Chemical bonding lecture 1

Question 1

what is the phenomenon known as glow discharge

Answer 1

it is a process in which soft and beautiful light is shown when electricity is forced through a gas at low pressure

Question 2

what is J.J Thomson’s contribution to the atom model

Answer 2

he discovered electrons and the charge-to-mass ratio of electrons by using the glow discharge phenomenon, where electrons being produced from the cathode accelerate towards the anode that had a hole causing the formation of a collimated electron beam that lit up the phosphorus gas when it hit it.

The electrical deflection plates used an electric field to bend the electron beam causing a displacement that can be measured by the distance between the different spots formed. The magnetic field would then be turned on to revert the bent electron beam back to its original position.

this proves the existence of electrons. Further more, by finding the displacement and the values of the electrical and magnetic force we can find the mass-to-charge ratio as seen in my note (check the notes)

Question 3

what is the equation for the force applied in a uniform electrical field

Answer 3

F = qE where both E and F are vectors that are positive when the particle is positively charged and is making its way to the positive terminal (+ —–> -)

Question 4

what was Millikan’s experiment and how did we achieve the electron charge from it

Answer 4

Millikan used an atomizer to spray very small droplets of oil into a chamber that contains electrically charged plates. Eventually, the droplets pick up a charge either by colliding with the ions in the air of the chamber or during the spray process. Once the droplets fall and get in between the charged plates a constant electrical field can be applied, this electric field can then be equated to the gravitational force acting on the oil droplet. This causes the oil droplet to remain stationary and allows for the Electrical force to equal the gravitational forces in play giving us Mg = Eq, then we solve for q.

Question 5

what are the properties of the “canal rays” that were found behind the cathode in the crooks tube

Answer 5

1. the deflection of the canal rays by a magnetic field showed that they are positively charged
2. the electric and magnetic fields required to deflect the canal rays were much larger than those used on electrons showing that it is a heavy particle
3. positively charged particles associated with different gases have different masses

Question 6

how were the canal rays analyzed and what was the conclusion

Answer 6

by using J.J Thomson’s apparatus used to find the mass-to-charge ratio of the electron but by making 3 changes which are:
1. the voltages of the anode and cathode are flipped
2. the magnetic field is placed parallel to the electric field
3. the front of the tube is fitted with a photographic plate

Then using the same previous method to find the mass-to-charge ration, it was concluded that the canal rays were positive ions

Question 7

what did Ernest Rutherford discover

Answer 7

in short, by shooting heavily dense positively charged alpha particles at a thin gold foil, he was surprised at how some of the alpha particles were reflected back, at the time this went against the accepted model of the atom for which Ernest then developed his own planetary model which state that the positive charge is concentrated at the very small nucleus which is being orbited by negative electrons.

Question 8

What are the different types of bonding

Answer 8

1. ionic - one or more electrons are completely transferred from one atom to the other
2. covalent - the electrons are shared more or less equally between two atoms comprising the bond
3. polar covalent - a partial transfer of charge from one atom to the other

Note: the type of bond is determined by the atoms electronegativity

Question 9

what is the condensed structural formula

Answer 9

it is a structural formula that represents which atom is bonded to which

for example CH3OH

Question 10

Explain the movement of an electron around the positive nucleus

Answer 10

The movement follows a circular orbit around the nucleus with a constant potential force acting in the direction of the positive nucleus where the force would constantly change the direction of the electron’s velocity but keeps its speed constant thus its kinetic energy as well as constant.

For visual representation please look at pdf page 104

Question 11

the electrostatic force between an electron and a nucleus can be represented by?

Answer 11

Coulomb’s aw

V = q1q2 / 4pier

Coulomb’s law is applied to all charged particles

Question 12

Explain the attraction between proton-proton and electron-proton

Answer 12

Following figure 3.6 (PDF page 105), we can see that as the distance between two protons increases the potential energy between them decreases and that is because as the are closer to each other they repeal more but as they get further away they start losing the potential energy between them and start gaining kinetic energy instead.

In the figure, it also shows that as the distance between the proton and electron increases the potential energy increases as well, and that is because they are oppositely charged therefore the more distance between them the more potential to attract each other.

The interactions can be further sustained by using the Coulomb’s law equation as with electron-proton a large value of r will lead to a smaller NEGATIVE value of V and vice versa.

Question 13

what is the relationship between the attractive forces and potential energy

Answer 13

as the attractive force between the particles increases then the potential energy between them decreases

Question 14

How can we measure the force exerted on particles by other particles

Answer 14

simply by taking the derivative of the graph of Potential energy against distance r (dV/dr) and then multiplying it by the distance vector r between the two particles.

note repulsive forces are negative derivatives and attractive forces have a positive derivative

for eq the force between a proton and an electron.

F = dV/dr x (the vector r between the two particles). sub V in as the Coulombs potential energy.

Question 15

what is the total energy of an electron orbiting around a positive nucleus

Answer 15

E(total) = E(Kinetic) - E(potentail)

Question 16

What is ionization energy and how do you calculate it

Answer 16

it is the minimum energy necessary to remove an electron from a neutrally charged atom in the gas phase to form a positively charged ion in the gas phase

[ENERGY OF THE PRODUCTS] - [ENERGY OF THE REACTANTS] = E (which for ionization is always positive)

Question 17

what is the trend of ionization energy as we move across the period

Answer 17

the ionization energy is increases until it reaches the noble gasses then abruptly decreases and that is due to the noble gass’s very stable structure

Question 18

what was concluded after taking successive ionization energy of an atom

Answer 18

it is concluded that there is a shell model for the atomic structure, where the electrons are grouped into different shells that are concentrated around the nucleus where as we go outward from the nucleus the ionization energy decreases.

This follows the 2,8,8 electronic configuration

Question 19

how can we calculate the total potential energy between all particles in an atom

Answer 19

By drawing a diagram representing all of the different interactions of the particles, for example, placing the positive nucleus in the middle and sounding it by electrons that are spatially arranged to give the lowest repulsion. After that use Coulomb’s law with V = e^2/4pie (sum of all the electrostatic interactions)

Question 20

what is electron affinity

Answer 20

it is the energy measured when an electron attaches itself to a neutral atom to form an anion in the gas phase

The energy measured is always negative unless it comes to second electron affinities then adding an electron to a negatively charged ion will require energy making the electron affinity potentially positive

Note in the book for some reason they define electron affinity as the energy measured when an electron detaches its self so be aware but idc that’s dumb

Question 21

What is the trend in electron affinity

Answer 21

it keeps increasing across the period and reaches a maximum at the halogens in group 7 then supply decreases at the noble gases due to their full valence shell

Question 22

What is the ionic character

Answer 22

it is the partial charge separation in a polar covalent bond

Question 23

What is Pauling eletronnegativity scale

Answer 23

On the notes and pdf number 122 the notes also mention the range of values Electronegativity difference when the bond is polar covalent.

Question 24

how do we set up a potential energy equation for bond formation and how can we then achieve the total energy required to make the bond

Answer 24

find the potential energy of all the particles that have attractive forces and then the potential energy of the particles that have repulsive energies.

Total energy = the kinetic energy of all particles + the potential energy calculated.

note: remember potential energy is negative for attractive forces and there can be more than one expression for the repulsion of particles

Question 25

Describe the graph of Veff plotted against the Distance between the nuclei as a diatomic molecule dissociates

Answer 25

At the largest distance between the nuclei, there will be effectively ‘zero’ Veff, when bond dissociation energy is applied to break the bond at first there will be a negative potential as the distance between the nuclei decreases as there is still attractive forces between the electrons and the positive nuclei, once we reach the equilibrium bond length (Re) of the system there will be no repulsive or attractive forces between the nuclei’s which then as the bond length gets smaller the repulsive forces between positive nuclei’s increases which causes an increase in Veff.

Question 26

describe the graph of the potential energy against the distance between K and F in the natural state

Answer 26

At first, when the K and F are very far from each other there is virtually zero potential energy between them, as they get closer the attractive forces pulls them together until the reach their natural state in which they would have no repulsions nor attractions, after that the repulsive forces will take over causing a dramatic increase in the potential energy

Question 27

describe the graph of the potential energy against the distance between K^+ and F^-

Answer 27

There is a starting positive potential for the bond formation of K^+ and F^- and that is because energy is required to remove an electron from K energy and energy is required to attach an electron to fluorine where the ionization is always greater than the electron affinity which always results in a positive value.

As the ions approach each other the attractive forces between them cause the potential energy to decrease as they approach the Re where there are no attractive or repulsive forces in play, after that repulsion occurs between the ions as they get closer causing an increase in the potential energy

Note

• That the distance between the ions at Re is the bond length.
• The energy of bond formation is the difference between the Re and the starting energy

Question 28

what is the relation between bond energy and bond length as we go down the group

Answer 28

bond length increase.

bond energy decreases.

Question 29

How do you calculate the dipole moment?

Answer 29

Check notes made.