Chem 373- Lecture 18: Orbitals of the Hydrogen Atom

8/3/2019 Chem 373- Lecture 18: Orbitals of the Hydrogen Atom

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Lecture 18: Orbitals of the Hydrogen

AtomThe material in this lecture covers the following in Atkins.

The structure and Spectra of Hydrogenic Atoms

13.2 Atomic orbitals and their energies(c) Shells and subshells(b) s-orbitals(c) Radial distribution function(d) p-orbitals(e) d-orbitals

Hydrogenic atomic orbitals

Lecture on-lineHydrogenic atomic orbitals (PDF Format)

Hydrogenic atomic orbitals (PowerPoint)

Handout for this lecture


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Audio-visuals on-line key hydrogen orbitals (PowerPoint)(From the Wilson Group,***)key hydrogen orbitals (PDF)(From the Wilson Group,***)

Vizualization of atomic hydrogen orbitals (PowerPoint)

(From the Wilson Group,***)Vizualization of atomic hydrogen orbitals (PDF)(From the Wilson Group,***)

Slides from the text book (From the CD included in Atkins ,**) Interactive Hydrogen Orbital Plots (For Mac users only)( Visualizes all the

angular and radial wavefunctions of the hydrogen atom, *****)


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Orbitals of Hydrogenic Atom

Thenlm

orbitals for the hydrogenic atom are given by(r, , R(r) Y ( , ; n = 1,2,3 ...l < n - 1; m = - l, - l+ 1, ....l -1, l

nl l,m ) )=

Wherel ml m

im

Y , ) =2l+14

P

are eigenfunctions to L and L

l,m l|m|

2 z

(( | !|( | !|)

(cos ) exp[ ] +

With the radial part given as

R r Nn

L r enl nll

n ln( ) ( ),

/ = 2


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Hydrogen Levels

nn l m=1,

(, ,l = 0

( , ) = R (r)Y ( , )nl l,m

R r e1 0

3 22

,

//

( ) =

2

Za o

No rnodes. R everywhere positive1,0 ( )

For l = 0 we have m = 0;

Yo,o =1

4

Value of Y is uniform oversphere

oo1s orbital


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Orbitals of Hydrogenic Atom...prop.dens.

The probability density(r, , ) = (r, , (r, ,nlm nlm nlm PP ) )*

A constant-volume electron-sensitivedetector (the small cube) gives its greatest

reading at the nucleus, and a smallerreading elsewhere. The same reading isobtained anywhere on a circle of given

radius: the s orbital is sphericallysymmetrical.

PP 100

3 22

3 22

3,

/ / *

/ /

( , . )r e Y

e Y e

oo

oo

=

=

2Z

a

2Z

a1 Z

a

o

o o


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Representations of the1 s and 2 s hydrogenicatomic orbitals in termsof their electron densities

(as represented by thedensity of shading).

The probability density

(r, , ) = r, , r, ,nlm nlm nlmPP

) )

PP

PP

1 00

3

200

32 22

12

,

/

( )

( ) ( )

r e

r e

=

=

1 Z

a

132

Za

o

o

Orbitals of Hydrogenic Atom...prop.dens.


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constant-volume electron-sensitiveetector (the small cube) gives its greatesteading at the nucleus, and a smallereading elsewhere. The same reading is

btained anywhere on a circle of givenadius: the s orbital is sphericallyymmetrical.

PP 1 00

3 22

3 22

3

,

//

//

( , . )r e Y

e Y e

oo

oo

=

=

2Z

a

2 Za

1 Za

o

o o

Radial probability density


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he radial distribution function P gives the probability that the electron

ill be found anywherein a shell of radius r. For a 1 s

lectron in hydrogen, P

is aaximum when r is equal tohe Bohr radius a 0. The valuef P is equivalent to the reading

hat a detector shaped like apherical shell would give asts radius was varied.



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The probability of finding the electronbetween r and r + dr is :

| R (

| R (

nlm

nlm

nl

nl

P r dr r

r r d d r dr

r Y d d r dr

r r dr

Thus

P

nl oo

oo

lmoo

( ) ( , , )

( , , )

( ) | ( , ) |

( ) |

= =

=

=

PP

PP

dv

sin

sin

1

the radial probability density is

2

22

2 2 22

2 2

nlnl r r r( ) ( ) |=| R (nl2 2



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The radial probability density is

most probable radius corrsponds

to the maxima for1s

r max

P r r r

The

P rFor

aZ

nl

nl

o

( ) ( ) |

( ).

=

=

| R (nl2 2



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Orbitals with increasing l are called

p - orbitals m = - 1, 0,1n1m =R r Yn m1 1( ) ( , )

n2m d - orbitals m = - 2, -1, 0,1, 2=R r Yn m2 2( ) ( , )

n3m f - orbitals m = - 3, -2, -1, 0,1, 2, 3=R r Yn m3 3( ) ( , )


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R

R (r) R (r)R (r) R (r)

nl

nl nlnl nl

is a solution to

+ + + + =h h2 2

2

2

222

41

2

{ ) {

( )}

r r rZe

rl lmr

Eo

Veff

As l increases the centrigugal term

becomes more repulsive near nucleus at small r.Hence R tend to zero as r o with increasingspead as l becomes larger

nl

h 2

21

2

l l

mrr

( )

( )

+

Orbital near nucleus


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Close to the nucleus, p orbitals are proportional to r ,d orbitals are proportional to r 2,and f orbitals are proportional to r 3.

Electrons are progressively excludedfrom the neighbourhood of the nucleusas l increases. An s orbital has afinite, nonzero value at the nucleus.

Orbital near nucleus


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he variation of the mean radius of aydrogenic atom with the principal andrbital angular momentum quantum

umbers. Note that the mean radius lies inhe order d < p < s for a given value of n.

Mean radius

The mean radius is given as theespectation value

< >= r r dvnlm nlm

For the hydrogenic atom

< r > nl = + +

nl l

naZo2

2112

11

(( )


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The boundary surface of an s orbital,within which there is a 90 per cent

probability of finding the electron.


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nlm (r, , R(r) Y ( ,n = 2 , l = 1

nl l,m) )=

211

1 238

(r, , R(r) Y ( , = -R(r)21 1,1 21) ) sin/

= e

i

m = 1 0 1, ,

2101 23

4(r, , R(r) Y ( , = R(r)21 1,0 21) ) cos

/

=

21 1

1 238

= (r, , R(r) Y ( , = R(r)21 1,-1 21) ) sin/

e i

Let us now look at the angular part of the p - functions

Angular part


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2101 23

4(r, , = R(r) 21) cos

/ l = 1 , m = 0

xy

z

L2 =h 2l(l +1) Lz = hm l

m=0

The electron has an angular momentum L

such that L L =z - component of L is L

2

z

r

hs +

=l l

The o( )1

Angular part


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211

1 23

8(r, , -R(r) 21) sin

/

= ei


x

y

z

L2 = h 2l(l +1)

Lz= h

m l

x

yz

r v

vL =

vr

vv

l = 1 , m = 1



2

z

r

hs

h

+=

l lThe

( )1

Angular part


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21 1

1 23

8 (r, , = R(r) 21) sin

/e i l = 1 , m = - 1

x

yz

L2 =h 2l(l +1)

Lz = hm lm=-1



2

z

r

hs

h +

= l l

The( )1

Angular part


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Plotting

p z

angular part of

2101 23

42(r, , = R(r) 21) cos

/

=


R =k cos +

-

Draw vector R of lengthkcos through each point

( , ) on sphere

r

Draw surface through allvectors R

Real orbitals


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2111 23

8(r, , -R(r) 21) sin [cos sin ]

/

= +i


21 11 23

8

(r, , = R(r) 21) sin [cos sin ]/

i

The remaining two orbitals are complex

However by linear combinations we can get the real orbitals

22

38211 21 1

1 2p

iy =

{ + } = R(r) 21 /

sin sin

212

38211 21 1

1 2p x =

{ - } = R(r)21 /

sin cos

Angular part


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22

3

8211 21 1

1 2p

iy =

{ + } = R(r) 21

/

sin sin

2 12 38211 21 11 2

p x = { - } = R(r)21 /

sin cos


The orbitals 2p and 2p have the same energies and eigenvalues to

L as andx y

2211 21-1 . However only the latter are eigenfunctions of L z

+

-

x

z

+-

y

2p y 2p x

Angular part


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The boundary surfacesof p orbitals. A nodalplane passes throughthe nucleus and

separates the two lobesof each orbital.The dark and light areas denoteregions of opposite signof the wavefunction.

Angular part


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Grotrian diagram that summarizes theppearance and analysis of the spectrum of tomic hydrogen. The thicker the line, theore intense the transition.

In transitions betweendifferent energy levels of

the hydrogenic atom

the following selectionrules apply

l = 1m = o, 1

Transitions


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What you need to learn fromthis lecture about the hydrogen atom

for the quizz and the final exam

The

R r

definition of the radialdistribution functionP(r) = r and how itis used to calculate the

expaction values < r

2

n

( )2

>

The rbehaviour of R nearthe nucleus (Fig 13.16)

nl ( )

Understand how the

balance between kineticenergy and potentialenergy shapes R

3.10nl

Fig.1

Selection rules for electronictransitions in thehydrogen atom

l = 1m = o, 1

Memorize the relation betweenreal p - orbitals (p and

imaginary p - orbitals (pUnderstand the physicaldifference. Understandplots of (p

xo,

x

, , )

, ).

, , ).

p p

p p

p p

y z

y z

1 1

Same for d and f great,but not required


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Appendix : Orbitals of Hydrogenic Atom..why nodes

R r e20

3 242

1

2( ) ( )

//=

1

2 2

Z

a o

n n on l m

= ,(, ,

l = 0

( , ) = R (r)Y ( , )nl l,m

R r e30

3 22 66 2

19

( ) ( )/

/=

+

19 3

Za

o

One node at

= 2Zr/anode : 2a

o

o

( )

/

2 12

0 =with

Z

two nodes at

= 2Zr/anodes : 1.9a

7.1Zr/a

o

o

o

( )

/

6 21

902 + =

withZ

and

For l = 0 we have m = 0;

Yo,o = 14


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n n on o oo

= ,(, ,

l = 0( , ) = R , (r)Y ( , )n o o,oo

Thus Y Y

R Y R Y dv

oo oo

n o oo n o oo

any two s - orbitals R and Rmust be orthorgonal

n'o no

' '000

2

0

=

Why has R n - 1 nodes ?nl ( )r

We have seen previously that

any two independent solutionsto the Schrdinger equation mustbe orthogonal :

i =j ijdv



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R Y R Y dv

Y Y d d R r R r r dr

n o oo n o oo

oooo

oo n oo

r

no

' '

'sin ( ) ( )

000

2

22

0

0

=

=

1

normalized

R r R r r dr

Y

n oo

r

no

oo

' ( ) ( ) =2 0

The

R r e

R function is positiveevery where

2Z

a

1o

o1 0

3 22

,

//( ) =


f


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Thus for R to be orthogonal

to R

must have positive andnegative regions

2o

1o

R r R r r dr

R

oo

r

o1 22

20

0 =( ) ( )

R r e20

3 242

12

( ) ( )/

/=

12 2

Za o

For R to be orthogonal toR and R two nodes arerequired etc...

3o

1o 20



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e1

e 2

e3e 4

e 5

Appendix : Drawing Orbitals of Hydrogenic Atom An asideHow do we plot Y ?lm

Consider a unit spherewithradius 1. Draw a large

number ofunit vectors fromthe origin ofthe sphere indifferent directions(e.i. different , ) e e ee e

1 2 3

4 n

r r r

r r

, , ,...

Calculateat

where R eDraw

n

n n

the value of Ythe position of each e

(e.i for each Constructthe vectors : R R R R

YR

lm

n

n

1 2 3 n

lm

n

( , )

, )., , , ..

: ( , )

r

rr=

R2

R 3

1R

R4

R 5


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An aside

R2

R 3

1R

R4

R 5

Draw

This

a surface through the

endpoints of all Rsurface represents Y

nlm

r

.( , )

Appendix : Drawing Orbitals of Hydrogenic Atom

A idd b l f d


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An aside

For Y we have

R

oo

n

( , )

r r= 1

2for all ne n

We

n n

have that Y is sphericale.u. the same for all

oo ( , ),

THus Y representing the angularpart of a ns function is representedby a sphere

oo

ns - orbital

Appendix : Drawing Orbitals of Hydrogenic Atom

Chem 373- Lecture 18: Orbitals of the Hydrogen Atom

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Transcript of Chem 373- Lecture 18: Orbitals of the Hydrogen Atom