CHEM1101
2014-J-6
June 2014
• A schematic representation of a p orbital is shown below. The central sphere (mostly
obscured) represents the atomic nucleus.
Marks
2
planar node
spherical
node
How many spherical and planar nodes does this orbital have? Label them on the
diagram above.
Number of spherical nodes: 1
Number of planar nodes: 1
What is the principal quantum number, n, of this orbital? Explain your answer.
n=3
The total number of nodes is 1 fewer than the principal quantum number. As the
total number of nodes is 1 + 1 = 2, the principal quantum number is 3.
• Shielding is important in multi-electron atoms. Briefly explain the concept of
shielding.
Electrons closer to the nucleus partially block the attractive force of the nucleus
on the electrons that are further away, resulting in a lowering of the effective
nuclear charge on such electrons.
Give one example of a consequence of shielding.
The elements in a group of the Periodic Table have similar reactivities but
ionisation energies decrease and sizes increase.
3
CHEM1101
2014-N-2
November 2014
• Consider the 4p orbital shown below. Note that, for clarity, the nucleus of the atom is
not shown.
How many spherical and planar nodes does this orbital have?
Number of spherical nodes: 2
Number of planar nodes: 1
Complete the following table to give a set of quantum numbers that describes an
electron in a 4p orbital.
Quantum number
n
l
ml
ms
Value
4
1
-1, 0 or +1
½ or -½
Marks
3
CHEM1101
2012-J-4
June 2012
• The “Paschen” series of emission lines corresponds to emission from higher lying
energy states to the n = 3 state in hydrogen-like atoms. Calculate the wavelength
(in nm) of the lowest energy “Paschen” emission line in Li2+.
The energy of an orbital in an 1-electron atom or ion is given by
En = –Z2 ER (1/n2)
The energy difference between two levels is therefore:
ΔE = En1 – En2 = [–Z2 ER (1/n12)] - [–Z2 ER (1/n22)] = Z2 ER (1/n22 - 1/n12)
The lowest energy line in the Paschen series corresponds to moving from n = 4 to
n =3. As Li2+ has Z = 3, the energy of this transition is therefore:
ΔE = (3)2 ER (1/32 - 1/42)
= 9.54× 10–19 J
Using E = hc / λ, this corresponds to a wavelength of:
λ = hc / E = (6.626 × 10–34 J s)(2.998 × 108 m s-1) / (9.54 × 10–19 J)
= 2.08× 10–7 m
= 208 nm
Answer: 208 nm
What are the possible l states for the n = 4 level of Li2+?
l = 0, 1, 2 and 3
Sketch the atomic orbital with n = 3 and the lowest value of l.
The orbital is 3s:
Marks
4
CHEM1101
2012-N-4
November 2012
e) The oxygen atom in the reaction in part d) is formed in its ground electronic state.
What is the ground state electronic configuration for O?
1s2 2s2 2p4
Draw an atomic orbital energy level diagram for the ground state O atom. Name the
orbitals and show all electrons.
Name and sketch the atomic orbitals for the highest occupied atomic orbital and the
lowest unoccupied atomic orbital in the ground state O atom. Make sure all nodes are
clearly identified in your sketch.
sketch of highest occupied orbital
Name: 2p orbital
sketch of lowest unoccupied orbital
Name: 3s orbital
Marks
5
CHEM1101
2010-J-3
June 2010
• Sketch the wavefunction of the 3s atomic orbital as described below. Clearly mark all
nodes and the relative sign (+ or –) of the wavefunction.
a) using lobe representations
b) by plotting wavefunction versus distance from the nucleus
ψ
r
Explain the significance of (a) the lobes, (b) the nodes and (c) the sign of the
wavefunction, in terms of the probability of finding an electron at a given point in
space relative to the nucleus.
The lobes define the volume within which there is a certain probability of finding
the electron (usually 95%).
The nodes represent surfaces where there is zero probability of finding the
electron.
The sign of the wavefunction is not relevant to the probability of finding the
electron. The probability distribution depends on the square of the wavefunction,
which is always positive.
CHEM1101
2009-J-4
June 2009
 Sketch the following wavefunctions using lobe representations. Clearly mark all
nodal surfaces, nuclear positions and the relative sign (+ or –) of the wavefunction
within the lobes.
a 3p atomic orbital
a 2s atomic orbital
nodal surf ace
nodal surf ace
_ +
nucleus
+
_
nodal plane
nucleus
+
_
Explain the significance of (a) the lobes, (b) the nodes and (c) the sign of the
wavefunction, in terms of the probability of finding an electron at a given point in
space relative to the nucleus.
(a) The lobes define the volume within which there is a certain probability
(e.g. 95 %) of finding the electron.
(b) The nodes represent surfaces where there is zero probability of finding
the electron. Alternatively (and equivalently), they are the surfaces
where the sign of the wavefunction changes.
(c) The sign of the wavefunction is not relevant to the probability of finding
the electron. The probability distribution depends on the square of the
wavefunction, which is always positive.
Marks
5
CHEM1101
2007-N-4
November 2007
• Calculate the energy (in J) and the wavelength (in nm) expected for an emission
associated with an electronic transition from n = 4 to n = 2 in the Be3+ ion.
Marks
3
Be3+ has one electron and so the equation below can be used with Z = 4 to
 1 
calculate the orbital energies, En = − Z 2 E R 

 n2 
The energies of the n = 4 and n = 2 levels are therefore,
 1 
En = 4 = − (4)2 E R   = ER
 42 
and
 1 
En = 2 = − (4)2 E R   = 4ER
 22 
The energy associated with emission from n = 4 to n = 2 is the difference
between the energy of these two levels:
∆E = En = 2 - En = 4 = 4ER – ER = 3ER = 3 × (2.18 × 10-18) = 6.54 × 10-18 J
The energy is related to the wavelength, λ, by E =
λ=
hc
λ
:
hc (6.634 × 10−34 ) × (2.998 × 108 )
=
= 3.04 × 10−8 m = 30.4nm
−18
E
6.54 × 10
Energy: 6.54 × 10-18 J
Wavelength: 3.04 × 10-8 m or 30.4 nm
• What two properties do electrons in atoms have which lead to discrete energy levels?
Explain your answer.
All particles have wave-like properties with momentum given by mv =
h
λ
2
.
The wavelengths that are possible for an electron are limited because the
electron has restricted motion due to its attraction to the positive nucleus.
• What is the % transmission of a sample measured in an atomic absorption
spectrometer to have an absorbance of 0.5?
The absorbance, A, is related to the intensity of the incident light, I0, and the
 I 
transmitted light, I, by A = − log10  
 I0 
If A = 0.5, then the fraction transmitted is
I
= 10−0.5 = 0.3 or 30%
I0
Answer: 30%
2