Chapter 5. Periodicity and the Periodic Table

Chapter 5. Periodicity and the Periodic Table

Many properties of the elements follow a regular pattern.

In this chapter, we will look at theory that has been developed to explain this periodicity

ionization energy: the amount of energy needed to remove an electron

Much of what we have learned about atomic and molecular structure, has come from our understanding of how matter interacts with light.

What is light?

The interaction of light with matter forms the foundation of ourunderstanding of atomic structure, molecular structure, and the structure of the universe!

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Light: What is it?

Light is referred to as electromagnetic radiation

Light has both wave properties and particle properties

All light, whether radiowaves or visible light, travels as the same speed, 3 *108 meters/sec

As a result, since the length of each wave decreases from left to right, the frequency of the peaks an troughs of the waves shown above must increase from left to right

Referring to light as a particle, known as a photon of light, the energy of each particle of light is also known to increase from left to right

The total energy of course depends on the frequency of the light and the number of particles

Light also has a magnetic field associated with. If varies in the same fashion as the electric field, traveling at the same speed but perpendicular to the electric field. Both the electric and magnetic fields are used in medicine and science

A MRI (magnetic resonance image ) of a heart and lungs using radiowavefrequencies in the presence of a strong external magnetic field

The interaction of light with matter forms the foundation of ourunderstanding of molecular structure. How so?

Wave properties of light

c = λν where c = 3x1010 cm/s speed of light in a vacuum

λ = wavelength of light and ν = frequency of light

Particle properties of light

E = hν where E is the energy of a photon of light and

h = 6.626 x 10-34 J.s

The total energy associated with light depends on the frequency and also the intensity (the number of photons per unit time)

Heisenberg uncertainty principle:

(uncertainty in position)(uncertainty in momentum (mv)) ≈ h/4π

How does the emission spectrum of black body look like?

White light that passes through a prismis separated into all colors that together are called a continuous spectrum gives the colors of a rainbow

When an element is heated, it gives off light. However the entire rainbow of colors is absent and only certain colors are present. Each element gives it own spectrum of color. Not all of the light is in the visible region. Depending on the temperature, the element, some light covering a large portion of the electromagnetic spectrum can be observed

What happens if we take a hydrogen atom and heat it up, or for that matter, any element?

The emission spectrum of H and sodium atoms in the visible region of the spectrum

The significance of observing discrete line may not be immediately apparent but these atomic species when heated do not give off all wavelengths of light but only discrete wavelengths

The interpretation of these observations is that upon heating H atoms, the hydrogen does not emit any light until a certain amount of energy is put into the atom. Since an atom of hydrogen consists of only a proton and an electron, it is believed that the emission of light by the hydrogen is due to excitation of the electron.

Unexcited hydrogen

Excited hydrogen

energy

The interpretation of these observations is that upon heating H atoms, the hydrogen does not emit any light until a certain amount of energy is put into the atom. Since an atom of hydrogen consists of only a proton and an electron, it is believed that the emission of light by the hydrogen is due to excitation of the electron. Alternatively, we can’t excite hydrogen electronically unless we put in the correct amount of energy.

Unexcited hydrogen

Excited hydrogen

energyhν given off E = h ν

Balmer equation1/ λ = R[1/22-1/n2] where R = 1.097x10-2 nm-1 and n is some integer > 2

The energy of the light observed in the visible region is only a portion of the light emitted by a hydrogen atom. This transition occurs in the ultraviolet region

A model has been devised to explain this phenomena

Think of the model as a bookcase with each succeeding shelf getting closer and closer

Depending on the element, different electronic transitions can be observed. The emphasis her is electronic, the electrons are being excited to different levels

An e-

ionization

potential I.E.

hydrogen

Balmer Rydberg equation

1/ λ = R[1/m2-1/n2] where R = 1.097x10-2 nm-1 and n is some integer > m. This equation accounts for the lines observed for hydrogen both in the visible region and elsewhere.

Why are these other lines also included?

Why are the orbital energies of hydrogen written as 1s, 2s, 2p, 3s, 3p, 3d….?

Also why the difference in energy between the 2s and 2p level, for example, in a multi-electron atom?

Many more emission lines are observed in multi electron atoms. These terms are used to describes the levels an electron can occupy

3d

3p

3s

2p

2s

1s

multi-electron atom

3d

3p

3s

2p

2s

1s

multi-electron states in a magnetic field

Energy

Absorption lines were observed to increase in magnetic field

States needed to explain emission lines in:

What is observed in the spectra of multielectron atoms are multiple lines closely spaced followed by big gaps.

The number of lines observed with other atoms are numerous and beyond our concern. We will be interested in summarizing the theory that has been developed to explain these emision lines.

Attempts to explain the emission (and absorption spectra) of atomic hydrogen and the other atoms, resulted in discovery/development of a mathematical equation with properties that mimicked the observedspectra of atoms.

Schroedinger Equation is a differential equation:

Properties of a differential equation:

1. the equation may have more than one solution.

2. any combination of solutions (sum or difference) is also a solution

Solutions to this equation are found only when certain terms in the equation have unique values: these terms have been called quantum numbers and have been given the symbols: n, l, m, and s.

The quantum numbers have names and also must have certain relationships between each other, otherwise the equation vanishes (has no solution)

n = principle quantum number, must have integer values of 1, 2, 3, …

L = is called the angular momentum quantum number and must have integer values from –(n-1), -(n-2), …0

m = is called the magnetic quantum number and can have values from

–L, (L-1),..0,..+L (cap L used here because lowercase l looks like the number 1)

s = is called the spin quantum number, must be +1/2 or –1/2

Each electron in an atom is assigned 4 quantum numbers; no two electrons can have the same 4 quantum numbers or the solution vanishes?

What do the solutions to the Schroedinger Equation look like and what information do they provide.

The solutions are mathematical equations often described in spherical coordinates.

What are spherical coordinates?

What are Cartesian coordinates?

Cartesian Coordinates

x

y

z. (x1,y1,z1)

Spherical Coordinates

x

y

z. (r,θ,ϕ)

θ

ϕ

Some solutions to the Schroedinger Equation

Solution to this equation are called Ψ (psi)

What do they look like:

Ψ1s = (1/πa3)(2.718)r/a where a is a constant 5.29*10-9 cm and r is the distance of the particle from the origin (n=1, l =0)

Ψ2s = 1/4(1/2πa3).5(2-r/a)(2.718)r/2a (n = 2, l=0)

Ψ2p = 1/4(1/2πa3).5(r/a)(2.718)r/2acos θ (n = 2, l=1)

…

What is the physical interpretation of the information they provide?

The functions Ψ (psi) are amplitude functions, when squared and multiplied by an element of volume, they provide the probability of finding an electron at some location ((r,θ,ϕ or x,y,z) in space.

What do these functions look like?

1s

n = 1, l =0

2s

n = 2, l = 0

3s

n = 3, l = 0

Ψ1sΨ2s

Ψ3sa node is a region where the function = 0

+

-

+

+-+

What do the p orbitals look like?

How do they compare in energy to s orbitals?

P orbitals

n = 2, l = 1, m = -1 n = 2, l = 1, m =0 n = 2, l = 1, m = 1

Ψ2p

- + + - -

+

What do the d orbitals look like and how many are there?

How do they compare in energy to p orbitals?

Ψ3d Ψ4f

Why are these orbitals significant:

These orbitals are solutions to the Schroedinger Equation for the hydrogen atom. However they are very useful because they provide a model to mimic the behavior observed for the remaining elements in the periodic table.

Rules for predicting the electronic properties of the remaining elements of the periodic table:

1. Electrons want to occupy orbitals with the lowest energy possible

2. No two electrons can have the same four quantum numbers

3. Electrons repel each other and prefer to go in orbitals of equal energy that are unoccupied; they prefer to go in with the same spin (Hund’srule)

4. A maximum of 2 electron are possible in any given orbital

H 1 proton and 1 electron

Designation: 1s1

Remember, if we excite hydrogen, we can excite it to a 2s level,3s level, 4s level, and then it can decay from any one of these leves to a lower level by emitting a specific wavelength of light. This model explains the observed spectra of hydrogen, both emission (light given off from an exited state to one of lower energy) or absorption (light absorbed in going from the ground state to an excited state)

He has 2 protons and 2 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 2 protons

Designation: 1s2

Also note that this fills the 1s level; the next level is much higher in energy

Li has 3 protons and 3 electrons; note that the orbital energy scale will change again because each electron will be attracted to a nucleus that has 3 protons

Designation: 1s2 2s1

Be has 4 protons and 4 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 4 protons

Designation: 1s2 2s2

B has 5 protons and 5 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 5 protons

Designation: 1s2 2s2 2p1

C has 6 protons and 6 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 6 protons

Designation: 1s2 2s2 2p2

Note Hund’s rule: electrons occupy different orbitals with the same spin

N 7 has protons and 7 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 7 protons

Designation: 1s2 2s2 2p3

O has 8 protons and 8 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 8 protons

Designation: 1s2 2s2 2p4

F has 9 protons and 9 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 9 protons

Designation: 1s2 2s2 2p5

Ne has 10 protons and 10 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 9 protons

Designation: 1s2 2s2 2p6

Also note that this fills this level

Na has 11 protons and 11 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 11 protons

Designation: 1s2 2s2 2p6 3s1

Name the element with the following electronic configurations

1s2 2s2 2p6 3s1 (Ne 3s1)Na

1s2 2s2 2p6 3s2

Mg1s2 2s2 2p6 3s2 3p6

Ar1s2 2s2 2p6 3s2 3p6 4s1

K1s2 2s2 2p6 3s2 3p6 4s23d5

Mn1s2 2s2 2p6 3s2 3p6 4s2 3d103p3

As

In a multi-electron atom, which orbital shape do you think best shields the nucleus to an electron further out in space (in a higher level)?

s orbital

p orbital

d orbital

f orbital

Some anomalous electron configurations

Stability associated with half filled and fully filled shells

Cr [Ar] 4s2 3d4 → [Ar] 4s1 3d5

Cu [Ar] 4s2 3d9 → [Ar] 4s1 3d10

Which of the following combination of quantum numbers can refer to any electron in a ground state Co atom (Z =27)?1. n = 3, l = 0, ml = 2 2. n = 4, l = 2 ml = -23. n = 3, l =1, ml = 0

What type of electron has the following quantum numbers?

n = 3, l =0

3s orbital,

n= 4, l =2

a 4d orbital, ml = -2, -1, 0, 1, 2

n= 3, l =1

a 3p orbital, ml = -1, 0 ,1

Which of the following electron configurations refer to an excited state of V?

[Ne]3s2 3p6 4s2 3d3

[Ne]3s2 3p6 4s2 3d2 3f1

[Ne]3s2 3p6 4s2 3d2 4p1

ground state V 1s2 2s2 2p6 3s2 3p6 4s2 3d3

[Ne]3s2 3p6 4s2 3d2 4p1

What is the electronic configuration of Se?

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4

How many unpaired electrons are there in

a. K?

b. Cr?

c. Fe?