons assigned to sub-shells of increasing ‘n
+ ℓ’ value
• For 2 sub-shells with the same ‘n + ℓ’ value,
electrons assigned first to sub-shell of lower n.
• e.g. e- assigned to 2s (n + ℓ = 2 + 0 = 2) before
2p (n + ℓ = 2 + 1 = 3)
• 3s (n + ℓ = 3 + 0 = 3) before 3p (n + ℓ = 3 + 1 =
4) before 3d (n + ℓ = 3 + 2 = 5)
Core electrons (i.e. e in previous E shell) can be
ignored when considering the chemistry of atom.
Remaining electrons i.e. 2s1 = valence electrons –
electrons that determine the properties of the
element.
• Position of Li on the periodic gives the
configuration → all group 1A elements have 1
valence electron in an s-orbital of the nth shell,
where n = number of period in which element is
found.
• e.g. Potassium (K), 4th period, 1st group:
• [Ar] 4s1
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Beryllium (Be) and group 2A
p+
e- )
• Be: Z = 4 (4
and 4
• 2 e-s in 1s and 2 e-s in 2s:
[He]2s2
• All elements in group 2A: [preceding noble gas]
ns2
• n = period where element is to be found
• Group 1A and Group 2A = s-block elements
• Boron (B) and group 3A
• B: Z = 5 (5 p+ and 5 e-)
1st element in block
on RHS
• 1s and 2s orbitals are filled, 5th e- in 2p orbital
3 valence e-s
Elements from group 3A → 8A = p-block elements
• General configurations ns2 npx, where x varies
from 1 – 6 (group number – 2)
• Carbon (C) and group 4A
• C: Z = 6 (6 p+ and 6 e-) 1s2 2s22p2
[He]2s22p2
• 2nd p-electron must be assigned to either of the
remaining p-orbitals and will have the same spin
as the 1st p-electron. Hund’s Rule
• All elements of group 4A have similar outer shell
configuration: [noble gas]ns2np2
Nitrogen (N) and group 5A
Fluorine (F) and group 7A (Halogens)
• N: Z = 7 (7 p+ and 7 e-)
• [He]2s22p3
All p-electrons in different p-orbitals
with same spin.
• All elements outer shell configuration [noble
gas]ns2np3, n = period in which element is located.
• F: Z = 9 (9 p+ and 9 e-)
• [He]2s22p5
• Outer shell: ns2np5
• Oxygen (O) and group 6A
• O: Z = 8 (8 p+ and 8 e-) 1s2 2s22p4 ….[He]2s22p4
• 4th 2p-electron must pair up with one already
present – must have opposite spin to electron
already there.
• Outer shell:
ns2np4
• Neon (Ne) and group 8A (Noble gases)
• Group 8 (except He) …8 e-s in shell of highest
value → ns2np6 i.e. all noble gases have filled ns
and np subshells → this explains their nearly
complete chemical inertness.
• Ne: Z = 10 (10 p+ and 10 e-) 1s2 2s22p6
[He]2s22p6
Period 3
Period 4:
• Similar electron configuration to 2nd period –
preceding noble gas neon and valence shell n =
3
• e.g. Si (group 4 A) – compare with C
1s2 2s22p6 3s23p2
or
[Ne]3s23p2
• Period 3 ends with Argon
• 1s2 2s22p6 3s23p6
• Next element, K (Z = 19), does the electron go
into 4s or 3d?
• 4s (n + ℓ = 4 + 0 = 4), 3d (n + ℓ = 3 + 2 = 5) → n
+ ℓ rule tell us 4s, also …
• Chemical properties of K indicate it is a member
of the alkali metal group → implies outermost
electron occupies an s-orbital and not 3d-orbital.
Q1. Give a set of quantum numbers for each of
the valence electrons of aluminium:
• Elements of 4th → 7th periods use d- and fsubshells in addition to s- and p-subshells.
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Elements filling d-subshells → transition elements
Exceptions:
• f-subshells → lanthanides (4f) and actinides (5f)
• Transition elements always preceded by 2 sblock elements, i.e. fill ns orbital and then (n –
1)d
• Period 4: Scandium (Sc) – 1st transition element
• Sc: [Ar]3d1 4s2 or [Ar] 4s2 3d1
• Ti: [Ar] 3d2 4s2 or [Ar] 4s2 3d2
• Use Hund’s rule – singly into each d-orbital until
all 5 have one electron each.
• All orbitals of given subshell i.e. 5 x 3d have the
same energy …degenerate
• Chromium (Cr) [Ar]3d5 4s1 not [Ar]3d4 4s2 – 3d
and 4s orbitals have approximately the same
energy in Cr → each of the 6 valence electrons
are assigned to one of the orbitals, same spin.
•
Q2. Give a set of quantum numbers for each of the
valence electrons for Arsenic.
Q3. a) Give the orbital energy diagram as well as
the spdf electron configuration for the nickel atom
in the ground state.
b) Give a full set of quantum numbers for an
electron in the highest subshell energy.
c) If one of these e-s is excited by absorbing a
photon, and 10 line spectra are observed, to which
subshell was the electron excited?
Energies of 4f and 5d orbitals are very similar →
occasional variations in occupation of 5d and 4f
orbitals followed by 6p orbitals.
• Ends with Radon (Rn) – the heaviest of known
nobel gases.
• 7th period – first fill the 7s orbitals
• also includes elements using f-orbitals, the
actinides, which begin with Actinium (Ac).
• Ac: [Rn] 6d1 7s2
• Next Thorium (Th): [Rn] 5f1 6d1 7s2 – first using
5f orbitals.
• Actinide elements are radioactive, most not
found in nature.
• Copper (Cu): [Ar] 3d10 4s1, more stable when all
d orbitals have paired electrons.
•
Lanthanides and Actinides
• 6th period includes the lanthanide series,
starting with Lanthanum (La):[Xe] 5d1 6s2 (or
[Xe] 6s2 5d1)
• La – 1st element in d-block
• Next: Cerium – in separate row → 7 degenerate
f-orbitals corresponding to 7 allowed mℓ values
from +3 to -3.
• Elements in this row, Ce → Lu are first to have
electrons assigned to f-orbitals.
• Ce: [Xe] 4f1 5d1 6s2
Q4: Using the periodic table, write condensed
electron configurations for:
(a) phosphorus, (b) cobalt, (c) tellurium, (d)
bismuth. How many valence electrons?
Paramagnetic or diamagnetic?
Q5. In the ground state of mercury, Hg
a) How many electrons occupy orbitals with n =3?
b) How many electrons occupy d atomic orbitals?
c) How many electrons occupy pz atomic orbitals?
d) How many electrons have spin “up” (ms = +1/2)
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Electron Configuration and Periodic
Table
• Electron configuration of elements related to its
position on the periodic table → elements with
same pattern of outer – shell (valence) electrons
arranged in groups (columns).
• e.g. Group 3A:
B:
[He] 2s2 2p1
•
Aℓ: [Ne] 3s2 3p1
•
Ga: [Ar] 3d10 4s2 4p1
•
In: [Kr] 4d10 5s2 5p1
•
Tℓ: [Xe] 4f14 5d10 6s2 6p1
Electron configuration of Ions ….
• Germanium: remove 2 x 4p electrons
Ge: [Ar] 3d10 4s2 4p2 → Ge2+: [Ar] 3d10 4s2 + 2e• Ge2+: [Ar] 3d10 4s2 → Ge2+: [Ar] 3d10 + 2 e• Same general rule applies to transition metals –
ns electrons always lost before (n-1)d electrons.
e.g. Fe: [Ar] 3d6 4s2
→ Fe2+: [Ar] 3d6 + 2eFe2+ [Ar] 3d6 →
Fe3+: [Ar] 3d5 + ecations formed have the general electron
configurations [noble gas] (n-1)dx
Effective Nuclear Charge (Z*)
• i.e. the nuclear charge experienced by a particular
electron in a multi-electron atom, modified by the
presence of other electrons.
• → In H: only 1 e-, 2s and 2p have the same Energy.
• → Li: 3 e-s, presence of 1s e-s alters the energy of 2s
and 2p subshells.
• Li: 3 protons in nucleus (Z = 3), nuclear charge = +3.
If third e- (2s) is very far from nucleus – experiences
+1 nuclear charge (screened from nucleus by 2 x 1s
e-s).
• Due to the penetration of the 2s orbital into 1s orbital
region, 2s e- experiences charge < +3 but > +1
Electron configuration of Ions
• To form a cation from a neutral atom, one or
more valence electrons are removed.
• electrons are always removed from electronshell of highest n.
• if several subshells are present within the nth
shell, electron/s of maximum ℓ are removed.
e.g. Na ion – remove 3s1 electron from Na atom.
Na: [Ne] 3s1 or
1s2 2s2 2p6 3s1 → Na+: 1s2 2s22p6 + 1 e-
Electron configuration of Ions
• magnetic properties of transition metals
determined by the number of unpaired electrons
in d-orbitals
• Fe3+ - ions are paramagnetic – 5 unpaired
electrons.
• e.g. Cu, Cu+ and Cu2+
Cu: [Ar] 3d10 4s1 →
Cu+: [Ar] 3d10 + eall e-s paired, diamagnetic
→
Cu2+: [Ar] 3d9 + 2eone unpaired e-, paramagnetic
• Z* is greater for s-electrons than p-electrons in
same energy shell
→ s-electrons have lower E than p- electrons
• difference becomes larger as n becomes larger,
i.e. down in a group.
• Penetrating power of sub-shells: s > p > d > f
• Effective nuclear charge experienced by orbitals:
ns > np> nd > nf within the same e- - shell
• Z increases from left to right across the period –
increased number of protons → value of Z* also
increases across the period.
• Z* = Z – S where S = screening constant, how
much the inner (core) electrons shield or screen
the outer electrons from the nucleus.
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Atomic Properties and Periodic Trends
• Similarities in properties of elements result from
similar valence shell electron configuration.
• Atomic electron configuration related to some of
the physical and chemical properties of
elements and why they change.
• Atomic Size
Main Group Elements: atomic radii generally
increase going down a group in the Period Table,
and decrease across a period.
• Size of atom determined by
• From top to bottom – outermost electrons assigned to
orbitals with higher value of n (principle quantum
number)
• Core electrons require space →
• Orbital has no sharp boundary
→ Distance between atoms in element used to
determine the radius.
• e.g. Cℓ – Cℓ bond distance 198 pm → radius of
Cℓ = 99 pm
• In a given period, n of valence electrons is the same.
→ effective nuclear charge, Z*, increases slightly from left
to right
→ attraction between nucleus and electrons increases and
radius decreases.
Transition Metal Atoms
Ionisation Energy
• Slight decrease initially from left to right, then
slight increase.
• Electrons added to d-orbitals, increase in Z*
causes the decrease – only slight due to e- – erepulsion.
• Slight increase due to increasing e- - e- repulsion
when d-electrons are pairing up.
• i.e. the energy required to remove an electron
from an atom in the gas phase.
• Atom in ground state (g) → Atom+(g) + e∆E ≡ ionisation energy, IE
• To separate the electron from atom, energy
must be supplied to overcome the attraction of
the nuclear charge, → IE positive.
• Atoms other than H have a series of IE’s, more
than 1 electron can be removed.
• Removing each subsequent electron requires
more energy due to electron being removed
from increasingly positive ion.
• e.g. 1st ionisation Mg (g)
IE1 = 738 kJ/mol
1s2 2s22p6 3s2
→
• 2nd ionisation
Mg+ (g)
IE2 = 1451 kJ/mol
1s2 2s22p6 3s1
→
• 3rd ionisation
Mg2+ (g)
IE3 = 7733 kJ/mol*
1s2 2s22p6
→
Mg+ (g) + e1s2 2s22p6 3s1
Mg2+ (g) + e1s2 2s22p6 3s0
Mg3+ (g) + e1s2 2s22p5
• *Removal of core electrons requires much more energy
than removal of valence electrons. Core electrons are
not lost in chemical reactions.
Main Groups (s- and p- block)
• First ionisation energies generally increase
across a period and decrease down in a group.
• Across period: increase in effective nuclear
charge, Z*, with increasing atomic number →
attracts the electrons more strongly → atomic
radius decreases → energy to remove an
electron increases.
• Variations due to e- – e- repulsions i.e. group 5
– 6, 4th e- in 2p makes it easier to remove.
• Down in group: electron being removed is
further from nucleus → less nuclear – electron
attractive force → less energy required to
remove electron.
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Electron Affinity (Electron Attachment
Enthalpy ∆EAH)
• Affinity = liking (for electrons), i.e. atoms acquire one or
more electrons to form a negative ion.
• EA : Energy change for a process in which an electron is
acquired by an atom in the gas phase.
A (g) + e- → A− (g) ∆U = ∆EAH
The greater the affinity an atom has for an electron, the
more E is released → the more negative the value of
∆EAH.
e.g. Fluorine: F (g) + e− → F− (g) ∆EAH = – 328 kJ/mol
• large value, reaction is exothermic and product –
favoured.
• Boron – much lower affinity for an electron,
∆EAH = – 26.7 kJ/mol
EA generally becomes less negative going down in
a group – more difficult to form neg. ion.
• Electrons added progressively further from
nucleus → decrease in attractive force between
nucleus and electrons.
• Does not apply to the 2nd period → smaller
volume of n = 2 shell causes significant electron
– electron repulsions.
• Few elements, e.g. N, noble gases – have no
affinity for electrons, ∆EAH = 0
Trends in Ion sizes
• down in a group – same as neutral atom
- positive and negative ions increase in size.
• Compare atom and ion sizes:
• electrons removed to form positive ions →
• nucleus has stronger hold on remaining
electrons →
• i.e. radius of cation smaller than neutral atom,
e.g. Li (152 pm); Li+ (78 pm).
∆EAH and IE represent energy involved in the gain
or loss of electrons by an atom – the formation of
negative and positive ions.
• Increase in Z* across a period → more difficult to
ionise the atom (lose e-)
→ increased attraction for additional electrons
→ element with high IE usually has a more
negative ∆EAH.
• Values for ∆EAH generally become more
negative moving L to R across period.
• Trend not smooth – group 2 lower → adding ewould start p-subshell, more difficult
– group 5 lower → adding e- would start pairing
p – electrons.
e.g. for chlorine:
• Ionisation energy:
• Cℓ (g)
→
Cℓ+ (g) + e∆E = 1251 kJ/mol
2
5
[Ne]3s 3p
[Ne]3s23p4
• Positive E value – E must be added to remove
the e• Electron Affinity:
Cℓ (g)
+ e- →
Cℓ- (g)
∆E = -349 kJ/mol
2
5
•
[Ne]3s 3p
[Ne]3s23p6 – noble gas
structure
• Negative E value → energy released → Cℓmore stable.
→ large decrease also expected when 2 or more
electrons are removed.
• e.g. Aℓ (143 pm); Aℓ3+ (57 pm) – loses n = 3 shell
• anions are larger than atoms from which they
are formed →
• When electron removed in Li, attractive force of
3 p+ exerted on only 2 electrons →
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• Compare isoelectric elements (same
number of electrons)
N3- O2- F- Na+ Mg2+
No. of electrons
10 10 10 10 10
No. of protons
7
8
9 11 12
Ionic radius (pm)
146 140 133 98 79
• → electron – proton attraction increases
→ radius decreases
• ← electron – electron repulsion increases
←
• Q6. Compare C, O and Si – place in order of
increasing atomic radius.
Which has the largest ionisation energy?
• Q7. Compare B, Aℓ and C for atomic radius and
ionisation energy.
• Q8. Why is fluorine a smaller atom than carbon,
but also smaller than chlorine? Use spdf
electron configurations of all elements to
explain.
• Q9. Give the trend in the sizes of the ions K+, S2and Cℓ-. Explain.
• Also Textbook No. 34, 41, 42, 48, 50 and 52.
Group 1 metals form compounds containing 1+ ions,
i.e. Na+, K+
2 Na (s) + Cℓ2 (g) → 2 NaCℓ (s), i.e. Na+ and Cℓ- ions
• 2 K (s) + 2 H2O (ℓ) → 2 KOH (aq) + H2 (g)
i.e. K+ and OH- ions
• Not NaCℓ2 or K(OH)2 as the 2+ ion is not
favourable
• C - does not have a favourable IE or ∆EAH does not easily form a cation or an anion →
don’t find many ionic compounds containing C,
C usually shares electrons with other elements,
CO2, CCℓ4
Q10. Using your knowledge of the trends in
element sizes, explain why the density of the
elements increases from K through V
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