chapter 5 preview questions are due on monday! section 5.1 the development of a new atomic model...

Chapter 5 Preview Questions are Due on Monday!

Section 5.1 The Development of a New Atomic Model

Previously, Rutherford reshaped our thoughts of the atom by showing the protons were located in the nucleus of the atom, but he could not model for us where the electrons were, other than outside the nucleus somewhere. Fortunately, studies into the properties of light and the effects of light on matter soon gave clues to where electrons actually are.

Light is a small part of all the radiation (something that spreads from a source) called electromagnetic radiation. Electromagnetic radiation is energy in the form of waves (of electric and magnetic fields). Electromagnetic radiation includes radio waves, microwaves, infrared, visible light, X-rays, and Gamma rays. All these together are considered the Electromagnetic Spectrum.

As all the forms of electromagnetic radiation are waves, they all have similar properties.

• All electromagnetic radiation travels at the speed of light (c), 299,792,458 m/s (2.998 x 108) in a vacuum

•The crest is the top of the waves, the trough is the bottom of the waves, and the amplitude is a measurement from the rest or zero line to a crest or trough

•The wavelength (λ – lambda) is the distance between successive crests/troughs and is measured in meters (often nm, and 1 x 109 nm = 1 m)

•The wavelength (λ – lambda) is the distance between successive crests/troughs and is measured in meters (often nm = 1 x 10-9 m)

•The frequency (ν – nu) is the number of waves that pass a point in one second and is measured

in (per second – can be written as s-1) or

Hz (Hertz)

1

s

How many hertz is the first wave?

How many hertz is the second wave?

How many hertz is the first wave?1 wave per second = 1 Hz

How many hertz is the second wave?2 waves per second = 2 Hz

The speed of a wave is directly proportional to the wavelength and the frequency; c = λν is the formula

c

λ ν

Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?


ν =c

λ


ν =c

λ

ν =2.998 × 108 m/s

413 nm


ν =c

λ

ν =2.998 × 108 m/s

413 nm

WAIT, This won’t work!


ν =c

λ

ν =2.998 × 108 m/s

413 nm

413 nm 1 m=


ν =c

λ

ν =2.998 × 108 m/s

413 nm 1 m=

1 x 109 nm


ν =c

λ

ν =2.998 × 108 m/s

4.13 x 10-7 m

413 nm 1 m= 4.13 x 10-7 m

1 x 109 nm


ν =c

λ

ν =2.998 × 108 m/s

4.13 × 10-7 m

ν = 7.26 × 1014 or Hz

Unfortunately, thinking of light as waves lead to a problem. It was noticed that if light strikes a metal, then sometimes it could cause electrons to be emitted (leave the atoms entirely – like in a solar panel); called the photoelectric effect. If light was a wave, then all amounts of light energy should cause this to happen, but this was not the case. It always took some minimum amount of energy to get the electrons to be emitted.

This lead Max Planck to theorize that light must carry energy in basic minimum amounts that he called quanta. Like a delivery person cannot correctly deliver half a box, the electrons in atoms cannot gain a fraction of a quantum of energy (it has to be in whole numbers).

He proposed that this energy was directly proportional to the frequency of the electromagnetic radiation and a constant, now called Planck’s constant. E = h ν

E = energy in Joules (J)

h = Planck’s constant = 6.626 × 10-34 Js

ν = frequency in Hz or 1/s

E

h ν

Example. What is the energy content of one quantum of the light with a wavelength of 413 nm?


Note: wavelength is not in the energy equation, but frequency is. So first, you must solve for the frequency.


Note: wavelength is not in the energy equation, but frequency is. So first, you must solve for the frequency. As seen in the earlier example, a wavelength of 413 nm gives a ν = 7.26 × 1014 Hz.


ν = 7.26 × 1014 Hz

E = h × ν

E = 6.626 × 10-34 Js × 7.26 × 1014 1/s


ν = 7.26 × 1014 Hz

E = h × ν

E = 6.626 × 10-34 Js × 7.26 × 1014 1/s

E = 4.81 × 10-19 J

In 1905 Einstein used Plancks work to propose that electromagnetic radiation had a dual wave-particle nature. As a particle, electromagnetic radiation carries a quantum of energy of energy, has no mass, and is called a photon.

So to get an electron to emit from a metal, it must be struck with a photon having quantum energy big enough, or nothing will happen. Each metal requires a different quantum energy, thus each metal can be identified by the frequency of light needed to emit electron.

This idea was expanded upon to develop an idea of where the electrons were in an atom. It was found that low pressure gases could be trapped in a tube and electrified, and would then glow a color particular to the gas inside.

Furthermore this light could be passed into a prism, and instead of getting the entire spectrum (rainbow) of colors, only certain wavelengths of light would be seen as small bars of color, called a line-emission spectrum.

This would indicated that the electrons in an atom were only absorbing specific amounts of energy from the electricity, causing the electrons to move from their ground state (normal position close to the nucleus) to an excited state (higher energy position further away from the nucleus). The electrons do not stay in the excited state for long and fall back to their ground state, losing the energy equal to what they gained.

Quanta Video

Niels Bohr used this to develop a model of the atom where the electrons could only be in certain, specific energy level (n) orbits around the nucleus. Just as you cannot go up half a rung on a ladder, the electron could not go up a partial energy level. The electrons gained or lost enough energy to move a whole number amount of energy levels (n) away from or closer to the nucleus, or it did not move.

He calculated the amount of energy needed for an electron of hydrogen to move between each energy level (n) (which was not constant) and his calculations agreed with experimental results.

The Balmer series of hydrogen spectral lines refer to the four lines seen in the visible light region (the four colored bars). If the electron was excited to energy level (n) 6, 5, 4, or 3 and fell to energy level (n) 2, the resulting energy given off would have a frequency in the visible region of electromagnetic radiation. (One line for dropping from 6 to 2, one for 5 to 2, one for 4 to 2, and one for 3 to 2).

However, there are other possibilities. If the electrons drop from n=6, 5, or 4 to n=3, then the energy given off is not big enough to be seen as it is in the infrared region. These three lines in the infrared region are referred to as the Paschen series. If the electrons drop to n=1, then the five lines given off are too high in energy to be seen, as they are in the ultraviolet region. These lines are referred to as the Lyman series.

•Turn your lab in to the white box, by tomorrow.

•You need destructible pens for tomorrow!

•Grab 4 different colors of pencils (or use your own)

Model of Atom Review:

1. Thomson’s Plum Pudding Model – the atom is a ball of evenly spread positive stuff with random negative particles (electrons).


2. Rutherford’s Nuclear Model – the atom has a central nucleus containing the positive particles (protons) with the electrons outside.


2. Rutherford’s Nuclear Model – the atom has a central nucleus containing the positive particles (protons) with the electrons outside.

3. Bohr’s Orbital Model – The electrons circle the nucleus in specific energy orbits, like the planets orbit the sun. Unfortunately this only works for atoms with one electron…

4. Quantum Mechanical Model – electrons are found in specific regions around the nucleus, but the exact location of the electrons inside the regions cannot be determined

The quantum mechanical model starts with a Principal Quantum Number (n), which is the basic energy level of an electron, and often matches the period number. Possible values (currently) are 1-7.

The quantum mechanical model starts with a Principal Quantum Number (n), which is the basic energy level of an electron, and often matches the period number. Possible values (currently) are 1-7.

Inside the principal quantum energy level are sublevels that correspond to different cloud shapes. The sublevels are designated as s (sharp), p (principal), d (diffuse), and f (fundamental).

Inside the sublevels are orbitals, specific regions with a 90% probability of finding electrons.

• s –orbitals are spherically shaped clouds around the nucleus

• p -orbitals are bar-bell shaped clouds with the nucleus between the lobes

• d and f are much more complex in shape

Each sublevel has room for a different amount of electrons, because an orbital can hold two electrons, then each sublevel has a different amount of orbitals

• s –sublevel can hold 2 electrons, so it has 1 orbital (shape)

• p –sublevel can hold 6 electrons, so it has 3 orbitals (shapes)

• d –sublevel can hold 10 electrons, so it has 5 orbitals (shapes)

• f –sublevel can hold 14 electrons, so it has 7 orbitals (shapes)

The s sublevel is simply a sphere centered on the nucleus.

The p sublevel has three orbitals. These are often referred to a dumbbell shape.

The d sublevel has five orbitals:

The f sublevel has seven orbitals

To know the maximum amount of electrons that could be in any principal quantum level (and the number of elements that could be represented) use the formula 2n2

if n=1, then


if n=1, then 2 electrons will fit

if n=4,

Color Now

Section 4.3 Electron Configurations

We don’t want to draw this, do we?

Section 4.3 Electron Configurations

In order to show on paper where electrons are likely to be located in an atom, orbital filling diagrams and electron configurations are drawn or written. When this is done, three rules must be followed:

1. Aufbau principle – electrons fill lower energy levels first, thus 1 before 2 and s before p, etc.

a. orbitals within a sublevel are equal in energy (called degenerate)

b. the principal energy levels often overlap, making them seem a little out of order

c. boxes are used to represent orbitals

Another way of writing the aufbau principle diagram:

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d

7s 7p

2. Pauli Exclusion principle – an orbital (box) can hold a maximum of two electrons (arrows)

a. for two electrons to fit, they have to have opposite spins

b. for one electron in the orbital

c. for two electrons in the orbital (opposite spins)

3. Hund’s Rule – when electrons occupy degenerate orbitals, one electron is placed into each orbital with parallel spins before doubling up

Ex. _____ _____ _____ NOT _____ _____ _____

3p 3p

1s

2s2p

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5dE

ner

gy a

nd

Dis

tan

ce f

rom

th

e N

ucl

eus

2s2p

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

H

2s2p

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

He

2p

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

Li

2p

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

Be

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

B

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

C

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

N

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

O

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

F

3d

4f

3s

4s

5s

6s

3p

4p

5p4d

5d

Ne

3d

4f

4s

5s

6s

3p

4p

5p4d

5d

Na

3d

4f

4s

5s

6s

3p

4p

5p4d

5d

Mg

3d

4f

4s

5s

6s

4p

5p4d

5d

Al

3d

4f

4s

5s

6s

4p

5p4d

5d

Si

3d

4f

4s

5s

6s

4p

5p4d

5d

P

3d

4f

4s

5s

6s

4p

5p4d

5d

S

3d

4f

4s

5s

6s

4p

5p4d

5d

Cl

3d

4f

4s

5s

6s

4p

5p4d

5d

Ar

3d

4f

5s

6s

4p

5p4d

5d

K

3d

4f

5s

6s

4p

5p4d

5d

Ca

4f

5s

6s

4p

5p4d

5d

Sc

4f

5s

6s

4p

5p4d

5d

Ti

4f

5s

6s

4p

5p4d

5d

V

4f

5s

6s

4p

5p4d

5d

Cr

4f

5s

6s

4p

5p4d

5d

Mn

4f

5s

6s

4p

5p4d

5d

Fe

4f

5s

6s

4p

5p4d

5d

Co

4f

5s

6s

4p

5p4d

5d

Ni

4f

5s

6s

4p

5p4d

5d

Cu

4f

5s

6s

4p

5p4d

5d

Zn

4f

5s

6s

5p4d

5d

Ga

4f

5s

6s

5p4d

5d

Ge

4f

5s

6s

5p4d

5d

As

4f

5s

6s

5p4d

5d

Se

4f

5s

6s

5p4d

5d

Br

4f

5s

6s

5p4d

5d

Kr

4f6s

5p4d

5d

Rb

4f6s

5p

5d

Sr

4f6s

5p

5d

Y

4f6s

5p

5d

Zr

4f6s

5p

5d

Nb

4f6s

5p

5d

Mo

4f6s

5p

5d

Tc

4f6s

5p

5d

Ru

4f6s

5p

5d

Rh

4f6s

5p

5d

Pd

4f6s

5p

5d

Ag

4f6s

5p

5d

Cd

4f6s

5d

In

4f6s

5d

Sn

4f6s

5d

Sb

4f6s

5d

Te

4f6s

5d

I

4f6s

5d

Xe

4f5d

Cs

4f5d

Ba

5d

La

5d

Ce

5d

Pr

5d

Nd

5d

Pm

5d

Sm

5d

Eu

5d

Gd

5d

Tb

5d

Dy

5d

Ho

5d

Er

5d

Tm

5d

Yb

Orbital Notation shows the arrows in the boxes to represent the electrons in an atom. To shorten this process, an electron configuration can be written. It leaves out the information about the number of orbitals in each sublevel, so it will be expect you remember that information.

It has the general form nl°n = principal quantum number (1-7…)

l = sublevel letter (s, p, d, or f)

° = number of e- in that orbital (1-14)

Ni = 28 e-

Ni = 28 e-

1s2

Ni = 28 e-

1s2 2s2

Ni = 28 e-

1s2 2s2 2p6

Ni = 28 e-

1s2 2s2 2p6 3s2

Ni = 28 e-

1s2 2s2 2p6 3s2 3p6

Ni = 28 e-

1s2 2s2 2p6 3s2 3p6 4s2

Ni = 28 e-

1s2 2s2 2p6 3s2 3p6 4s2 3d8

Ni = 28 e-

1s2 2s2 2p6 3s2 3p6 4s2 3d8

2 + 2 + 6 + 2 + 6 + 2 + 8 = 28

Sn = 50 e-

Your Turn to Try

Sn = 50 e-

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2

If writing out the entire electron configuration is too much, we can use the previous (in the periodic table) noble gas to take the place of part of the electron configuration:

Polonium:1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p4

Xenon:1s22s22p63s23p64s23d104p65s24d105p6

Polonium: [Xe] 6s24f145d106p4

When the electron configuration is written for an element using the noble gas configuration the electrons written after the noble gas are the ones that appear on the outside of the atom, called valence electrons..

When elements bond to form compounds, it is these electrons that are involved. The amount of valence electrons makes a big difference in how the element will bond, so to make it easy to predict, we draw electron dot diagrams.

A) In an electron dot diagram, we use the symbol of the element and dots to represent the number of valence electrons.

B) Only s and p electrons with the highest quantum number count for dot diagrams, even if there are d and f electrons after the noble gas.

Lithium = 1s22s1

So Li

Beryllium = 1s22s2

So Be

Boron = 1s22s22p1

So B

Carbon = 1s22s22p2

So C

Nitrogen = 1s22s22p3

So N

Oxygen = 1s22s22p4

So O

Fluorine = 1s22s22p5

So F

Neon = 1s22s22p6

So Ne

QUIZ TIME!

Strontium = 1s22s22p63s23p64s23d104p65s2

Strontium = 1s22s22p63s23p64s23d104p65s2

So Sr

Cobalt = 1s22s22p63s23p64s23d7

Cobalt = 1s22s22p63s23p64s23d7

So Co

Bromine = 1s22s22p63s23p64s23d104p5

Bromine = 1s22s22p63s23p64s23d104p5

So Br

chapter 5 preview questions are due on monday! section 5.1 the development of a new atomic model...

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