Unit Four: The Periodic Table

Question 1

Where are Groups #1-18 on the Periodic Table?

Answer 1

Just label the top of the groups with the numbers 1 through 18.

Question 2

Where are Periods #1-7 on the PT?T

Answer 2

Rows 1 through 7 on the periodic table.

http://www.green-planet-solar-energy.com/images/PT-blank-2.gif

Question 3

What are the 8 diatomic atoms?

Answer 3

1. Hydrogen
2. Oxygen
3. Nitrogen
4. Chlorine
5. Fluorine
6. Bromine
7. Iodine
8. AStatine (radioactive)

Question 4

Define atomic radius

Answer 4

Distance from center of the nucleus to the outermost enegy shell.

Half the distance between the nucleii of two identical, neutral atoms bonded together.

Question 5

Define ionization energy.

Answer 5

The amount of energy required to remove a valence electron from an atom in gaseuous state.

Question 6

Define Electronegativity

Answer 6

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.

Question 7

Electron affinity

Answer 7

The energy change associated when a neutral atom gains an electron

Question 8

Which element has the highest atomic radius and the lowest ionization energy and the lowest electro negativity?

Why?

Answer 8

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.

Question 9

Atomic radius trends: left to right, top to bottom?

Answer 9

Across a period (left to right), atomic radius decreases.

Down a group, it increases.

Question 10

Ionization energy trends

Answer 10

Across a period, increases.

Down a group, decreases.

Question 11

Effective Nuclear Charge.

Answer 11

Z_eff=# protons - #shielding electrons.

Question 12

Core electrons trend across period

Answer 12

Doesn’t. change.

Question 13

Effective nucleur charge experienced by valence electrons trend across a period?

Answer 13

A valence electron’s experienced Z_eff increases, b/c number of protons increases, but shielding effect does not.

Question 14

Z_eff relationship w/ atomic radius?

Answer 14

Each of the following examples has the same number of electrons, but the Zeff changes b/c # of protons differs.

Z_eff(F^-) = 9 - 2 = 7+

Z_eff(Ne) = 10 - 2 = 8+

Z_eff(Na⁺) = 11 - 2 = 9+

So the sodium cation has the largest effective nuclear charge, and thus the smallest radius.

Question 15

Zeff relationship with atomic radius trend down a group?

Answer 15

Zeff decreases, consequently atomic radius increases.

Question 16

Cation

Answer 16

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

Question 17

Cation radii are greater than neutral atom.

True or False?

Answer 17

False. Zeff is greater, therefore atomic radius will be smaller.

Question 18

Anion

Answer 18

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]

Question 19

Atomic radius of an anion is greater than that of the neutral atom.

True or False?

Answer 19

True, More electrons means more shielding, therefore smaller Zeff, therefore greater atomic radius.

Question 20

Isoelectronic

Answer 20

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.

Question 21

2 cations, 2 anions, isoelectronic w/ Neon.

Answer 21

1. Na⁺
2. Mg²⁺
3. F^-
4. O^2-

Question 22

Predict the charge of Rb

Answer 22

Rb⁺

Question 23

Predict the charge of Cs

Answer 23

Cs⁺

Question 24

Predicted Charge:

Ga

At

Se

Answer 24

Ga³⁺

At^-

Se^2-

Question 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

Answer 25

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.

Question 26

Ionization energy sodium: equation

Answer 26

Na + energy → Na⁺ + e^-

Question 27

What are Fajans’ rules?

Answer 27

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.

Question 28

Two factors controlling: first ionization energy

Answer 28

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

Question 29

Ionization energy increases going from Lithium ot Neon. Why?

Answer 29

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.

Question 30

He has largest ionization energy of all noble gases. why?

Answer 30

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

Question 31

Ionization Energy Equation

Answer 31

I=R_H*(Z_eff/n)²

1. Across a period, Z_effincreases and n (principal quantum number) remains the same, so the ionization energy increases.
2. Down a group, n increases and Z_effincreases slightly; the ionization energy decreases.

Question 32

Which element is expected to have the higher first ionization energies:

1. Ca or Be?
2. Na or Ar?

Answer 32

Figure out answer from table below.

Question 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?

Answer 33

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.

Question 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

Answer 34

Looks like Aluminium, in group 13. Because magnitude 5x between 3rd and 4th IE. (Found by scrounging around on the Internet).

Question 35

Predict formula with chlorine:

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

Answer 35

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 :

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.

Question 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]

Answer 36

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.

Question 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

Answer 37

5 (or 4)?

Question 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?

Answer 38

Group 2, as can e seen by the jump from IE 2 to IE 3.

Question 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

Answer 39

There are only 3 electrons.

Question 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

Answer 40

Y²⁺

Question 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

Answer 41

YO

Question 42

Arrange these species in order of increasing size: Rb⁺, Y³⁺, B^r-, Kr, Sr²⁺, Se^2-

Answer 42

Use the following table:

Question 43

Arrange these species in order of increasing size: K⁺, Ca²⁺, S^2-, Cl^-

Answer 43

An exercise left to your delectation.

Question 44

Expected sizes of the hydrogen ion (H+) and hydride (H-) ion relative to hydrogen atom (and why?)

Answer 44

Indeed?

Question 45

Why do elements in the same group react in similar ways?

Answer 45

Valence electrons

Question 46

Rank in order of increasing atomic radius and explain: O, S, F

Answer 46

F < S < O,

b/c this is the order of EN, which is proportional to Zeff

Question 47

Largest first ionization energy: Si, S or Se

Answer 47

S, Se, Si

Question 48

Increasing first ionization energies: Cs, Sr, Ba

Answer 48

Cs, Ba, Sr

Question 49

Element with largest jump between IE2 and IE3? K, O, Mg, Fe

Why?

Answer 49

Mg, b/c it goes down an energy level

Question 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?

Answer 50

to do

Question 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?

Answer 51

a = calcium,

b = bromine

2. B
3. B

Question 52

Configuration of an element is: [Ar]3d³4s²

Identify

Answer 52

Vanadium

Question 53

Configuration of an element is: [Ar]3d³4s²

Which group and period?

Answer 53

Group 5, Period 4

Question 54

Configuration of an element is: [Ar]3d³4s²

Identify type(?): nonmetal, main group, transition, lanthanide or actinide?

Answer 54

Transition element

Question 55

Configuration of an element is: [Ar]3d³4s²

Diamagnetic or paramagnetic? If paramagnetic, number of unpaired electrons?

Answer 55

Paramegnetic; 3 unpaired electrons.

Question 56

Which electrons experience greater effective nuclear charge in a Be atom: 1s or 2s electrons? Why

Answer 56

1s because of larger Zeff from the positive pull of the protons;

less shielding?

Question 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

Answer 57

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.

Question 58

Name neutral atoms isoelectronic with ion:

Cl^-

Mg²⁺

Se^2-

Answer 58

to do ..

Question 59

Use electron configurations to explain why H exhibits properties similar to both Li and F.

Answer 59

H could lose one e like lithium to become 1s0, or gain 1 electron like F to become 1s2

Question 60

Why is chlorine generally more reactive than bromine?

Answer 60

Higher EN and IE, greater Zeff

Question 61

Why is cesium more reactive towards water than lithium?

Answer 61

Cesium has a higher ractivity because it loses electrons easily, b/c of its large atomic radius.

Question 62

Phase of astatine at room temperature?

Answer 62

solid

Question 63

Astatine: Chemical formula of compound with Na

Answer 63

NaAt, chemically similar to Chlorine

Question 64

Electronegativity

Answer 64

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.

Question 65

Electron affinity

Answer 65

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.

Question 66

Which element has the highest atomic radius, lowest IE and lowest electronegativity and why?

Answer 66

.. to do ..

Question 67

Smallest to largest ionization energy (Al, O, Ba and Fe)

Answer 67

.. to do ..

Question 68

Largest to smallest atomic radius (Al, O, Ba and Fe)

Answer 68

Ba, Fe, Al, O

Question 69

8 metalloids on the periodic table: Locate

Answer 69

On a diagonal sloping to the right starting at Group 13, first row.

Boron, Silicon, Germanium, Arsenic, Antimony, Tellurium, Polonium and Astatine

Question 70

Alkali metals: location

Answer 70

Group 1

Question 71

Alkaline earth metals: Location

Answer 71

Group 2

Question 72

Transition metals

Answer 72

d block and f blocks.

Question 73

Halogens

Answer 73

Group 17 or VII

Question 74

8 diatomic atoms

Answer 74

Hydrogen, Nitrogen, Oxygen, Fluorine, Chlorine, Bromine, Iodine and Astatine

Question 75

Smallest to largest ionic radius: O, Al, Mg and N

Answer 75

Al+3, Mg+2, O-2, N-3

Question 76

Largest to smallest EN (Cl, S, Rb and K)

Answer 76

Cl > S > K > Rb

Question 77

Lowest AR, highest IE and highest EN? Why

Answer 77

Helium has the smallest AR, highest IE because largest Zeff

Fluorine has the highest EN because He does not form bonds.

Question 78

Smallest to largest IE: (Al, O, Ba and FE)

Answer 78

Ba < Fe < Al < O

Question 79

Smallest to largest AR: Al, O, Ba and Fe

Answer 79

O < Al < Fe < Ba

Question 80

IE trends increasing accross with exceptions. Be vs B, N vs O. Use electron configurations to explain anomaly.

Answer 80

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.

Question 81

What phase would you expect Astatine to be at STP and why?

Answer 81

Solid, b/c iodine is also solid.