For online help with
this topic, see the chemcal module "Atomic
and Nuclear Structure" which deals with the topics: Fundamental
concepts; nuclear equations; electromagnetic radiation; waveparticle
duality; Bohr model.
More advanced topics
will be found in the chemcal module "Atoms
, Electrons and Orbitals" which deals with the topics:
Atomic orbitals  shapes, quantum numbers; subshell structures of
atoms and ions.
Atomic Structure (General questions)
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Shortcut to Questions
Q: 1 2 3
4 5
6 7 8 9
10

1

Give the symbols and
ground state electronic orbital configuration of the atoms having
the following numbers of electrons:
(1)
7
(2)
11
(3)
16
(4)
25


2 
An atom of an element has two electrons in the n=1 shell,
eight electrons in the n=2 shell, and five electrons in the
n=3 shell. From this information, give for the element
(1)
its atomic number
(2)
its approximate atomic weight
(3)
the total number of s electrons in its atom
(4)
the total number of d electrons in its atom
(5)
the name of the element
(6)
two common oxidation states of the element in its compounds


3 
List the atomic numbers of the following elements:
(1)
Li
(2)
O
(3)
N
(4)
Mg
(5)
Ne
(6)
Br


4 
For one atom of the isotope ^{19}E (atomic number 9) give
(1)
the number of protons
(2)
the number of electrons
(3)
the electronic configuration of the atom in the ground state
(4)
an equation for one typical reaction of the element


5 
Give the names and symbols (including mass numbers) for the isotopes
of hydrogen.


6 
An energy of 419 kJ mol^{1} is required to convert (in
the gas phase) potassium atoms to potassium ions and electrons.
Given that 1 kJ mol^{1} equals 1.67 x 10^{21}
joule per atom, calculate the longest possible wave length of light
which is capable of ionising potassium atoms.


7 
Calculate (1)
the frequency and (2)
the energy per quantum for electromagnetic radiation having a wavelength
of 580 nm.


8 
(1)
Calculate the wavelength of a photon that has a frequency of 1.20
x 10^{15} Hz
(2)
What is the energy of the photon in joules per photon?
(3)
What is the energy in kJ mol^{1} of photons?
(4)
What is the name usually given to such radiation? (ie what region
of the electromagnetic spectrum does it belong to?)


9 
Calculate for 1.00 MHz broadcast band radio waves:
(1)
The energy of photons in joules per photon and (2)
in kJ mol^{1}.
(3)
What is the wavelength of such photons?
(4)
How does the energy compare with that for a CC single bond? Would
you expect radio waves to be able to produce chemical reactions?


10 
(Advanced question) Find the energy difference (in kJ mol^{1})
between the ground and first excited states of the hydrogen atom.



Atomic Structure (General answers)


1

The number of electrons
in a neutral atom equals the number of protons, so the number of
electrons will be equal to the atomic number.
(1)
Atomic number 7, nitrogen (N):
1s^{2} 2s^{2} 2p^{3}
(2)
Atomic number 11, sodium (Na)
1s^{2} 2s^{2} 2p^{6} 3s^{1}
(3)
Atomic number 16, sulfur (S):
1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{4}
(4)
Atomic number 25, manganese (Mn):
1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}
3d^{5} 4s^{2}


2 
(1)
Counting the total number of electrons across all the shells we
obtain 15. This must equal the number of protons in the neutral
atom, so the atomic number is 15.
(2)
An element's atomic mass is determined by the number of protons
and neutrons, since they contribute practically all mass to the
atom. There are approximately equal numbers of protons and neutrons
present in atoms, so the approximate mass will be 2 x 15
= 30.
(3)
(4)
From the information given we can work out the element's atomic
configuration:
Two first shell electrons: 1s^{2}
Eight second shell electrons: 2s^{2} 2p^{6}
Five third shell electrons: 3s^{2} 3p^{3}
Therefore there are a total of six s electrons, and no d
electrons.
(5)
From the atomic number, 15, the element is identified as phosphorus.
(6)
Phosphorus is a nonmetal, and so will form anions in ionic bonds
to metals and covalent bonds to nonmetals. Gaining 3 electrons
(or a share of 3 electrons) fills the 3p orbital making it isoelectronic
with Ar. Thus the III state is common, and shown in the P^{3}
ion, and in covalent molecules such as PH_{3}. Phosphorus
can also have more than 8 electrons in its outer level by incorporating
the adjacent empty 3d orbitals into the valence level to form 5
covalent bonds, such as in PF_{5} and PCl_{5} where
the oxidation state of P is +V.


3 
The atomic number of
an element is equal to the number of protons in the nucleus. In
a neutral atom the protons (+1 charge) and electrons (1 charge)
are present in equal numbers, so their charges balance. The atomic
number of an element is readily determined by consulting a periodic
table.
(1)
Li has an atomic number of 3
(2)
O has an atomic number of 8
(3)
N has an atomic number of 7
(4)
Mg has an atomic number of 12
(5)
Ne has an atomic number of 10
(6)
Br has an atomic number of 35


4

(1)
The number of protons equals atomic number, which is 9.
(2)
We assume here that the atom in question is neutral, since no charges
have been explicitly mentioned. In a neutral atom the number of
electrons equals the number of protons, which is 9.
(3)
The electronic configuration is found by assigning all available
electrons to orbitals, starting with 1s.
The configuration for 9 electrons will be 1s^{2}
2s^{2} 2p^{5}.
(4)
Noting that the outermost orbital (the 2p orbital) is 1 electron
short of being filled, the element will assume the I oxidation
state (ie will very readily accept 1 electron) to form stable compounds.
Indeed, we can now identify "E" as flourine, which occurs as the
F_{2} molecule. A typical reaction would be the reaction
of fluorine with a Group 1 metal such as lithium:
2Li(s) + F_{2}(g) ® 2LiF(s)


5 
hydrogen : ^{1}H
(this isotope is also known as protium)
deuterium : ^{2}H
tritium : ^{3}H


6 
The energy required to
ionise one potassium atom = 419 x 1.67 x 10^{21} joule
per atom = 7.00 x 10^{19} joule per atom. This is the minimum
amount of energy required from a single photon of electromagnetic
radiation. Converting this energy into a wavelength:
E = h c / l
l =
h c / E
= (6.626 x 10^{34}
J s) x (3.00 x 10^{8} m s^{1}) / (7.00 x 10^{19}
J)
= 2.84 x 10^{7}
m
In the above equation
l (lambda)
is wavelength (in metres), h is Planck's constant (6.626
x 10^{34} Joule seconds), c is speed of light in
vacuum (3.00 x 10^{8} m s^{1}), and E is energy
(in Joules).
The minimum wavelength
of electromagnetic radiation required to ionise potassium is therefore
284 nm.


7 
(1)
n = c / l
= (3.00 x 10^{8} m s^{1}) / (580 x 10^{9}
m)
= 5.17 x 10^{14} Hz
Note that frequency
is denoted by the symbol n ("
nu "), and the units are Hertz which are equivalent
to s^{1}. Also take note of the units used for length which
must be in metres to be consistent with energy in Joules.
(2)
E = h c / l
= (6.626 x 10^{34} J s) x (3.00 x 10^{8}
m s^{1}) / (580 x 10^{9} m)
= 3.43 x 10^{19} J


8 
(1)
l = c / n
= (3.00 x 10^{8} m s^{1}) / (1.20 x 10^{15}
s^{1}) = 2.50 x 10^{7} m
(2)
E = h n = (6.626 x 10^{34}
J s) x (1.20 x 10^{15} s^{1}) = 7.95 x 10^{19}
J
(3)
There are N_{A} photons in 1 mole = 6.022 x 10^{23}.
Therefore the energy of 1 mole of photons = (6.022 x 10^{23}
mol^{1}) x (7.95 x 10^{19} J)
= 479 kJ mol^{1}.
(4)
A wavelength of 250 nm corresponds to light in the ultraviolet region.


9 
(1)
E = h n = (6.626 x 10^{34}
J s) x (1.00 x 10^{6} s^{1})
= 6.626 x 10^{28} J per photon.
(2)
Energy per mole photons = energy per photon x N_{A}
= (6.626 x 10^{28} J s) x (6.022 x
10^{23} mol^{1})
= 3.99 x 10^{7} kJ mol^{1}
(3)
l = h c / E
= (6.626 x 10^{34} J s) x (3.00 x 10^{8}
ms^{1}) / (6.626 x 10^{28} J)
= 300 m.
(4)
The energy of radio waves (3.99 x 10^{7} kJ mol^{1})
is much less than that of a CC bond (3.48 x 10^{2}
kJ mol^{1}), therefore the waves would not be expected
to produce a chemical reaction.


10 
The Bohr model of the hydrogen atom can be used
to predict the energy of various electronic transitions within the
hydrogen atom. In this model, electrons are assumed to exist within
certain "orbits" (designated by the symbol n) each of which
has an associated discrete energy.
The energy of an electron in a particular orbit
n = E(n)
= (2.18 x 10^{18} J) x (1 / n^{2}),
for the hydrogen atom only.
For the n = 2 ® 1 transition,
n_{final} = 1, and n_{initial} = 2,
so E(n = 2 ® 1) = (2.18
x 10^{18} J) x ((1 / 1^{2})  (1 / 2^{2}))
= 1.64 x 10^{18} J.
This is the energy difference per atom, finding the energy in kJ
mol^{1}:
E per mole = N_{A } x E(n = 2 ®
1)
= (6.022 x 10^{23} mol^{1}) x (1.64 x 10^{18}
J)
= 988 kJ mol^{1}.


