Chapter 1: Atomic structure

Scheduled Lectures: 4 (max)

 Atomic Structure

Atomic structure: Overview of Bohr’s atomic model,

Schrodinger wave equation, Interpretation of wave
function (radial and angular), Hydrogen like atom,
Concept of atomic orbitals.

Books:
1. Atkins’ Physical Chemistry by Peter Atkins and Julio de Paula
2. Inorganic Chemistry by Gary L. Miessler and Donald A. Tarr
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 Atoms
• Two regions.
• Nucleus- protons and neutrons. (Occupies very
small space)
• Electron cloud- region where you might find
an electron. [Occupies large (most of the)
space]
Subatomic particles

Name Symbol Discovered by, year

Electron e- J. J. Thomson, 1897

Nucleus n+ p E. Rutherford, 1911

Proton p+ H.G.J. Moseley, 1913 3

 Atomic Structure

Bohr’s Model (planetary Schrödinger Model (electron

model)-1914 cloud model)- 1926
Assumes electron as particle Assumes electron as wave

Got Nobel Prize in 1922 for Got Nobel Prize in 1933

foundational contribution to for his Schrödinger wave
understand atomic structure. equation for H-like atoms.

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Line Spectram Black Body Radiation

The kinetic energy of the emitted electron Ek = h −  = h( − 0)

 is the work function (minimum energy required to remove an electron from the metal surface )
 = h0 , 0 is the threshold frequency and for radiation with  > 0, electron emitted.
Rutherford Atomic Model
 Bohr’s Model

➢ Electrons revolves around nucleus (positively charged) in well

defined orbits, which are distinctly separated from each other.

➢ Spiraling of electron does not occur to nucleus because electron can

not exist in between the allowed orbits.

➢ This model is very similar to our planetary model.

The orbits are discrete, so the energy of an electron in an orbit is quantized.

mvr = nh/2π

𝑛2 ℎ 2
v2 = 4π2 𝑚2𝑟² ……(1)

The electrostatic attraction between electron and nucleus must be equal to

centrifugal force.
𝑚𝑣2 𝑍𝑒 2 𝑍𝑒 2
2v ………..(2)
= 2 =
𝑟 4πεₒ𝑟 4πεₒ𝑚𝑟

Combining (1) and (2)

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 Bohr’s Model ∞ 0.00 eV

Combining (1) and (2) 5 −0.54 eV

4 −0.85 eV
−1.51 eV
r = εon2h2/πme2Z …….. (3) 3

The energy of an electron when it jumps 2 −3.4 eV

Increasing energy
from one orbit to other is
E = −1/2mv2, E = −Ze2/8πεₒr

By putting the value of r in the above equation,

E = −Z2e4m/8εₒ2n2h2 ……. (4)

1 −13.6 eV
Or, E = −13.6/n2 eV ….… (5)
m = mass of electron
Z = Charge of nucleus or atomic number
e = electronic charge
h = Planck’s constant
n = principal quantum number
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 Drawbacks of Bohr model

➢ The model only worked for hydrogen-like atoms.

➢ Could not explain why the intensity of the spectra lines were NOT all equal.
➢ The existence of fine and hyperfine structure in spectral lines.
➢ The Zeeman effect - changes in spectral lines due to external magnetic fields.

➢ Wave nature of electron was not considered in Bohr’s model

Information gained from Bohr’s model

It provide foundation to understand atomic structure.

The total energy of electron in H-atom it predicts is exactly agree with these
obtained from Schrodinger equation (discussed later).
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 de Broglie equation (1924)
(Wave-particle duality)

All the moving particles have wave properties. So electrons

have dual character.

The wave length of a particle moving with a velocity ‘v’ is

given by:

λ = h/mv

Q. Calculate the de Broglie wave length of an electron moving with velocity

104 m/sec. [mass of e = 9.1 x 10−31 kg]

Ans. 72 nm
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 Heisenberg’s uncertainty principle (1927)

The more precisely we can define the position of an electron

the less certainly we are able to define its velocity, and vice
versa.

Δx . ΔPx ≥ h/4π

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 Schrödinger's equation
The Schrödinger equation describes the wave properties of an
electron in terms of position, mass, energy.

The simplest form of time independent Schrodinger equation:

H=E
H = Hamilton operator
E = Total energy of electron
 = wave function of electron

H = −h2/8π2m 2 + V
V = potential energy, (V = − Ze2/4πor)

[−h2/8π2m 2 +V]  = E 
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 Understanding the Schrödinger equation

(Hamiltonian operator) (Eigenfunction) = (Eigenvalue) (Eigenfunction)

H=E
➢ Eigenfunction is the wave function of an electron corresponding to
the energy E.

➢ Eigenfunction is different for each eigenvalue.

➢ By solving Schrödinger equation one can find the wave functions

(eigenfuncctions) and the corresponding allowed energies
(eigenvalues).
(operator corresponding to an observable) () = (value of observable) ()

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 Understanding the Schrödinger equation

The probability of finding the particle in an infinitesimal

volume, dV, about a given point is proportional to 2 dV.

Normalization
the probability of existence of the particle in
the entire space should be 1.

 dxdydz = 1
 

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 Solutions of Schrödinger equation
[−h2/8π2m 2 +V]  = E 
There are many solutions to the above equation. However, the
acceptable solutions must satisfy following conditions.
1.  must be single valued.
2.  must be continuous.
3.  must be finite.
➢ For an atom several wave functions (1, 2, 3) will satisfy
these conditions and each of these has a corresponding energy
(E1, E2, E3)

➢ These wave functions are called as orbitals analogy to orbits in

Bohr’s theory.

➢ Each orbital is described uniquely by a set of three quantum

numbers, n, l, ml.
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 Identifying an eigenfunction
Q1. Show that eax is an eigenfunction of the operator d/dx,
and find the corresponding eigenvalue.

Q2. Show that eax² is not an eigenfunction of d/dx.

d/dx(eax²) = 2ax (eax²)

This is not an eigenvalue even though the same function

occurs on the right, because the function is now multiplied by
a variable factor (2ax), not a constant factor.
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 Quantum Numbers
There are four quantum numbers.

Orbital is defined by three quantum numbers (n, l, ml).

An electron is defined by four quantum numbers (n, l, ml, ms).

Principal quantum number (n):

Determines the total energy of an electron.
E = −1/n2 (Z2e4μ/8εₒ2h2)
Or, E = 1/n2 (−13.6) eV
Has value 1, 2, 3, …………∞

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 Quantum Numbers

Q. Why the value of ‘l’ is always less than ‘n’?

Magnetic quantum number (ml):
Describe orientation of orbital in space.
It is the z-component of angular momentum.
(so it never be larger than the ‘l’ value)
It may have +ve z-component or
–ve z-component
So the value ranges from –l, ……0, …….+l
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 Quantum Numbers

Spin quantum number (ms):

Describes orientation of electron spin in a magnetic field.

The value of ms either +1/2 (in the direction of the field)

or −1/2 (opposed to it).

Arises due to spinning of electron.

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 Radial and Angular part of wave function 

In 3D,  may be expressed either in Cartesian coordinates

(x,y,z) or in spherical coordinates (r,θ,ϕ).

r = represents the distance from the nucleus.

θ = is the angle from z-axis, it varies from 0-π.
ϕ = is the angle from x-axis, it varies from 0-2π.

(x,y,z) can be converted to (r,θ,ϕ) using:

x = r sinθ cosϕ
y = r sinθ sinϕ
z = r cosθ

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 Radial and Angular part of wave function 

 may be separated into radial component and two angular

components.

(r,θ,ϕ) = R(r) (θ ) (ϕ)

Combining the two angular components

(r,θ,ϕ) = Rn,l(r) Yl,m(θ,ϕ)

radial
angular
function
function

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 Radial and Angular part of wave function 

The radial function, Rn,l(r) :

The radial function is determined by the quantum numbers n

and l.

Information it gives
The size of the orbital is determined by Rn,l(r).

4πr2R2 = The radial probability function. It describes the probability of

finding the electron at a given distance from the nucleus.

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 Radial probability function
Orbital Schrodinger model Bohr Model
1s (1 peak) rmax = ao r = ao
2s (2 peaks) rmax ≈ 5 ao r = 4 ao
3s (3 peaks) rmax ≈ 13 ao r = 9 ao

1s

2s
3s

Maximum correspond to distance from nucleus at which the electron has

highest probability of being found.
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 Radial and Angular part of wave function 

The angular function (Yl,m(θ,ϕ) ):

The angular function is determined by the quantum numbers

l and ml.

The angular function, Yl,m(θ,ϕ), describes shape of the

orbitals and their orientation in space.

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 Hydrogen atom

Hydrogen has special significance

•No approximation is required in solution of Schrödinger equation
•Can get expression for energy levels

For H atom the Schrödinger wave equation can be written as

Ĥψ=Eψ

−(ħ2/2m 2 +Ze2/r ) ψ = E ψ
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 Solution of the Schrödinger wave equation
ψn,l,ml(r,,) = Rn,l(r)Yl,ml(,)
1s atomic orbital (n = 1, l = 0, ml = 0
3 1
 Z  2 − 2  1  2
 1s = 2  e  
 a0   4 

Radial part Z = effective nuclear charge Angular part

ao = Bohr radius
ρ = 2Z r/n ao
r = distance from the nucleus
n = principal quantum number
3 1
 Z  − 2
2
 1  2
R n, l = 2  e Y l, ml (,) =  
 a0   4  28
 1s atomic orbital (n = 1, l = 0, ml = 0
For H atom Z = 1, so
3 1
1 2
−  1  2
 1s = 2  e 2
 
 a0   4 

➢ Y is a constant and does not depend on 

and  , s-orbitals are spherically symmetrical.

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 2p atomic orbitals (n = 2, l = 1, ml = +1, 0, −1

3
1  Z  2 − / 2  3  2  z 
1

 2pz =    e    
2 6  a0   4   r 

Radial part (R n, l) Angular part (Y l, ml (,))

Note: Z is the atomic number and z is the z-direction/component

In polar coordinate the above equation becomes

3
1  Z  2 −  / 2 3  2
1

 2pz =    e   cos (Since, z = r cosθ)

2 6  a0   4 
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 2p atomic orbitals (n = 2, l = 1, ml = +1, 0, −1

For H atom Z = 1, so
3
1  1  2 −  / 2 3  2
1

 2pz =    e   cos
2 6 0
a  4 

➢ The angular variation of wave function depend

on cos .
➢ The probability density is proportional to cos2.
➢ The probability density has maximum value
along an arbitary axis (z-axis) on either side of the
nucleus ( at = 0 and 180o)

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 2p atomic orbitals
Similarly the wave functions (real part of wave functions) of
other two p-orbitals are given below:
3
1  1  2 − / 2  3  2
1

 2px =    e   sin  cos (Since, x/r = sinθcos)

2 6  a0   4 
3
1  1  2 −  / 2 3  2
1

 2py =    e   sin  sin  (Since, y/r = sinθsin)

2 6  a0   4 

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3d atomic orbitals (n = 3, l = 2, ml = +2, +1, 0, −1, −2)
(dxy, dxz, dyz, dx²-y², dz²)

1
1  15  2
xy Similarly the dxz, dyz
 3dxy = R n, l  
4   r2
Y (θ, )
1
1  15  2
x2− y 2
 3dx²-y² = R n, l   Since the angular part contains
4   r2 two or more variables, so these
orbitals have shapes in two
1
axial directions or more.
1 5 2
2 z2− x 2− y 2
 3dz² = R n, l  
4   r2

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 Nodal surface(s) of atomic orbitals
node
Nodal surface is a surface with zero electron density.

Nodes appear naturally as a result of the wave nature of the electron

At the nodal surface the wave function changes its sign.

At nodal surface,  = 0
either R(r) = 0
or Y(,) = 0
Angular node
Radial node

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 Angular nodal surface

If Y(,) = 0, angular nodes result.

Angular node
Angular nodes are planar or conical.

Number of angular nodes = l

Orbital No. of angular nodes

s- orbital 0
p-orbital 1
d-orbital 2
f-orbital 3

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 Radial or Spherical nodal surface

If R(r) = 0, radial nodes or spherical nodes result.

Number of radial nodes = n−l −1

Orbital Radial Orbital Radial Orbital Radial

nodes nodes nodes
1s 0 2p 0 3d 0
2s 1 3p 1 4d 1
3s 2 4p 2 5d 2
Radial node

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Q. Describe the angular nodal surfaces for a 3dxy orbital, whose
angular wave function is
1
1  15  2
xy
Y = 4  
  r2

xz -plane
Ans. In nodal surfaces, Y = 0

Y = 0, when either x = 0 or y = 0
So the nodal surfaces are
x = 0 (yz plane)
and y = 0 (xz plane) yz -plane

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Q. Why the lobes of dxy, dyz and dzx orbitals are in between
axes and the lobes of dx2−y2 is along the axes?

Ans: The nodal surfaces of dxy, dyz and dzx orbitals contains the
axes and hence their lobes lie in between axes. However, the
nodal surfaces of dx2−y2 orbital are x = y and x = −y. So the lobes
of dx2−y2 lie along the axial direction.

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Q. Describe the spherical nodal surface of 2s orbital of hydrogen
atom. The radial component of 2s atomic orbital is:
3
Z 2 −
R2s = 2  (2 −  )e 2 (ρ = 2Zr / nao)
 a0 

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 What we learn in atomic structure?

➢ Bohr’s model of atomic structure.

➢ Schrodinger atomic model or wave equation.

➢ Quantum numbers.

➢ Solutions (only wave function part) of Schrodinger equation.

➢ Nodal surfaces and sign of atomic orbitals.

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