Unit Four: The Periodic Table Flashcards
1
Q
Where are Groups #1-18 on the Periodic Table?
A
Just label the top of the groups with the numbers 1 through 18.
2
Q
Where are Periods #1-7 on the PT?T
A
Rows 1 through 7 on the periodic table.
http://www.green-planet-solar-energy.com/images/PT-blank-2.gif
3
Q
What are the 8 diatomic atoms?
A
- Hydrogen
- Oxygen
- Nitrogen
- Chlorine
- Fluorine
- Bromine
- Iodine
- AStatine (radioactive)
4
Q
Define atomic radius
A
Distance from center of the nucleus to the outermost enegy shell.
Half the distance between the nucleii of two identical, neutral atoms bonded together.
5
Q
Define ionization energy.
A
The amount of energy required to remove a valence electron from an atom in gaseuous state.
6
Q
Define Electronegativity
A
Tendency of attracting electrons in a bond.
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values range down to caesium and francium which are the leastelectronegative at 0.7.
7
Q
Electron affinity
A
The energy change associated when a neutral atom gains an electron
8
Q
Which element has the highest atomic radius and the lowest ionization energy and the lowest electro negativity?
Why?
A
Francium
Francium has the smallest Zeff; highest atomic radius and lowest IE
because
its valence electron is in the
7th energy level, farthest from the nucleus
Lots of shielding effect which lowers IE and electronegativity.
9
Q
Atomic radius trends: left to right, top to bottom?
A
Across a period (left to right), atomic radius decreases.
Down a group, it increases.
10
Q
Ionization energy trends
A
Across a period, increases.
Down a group, decreases.
11
Q
Effective Nuclear Charge.
A
Zeff=# protons - #shielding electrons.
12
Q
Core electrons trend across period
A
Doesn’t. change.
13
Q
Effective nucleur charge experienced by valence electrons trend across a period?
A
A valence electron’s experienced Zeff increases, b/c number of protons increases, but shielding effect does not.
14
Q
Zeff relationship w/ atomic radius?
A
Each of the following examples has the same number of electrons, but the Zeff changes b/c # of protons differs.
Zeff(F-) = 9 - 2 = 7+
Zeff(Ne) = 10 - 2 = 8+
Zeff(Na+) = 11 - 2 = 9+
So the sodium cation has the largest effective nuclear charge, and thus the smallest radius.
15
Q
Zeff relationship with atomic radius trend down a group?
A
Zeff decreases, consequently atomic radius increases.
16
Q
Cation
A
A cation (+) (/ˈkæt.aɪ.ən/ kat-eye-ən), from the Greek word κατά (katá), meaning “down”,[7] is an ion with fewer electrons than protons, giving it a positive charge
17
Q
Cation radii are greater than neutral atom.
True or False?
A
False. Zeff is greater, therefore atomic radius will be smaller.
18
Q
Anion
A
An anion (−) (/ˈæn.aɪ.ən/ an-eye-ən), from the Greek word ἄνω (ánō), meaning “up”,[5] is an ion with more electrons than protons, giving it a net negative charge (since electrons are negatively charged and protons are positively charged).[6]
19
Q
Atomic radius of an anion is greater than that of the neutral atom.
True or False?
A
True, More electrons means more shielding, therefore smaller Zeff, therefore greater atomic radius.
20
Q
Isoelectronic
A
Having the same # of electrons
Two or more molecular entities (atoms, molecules, or ions) are described as being isoelectronic with each other if they have the same number of electrons[1] or a similar electron configuration[2] and the same structure (number and connectivity of atoms), regardless of the nature of the elements involved.
21
Q
2 cations, 2 anions, isoelectronic w/ Neon.
A
- Na+
- Mg2+
- F-
- O2-
22
Q
Predict the charge of Rb
A
Rb+
23
Q
Predict the charge of Cs
A
Cs+
24
Q
Predicted Charge:
Ga
At
Se
A
Ga3+
At-
Se2-
25
An element's most stable ion forms an ionic compound with bromine, having the
formula XBr2. If the ion of element X has a mass number of 230 and has 86 electrons,
what is the identity of the element? How many neutrons does it have?
\*from:
http://depts.washington.edu/chemcrs/bulkdisk/chem142A\_win07/quiz\_answers\_Homework\_Solution\_02.pdf
1. Solution: first, in order to determine the identity of element X, need to determine the atomic number.
2. This can be done by using the charge on the bromide ion to determine what the positive charge on X is.
3. Then using the number of electrons, we can determine the atomic number, which will tell us what element it is.
4. So, Br has a charge of -1, and since there are two of them, X has a charge of +2. The atomic number then will be 86+2=88, which corresponds to radium (Ra).
5. Since the mass number is 230, the # of neutrons is going to be 230-88=142.
26
Ionization energy sodium: equation
Na + energy → Na+ + *e-*
27
What are Fajans' rules?
```
Ionic Covalent Low positive charge High positive charge Large cation Small cation Small anion Large anion
```
Fajans' Rules, formulated by Kazimierz Fajans in 1923,[1][2][3] are used to predict whether a chemical bond will be covalent or ionic, and depend on the charge on the cation and the relative sizes of the cation and anion.
Thus sodium chloride (with a low positive charge (+1), a fairly large cation (~1 Å) and relatively small anion (2 Å) is ionic; but aluminium iodide (AlI3) (with a high positive charge (+3) and a large anion) is covalent.
28
Two factors controlling: first ionization energy
1. **Valence shell -- filled or unfilled**. Generally, elements on the right side of the periodic table have a higher ionization energy because their valence shell is nearly filled. Elements on the left side of the periodic table have low ionization energies because of their willingness to lose electrons and become cations. Thus, ionization energy increases from left to right on the periodic table
2. **Electron shielding**: the ability of an atom's inner electrons to shield its positively-charged nucleus from its valence electrons. When moving to the right of a period, the number of electrons increases and the strength of shielding increases. As a result, it is easier for valence shell electrons to ionize.
*Best answer I have found is the Ionization energy equation which boils it down to Z\_eff and n (the principal quantum number, whatever that is): Ionization\_energy = R\_h \* (Z\_eff/n)^2*
29
Ionization energy increases going from Lithium ot Neon. Why?
1. **Increasing number of protons** in the nucleus as you go from lithium to neon. Causes greater attraction between the nucleus and the electrons and so increases the ionisation energies.
2. Increasing nuclear charge **drags the outer electrons in closer** to the nucleus. That increases ionisation energies still more as you go across the period.
3. Other factors remain the same in Period 2: screening effect is constant, outer electrons are in 2-level orbitals - 2s or 2p.
30
He has largest ionization energy of all noble gases. why?
The ionization energy of the elements within a group generally decreases from top to bottom because of **electron shielding**. Since Helium is at the top of its group, it will have the highest ionization energy.
***helium has no electron shielding, therefore the zeff on each electron is very high***
http://chemwiki.ucdavis.edu/Inorganic\_Chemistry/Descriptive\_Chemistry/Periodic\_Trends\_of\_Elemental\_Properties/Periodic\_Trends
31
Ionization Energy Equation
[I=RH*(Zeff/n)2](http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/Periodic_Trends)
1. Across a period, Zeff increases and *n* (principal quantum number) remains the same, so the ionization energy increases.
2. Down a group, *n* increases and Zeff increases slightly; the ionization energy decreases.
32
Which element is expected to have the higher first ionization energies:
1. Ca or Be?
2. Na or Ar?
Figure out answer from table below.
![]()
33
First four ionization energies of for an element A are:
1st: 578 kJ/mol
2nd: 1817 kJ/mol
3rd: 2747 kJ/mol
4th: 11580 kJ/mol
Why do they increase?
1. Once you have removed the first electron you are left with a positive ion. (Charge inequality).
2. Trying to remove a negative electron from a positive ion is going to be more difficult than removing it from a neutral atom.
3. Removing an electron from a 2+ or 3+ (etc) ion is going to be progressively more difficult.
34
Predict group and identity: First four ionization energies of for an element A are:
1st: 578 kJ/mol
2nd: 1817 kJ/mol
3rd: 2747 kJ/mol
4th: 11580 kJ/mol
Looks like Aluminium, in group 13. Because magnitude 5x between 3rd and 4th IE. *(Found by [scrounging around on the Internet](https://chemistry.osu.edu/~woodward/ch121/ch7_ie.htm)).*
35
Predict formula with chlorine:
## Footnote
First four ionization energies of for an element A are:
1st: 578 kJ/mol
2nd: 1817 kJ/mol
3rd: 2747 kJ/mol
4th: 11580 kJ/mol
1. The big leap in the ionization energies is from the 3rd to the 4th.
2. This implies that once 3 electrons have been removed, we have dropped down to the lower shell.
3. This implies that it has 3 electrons in its outer shell
4. This implies A → A3+ + 3e-
5. Chlorine forms the anion: Cl-
6. From (4) and (5) we can say that the compound will be: **ACl3**
We get a different answer from [chemguide in England](http://www.chemguide.co.uk/atoms/properties/moreies.html) :
The first four ionisation energies of aluminium, for example, are given by
1st I.E. = 577 kJ mol-1
2nd I.E. = 1820 kJ mol-1
3rd I.E. = 2740 kJ mol-1
4th I.E. = 11600 kJ mol-1
In order to form an Al3+(g) ion from Al(g) you would have to supply:
577 + 1820 + 2740 = 5137 kJ mol-1
1. That's a lot of energy. Why, then, does aluminium form Al3+ ions?
2. It can only form them if it can get that energy back from somewhere, and whether that's feasible depends on what it is reacting with.
3. For example, if aluminium reacts with fluorine or oxygen, it can recover that energy in various changes involving the fluorine or oxygen - and so aluminium fluoride or aluminium oxide contain Al3+ ions.
4. If it reacts with chlorine, it can't recover sufficient energy, and so solid anhydrous aluminium chloride isn't actually ionic - instead, it forms covalent bonds
5. Why doesn't aluminium form an Al4+ ion? The fourth ionisation energy is huge compared with the first three, and there is nothing that aluminium can react with which would enable it to recover that amount of extra energy.
36
Why is the 4th ionization energy of aluminum so much larger than the others?
Al (g) � Al+ (g) + e - [1st IE = 578 kJ/mol]
Al+(g) � Al2+(g) + e- [2nd IE = 1820 kJ/mol]
Al2 +(g) � Al3+ (g) + e- [3rd IE = 2750 kJ/mol]
Al3 +(g) � Al4+ (g) + e- [4th IE = 11,600 kJ/mol]
Consider the electron configuration of aluminum, [Ne] 3s23p1, after aluminum has lost three electrons to form Al3+ it has an electron configuration identical to neon. Since the noble gas electron configurations are particularly stable it takes a lot of energy to change the electron configuration ....
From [Ohio State](https://chemistry.osu.edu/~woodward/ch121/ch7_ie.htm).
37
What is the minimum number of electrons that A must have, given the following first four ionization energies of for an element A are:
1st: 578 kJ/mol
2nd: 1817 kJ/mol
3rd: 2747 kJ/mol
4th: 11580 kJ/mol
*5 (or 4)?*
38
Consider the following ionization energies (kJ/mol) of elements X and Y:
```
1 2 3 4 5 6 X 513 7298 11814 Y 737 1450 7732 10540 13360 17995
```
Which group is X in?
Group 2, as can e seen by the jump from IE 2 to IE 3.
39
Why does element X only have values for the first three ionization energies?
Ionization Energies in kJ/mol:
```
1 2 3 4 5 6 X 513 7298 11814 Y 737 1450 7732 10540 13360 17995
```
*There are only 3 electrons.*
40
Most likely charge on element Y, given:
Ionization Energies in kJ/mol
```
1 2 3 4 5 6 X 513 7298 11814 Y 737 1450 7732 10540 13360 17995
```
*Y2+*
41
Most likely formula of element Y's oxide, given:
Ionization Energies in kJ/mol
```
1 2 3 4 5 6 X 513 7298 11814 Y 737 1450 7732 10540 13360 17995
```
*YO*
42
Arrange these species in order of increasing size: Rb+, Y3+, Br-, Kr, Sr2+, Se2-
Use the following table:
43
Arrange these species in order of increasing size: K+, Ca2+, S2-, Cl-
*An exercise left to your delectation.*
44
Expected sizes of the hydrogen ion (H+) and hydride (H-) ion relative to hydrogen atom (and why?)
*Indeed?*
45
Why do elements in the same group react in similar ways?
*Valence electrons*
46
Rank in order of increasing atomic radius and explain: O, S, F
F \< S \< O,
b/c this is the order of EN, which is proportional to Zeff
47
Largest first ionization energy: Si, S or Se
S, Se, Si
48
Increasing first ionization energies: Cs, Sr, Ba
Cs, Ba, Sr
49
Element with largest jump between IE2 and IE3? K, O, Mg, Fe
Why?
Mg, b/c it goes down an energy level
50
A=[Ar]4s2, B=[Ar]3d104s24p5
1. A: metal, metalloid or non-metal
2. B: as above
3. Identify A and B
4. Which is expected to have the larger ionization energy?
5. Which element is expected to have the smaller atomic radius?
to do
51
A=[Ar]4s2, B=[Ar]3d104s24p5
1. Identify A and B
2. Which is expected to have the larger ionization energy?
3. Which element is expected to have the smaller atomic radius?
a = calcium,
b = bromine
2. B
3. B
52
Configuration of an element is: [Ar]3d34s2
Identify
Vanadium
53
Configuration of an element is: [Ar]3d34s2
Which group and period?
Group 5, Period 4
54
Configuration of an element is: [Ar]3d34s2
Identify type(?): nonmetal, main group, transition, lanthanide or actinide?
Transition element
55
Configuration of an element is: [Ar]3d34s2
Diamagnetic or paramagnetic? If paramagnetic, number of unpaired electrons?
Paramegnetic; 3 unpaired electrons.
56
Which electrons experience greater effective nuclear charge in a Be atom: 1s or 2s electrons? Why
1s because of larger Zeff from the positive pull of the protons;
less shielding?
57
You are shown a diagram illustrating a reaction. On the left, we see two balls of equal medium size; one grey and one white. On the right, we see a small grey and a big white ball. Which sphere (grey or white) represents a metal? Explain
Grey is the metal, b/c its atomic radius goes down. This is because it loses electrons, and the Zeff increases The white sphere grows larger because it gains electrons, becomes an anion, and Zeff decreases.
58
Name neutral atoms isoelectronic with ion:
Cl-
Mg2+
Se2-
to do ..
59
Use electron configurations to explain why H exhibits properties similar to both Li and F.
H could lose one e like lithium to become 1s0, or gain 1 electron like F to become 1s2
60
Why is chlorine generally more reactive than bromine?
Higher EN and IE, greater Zeff
61
Why is cesium more reactive towards water than lithium?
Cesium has a higher ractivity because it loses electrons easily, b/c of its large atomic radius.
62
Phase of astatine at room temperature?
solid
63
Astatine: Chemical formula of compound with Na
NaAt, chemically similar to Chlorine
64
Electronegativity
Electronegativity can be understood as a chemical property describing an atom's ability to attract and bind with electrons. Because electronegativity is a qualitative property, there is no standardized method for calculating electronegativity.
65
Electron affinity
Energy change when an atom accepts an electron.
Unlike electronegativity, electron affinity is a quantitative measure that measures the energy change that occurs when an electron is added to a neutral gas atom. When measuring electron affinity, the more negative the value, the higher an atom's affinity for electrons.
66
Which element has the highest atomic radius, lowest IE and lowest electronegativity and why?
.. to do ..
67
Smallest to largest ionization energy (Al, O, Ba and Fe)
.. to do ..
68
Largest to smallest atomic radius (Al, O, Ba and Fe)
Ba, Fe, Al, O
69
8 metalloids on the periodic table: Locate
On a diagonal sloping to the right starting at Group 13, first row.
Boron, Silicon, Germanium, Arsenic, Antimony, Tellurium, Polonium and Astatine
70
Alkali metals: location
Group 1
71
Alkaline earth metals: Location
Group 2
72
Transition metals
d block and f blocks.
73
Halogens
Group 17 or VII
74
8 diatomic atoms
Hydrogen, Nitrogen, Oxygen, Fluorine, Chlorine, Bromine, Iodine and Astatine
75
Smallest to largest ionic radius: O, Al, Mg and N
Al+3, Mg+2, O-2, N-3
76
Largest to smallest EN (Cl, S, Rb and K)
Cl \> S \> K \> Rb
77
Lowest AR, highest IE and highest EN? Why
Helium has the smallest AR, highest IE because largest Zeff
Fluorine has the highest EN because He does not form bonds.
78
Smallest to largest IE: (Al, O, Ba and FE)
Ba \< Fe \< Al \< O
79
Smallest to largest AR: Al, O, Ba and Fe
O \< Al \< Fe \< Ba
80
IE trends increasing accross with exceptions. Be vs B, N vs O. Use electron configurations to explain anomaly.
Be vs B: entire new sub-energy level 2p in B. Outermost 2p has more energy, more shielding, less Zeff, lower first IE.
N vs O: 3 electrons in 2p in N have their own orientations(px, py and pz): the extra electron from O pairs up into one orientation, causing the configuration to be less stable.
81
What phase would you expect Astatine to be at STP and why?
Solid, b/c iodine is also solid.
