CHEM1011 Revision Lecture

8/13/2019 CHEM1011 Revision Lecture

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CHEM1011Essentials of Chemistry 1A

Revision lecture

A/Prof. John A. StrideDalton 131

[email protected]


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Questions.

question in the tutorial set 3, for number 2 about the wavelength lightemitted by atomic hydrogen. Can you please explain it in the lecture?and also electron affinity.

calculations on light and it's energies with use of the various equations,

a) Give a valid set of four quantum numbers for each of the electrons inthe nitrogen atom (in the ground state)

b) Show how the Pauli principle and Hund's rule are involved in youranswer to part (a)

quantum numbers and atomic orbitals.

electron configuration of atoms


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Electromagnetic radiation: light

Is a self-sustaining oscillation of electric andmagnetic fields moving through space.

Is characterised by its frequency (!) or wavelength(").


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Electromagnetic radiation units of frequency are time-1(usually sec-1or

Hz, where 1 sec-1= 1 Hz).

units of wavelength are length (usually the SIunit: metres), with smaller units for shortwavelengths: m (= 106m); nm (= 109m);

(= 1010

m).

Its speed of propagation depends upon thedensity of the substance passing the light.For vacuum, c= 2.998 x 108m s-1.

c ="#


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Question?

Calculate the wavelength of electromagneticradiation of frequency != 104 MHz?Assume vacuum is the medium, so c= 2.998 x 108 m s-1.

Rearrange equation in terms of "; "=c

#

Thus,"=

2.998$108 ms%1

104 $106 s%1=

2.88m


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visible light The range of the electromagnetic waves that we call visible

lightis a very small fraction of the completeelectromagnetic spectrum

ROYGBIV


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visible light The range of the electromagnetic waves that we call visible

lightis a very small fraction of the completeelectromagnetic spectrum

ROYGBIV

Monochromatic radiation:a selection of one frequency of e.m. radiation (in practical

terms, a narrow band of frequencies) which can be used for various scientific

measurements etc.

Polychromatic radiation:radiation consisting of many frequencies - such as the light

we receive from the sun.


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Question? Calculate the energy of red light (wavelength = 700 nm).

Compare it with blue light (wavelength 470 nm).

From the formula c ="#;

" =c

#= 2.998

$

10

8

ms

%1

700 $10%9 m

=4.28 $1014 s-1

Then E =h" = 6.626 $10%34 Js( )4.28 $1014 Js( ) =2.84 $10%19 J

For blue light (shortcut);

E =h" =h $c

#=

6.626 $10%34 Js( )2.998 $108 ms%1( )470 $10

%9 m( )=4.23 $10

%19 J

blue light has a higher energy than red


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Quantisation of energy

White (polychromatic) light passing througha gas composed of single atoms has lines -or specific frequencies - removed

the lines represent the absorptionspectrum ofthe specific gas


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A similar experiment can be performed inemission: the same gases heated by an electrical

discharge (passage of current) produces lightconsisting of a series of lines

emissionspectrum


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The line spectra vary with the gas:

the spectra also vary with the pressure of the gas


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Recall that a bound electron has negativepotential energy:


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In fact the bound electrons can occupyvarious discrete states

Electrons can undergo transitions between

these states, rather like a ball on a staircase


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the hydrogen atom

UV: Lyman seriesvis: Balmer series

IR: Paschen series

1

"= R

H

1

n1

2#

1

n2

2

$

%&

'

()


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the hydrogen atom

1

"= R

H

1

n1

2#

1

n2

2

$

%&

'

()

The value of nfor each level is called its principalquantum number.

An atom in its lowest energy level is said to be in itsground state.

Atoms can undergo transitions to higher energy levels(excited states) by being heated strongly, or bycolliding energetically with other bodies.

Excited states are unstable, and such states undergo

transitions to lower energy levels (eventually to theground state) by loss of energy in the form of photons.

E= h"


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the hydrogen atom

1

"= R

H

1

n1

2#

1

n2

2

$

%&

'

()

E= h"


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Question? One specific calculation of chemical interest is how much

energy it takes to completely remove an electron from ahydrogen atom in its ground state (this is called the ionisationenergy of the H atom).

"Complete removal" of an electron means that the electron has

been moved to the n=!

energy level.For one H atom, "E=Efinal # Einitial

"E=# 2.18 $10#18 J( )$ 1

nfinal2 #

1

ninitial2

%

&''

(

)**=# 2.18 $10

#18 J( )$ 0 #1( )

"E=2.18 $10#18 JFor a mole of H atoms, the molar ionisation energy would be :

"E= 2.18 $10#18 J( )$ 6.023$1023 mole#1( ) =1.31$106 Jmol#1


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Question: (tut. 3, Q.2)The wavelength of light emitted by atomic hydrogen is given by the equation:

WhereRHis the Rydberg constant for hydrogen with the valueRH= 1.096776 "107m-1.

(a) The Lyman series in the spectrum of the hydrogen atom originate from transitions for which the lower

energy level has n= 1. Calculate the wavenumber (either m-1or cm-1), wavelength (either in nm or m)and the frequency (s-1or Hz) of the two longest wavelength lines in the Lyman series. Indicate theregion of the spectrum corresponding to these transitions.

(b) If the electron of the hydrogen atom is raised from the lowest level (n= 1) to the energy level n= !, theelectron becomes free of the atom and the H+(g) ion is produced. Calculate the energy required for thisprocess (in J atom-1and kJ mol-1)

1

"= R

H

1

n1

2# 1

n2

2

$

%& '

()


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Question: (tut. 3, Q.2)The wavelength of light emitted by atomic hydrogen is given by the equation:

WhereRHis the Rydberg constant for hydrogen with the valueRH= 1.096776 "107m-1.

(b) If the electron of the hydrogen atom is raised from the lowest level (n= 1) to the energy level n= !, the

electron becomes free of the atom and the H+(g) ion is produced. Calculate the energy required for thisprocess (in J atom-1and kJ mol-1)

1

"= R

H

1

n1

2# 1

n2

2

$

%& '

()

1

"= R

H

1

12#

1

$2

%

&''

(

)**=1.096776+10

7 1

1# 0

%

&'

(

)*=1.096776+107 m#1

E = h, and c =,": -, =c

"

and so E = hc

"

=6.626068+10-34 + 3.0+108 +1.096776+107 =2.178721+10#18 Jatom#1

2.178721+10#18 Jatom#1 = 6.022045+1023 +2.178721+10#18 Jmol#1 = 1312036J mol#1 =1312kJmol#1


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QUANTUM MECHANICS OF THE H ATOM

solutions to the Schrdinger equation:

have exact analytical forms for the 1 electron H atom.

the electron density at any point is given by the square of thewavefunction:

however, the Heisenberg uncertainty principle limits theability to know both the position and energy of a quantum

particle such as an electron.

H"=E"

"2=""

#

We therefore need to think about electronprobability densities.


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quantum numbers

principal quantum number, n% n= 1, 2, 3, 4,., !

azimuthal (angluar mometum) quantumnumber, l% l= 0, 1, 2, 3,., (n- 1)

magnetic quantum number, ml% ml= (-l), (-l+ 1),., 0,., (l- 1), (l)

spin quantum number,s% s= +1/2 or -1/2


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Question?

For an orbital with n= 3, what values of land mlare allowed?

Values of ldepend upon the nvalue, ranging from 0 up to n 1 in

integer steps:

thus l = 0, 1 and 2.

In turn, the mlvalues are dependent upon the lvalues.

For each lvalue, ml= +l,...0-l by integer steps:

For l = 0, ml

can only take the value of 0.

For l= 1, ml takes the values ml = +1, 0 and 1.

For l= 2, ml takes the values ml = +2, +1, 0, 1, and 2.


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SHELL

A set of orbitals with the same value ofn

are called a shell.For the hydrogen atom, with its single electron, all orbitals inthe same shell have the same energy.

ORBITAL SYMBOLS

A set of orbitals with the same nand lis called a sub!shell.

For convenience these are labelled with letters:

quantum numbers

If l= 0, the orbitals are called sorbitals,If l= 1, the orbitals are called porbitals,If l= 2, the orbitals are called dorbitals,If l= 3, the orbitals are called forbitals.

Orbitals are labelled with their principalquantum number,n,followed by the orbital symbol (equivalent tol).e.g. a "2s" orbital has n= 2 and l= 0.


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porbitals(l= 1)

Eachporbital consists of two lobes of electron density onopposite sides of the nucleus, with a nodal plane (zeroelectron density) between the lobes.

Since aporbital has l= 1, mlcan have the values of +l, 0and l, thereby giving rise to the threeporbitals:px,pyandpz.

The threeporbitals have their lobes at right angles to one-another

and can be pictured to lie along a set of x,y,zCartesian axes.


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dorbitals(l= 2)

dorbitals come in sets of 5, since dorbitals have l= 2, which gives the

values of ml= +2, +1, 0, 1 and 2, (i.e. five possible values). Unlike the porbitals, the dorbitals are not simple 90 image rotations of one-another.

Four of them consist of four lobes of electron density arranged in a quadrantabout the nucleus, with two nodal planes between the lobes. Of these, onehas its four lobes along thexandyaxes of a set of Cartesian axes. Relative

to this orbital, the other three have their lobes between the axes, in each ofthe planes defined by the pairs of axes.

The other dorbital has just two main opposed lobes of density along the zaxis, with a small torus of density at the waist.


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atomic orbitals

atoms can absorb energy(photons) to promote electronsto excited states

the Coulombic attractionbetween nucleus & electronslead to a contractionof shellswith increasing nuclearcharge


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multi-electron atoms the electrons around a multi-electron atom repel one-

another in a complicated way meaning that it isdifficult to obtain analytical solutions to theSchrdinger equation.

a useful approximation is that the orbitals resemblethose of the hydrogen atom

two factors come in to play in these atoms: increased nuclear charge

electron-electron repulsions%shielding


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multi-electron atoms


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electronic shielding

Shielding does not affect the electrons of anouter shell equally. Theselectrons of an outer shell have one or more

small lobes of density close to the nucleus (insidethe region of shielding electrons), and thus are lessaffected by shielding.

This effect is smaller forporbitals (l= 1) andprogressively less for d(l= 2) andf(l= 3) orbitals.


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electronic shielding

To summarise, the degree of "penetration" to the nucleus iss>p> d>f. As a result of the off-setting of shielding by

different amounts of penetration, the energies of orbitals ineach shell follow the orders


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orbital filling

orbitals are filled from the lowest energylevels up..

a simple way to remember the ordering:


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Questiona) Give a valid set of four quantum numbers for each of the electrons in the

nitrogen atom (in the ground state)

b) Show how the Pauli principle and Hund's rule are involved in your answerto part (a)

Nitrogen has the ground state electronic configuration: 1s22s22p3:

1selectrons: n= 1, l= 0, ml= 0, s= +1/2& -1/22selectrons: n= 2, l= 0, ml= 0, s= +1/2& -1/22pelectrons: n= 2, l= 1, ml= -1, 0, +1, s= +1/2(for each of the 3 electrons)

Pauli Principle: no more than 2 electrons in each orbital

each of the s obitals has an occupancy of 2, despite being lower in

energy than the 2p orbitals.

Hunds Rule: maximise the multiplicity when filling orbitalsthe electrons in the 2p orbitals populate each of the px, py& pzwith one

electron (3 values of ml) rather than 2 electrons entering any one


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GOOD LUCK!

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CHEM1011 Revision Lecture