ch 301 unit 1 exam 2 review notes 7. quantum mechanics

CH 301 Unit 1 Exam 2 Review Notes

7. Quantum mechanics applications – H atom – easy

Be able to explain and distinguish T/F questions.• Schrodinger developed quantum mechanics. Explains wave

behavior of electrons.• Can’t know exact location, but we can figure out the most probable

location.• Not all locations are possible. There are rules for where electrons

are found. These are quantum rules.• n – energy level, distance from the nucleus, n = 1, 2, 3, . . . • l – shape of energy level, l = 0, 1, 2, 3, . . . , n – 1 • ml – orientation of the shape in space, ml = - l, . . . 0 . . . , +l

• ms – the spin of the electron, ms = !"# ,

%"#

s sub-shell ( l = 0 ) d sub-shell ( l = 2 ) p sub-shell ( l = 1 )

These are the most likely location of the electron.

• s has 1 orientation• ms = 0

• p has 3 possible orientations• ms = -1, 0, +1

• d has 5 possible orientations• ms = -2, -1, 0, 1, 2

ms = !"# ms =

%"#

Practice Exam Problem!

The size of an atomic orbital is determined by which quantum number?

1. ml

2. l3. ms

4. nAnswer: 4. n

All of these numbers translates to the periodic table! So you don’t really need to memorize everything. Just remember

that . . .

• n = shell ( 1, 2, 3 )• l = subshell ( s, p, d )• ml = orbital • 2 electrons per orbital!

Here are the skills and knowledge you need to answer this question correctly on the exam:

8. Quantum number boundary conditions (Two questions on exam) - medium/hard


Given boundary conditions, be able to apply it to the multielectron system to explain how many possible electrons fit the condition.

Remembern = 1, 2, 3, . . .l = 0, 1, 2, . . . , n-1 ml = -l, . . . 0 . . . , +l

ms = !"# ,

%"#Example: How many electrons are in the n = 1 shell?

Answer: Only 2

( n = 1 so l = only 0 and so ml = only 0 and since ms = !"# ,

%"# , there are

only 2 e- in the n = 1 shell )

Example: How many electrons in n = 1 through 3 are p?Answer: Only 12 ( When n = 1 there are no p orbitals, when n = 2 there are 3 p orbitals, and when n = 3 there are 3 p orbitals. Giving us a total of 6 p orbitals with 2 e- in each of them, meaning that there are 12 e- total! )


In an atom, what would be the maximum number of electrons having the quantum numbers n = 6 and l = 2?

1. 722. 63. 54. 105. 8 Answer: 4. 10

Example: Is this a possible combination of quantum numbers?

n = 1 l = 1 ml = 0 ms = !"#

Answer: No ( l must be a number that satisfies the condition n – 1, so l cannot be equal to n, which in this case it is)


n = 4 l = 2 ml = -2 ms = !"#

Answer: Yes (these numbers satisfy the quantum number rules)


n = 3 l = 2 ml = -3 ms = %"#

Answer: No ( ml must stay in the values of -l, … 0 …, l )


Which set of quantum numbers does notprovide a satisfactory solution to the wave equation?

1. n = 2, l = 0, ml = -12. n = 1, l = 0, ml = 03. n = 4, l = 2, ml = +24. n = 5, l = 3, ml = -35. n = 3, l = 2, ml = -1 Answer: 1



9. Periodic table nomenclature (Two questions on exam) – easy

You will have a periodic table on the exam. It tells you so much if you know

how to read it.

• It lets you calculate molar mass

• It lets you know the number of electrons, protons, and neutrons

• It tells you the number of valence electrons

• It answers quantum rules about n, l , ml, m

s

• It tells you energy levels of electronic configurations

• It tells you the state of the elemental form


Fill in the blanks: potassium is one of the most

well-known elements in the alkali metal ____. It

is in the ____ which makes it a ____ element. Its

single valence electron is in the ____ subshell of

the ____ shell, making it very reactive. It reacts

readily with non-metals to form ____.

1. series; s block; common; l = 1; n = 4;

networks

2. row; d block; main group; l = 1; n = 4; salts

3. family; s block; main group; l = 0; n = 4; salts

4. row; d block; non-metal; l = 1; n = 3; alloys

5. family; s block; reactive; l = 0; n = 3; alloys

Answer: 3.

What else? Know the definitions of:

Period, group, family, block, main group,

transition metal, lanthanum, actinium series,

alkali metal, alkali earth metals, halogens,

noble gases, shell, subshell, orbital, non-

metals, metals, metalloids.


Remember the quantum number ms: which only has

two numbers that it could be !"# , %&'

("# .

Pauli:No 2 electrons have the same set of quantum numbers.This translates as ”no more than 2 e- per orbital”

3p ↑ ↑ ↑3s ↑↓2p ↑↓ ↑↓ ↑↓2s ↑↓1s ↑↓

3p ↑ ↑ ↑ 3s ↑↓↑2p ↑↓ ↑↓ ↑↓2s ↑↓1s ↑↓

3p ↑↓ ↑ ↑3s ↑↓2p ↑↓ ↑↓ ↑↓2s ↑↓1s ↑↓

Yes!

No! You can’t have 3 electrons in an orbital.

Yes!


10. Aufbau, Hund, Pauli theory (Two questions on exam) – easyHere are the skills and knowledge you need to answer this question correctly on the exam:

In addition to knowing quantum rules for n, l, ml, ms when filling electron configuration for multiple electron, there are also rules for how they fill.

Aufbau:Electrons fill from the lowest to the highest.

Hund:Spread out electrons when they have the same energy levels in p and d orbitals

3p ↑ ↑ ↑3s ↑2p ↑↓ ↑↓ ↑↓2s ↑↓1s ↑↓

3p ↑ ↑ 3s ↑↓2p ↑↓ ↑↓ ↑↓2s ↑↓1s ↑↓

3p ↑↓ ↑3s ↑↓2p ↑↓ ↑↓ ↑↓2s ↑↓1s ↑↓

No! We need to fill the 2s before we fill the 3p.

Yes!

No! the electrons aren’t spread out.


One of the following does NOT represent the ground state electron configuration for an atom. Which one?

1. [Ne] 3s2 3p5

2. [Ne] 3s1 3p3

3. [Ne] 3s1

4. [Ne] 3s2

Answer: 2.


11. Electron configurations of atoms: no exceptions (Two questions on exam) – medium


This is a mechanical process to assign electronic configuration.

• Determine the number of electrons. For neutral atoms, it is the same as the atomic number.

For example, there are 6 for C and 17 for Cl

• Fill up

4p __ __ __

3d __ __ __ __ __

4s __

3p __ __ __

3s __

2p __ __ __

2s __

1s __

You don’t need to draw out

this every time, just read off

of the periodic table

1s2 2s2 2p6 3s2 3p6 …

until you add the number of

e- to the atomic number.

Example:

C has 6 e-

So its electronic configuration is 1s2 2s2 2p2

Cl has 17 e-

So its electronic configuration is 1s2 2s2 2p6 3s2 3p5

But wait, in large atoms you don’t want to keep

writing 1s2 2s2 2p6 3s2 3p6 every time.

What you can do instead is write the previous

noble gas in brackets and continue from that

point.

So another way to write the electronic

configuration of Ti is [Ar] 4s2 3d2

This is called shorthand notation.

If you add up the

superscript numbers you

should get the total

number of electrons!


Which of the following atoms has an s2p3 valence electron

configuration?

1. Ga

2. Ge

3. As

4. Se

Answer: 3.

(As has the electronic configuration [Ar] 4s2 3d10 3p3)


12. Electron configurations of ions: no exceptions (Two questions on exam) – medium


Just like problem 11 • Count the number of e-• Fill 1s2 2s2 2p6 3s2 etc. • Use shorthand notation when you find your electronic configuration too long.

However, ions makes the first step just a little bit tricky.With ions we have to add or subtract electrons from the original number of e- count.

Remember:that a ‘ – ‘ superscript means that the atoms is more negative, so you have to add an electron,

and a ‘ + ‘ means that the atom is more positive, so you have to subtract an electron.

Na has 11 e- so it’s electron configuration is 1s2 2s2 2p6 3s1

Na+ has 10 e- so it’s electron configuration is 1s2 2s2 2p6

Cl has 17 e- so it’s electron configuration is [Ne] 3s2 3p5

Cl- has 17 e- so it’s electron configuration is [Ne] 3s2 3p6 ≡ [Ar]


How many electrons are in the 4s orbital of a Ca2+ ion?

1. 02. 23. –24. 35. 1

Answer: 1. (The Ca2+ ion has lost 2 electrons, so it does not have any electrons in the 4s orbital)

More examples!

Cr3+ has 24 – 3 = 21 e- so its electron configuration is [Ar] 4s2 3d1

What about Ur91+? This ion has 1 e- so its electron configuration is just 1s1.

This is a silly example but the point is to just add or subtract 3- and then treat them like the neutral atom that corresponds.


13. Electron configurations of atoms: d block exceptions (Two questions on exam) – medium


In creating electronic configurations for atoms and ions that normally have s2 d4 or s2 d9 (following the Aufbau principle), we have to take into consideration the stability of

filled and half-filled stability of the 2 subshells. So when writing the electron configurations there is a promotion of ‘s’ electrons to the ‘d’ subshell.

Some things to remember!

• Atoms and ions in the same column as Cr will have a s1 d5 at

the end of their electron configuration

• Examples: Cr, Fe+2, Ta-

• Atoms and ions in the same column as Cu will have a s1 d10 at

the end of their electron configuration

• Examples: Cu, Zn+, Co-2

What’s the reasoning behind the exceptions?

Look at the diagrams below:

The hard part is distinguishing the atoms and ions in group 11 and 12 that are not

exceptions from the ones in group 13 that are.

d ↑ ↑ ↑ ↑

s ↑↓

d ↑ ↑ ↑ ↑ ↑

s ↑

d ↑↓↑↓↑↓↑↓↑

s ↑↓

d ↑↓↑↓↑↓↑↓↑↓

s ↑

s2d4 s2d5

s2d9 s2d10


A comparison of the electron configurations of

nickel (Ni) and copper (cu) indicates that

1. Cu has two more d electrons and one less s

electron than Ni.

2. Cu has one more d electrons and one more s

electron than Ni.

3. Cu has one more d electrons and the same

number of s electrons than Ni.

4. Cu has two more d electrons and the same

number of s electrons than Ni.

5. Cu has one more d electrons and one less s

electron than Ni.

Answer: 1. (Cu’s configuration is [Ar] 4s1 3d10

while Ni’s is [Ar] 4s2 3d8 )

Which of the following would be the

correct electron configuration for

divalent ruthenium, Ru(II)?

1. [Kr] 4s2 4d4

2. [Kr] 5s1 4d5

3. [Kr] 5s2 4d4

4. [Kr] 4s1 4d5

5. [Kr] 5s1 5d5

Answer: 2. ( Ru(II) is the same as

Ru2+ placing it in the same column

as Cr, meaning we have to watch out

for the exception rule)


14. Electron configurations and magnetism – medium


4p 3d ↑ ↑ ↑ ↑ ↑4s ↑3p ↑↓ ↑↓ ↑↓3s ↑↓2p ↑↓ ↑↓ ↑↓2s ↑↓1s ↑↓

Be able to create a drawn out electronic configuration for an atom and ion.You would need to be able to draw out the entire “picture”, not just the shorthand notation.This will help you distinguish different strengths of magnetism between different ions and atoms.

Here is the electron configuration drawn out for Fe2+ which has 24 e-

Note the 5 unpaired electrons

The more unpaired electrons, means it is a

stronger magnet!


Based on electronic configurations, which of the following species would you predict to be most magnetic?

1. Sc2. Ti3. V4. Cr

Answer: 4. ( Sc has 1 unpaired e-, Ti has 2, V has 3, and Cr (like Fe2+) will have 5. Cr has the most unpaired e- so it’s the most magnetic. )

4p 3d ↑↓↑↓↑↓ ↑↓ ↑↓4s ↑↓3p ↑↓ ↑↓ ↑↓3s ↑↓2p ↑↓ ↑↓ ↑↓2s ↑↓1s ↑↓

Here is the electron configuration drawn out for Zn which has 30 e-

No unpaired electrons

So from drawing out the electron configuration you can see that

Fe2+ is magnetic, and zinc is not!


15. Periodicity theory: effective nuclear charge & shielding (Two questions on exam) – medium

Here are the skills and knowledge you need to answer this question correctly on the exam:One question will ask you to reason why ENC and shielding influence

periodic trends and the other will ask you to calculate ENC for e- in n = 1,

2, or 3 shell.

ENC = # protons - # shielding e-

For e- in n = 1 shell there is no shielding

For e- in n = 2 shell there is 2 e- shield

For e- in n = 3 shell there is (2+8) = 10 e- shield

So..

• an e- in n = 1 shell for Mg sees ENC = 12 – 0 = 12

• an e- in n = 3 shell for Mg sees ENC = 12 – 10 = 2


As you move down a family on the periodic table, the

increase in atomic radius can be explained by the ___.

1. increase in number of shielding electrons

2. increase in effective nuclear charge

3. decrease in number of shielding electrons

4. decrease in effective nuclear charge

Answer: 1. (As you move down a family on the periodic table,

the increase in atomic radius can be explained by the increase

in number of shielding electrons. Effective nuclear charge

(ENC) is constant within a given family on the periodic table;

changes in ENC can be used to explain periodic trends across

a row.)Explanation:

Shielding increases heading down table.

+

-

-

This electron ‘feels’

the entire nucleus

charge.

This electron is shielded

by the electrons before

it, so it ’feels’ less and is

therefore farther away

and easily ionized.

What is the effective nuclear charge of a valence electron

in the Al+ cation?

1. 2

2. 3

3. 12

4. 13

Answer: 2. (The ENC of Al- would be 13 protons – 10

shielding electrons which is 3.)

Why is Li larger than F?

+3 +9

- -

-

- - --

- - --

-

Sees only

ENC =1

Sees ENC =7! The

other e- in n = 2

have a negligible

effect on each

other.

Fluorine’s outer most

electrons feel a stronger

nuclear charge than lithium

does, so its atomic radius is

effectively pulled in.


16. Ranking periodic trends without exceptions, atomic radius – easyHere are the skills and knowledge you need to answer this question correctly on the exam:


Rank the following atoms in terms of decreasing atomic radius.

1. Na, N, O, Mg, F2. F, O, N, Na, Mg3. F, O, N, Mg, Na4. F, Mg, Na, O, N5. Na, Mg, N, O, F

Answer: 5.

Lilarge

Csvery large

Fvery small

smaller

The trend for atomic radius is very smooth.

You will be given a collection of atoms and asked to rank from largest to smallest of smallest to largest.

All you need to so is to identify them on the periodic table and rank them along the diagonal.

C N

Si P

Don’t worry we won’t try to trick you, for the case between these four atoms is a little complicated.

So we won’t ask you about these.


17. Ranking periodic trends without exceptions, ionic radius – mediumHere are the skills and knowledge you need to answer this question correctly on the exam:

Ionic radius also follows ENC smoothly.But you have to create an isoelectric table to compare.


Rank the following atoms and ions Li+, Be2+, He, H-, B3+

in order of decreasing size.

1. H-, Li+, He, Be2+, B3+

2. B3+, Be2+, He, Li+, H-

3. B3+, Be2+, Li+, He, H-

4. H-, He, Li+, Be2+, B3+

5. He, H-, Li+, Be2+, B3+

Answer: 4.

Isoelectric table around He (2e-)

Isoelectric table around Ne (10e-)

H- He Li+ Be2+

# e- 2 2 2 2

# p+ 1 2 3 4

ENC smallest ⟶ largest

Ionic radius largest ⟵ smallest

O2- F- Ne Na+ Mg2+

# e- 10 10 10 10 10

# p+ 8 9 10 11 12

ENC smallest ⟶ largest

Ionic radius largest ⟵ smallest


18. Periodicity theory: islands of stability (Two questions on exam) – medium


One question asks you to identify

islands of stability on the periodic table.A second question asks you to identify situations in which this

enhanced stability influences trends.

Here are a couple examples to remember.

Aufbau exception:

1. s2d4 à s1d5

2. s2d9 à s1d10

ENC exception

Ionization energy doesn’t follow trend so…

I.E.

The prediction based on ENC

d1

s2

d5

d6

d10

p1

p3

p4

s1

p6

Practice Exam Problems!

Which of the following trends is not solely explained by

effective nuclear charge and shielding?

1. Metallic character

2. Ionization energy

3. Atomic radius

4. Ionic radius

Answer: 2.

Which of the following valence electron configurations

does not exhibit enhanced stability?

1. s2

2. d5

3. d10

4. p5

Answer: 4.

s2 ↑↓

d5 ↑ ↑ ↑ ↑ ↑

d10 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

p3 ↑ ↑ ↑

p6 ↑↓ ↑↓ ↑↓

These families have enhanced

stability similar to the noble gases

electron configurations.


19. Ranking periodic trends, exceptions, 1st ionization energy (Two questions on exam) – medium


One question asks you to rank I.E. exception to ENC because of of the islands of stability. Note that given a series of atoms, general trend is lower left (Cs) is easiest because of shielding and low ENC, while the highest ionization energy is F.

But along this trend there are exceptionsd1 is lower than s2

d6 is lower than d5

p1 is lower than d10

p4 is lower than p3

These want to lose e- to be

either filled or half-filled.

You should also be able to draw the electron confirmation around p3 to explain it.

p4 ↑↓↑ ↑

p3 ↑ ↑ ↑

p2 ↑ ↑

The paired e- wants to leave to become like p3 even though the ENC for p4 is larger.

So…C < O < NSi < S < P


Rank the following from least to greatest ionization energy: silicon (Si), phosphorous (P), sulfur (S).

1. P < S < Si2. Si < P < S3. S < Si < P4. Si < S < P5. P < Si < S6. S < P < Si

Answer: 4. (Ionization energy increases across a given row, but because of the added stability of a half-filled p subshell, sulfur has a lower ionization energy than what would have been simply predicted based on effective nuclear charge arguments.

The ionization energy of an Oxygen atom (O) is (equal to/ greater than/ less than) what you would predict based on simple effective nuclear charge arguments because the half-filled 2p orbital for O+ is (more/less) stable.

1. equal to, less2. greater than, more3. equal to, more4. greater, than, less5. less than, more6. less than, less

Answer: 5.


20. Ranking electronegativity (Two questions on exam) – easy/mediumHere are the skills and knowledge you need to answer this question correctly on the exam:

One question will ask you to rank a series of atoms.They follow the same smooth trend as atomic radius but electronegativity gets larger, not smaller as you go towards F.

F

CsElectronegativity Increases

Li1

Be1.5

B2

C2.5

N3

O3.5

F4

H2.1

HeX

Another question will ask you to calculate !EN for elements in n = 1 and 2. So memorizing this table will be really helpful.


What is the correct order of increasing electronegativity for N, O, P, and K?

1. P, K, N, O2. K, O, N, P3. K, P, N, O4. O, N, P, K5. N, O, P, K

Answer: 3.

What is the difference in electronegativity between C and O atoms?

1. 0.52. 1.03. 1.54. 0.4

Answer: 2.

So !EN is a simple

subtraction.!EN for HF is

4.0 – 2.1 = 1.9

This is actually one of the

most important things you

will learn to do in all of unit 1!

ch 301 unit 1 exam 2 review notes 7. quantum mechanics

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