Atomic Structure Dalton’s Atomic Theory of Matter

					Dalton’s Atomic Theory of Matter
In 1803, John Dalton proposed the following explainations regarding the makeup of matter.

 All matter is made up of indivisible and indestructible atoms. (Atom is smallest unit of matter)  All atoms of a given element are identical in their physical and chemical properties.  Atoms of different elements differ in their chemical and physical properties.

Dalton’s Atomic Theory
 Atoms of different elements combine in simple whole-number ratios to form compounds.

 Matter cannot be created nor destroyed. Chemical reactions consist of the combination, separation, and rearrangement of atoms. (Law of Conservation of Mass)

Dalton’s Atomic Theory (cont.)
Law of Definite Proportions
Compounds always exist in the same, definite, whole # ratios. Ex: CO2 always 1 C, 2O
H2O always 2 H, 1 O

Law of Multiple Proportions
The same elements can combine in multiple ratios to form diffferent compounds.
Ex: CuCl and CuCl2 or MnO and MnO2

Dalton’s Model of the Atom
Since Dalton did not consider the atom to consist of smaller units, the model he proposed was called the hard sphere model.
The hard sphere model is essentially just “an atom”, no protons, neutrons, or electrons were accounted for.

Atomic Structure
Until approximately 1900, scientists referred to Dalton’s theory as an explanation of the makeup of matter (the atom as the smallest unit of matter). However, there are even smaller units of matter called subatomic particles. The work of JJ Thomson and Ernest Rutherford led to the discovery of these subatomic particles.

JJ Thomson’s Plum Pudding Model

JJ Thomson experimented with electricity and first discovered subatomic particles. Using a cathode ray tube, Thomson observed that tiny particles were being sent between metal plates (electrodes) from the cathode (-) to the anode (+). The reaction of the rays to magnets proved they were negatively charged. Thomson called these newly discovered particles electrons.

Thomson’s Cathode Ray Tube Experiment At very low pressure and very high voltage a beam appears as shown in diagram

Cathode Ray Tube

HIGH VOLTAGE SOURCE

JJ Thomson’s Plum Pudding Model (1898)

Thomson proposed that the atom consisted of negatively charged particles embedded in a ball of positive charge.

PLUM PUDDING MODEL

Ramifications of Thomson’s Experiment

Thomson’s proof that there were electrons disproved the part of Dalton’s theory that said atoms were indivisible and indestructible. Atoms were no longer considered to be the smallest units of matter. Scientists knew already that atoms were overall electrically neutral so there must be some other particles in atoms that balance the electrons.

Rutherford’s Gold Foil Experiments Rutherford set out to test Thomson’s plum pudding model. Rutherford shot alpha particles (+ charge) at a thin sheet of gold foil. He expected them to go straight through with perhaps a minor deflection. (hypothesis) Most did go straight through, but to his surprise some particles bounced directly off the gold sheet!
 Animated Rutherford Experiment

Rutherford’s Gold Foil Experiments Rutherford Concluded (1911):

1. The atom is mostly empty space since the majority of the alpha particles passed straight through. 2. The center of the atom is positively charged. He concluded this because the center is where he saw the positive alpha particles deflected. He called this positively charged, very dense center of the atom the nucleus.

Subatomic Particles
 After Rutherford and Thomson’s conclusions, scientists knew that atoms consisted of: 1. A positively charged, dense, central nucleus: - the proton is the positive subatomic particle located in the nucleus. (proved by Rutherford) 2. Negatively charged electrons that surrounded the nucleus. (Proved by Thomson and Rutherford)

Subatomic Particles

Since atoms must be electrically neutral, scientists reasoned that an atom must contain an equal number of positively charged particles (protons) and negatively charged particles (electrons). However, when the masses of protons (1 amu) and electrons (1/2000 amu) were added together the result was less than the actual mass of the atom.

Subatomic Particles Scientists concluded that there must be a third subatomic particle with a mass of 1 amu (equal to a proton) and no charge. They called this particle a neutron. Subatomic Particle Summary Protons- (+) charge. Located in the nucleus. Mass =1 amu Neutrons- no charge. Located in the nucleus. Mass = 1 amu Electrons- (-) charge. Surround the nucleus. Mass = 1/1836 amu

Limitations of the Rutherford Model
 Rutherford’s model could not explain: 1. Why elements radiate light when electricity is passed through them. 2. Why negatively charged electrons were not “sucked” into the oppositely charged central nucleus.

These questions that arose in Rutherford’s model led to its modification by Neils Bohr in 1913.

Bohr Model of the Atom
The Bohr model proposed a “planetary” model of the atom with the positive nucleus in the center and the negative electrons in orbits (energy shells) surrounding it. The further an orbit (shell) was from the nucleus, the higher its energy level.

Bohr Models Ultimately, Bohr concluded that the following number of electrons could fit into each energy shell: 1st : 2 electrons 2nd: 8 electrons 3rd: 18 electrons 4th: 32 electrons

Ground State vs Excited State

When an atom is in its lowest energy state, it is said to be in its ground state. When an atom absorbs a specific amount of energy, one of its electrons can “jump” into the next energy level. When this happens, the atom is said to be in the excited state. Atoms emit light as this excited electron returns to the ground state.

Ground State vs Excited State
Electron configurations on the periodic table are given in the ground state.
Ex: Na: 2-8-1 Ca: 2-8-8-2

Excited state can be recognized because one of the electrons jumps to a higher energy level.
Ex: Na: 2-7-2 Ca: 2-8-7-3

Properties of Atoms
Atomic Number- The number of protons in the nucleus of a specific type of atom. Mass Number- The number of protons plus neutrons in the nucleus of a specific type of atom. Also referred to as atomic mass.

To figure out the number of neutrons in an atom, subtract the atomic number from the mass number. Example: Carbon has an atomic # of 6 and a mass number of 12. This means carbon has 6 protons (A #) and 6 neutrons (M # - A #)

Symbols

The symbol tells you the number of each subatomic particle in one atom of the element. The number in the top left is the mass #. - The number of protons plus neutrons. The number in the bottom left is the atomic #. - The number of protons.

Symbols

The number in the top right is the charge. - If the charge is 0, the # of electrons equals the number of protons - If the charge is positive, there are more protons than electrons. - If the charge is negative, there are more electrons that protons.

Atomic Symbols
208 82

Pb

2

51 23

V

17 8

O

2

82 protons 126 neutrons 80 electrons

23 protons 28 neutrons 23 electrons

8 protons 9 neutrons 10 electrons

Formation of Ions
 Ions = an atom with a charge due to loss or gain of electrons  METALS ARE LOSERS!!!!  Metals lose electrons to form (+) ions that are smaller than the neutral atom. Ex: Na: 2-8-1 Na+ : 2-8 Ca: 2-8-8-2 Ca+2 : 2-8-8  Nonmetals gain electrons to form (-) ions that are larger than the neutral atom. Ex: O : 2-6 O-2 : 2-8 P: 2-5 p-3 : 2-8 O-2 and P-3 are isoelectronic (same # electrons)

Isotopes
Isotopes are atoms of the same element that have different numbers of neutrons. The additional neutrons add mass but not charge to the atom. The atomic mass that is on the reference tables is actually an average of the masses of all of each element’s isotopes.

Isotopic Mixtures

To find an atom’s average atomic mass, you must:  First, convert the % abundance of each isotope into decimal form. (ie: 95.0%  0.95). Next, multiply the abundance by the mass of each isotope and add up the products. This is your average atomic mass.

Isotopic Mixtures Example
Three isotopes of Argon occur in nature: Argon-38 (35.97 amu;0.337%) Argon-39 (37.96 amu; 0.063%) Argon-40 (39.96 amu; 99.600%) What is the average atomic mass? (35.97)(.00337)= 0.1212 (37.96)(.00063)= 0.02391 (39.96)(.99600)= 39.80 (0.1212) + (0.02391) + (39.80) = 39.95 AMU