Transcript Tutorial 3 - Atomic Theory

JF Basic Chemistry Tutorial : Atomic Theory
Shane Plunkett
[email protected]
Matter
-anything that had mass and takes up space
-classified into 3 types: elements, compounds
and mixtures
Element
-consists of only one kind of atom. So, an
element is a pure substance.
Atom
-the smallest particle of an element that can exist.
-made up of three subatomic particles protons,
neutrons and electrons.
How do we know this?
Theories through time….
1890’s
J.J. Thompson - The cathode ray experiment
negative particles discharged were named electrons and denoted
eEarly 20th Century
The Plum Pudding model!
atoms made up of blobs of a positively charged jelly with electrons
suspended in it
1908
Ernest Rutherford – The Gold Leaf Experiment
2 students Geiger and Marsden shot α-particles (positively charged
particles) at a thin piece of gold foil. Most passed through, some
bounced back.
Conclusions: Can’t be a plum pudding structure!
- central positively charged point and a large volume of empty space
Central point called the atomic nucleus and this is where all positively
charged protons are located.
Atoms do not have an overall charge. There must be the same number
of electrons as there are protons.
Definition:
The number of protons in an atomic nucleus is called
the atomic number (Z) of the element.
Note: the atomic number of an element is the smaller number
If we take a look at the periodic table, we see that it is made up of
vertical blocks (these are called groups) and horizontal rows
(these are called periods).
Mass Number or Atomic Mass (A)
Element symbol
Atomic Number (Z)
16.00
O
8
e.g. look at Oxygen in the Periodic Table
Questions
Give the name, atomic number and group number of the
element
with the following Z value:
(a) Z=16
Sulfur
Group VI
(b) Z=8
Oxygen
Group VI
(c) Z=19
Potassium
Group I
(d) Z=13
Aluminium
Group III
(e) Z=32
Germanium
Group IV
What about neutrons?
A mass spectrometer is an instrument used by chemists to determine
the mass of a given element. Not all atoms of a particular element
have the same mass.
Example: Neon (Ne) gives 3 types of atoms, each heavier than the
last.
Explanation: There must be a third type of subatomic particle that
contributes to this change in mass
Neutrons (n) have no electric charge, but have the same mass as
protons.
Definition:
Atoms of a single element that differ from
each other in mass are called isotopes
Isotopes have the same atomic number but different atomic masses
They must have the same number of electrons and protons but a
different number of neutrons
Definition:
The mass number or atomic mass (A) is the total
number of protons and neutrons in a nucleus of an atom.
Note: the mass number or atomic mass of an element is the larger
number
As we go across the rows of the periodic table, the mass of the nucleus
increases.
Example
Hydrogen: Atomic number = 1
This means a hydrogen atom possesses one electron and one proton
Mass number = 1.0079
No. of neutrons in the hydrogen atom = Mass no. – Atomic no.
= 1.0079 – 1  0
This means that the hydrogen atom has one electron, one proton and
no neutrons.
If we move onto the next element, Helium, we find:
Atomic number = 2, i.e. 2 electrons and 2 protons in each He atom
Mass number = 4.00
No. of neutrons in the helium atom = Mass no. – Atomic no. = 2
Each helium atom has 2 electrons, 2 protons and 2 neutrons
Sample Question
Given Z (the atomic number), how many protons, neutrons, and
electrons are in the following elements?
(a)
Z=6 (Carbon)
 have 6 protons and 6 electrons
number of neutrons = mass number – atomic number
= 12 – 6 = 6 neutrons
(b)
Z=12 (Magnesium)
 have 12 protons and 12 electrons
number of neutrons = 24 – 12 = 12 neutrons
(c)
Z=17 (Chlorine)
 have 17 protons and 17 electrons
number of neutrons = 35 – 17 = 18 neutrons
Now you try! Z = 30
Z=16
Z = 30, Ans.: 30 electrons, 30 protons, 35 neutrons
Z = 16, Ans.: 16 electrons, 16 protons, 16 neutrons
Z = 35, Ans.: 35 electrons, 35 protons, 45 neutrons
Z=35
Sample Question
The three naturally occuring isotopes of argon are 36Ar, 38Ar and 40Ar.
How many protons, neutrons and electrons are present in each?
Note that in isotopes, the mass numbers can change, but
the atomic number always remains the same. It is only
the number of neutrons that can change.
Mass numbers are 36, 38 and 40 respectively. From the Periodic Table,
we know that the atomic number of argon is 18. This means that each
isotope has 18 protons and 18 electrons.
Number of neutrons = mass number – atomic number
36Ar:
no. of neutrons = 36 – 18 = 18 neutrons
38Ar:
no. of neutrons = 38 – 18 = 20 neutrons
40Ar:
no. of neutrons = 40 – 18 = 22 neutrons
We have so far built up a picture of the atom that looks like this:
Atomic Nucleus
-contains the
protons and
neutrons
Empty space
Electrons are found
somewhere in here
We know that electrons orbit the atomic nucleus.
Can we be more definite about where electrons can be found?
YES!
There is a probability associated with finding an electron in a
particular location around the nucleus.
Definition:
An orbital is a region of space where the probability
of finding an electron is large.
Four types of orbitals: s, p, d and f. We will look at the first 3.
Quantum Numbers of Atomic Orbitals
Atomic orbitals have 4 quantum numbers associated with them
1. The Prinicipal Quantum Number (n)
-positive, whole number (n=1, 2, 3…)
-tells you about the size of the orbital, i.e., the distance from the nucleus
-tells you about the energy of the orbital; the bigger the number, the
higher the energy level
-the orbitals form a series of shells (like the layers of an onion).
Shells of higher n surround shells of lower n.
2. The Angular Momentum Quantum Number (l)
-can be a whole number with values from 0 to n-1
-tells you about the shape of the orbital, i.e. if it is an s, p or d orbital
-you can get its values when you know n,
e.g. if n=1, then l can only equal 0, so you have an s-orbital
if n=2, then l can equal 0 or 1, so you have s- and p-orbitals
if n=3, then l can equal 0, 1 or 2, so can have s-, p- and d-orbitals
3. The Magnetic Quantum Number (ml)
-whole number from –l through 0 to +l
- labels the orbitals and tells you how many of each orbital type you
have
4. The Spin Magnetic Quantum Number (ms)
-electrons behave like spinning spheres and this behaviour is called
spin
-spin can be: clockwise (usually represented as an arrow pointing
upwards (↑) and is given the value +½
or:
anticlockwise represented as an arrow pointing
downwards (↓) and is given the value -½
Sample Question
Determine the total number of orbitals in a shell with a principle quantum
number n = 3.
Step 1: Determine the angular momentum quantum numbers associated
with n = 3
Angular momentum quantum number, l = 0, 1, 2, …, n -1
For n =3, there are l = 0, 1, 2 subshells
For l =0, we have an s-orbital
l =1, we have a p-orbital
l =2, we have a d-orbital
To find out how many of each orbital we have, use the
magnetic
quantum number, ml
Step 2: Use the Magnetic Quantum Number to determine the number
of each type of orbital we have
Magnetic quantum number, ml= l, l -1, l -2, …, -l
For l=0;
ml=0
This means we have one s-orbital
For l=1;
ml=1, 1-1, -1
=1, 0, -1
This means we have three p-orbitals
For l=2:
ml= 2, 2-1, 2-2, 2-3, -2
=2, 1, 0, -1, -2
This means we have five d-orbitals
So, in total, we have 1+3+5 = 9 orbitals in the shell with n=3.
Now you try!
How many orbitals are there in the shell with n=2?
Ans.: 4 orbitals (1s-orbital and 3 p-orbitals)
What do these orbitals look like?
s-orbitals
-spherical in shape
p-orbitals
-a cloud with two lobes on opposite sides of the nucleus.
-the two lobes are separated by a planar region called a nodal plane
-there are 3 p-orbitals which point along the x, y, and z axes respectively
z
y
x
pz
z
x
py
x
px
d-orbitals
-there are five d-orbitals named after the directions in which they
point
x
z
y
x
y
dx2 + y2
x
dxz
dxy
z
z
y
dyz
dz2
There are rules for how electrons will accommodate these orbitals.
Rule 1: Electrons will occupy the orbital of lowest available energy.
The energy of orbitals increases in the order s<p<d,f
Rule 2: Each orbital can accommodate a maximum of two
electrons.
The two electrons in an orbital must have opposite spins
(i.e. one must be clockwise and the other must be
anticlockwise).
This is called the Pauli Exclusion Principle.
Example:
Helium, He
Atomic number = 2
Lowest energy orbital available is first s-orbital, denoted the 1s orbital
Because we have 2 electrons and each orbital can have 2 electrons, we
can place both electrons of He into the 1s orbital:
Electron
1s orbital
Nucleus containing
2 protons and 2
neutrons


Electron
The name we give to the build-up of these electrons in orbitals is
the electronic configuration
So, the electronic configuration of the He atom is 1s2, where the
superscript 2 indicates the number of electrons present in the orbital.
Question: How many electrons are present in the hydrogen atom
and how would you write its electronic configuration?
Answer: Hydrogen has one electron and an electronic configuration of 1s1
After the 1s orbital, next comes the 2s orbital. Again this can hold 2
electrons
Question: How many electrons are present in the lithium atom and
how would you write its electronic configuration?
Answer: Lithium has 3 electrons and an electronic configuration of 1s2 2s1
Once the 2s orbital has been filled, the 2p orbital is now filled
Example:
Boron, B
Atomic number = 5
First, the 1s and 2s orbitals are filled. This accounts for 4 out of the 5
electrons.
Because the 3s orbital is too high in energy and we can only place two
electrons in each orbital, the final electron goes into the 2p orbital
Hence, the electronic configuration for the ground state boron atom is
1s2 2s2 2p1
The third rule we need to know is called Hund’s Rule and states that
electrons will occupy degenerate orbitals (those of the same energy)
singly before pairing (happens in buses).
Example:
The atomic number of carbon is 6 and the electronic configuration is
1s2 2s2 2p2
The 2p orbital fills up as
follows:


1s2
2s2



2p2
Note! The 2p orbitals have been filled up singly!


1s2
2s2


2p2
Question
What are the electronic configurations of the following elements (a)
Nitrogen, (b) Magnesium, (c) Chlorine, (d) Silicon, (e) Potassium?
The elements we have dealt with so far have only contained s and p
orbitals
These elements make up the s and p blocks of the periodic table
The first two groups are called the s-block and the last six groups are
called the p-block
In between these, we find the d-block or the transition metal elements
The d-block elements are so called because this is where we begin to fill
the d-orbitals
5 d-orbitals  can accommodate 10 electrons
Question
What are the electronic configurations of the following elements
(a) Nitrogen, (b) Magnesium, (c) Chlorine, (d) Silicon, (e)
Potassium?
Answer:
(a) Nitrogen has 7 electrons and an electronic configuration
of 1s2 2s2 2p3
(b) Magnesium has 12 electrons and an electronic
configuration of 1s2 2s2 2p6 3s2
(c) Chlorine has 17 electrons and an electronic
configuration of 1s2 2s2 2p6 3s2 3p5
(d) Silicon has 14 electrons and an electronic configuration
of 1s2 2s2 2p6 3s2 3p2
(e) Potassium has 19 electrons and an electronic
configuration of 1s2 2s2 2p6 3s2 3p6 4s1
Example
The atomic number of Iron is 26 we have 26 electrons
We begin to fill up the orbitals as follows:

1s2
   
2s2
2p6
   
3s2
3p6
We have filled 18 out of 26 electrons
Now, we face a problem!
As n increases, the sublevel energies get closer together. This results
in the overlap of some sublevels and here we have such a case.
The 4s sublevel is slightly lower in energy that the 3d orbital, so it is
filled first.

4s2
 



3d6
Question
What is the electronic configuration of (a) Vanadium, (b) Cobalt,
(c) Zinc?
Answer:
(a) Vanadium has 23 electrons and an electronic configuration of 1s2 2s2
2p6 3s2 3p6 4s2 3d3
(b) Cobalt has 27 electrons and an electronic configuration of 1s2 2s2 2p6
3s2 3p6 4s2 3d7
(c) Zinc has 30 electrons and an electronic configuration of 1s2 2s2 2p6 3s2
3p6 4s2 3d10
The nature of light and Atomic Spectra
Light is emitted in discrete or definite packets called quanta or phonons
Frequency of light, ν, increases proportionally with increase in energy, E:
ΔE = hν
where h is Planck’s constant, 6.63 × 1034Js
Electrons possess both kinetic and potential energy
Electrons can behave like rocks on a cliff – when the rock falls it gives up
potential energy

If electron falls towards nucleus, they give up potential energy
When excited electrons in atoms fall from a high energy level to a low
energy state, light is emitted with a specific frequency or colour
Electrons must occupy certain energy levels – like steps on a ladder
If an electron absorbs a photon or quantum of light, it is elevated to a
higher energy level – an excited state
When the electron falls to lower energy state, it gives up this energy in
specific quanta
Atoms with all electrons in their lowest energy levels are said to
be in their ground state
The energy difference between any two levels may be found by using
Rydberg’s constant:
ΔE = Efinal – Einitial = 2.18 × 10-18 Jx  1
1 

 2
 n2

n
final
initial


where 2.18 × 10-18J is the Rydberg constant and n is a positive integer
representing the shell number
We can convert the units of the Rydberg constant from Joules to metres -1
ΔE = hν
ΔE = 2.18 × 10-18 Jx 1
ν = c/λ
 n2
 final
1 
 2
ninitial 
Combine these to give:
ΔE = hc = 2.18 × 10-18 Jx 1
λ

 n2
 final
1  h = Planck’s constant
 2
c = Speed of light
ninitial 
λ = wavelength
 1

1
x
 2 
2
n

n
final
initial




=
(2.18 × 10-18 J)
x 1  1 
2
 n2

n
-34
8
final
initial
(6.626 × 10 Js)(3.00 × 10 ms 

 1 = 2.18 × 1018J
λ
hc
.
1)
= 1.0967 × 107 m-1 x 

1
1

 2 
2
n

n
final
initial


Example
What is the wavelength of a photon emitted during a transition from the
ninitial = 5 state to the nfinal = 2 state in the hydrogen atom?
ΔE = R  1
 n2
 final

 2 
ninitial 
1
= 2.18 × 10-18 J (1/4 – 1/25)
= 4.58 × 10-19 J
To calculate the wavelength of the photon:
λ = c/ν = ch/ΔE
= (3.00 × 108 ms-1)(6.63 × 10-34Js)
4.58 × 10-19 J
= 4.34 × 10-7m
For wavelengths, must convert to nanometres:
= 4.34 × 10-7 m × (1 × 109 nm)/(1 m)
= 434 nm
Question:
What is the wavelength of a photon emitted during a transition from
ninitial = 4 to nfinal = 2 state in the H atom?
Answer: 487 nm