Chem 373- Orbitals of Hydrogenic Atom

8/3/2019 Chem 373- Orbitals of Hydrogenic Atom

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Orbitals of Hydrogenic Atom

The orbitals for the hydrogenic atom are given by

(r, , ) R(r)nlYl,m( , ) ; n = 1, 2, 3 ...

l < n - 1; m = - l, - l+1, ....l - 1, l

and Rnl solution to

h

2

2{

2

Rnl(r)2r

2r

Rnl(r)r

) { Ze4 or

h

2

l(l 1)2 mr2

}Rnl(r) EnRnl(r)

Where

( , ) = Yl,m(( , ) =2l+1

4

(l | m! |

(l | m!|)

Pl|m| (cos ) exp[im ]

are eigenfunctions to L2 and Lz


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Hydrogen Levels

R1,0(r) 2Z

ao

3 / 2

e / 2

n 1, l = 0

n,l,m ( , ) = Rnl(r)Yl,m(( , )

R1,0(r) 2Z

ao

3 / 2

e / 2

No nodes. R1,0

(r) everywhere positive

For l = 0 we have m = 0;

Yo,o1

4

Value of Yoo is uniform over

sphere

1s orbital

O bi l f H d i A


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The balance of kinetic and potentialenergies that accounts for thestructure of the ground state ofh dro en (and similar atoms).

(a) The sharply curved but localizedorbital has high mean kinetic energy,but low mean potential energy

(b) the mean kinetic energy is low,but the potential energy is not very

favourable;

; (c) the compromise of moderatekinetic energy and moderatelyfavourable potential energy.


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R20 (r)1

2 2

Z

ao

3 / 2

(21

2)e / 4

n no, l = 0

n,l,m ( , ) = Rnl(r)Yl,m(( , )

R30 (r) 19 3

Zao

3 / 2

(6 2 19

2 )e / 6

One node at

(21

2) 0

with = 2Zr/ao; node : 2a

o

/ Z

two nodes at

(6 21

9

2

) 0

with = 2Zr/ao; nodes : 1.9ao / Z

and 7.1Zr/ao

For l = 0 we have m = 0;

Yo,o1

4


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n no, l = 0

no ,o,o( , ) = Rno ,o (r)Yo,o(( , )

Thus any two s - orbitals Rn'oYoo and RnoYoomust be orthorgonal

Rn'oYoo000

2Rn'oYoodv 0

Why has Rnl(r) n-1 nodes ?

We have seen previously that

any two independent solutions

to the Schrdinger equation must

be orthogonal :

i jdv ij


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Rn'oYoo000

2

Rn'oYoodv 0

Yoooo

2

Yoo sin d d Rn'oo

r

(r)Rno(r)r2dr 0

1 Rn'oo

r

(r)Rno(r)r2dr 0

Yoo normalized

The R1o function is positive

every where

R1,0(r) 2Z

ao

3 / 2

e / 2


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Thus for R2o to be orthogonal

to R1o

R1oo

r

(r)R 2o (r)r2dr 0

R20 must have positive and

negative regions

R20 (r)1

2 2

Z

ao

3 / 2

(21

2)e / 4

For R3o to be orthogonal to

R1o and R20 two nodes arerequired etc...


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The probability density

Pnlm(r, , ) = nlmr, , ) nlmr, , )

A constant-volume electron-sensitive

detector (the small cube) gives its greatest

reading at the nucleus, and a smallerreading elsewhere. The same reading is

obtained anywhere on a circle of given

radius: thes orbital is spherically

symmetrical.

P1,00 (r, . ) 2Z

ao

3 / 2

e / 2Yoo

2

Z

ao

3 / 2

e/ 2

Yoo1 Z

ao

3

e


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Representations of the1s and 2s hydrogenic

atomic orbitals in termsof their electron densities(as represented by thedensity of shading).

The probability densityPnlm(r, , ) = nlmr, , ) nlmr, , )

P1,00 (r)1 Z

ao

3

e

P2o0 (r)1

32

Z

ao

3

(21

2)2 e / 2


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e1

e2

e3

e4

e5

Orbitals of Hydrogenic Atom An asideHow do we plot Ylm ?

Consider a unit sphere

withradius 1. Draw a large

number ofunit vectors fromthe origin ofthe sphere in

different directions

(e.i. different , )re1 ,

re2,

re3,

re4 ...ren

Calculate the value of Ylm( , )

at the position of eachren

(e.i for each n, n). Constructthe vectors : R1 , R2, R3,..Rn

where :rRn

renYlm( , )

Draw Rn

R2

R3

1R

R4

R5

O bi l f H d i A A id


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Orbitals of Hydrogenic Atom An aside

R2

R

31

R

R4

R5

Draw a surface through the

endpoints of all rRn .

This surface represents Ylm( , )

O bit l f H d i At An aside


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Orbitals of Hydrogenic Atom An aside

For Yoo( , ) we have

rRn

1

2

ren for all n

We have that Yoo( , ) is spherical

e.u. the same for all n, n

THus Yoo representing the angular

part of a ns function is representedby a sphere

ns - orbital

O bi l f H d i A


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

probability of finding the electron.

Orbitals of Hydrogenic Atom M di


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The variation of the mean radius of a

hydrogenic atom with the principal and

orbital angular momentum quantum

numbers. Note that the mean radius lies inthe order d p sfor a given value ofn.

Mean radius

The mean radius is given as the

espectation value

r nlmr nlmdv

For the hydrogenic atom

< r >nl n2 1

1

2(1

l(l 1)

n2aoZ

O bit l f H d i At


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A constant-volume electron-sensitive

detector (the small cube) gives its greatest

reading at the nucleus, and a smaller

reading elsewhere. The same reading is

obtained anywhere on a circle of given

radius: the s orbital is spherically

symmetrical.

P1,00 (r, . ) 2Z

ao

3 / 2

e / 2Yoo

2 Z

ao

3 / 2e / 2Yoo

1 Z

ao

3e



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The radial distribution function

P gives the probability that the electron

will be found anywhere

in a shell of radiusr. For a 1s

electron in hydrogen,P is a

maximum whenr is equal to

the Bohr radiusa0. The value

ofP is equivalent to the readingthat a detector shaped like a

spherical shell would give as

its radius was varied.

Orbitals of Hydrogenic Atom Radial probability density


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The probability of finding the electron

between r and r + dr is :

Pnl(r)dr Pnlm(r, , )dvoo

2

P

nlm

(r, , )rsin d d r2droo

2

|Rnl((r) |2 Ylm( , ) |

2 sin d d r2droo

2

1 |Rnl

((r) |2 r2dr

Thus the radial probability density is

Pnl(r) | Rnl((r)|

2

r

2

Radial probability density

Orbitals of Hydrogenic Atom Radial probability density


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The radial probability density is

Pnl(r) | Rnl((r)|2 r2

The most probable radius corrsponds

to the maxima for Pnl(r).

For 1s

rraoZ

Radial probability density



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Orbitals with increasing l are called

n1m Rn1(r)Y1m ( , ) p - orbitals m = -1,0,1

n2m Rn2(r)Y2m( , ) d - orbitals m = - 2, -1, 0,1,2

n3m Rn3(r)Y3m( , ) f - orbitals m = - 3, -2, -1,0,1, 2, 3


Structure of Hydrogenic Atoms


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Orbitals of Hydrogenic AtomStructure of Hydrogenic Atoms

Rnl is a solution to

h2

2

{2Rnl(r)

2

r

2

r

Rnl(r)

r

) {Ze

4 or

h2l(l 1)

2 mr2

}Rnl(r) ERnl(r)

Veff

As l increases the ve centrigugal termh

2l(l 1)

2 mr2

becomes more repulsive near nucleus at small r.Hence Rnl(r) tend to zero as r o with increasing

spead as l becomes larger



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R1,0(r) 2Z

ao

3 / 2

e Zr/r

R21(r)1

2 6

Z

ao

5 / 2

reZr / 2ao

R3,2 (r)4

81 30

Z

ao

7 / 2

r2eZr / 4ao

for r o ; 1s 2Z

ao

3 / 2

for r o ; 2p r

for r o ; 3p r2

The asymptotic behaviour of the radial s,p, and d functions

differ as a result of the centrifugal term in the radial Schrdinger

equation



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Close to the nucleus,p orbitals are proportional to r,d orbitals are proportional to r2,and f orbitals are proportional to r3.Electrons are progressively excludedfrom the neighbourhood of the nucleusas l increases. An s orbital has a

finite, nonzero value at the nucleus.



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nlm(r, , ) R(r)

nlY

l,m( , )

n = 2 , l = 1

211(r, , ) R(r)21Y1,1( , ) = -R(r)213

8

1/ 2

sin ei

m 1,0,1

210(r, , ) R(r)21Y1,0( , ) = R(r)213

4

1/ 2

cos

21 1(r, , ) R(r)21Y1,-1( , ) = R(r)213

8

1/2

sin e i

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



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211(r, , ) -R(r)213

8

1/ 2

sin ei


x

y

z

L2

h2l(l 1)

Lz hml

x

yz

r v

vL

vr

vv

l = 1 , m = 1

The electron has an angular momentumrL

such that L L = h2l(l 1)

The z - component ofsL is Lz h



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210(r, , ) = R(r)213

4

1/ 2

cos l = 1 , m = 0

x

y

z

L2

h2l(l 1)

Lz hml

m=0

The electron has an angular momentum rL


The z - component ofsL is Lz o



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21 1(r, , ) = R(r)213

8

1/ 2

sin e i l = 1, m = -1

x

yz

L

2

h

2

l(l 1)

Lz hml

m=-1

The electron has an angular momentum rL


The z - component ofsL is Lz h



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Plotting angular part of

210(r, , ) = R(r)213

4

1/ 2

cos 2pz

R k cos

+

-

Draw vectorrR of length

kcos through each point

( , ) on sphere

Draw surface through allvectors R



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211(r, , ) -R(r)213

8

1/2

sin [cos isin ]

21 1(r, , ) = R(r)213

8

1/ 2

sin [cos isin ]

The remaining two orbitals are complex

However by linear combinations we can get the real orbitals

2pyi

2{ 211 + 21 1} = R(r)21

3

8

1/2

sin sin

2px1

2{ 211 - 21 1} = R(r)21

3

8

1/ 2

sin cos



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2pyi

2{ 211 + 21 1} = R(r)21

3

8

1/2

sin sin

2px1

2{ 211 - 21 1} = R(r)21

3

8

1/ 2

sin cos

The orbitals 2px and 2py have the same energies and eigenvalues to

L2 as 211 and 21-1 . However only the latter are eigenfunctions of Lz

+

-

x

z

+-

y

2py 2px



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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 sign

of the wavefunction.



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A Grotrian diagram that summarizes the

appearance and analysis of the spectrum of

atomic hydrogen. The thicker the line, the

more intense the transition.

In transitions between

different energy levels of

the hydrogenic atomthe following selection

rules apply

l = 1m = o, 1

Chem 373- Orbitals of Hydrogenic Atom

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