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Merlin's Questions for EARTH

Q 1.

What was Newlands describing in his "Law of Octaves"?

What is the "rule" that Newlands found and what is its modern equivalent that we use in bonding?

What is the importance of the number "eight"?

And the answer is....

Q 2.

Mendeleev developed the first Periodic Table of the elements and it described the behavior of most of the smaller elements but not all of the elements. Moseley found a different way to arrange the elements.

What had Mendeleev done wrong and what was Moseley's discovery that set the Table right?

And the answer is....

Q 3.

What are the first 20 elements in order of increasing atomic number?
Include their symbols.

And the answer is....

Q 4.

In the Periodic Table, what is a Group and what is a Period?

What do elements in the same Group have in common?

What do elements in the same Period have in common?

And the answer is....

Q 5.

What is a "metalloid"?

And the answer is....

Q 6.

What name is used to describe the three forms of pure carbon (graphite, diamonds and fullerenes)?

And the answer is....

Q 7.

Name an element with two gaseous allotropes.

And the answer is....

Q 8.

Here's the atomic radii Table again.

I didn't give you a value for some of the elements, but you should be able to make a "guess".

Using this Table and the values that are shown, tell me the largest possible radius for the noble gases; helium (He), neon (Ne), argon (Ar), krypton (Kr) and xenon (Xe).

And the answer is....

Q 9.

What can you tell me about the atomic radii of francium (Fr), radium (Ra), polonium (Po) and astatine (At)?

And the answer is....

Q 10.

From the answers you calculated in questions 8 and 9 above, you might be able to give a maximum AND minimum size for radium (Ra).

This is a more complex puzzle. It requires you to think about the two different changes that occur as you move around the Table. Give it some thought, but don't be disappointed if you can't get a specific answer for the maximum size of radium (Ra).

And the answer is....

Q 11.

What is the size range (maximum and minimum) of radon (Rn)?

Don't forget you found some radii earlier.

Now that you did Q 10, this should be easier. Also, don't spend hours on this. It's another one meant to get you thinking about atomic radii.
It's a "trick question"!

And the answer is....

Q 12.

Alchemists use X-rays to measure the sizes of atoms. (You knew that.)
The Table you saw in the section on atomic radii showed that sodium (Na) has an atomic radius of 1.91 angstroms and chlorine has an atomic radius of 1.00 angstroms.
BUT, when Alchemists put their X-rays to crystals of sodium chloride (NaCl) they found that the sodiums had a radius of only 1.02 angstroms and the chlorine had a radius of 1.81 angstroms.

Why are the atoms in table salt (NaCl) of different sizes to those listed as atomic radii?
Why is sodium smaller but chlorine bigger in NaCl than in Na or Cl?

And the answer is....

Q 13.

Assume you had an element able to make covalent bonds and ionic bonds of either ion, as well as being all alone.

List, from smallest to largest, how the size of that atom would change for each type of atom and bond.
Or to put it another way, arrange its sizes from smallest to largest, for these four types of the atom; atomic, covalent, cation and anion.

And the answer is....

Q 14.

What atom (element) has all the properties in question 13? That is, what element can form covalent bonds, ionic bonds of either ion and also be atomic?

And the answer is....

Q 15.

Which Table would you use to calculate the energy needed to make Na+ from Na?

And the answer is....

Q 16.

Here's a piece of the Table showing the first ionization energy of the Group I and Group II elements. There are no values given for francium (Fr) or radium (Ra).

What do you think would be the maximum energy needed to remove an electron from an atom of francium?

What would be the maximum energy needed to remove an electron from an atom of radium?

And the answer is....

Q 17.

Is 15 eV enough energy to turn one Ca atom into one Ca+2 ion?

And the answer is....

Q 18.

Here's (part of) a Table of the SECOND ionization energy - the energy needed to remove the second electron once the first electron is gone.

In other words, this is the energy needed to make A+2 from A+1.
(That "A" means no element in particular.)

Do you see any trends? Can you explain them?
Note that the second ionization energies for the Group II elements are much lower than for Group I. Try to explain that.
(Hint: Think about an atom's "desire" for complete shells.)

And the answer is....

Q 19.

If you had 10 eV could you remove the second electron from the francium cation (Fr+)?

How about the second electron from the radium cation (Ra+)?

And the answer is....

Q 20.

Here's the first and second ionization energies. Notice the BIG difference in energy needed to remove the second electron. It takes a lot of eV to get that second electron off the already positively charged cation!

While you are looking at these two Tables together, remind yourself why the first ionization energy increases from Group I to Group II, but the second ionization energy decreases from Group I to Group II.

Now to the question. How many electron volts do you need to change one atom of Ca to Ca+2?

And the answer is....

Q 21.

What do you think would happen if you put 15 eV into three calcium atoms?

And the answer is....

Q 22.

What do you think would happen if you put 25 eV into three calcium atoms?

(Hint: There are two possibilities. Try to find both of them.)

And the answer is....

Q 23.

Describe "electron affinity".

And the answer is....

Q 24.

The noble elements (Group VIII) have high ionization energies and low electron affinities. What does that tell you about their ability to form ions?
Also, explain what that means in terms of their outer shell and position in the Periodic Table.

And the answer is....

Q 25.

What is "electronegativity" and how is it used by Alchemists?

And the answer is....

Q 26.

Fluorine (F2) is an extremely nasty gas causing sever chemical "burns" if it makes contact with your skin. Why?

And the answer is....

Q 27.

Here's the Table of electronegativities. Use it to predict and explain what would happen when hydrogen gas (H2) is mixed with fluorine gas (F2).

And the answer is....

Q 28.

The difference in electronegativity between hydrogen and fluorine is 1.78.

Would you expect that HF to be held together by covalent bonds or electrovalent bonds?

And the answer is....

Q 29.

You may have been surprised by that last answer. (But I bet you get this one!)

Is the H-F bond a polar bond? Can HF make hydrogen bonds?

And the answer is....

Q 30

Hydrogen fluoride (HF) is a "hydrogen halide", meaning a compound of hydrogen and a halogen (Group VII).

Name the other hydrogen halides and briefly explain how they compare to each other in terms of their polarized bonds.

And the answer is....

Q 31.

When you add hydrogen halides to water (by bubbling the gas through water) they break up into ions, forming "hydrohalic acids". The extra hydrogen ions added to the water cause the water to be described as an acid. (Acids have excessive amounts of hydrogen ions.)

An Alchemist would write that as
HF + H2O ----> H+ + F- + H2O
The reason the hydrogen halides break up in water is because water's two polarized bonds are able to pull apart the bonds thathold the hydrogen halides together. Then the water molecules solvate each ion. (Recall, that means that the water molecules form solvation shields around them.)

Which hydrogen halide do you think would be the last one to release its hydrogen ion to the water?

And the answer is....

Q32.

Freon is a covalent molecule with the formula CCl2F2.

Would you expect freon to have polarized bonds? Explain your answer and the way the electronegativity affects freon's overall partial charges.

And the answer is....

Q 33.

Freon is usually a gas except when the molecules are pushed very close to each other. That is, freon becomes a liquid only if you push the molecules together with a lot of pressure. As soon as the pressure is released, the freon molecules fly away from each other and return to their natural state, a gas.

Why? Why is freon so reluctant to form the intermolecular bonds with other freon atoms which would cause it to stay a liquid? What causes molecules of freon to push away from each other?

And the answer is....

Q 34.

Hexane is an organic molecule made of nothing but C-H bonds and it is a liquid at room temperature, like water. But the C-H bonds have an electronegativity difference of only 0.35. (Check the Table to be sure you know how I got that.)
We say that water is a polar liquid and hexane is a "non-polar" liquid.

Table salt (NaCl) dissolves in water but not in hexane. Why?

And the answer is....

Q 35.

Fingernail polish and most paints are made of non-polar compounds (with lots of C-H bonds). They don't wash off in water (easily), but they will come off if scrubbed with hexane. Why?

And the answer is....

Q 36.

Explain how sodium hydride (NaH) can be formed.

And the answer is....

Q 37.

Does the electronegativity of the Group I elements increase, decrease or remain the same when you go down the Group? And why?

And the answer is....

Q 38.

Does the reactivity of the Group I elements with water increase, decrease or remain the same when you go down the Group? And why?

And the answer is....

Q 39.

How do the electronegativity and reactivity with water of the Group I and Group II elements compare? And why?

And the answer is....

Q 40.

Describe the properties of the alkaline metals that gives them their name?

And the answer is....

Q 41.

Are the alkaline metals more alkaline than the alkali metals?

And the answer is....

Q 42.

Do the metallic properties of the Group I and II metals decrease down the Groups? And why (not)?

And the answer is....

Q 43.

Does the electronegativity of the halides (Group VII) increase or decrease as you go down the Group?

And the answer is....

Q 44.

Does the reactivity of the halides, as elemental molecules, increase or decrease down the Group? And why?

And the answer is....

Q 45.

If electronegativity decreases down all Groups,
why do the Group I elements become more reactive in water,
but the Group VII elements become less reactive in water,
as you move down those Groups?

And the answer is....

Q 46.

Do semimetals make semimetal bonds? Explain.

And the answer is....

Q 47.

Why does the Periodic Table display a "staircase" pattern to the metals (and semimetals)?

And the answer is....

Q 48.

Here's another look at the complete Periodic Table of the Elements.

You will recall that the K-shell holds only 2 electrons while the L-shell can hold 8 electrons.

How many electrons should the M, N, O, and P-shells be able to hold?

And the answer is....

Q 49.

A convenient and useful way to describe an atom is by the number of its protons, neutrons and electrons, with the electrons arranged into their correct shells. Also, because the outer shell is so important, it is useful to include a description of how its electrons are distributed among its orbitals.

For example, 16O would be described as:
8 protons, 8 neutrons, with its electrons arranged as 2 (in the K-shell) and 6 (in the L-shell). The L-shell electrons are the outer electrons and they will be arranged in orbitals as s2 and p4. (We'll ignore exactly which types of p orbitals the 4 electrons can be placed in.)

Use this method and this Table to describe 12C.
Then tell me what Group it is in and why.
Add a wee comment about the outer shell and what that might tell you about the atom's properties.

And the answer is....

Q 50.

Do the same (as in Q 49) for 40Ar.

And the answer is....

Q 51.

Calcium (Ca) is element number 20 and the last "typical" element before the transition elements start to make things complex. Scandium (Sc) is the first transition element (atomic number 21). Tests done on scandium prove that it has 2 electrons in its outer (N) shell, just like calcium!

Describe what the electron arrangement of these two atoms (Ca and Sc) should be. Then try to figure out how scandium could have 2 electrons in its outer shell. Where does the extra electron go?
(Note: Scandium does not lose it. Scandium keeps all 21 electrons.)

And the answer is....

Q 52.

What are the number of protons and neutrons in 137Ba? Also, how are barium's electrons distributed among its shells?
You'll need to see the Tables clearly, so I've shown you two of them.



And the answer is....

Q 53.

Do the same (as in Q 53) for 197Au. You'll need to see the Tables clearly, so I've shown you two of them.



Take your time with this puzzle. Gold is an inner transition metal, so its electron structure is complicated, but try to predict its orbital configuration anyway. (It's good practice.)

And the answer is....

Q 54.

Let's take a break from discussing electrons (phew) and think a wee bit about neutrons.
The number of neutrons in an atom are easy to calculate by subtracting the number of protons (the atom's atomic number) from its total atomic mass (of all nucleons).

Here's a reminder of the atoms and isotopes we've discussed recently.
Carbon has 6 protons and its common isotope has 6 neutrons (an atomic mass of 12).
Oxygen has 8 protons and its common isotope has 8 neutrons (an atomic mass of 16).
Argon has 18 protons and its common isotope has 22 neutrons (an atomic mass of 40).
Barium has 56 protons and its common isotope has 81 neutrons (an atomic mass of 137).
Gold has 79 protons and its common isotope has 108 neutrons (an atomic mass of 187).
These are all stable isotopes of that element. (So they are not radioactive.)
Look at their proton to neutron ratios (# of protons to # of neutrons).

Do you see a progression? Describe how the ratio of protons to neutrons changes as the atoms get larger.

And the answer is....

Q 55.

Alchemists sometimes include the atomic number and atomic mass with the atom's symbol in order to give the maximum amount of information. Alchemists use a "Z" to stand for atomic number (the number of protons) and place it as a subscript before the symbol. They also use an "A" for the atomic mass (the number of nucleons) and include it as a superscript before the symbol (like you learned earlier).

If the element has the symbol "X" then the entire atom is written as AZX.

Rewrite the atoms from question 54 in this manner so you can see their atomic mass (A) as a superscript and atomic number (Z) as a subscript to each symbol.
Here they are again.
Carbon has 6 protons and its common isotope has 6 neutrons (an atomic mass of 12).
Oxygen has 8 protons and its common isotope has 8 neutrons (an atomic mass of 16).
Argon has 18 protons and its common isotope has 22 neutrons (an atomic mass of 40).
Barium has 56 protons and its common isotope has 81 neutrons (an atomic mass of 137).
Gold has 79 protons and its common isotope has 108 neutrons (an atomic mass of 187).

And the answer is....

Q 56.

OK, back to electron configuration (ugh).

As you now know it is very difficult to keep track of the electronic configuration of an atom. Some Alchemists have to figure our the electronic configuration of all the inner shells as well as the outer shells! They don't always have a book nearby to help them. Sometimes they don't even have a Table to guide them!
So they developed another way to track and assign electrons to orbitals and shells. It is simple to learn and use this method.

First, let's recall that the shells, which we've been calling "K-shell", "L-shell", etc. are really based upon the first quantum number.

What is the (first) quantum number for each shell?
(This is easy. Just write down the order of the shells and number them.)

And the answer is....

Q 57.

Take a look at this chart. It is a simple list of the various orbitals for each shell. Notice I am now using the principle quantum number for the shells instead of their letters.

The first shell (K-shell) is shell number 1 and it has a single s orbital.
The next shell (L-shell) is number 2 and it can have an s orbital and p orbitals.
The next shell (M-shell) is number 3 and it can have an s orbital, p orbitals and d orbitals.
The next shell (N-shell) is number 4 and it can have an s orbital, p orbitals, d orbitals and f orbitals.
The remaining shells (O, P and Q-shells) are number 5, 6 and 7 they can also can have an s orbital, p orbitals, d orbitals and f orbitals.

Draw this chart into your notebook. The first column lists all the s orbitals with their principal quantum number in front of each. The second column is a list of all the p orbitals also with the principal quantum number preceding it. (Notice that we start the second column with principal quantum number 2 because the second shell, the L-shell, is the first to have p orbitals.) The third column lists the d orbitals starting with principal quantum number 3 (because d orbitals start with the M-shell). The chart ends with the fourth column where you list the f orbitals starting with principal quantum number 4.

Draw the chart carefully, following the logic I explained above.
Do you see any useful pattern here? (Think diagonally.)

And the answer is....

Q 58.

Assume you have a "monster" of an atom with 118 protons!
(In fact, an atom this big has never been created.)
Assume this monster atom is not an ion so it also has 118 electrons.

Use this table to write out the entire series of orbitals in this monster atom in order of increasing energy.
Then assign electrons to each orbital and tell me what its outer shell is like and what chemical properties it might have.
Reminder: Each s orbital gets two electrons so it is written as s2 if that orbital is full.
Similarly full orbitals are p6, d10 and f14.


(Once you have this all written out you will be surprised how easy this is.)

And the answer is....

Q 59.

Now that you have your orbital series worked out (that's the name for the sequence of orbitals all arranged in increasing order) you can easily work out the orbital configuration of any atom without looking at the Periodic Table or worrying about the complexities of "basement" shells.

Use your newly gained ability to work out the complete electron orbitals to all these atoms you did earlier.
8O, 6C, 18Ar, 56Ba, 79Au.

(Notice I've included their proton counts, Z number, as a superscript so you don't need to consult a Table.)

And the answer is....

Q 60.

Let's find the electron configuration of another three elements using the orbital energy series. Try
Nickel - 28Ni,
Copper - 29Cu,
and Zinc - 30Zn.

And the answer is....

Q 61

Alchemists use a shorthand to write electron orbitals. They use noble elements to keep track of things. That is, they compare the electron configurations to that of the noble element that preceded it. (Note, that means we are talking about the Group VIII element that is a Period BEFORE the atoms we are concerned with.)

Find the electronic configuration of the noble element that precedes nickel, copper and zinc and use it as a foundation on which to "build" the nickel, copper and zinc electronic configurations. (Use copper's correct configuration.)

And the answer is....

Q 62.

What are the "transuranic elements"?

Where are they hidden in the Table?

And what is special (unusual) about them?

And the answer is....

Q 63.

List the four kinds of orbitals,
how many types of each orbital there are,
which parts of the Table have them as their outermost orbitals,
and how many electrons can fit in all of them.

And the answer is....

Arthur's (and Merlin's) Answers

A 1.

Newlands noticed that when (some) of the elements are listed in ever increasing mass, they tended to repeat their properties every eighth time. He called this observation his "Law of Octaves". This is the evidence behind Lewis's "octet rule". We use Lewis's octet rule to draw Lewis structures in order to understand bonding.

Atoms can hold eight electrons in their outermost shell. It is this desire to have a complete outer shell (that is an outer shell with eight electrons) that gives rise to most bonding and chemical behavior. Indeed, there can never be more than 8 electrons in the outermost shell. (Transition and inner transition elements hide their extra electrons in lower shells.)

Back to the question.

A 2.

Mendeleev developed a theory about the chemical periodicity, but he arranged the elements according to their atomic mass, instead of their atomic number. This is pretty good for the smaller elements. Moseley's X-ray work allowed him to count the protons and arrange the elements into increasing atomic number. When Alchemists arranged the elements according to Moseley's atomic NUMBER (instead of atomic mass), the elements with similar electronic configuration lined up into patterns whose behavior repeated every eighth time.

Back to the question.

A 3.

Perhaps you felt this was a bit unfair because of all the memorization it requires but most Alchemists have in their heads the first 20 (or 40!) elements and so should you.

The first 20 elements can be remembered by learning this (stupid) poem:
"Hi. Here Little Beggar Boys Catch Newts Or Fishes. Never Nab Maggots Alone Since Poisonous Substances Clog Arteries. Keeping Calm."

This gives you a hint to their symbols but, of course, it is only a learning tool and you might have created a better "poem". Some of the symbols are not easy to remember because they are based on other languages (like Na for sodium or K for potassium) or require that you use a second letter because another element has used the first letter (like nitrogen using N so neon must use Ne).

The first 20 elements in order of increasing atomic number are:
  1. Hydrogen (H)
  2. Helium (He)
  3. Lithium (Li)
  4. Beryllium (Be)
  5. Boron (B)
  6. Carbon (C)
  7. Nitrogen (N)
  8. Oxygen (O)
  9. Fluorine (F)
  10. Neon (Ne)
  11. Sodium (Na)
  12. Magnesium (Mg)
  13. Aluminum (Al)
  14. Silicon (Si)
  15. Phosphorous (P)
  16. Sulfur (S)
  17. Chlorine (Cl)
  18. Argon (Ar)
  19. Potassium (K)
  20. Calcium (Ca)

Back to the question.

A 4.

In the Periodic Table each column (running from top to bottom) is called a Group and each row (running across) is a Period.

Elements in the same Group have the same type of outer shell. Note that they do not necessarily have the same number of electrons in the outer shell (although they often do). Elements in the same Group have the same needs to complete their outer shell. Therefore elements within a Group often have similar chemical properties. By knowing the behavior of one element in a Group, you can make a good guess that the others will behave the same.

Periods run from left to right in rows, each row repeating when the outer shell is full. Elements in the same Period have the same outer shell BUT not the same shell type. For example, the first row, or "Period", represents the K-shell and both elements in the first Period (hydrogen and helium) have a K-shell as their outer shell. (That's assuming they are not ionized.). The second period is the L-shell and all elements in the second Period have an L-shell as their outermost shell.

Back to the question.

A 5.

A metalloid is a semimetal displaying a chemical property of a metal, like the ability to turn water alkaline.

Back to the question.

A 6.

Graphite, diamonds and fullerenes are allotropes of carbon.
They differ only in the arrangement of their atoms (all of which are carbon).

Back to the question.

A 7.

Oxygen has two allotropes, "normal" oxygen (O2) which we need to breathe and ozone (O3) which we don't want to breathe, but which protects us from the harmful ultraviolet light.

Back to the question.

A 8.

Atoms get smaller as you move from left to right across the Periods. That means atoms are smaller than their neighbor to the left. So...
Helium must be smaller than 0.37 angstroms (the radius of hydrogen) because helium's extra proton pulls the K-shell a wee bit closer.
Neon must be smaller than 0.71 angstroms (the radius of fluorine) because neon's extra proton pulls the L-shell a wee bit closer.
Argon must be smaller than 1.00 angstroms (the radius of chlorine), krypton must be smaller than bromine (1.14 angstrom) and xenon must be smaller than iodine (1.34 angstrom) for the same reasons.
If you thought (for example) that neon's radius must lie between fluorine and sodium, you made a serious error in your view of the Periodic Table. The move from fluorine to neon added an extra proton which drew the shell closer, making it smaller. But the move from neon to sodium adds a whole new shell and there is a BIG increase in size because you have started a new Period (shell).

Back to the question.

A 9.

All these elements will have atomic radii larger than the elements immediately above them because they have an extra shell.
So, francium must be larger than 2.72 angstroms (the atomic radii of cesium right above it). Radium must be larger than 2.24 angstroms (because that's barium's size). Polonium must be bigger than 1.40 angstrom and astatine must be bigger than 1.34 angstroms.

Back to the question.

A 10.

Radium is directly below barium (which has a size of 2.24 angstroms) and immediately to the right of francium (which must have a radius bigger than 2.72 angstroms. Right?).
Radium must be bigger than barium, because it is below barium (and thus has an extra shell). And radon (Rn) must be smaller than francium (Fr), because it holds its outer shell closer.
Therefore, radium is definitely larger than 2.24 angstroms in size. It is definitely smaller than francium, but all we know about francium is that it is larger than 2.72 angstroms.
This problem may have been an irritation to you. The point of this problem is not to get a specific answer. It is really meant to exercise your understanding of how the radius changes.

Back to the question.

A 11.

This is another difficult problem and you may have been a bit frustrated with it. But I bet you learned something!
Radon (Rn) is below xenon (Xe) and to the right of astatine (At). Therefore, radon must be larger than xenon (because it has an extra shell) but smaller than astatine (because it has an extra proton to pull in the outer shell).
Xenon must be smaller than iodine (1.34 angstroms) as you know from question 8. Also, astatine must be larger than iodine as you know from question 9.
With this in mind you would be correct to come to this very silly conclusion....
"Radon must be bigger than a value smaller than 1.34, because xenon must be smaller than 1.34. And radon must be smaller than a value bigger than 1.34, because astatine must be bigger than 1.34."
Huh???!!

You have every reason to be confused. Out of frustration you may have decided that radon must be 1.34 angstroms. But that is not the correct way to look at it.
For example, xenon could be 1.30 angstroms and it would fulfill the requirement that it be smaller than iodine. Right? And astatine could be 1.50 angstroms and it would fulfill the requirement that it be larger than iodine. Right? So, radon could be 1.40 angstroms and thus fulfill its requirement that it be bigger than xenon (which in this example I'm assuming is 1.30 angstroms) and it would also be smaller than astatine (which in this example I'm assuming is 1.50 angstroms).
Notice that if radon really is 1.40 angstroms it fulfills the "silly conclusion" above. You see the "silly conclusion" is logical and correct but it is useless! Many values would satisfy the requirements of that silly conclusion.
If you had a size for both xenon and astatine you would be able to work out the range (maximum and minimum) of radon. The next best thing is to look at the values that ARE in the Table. Iodine is 1.34 angstroms and bismuth (Bi) is 1.71 angstroms. Because of the position of radon in the Table, you can safely say that radon will be smaller than 1.71 angstroms.

"So, what is the size of radon?", you ask? (Perhaps shouting!). I don't know. No one really knows for sure. Even the sizes of xenon and astatine are argued about. The reason for this disagreement is that the lower right corner of the Periodic Table includes elements that are very difficult to purify and measure. Many of these elements are rare and unstable (radioactive). Some books give sizes for these atoms but another book will give different sizes! That is because of differences in the methods they use to prepare and measure the elements.

If you found this problem irritating, I am sorry. But I think you will have learned that some things are not as straight forward as we all would like them to be. However, the rules about how the size of atoms change as you move around the Table still apply.

Back to the question.

A 12.

The atomic radii are determined using "pure" atoms of the element that are NOT involved in ionic bonds. When atoms undergo bonding their outer shells often change in size. Recall that covalently linked atoms share their outer shell electrons and are thus able to pull closer to one another, so they appear to have smaller radii.

NaCl is an ionic compound made of one cation (Na+) and one anion (Cl-). The sodium atom releases its (only) outer electron to become Na+. By doing that it sheds an entire shell, so it becomes smaller. The chlorine atom gains an electron so its outer shell grows larger because that extra electron cannot be held as tightly as the normal electrons. (That's because there are now more electrons than protons, so the nucleus cannot draw that extra electron in tightly.)

In the future, as you read Tables with values for atoms' sizes, keep in mind that bonds can dramatically change them. Especially ionic bonds.

Back to the question.

A 13.

The smallest form of any atom is the cation form. To become a cation the atom must lose an electron. (Right?). If that is the only electron in its outer shell, then the atom will shrink to the size of the next smallest shell (approximately). If there are still electrons remaining in the outer shell after the one electron leaves, the atom will still be smaller! That's because the cation will have more positive charge to it and will pull the shell a wee bit closer. (Electrostatic attraction, remember?) So all cations are smaller than the "normal" atom because they either have lost their outer shell completely or they pull the remaining electrons in the shell closer.

Slightly larger in size would be the atom involved in a covalent bond. This is due to the sharing of the outer shell. This makes it slightly smaller than the "normal" atom.

Next in size is the atomic or "normal" atom. It has no bonds and therefore it has a "normal" shell and "normal" size. This is the atom we think of when we start to think of an atom's size.

The largest type of atom would be an anion. The extra electron may require a new shell to be formed, thus making the atom larger than "normal". If there is room in the outer shell for the extra electron, that shell will grow a wee bit larger because there are fewer protons than electrons, so the electrostatic attraction can't hold all the electrons as close.

Summary: from smallest to largest; cation, covalent, "normal", and then anion.

Back to the question.

A 14.

Hydrogen! (Of course!).
The hydrogen cation (H+) is VERY small, just a proton! (Even a cation of deuterium or tritium would be tiny.)
Hydrogen commonly forms covalent bonds (O-H, C-H, etc.)
Hydrogen, of course, has an atomic size when all alone.
And, finally, the hydrogen anion, the hydride (H-), can be made.
Which kind of hydrogen you have depends upon the environment (the other elements around it and the conditions).

Note: the hydride (H-) will have the electronic shell of helium (a full K-shell), but will be larger than helium. That is because the hydride has only one proton pulling in two electrons and it doesn't do that very well. So the hydride has a bigger (K) shell than helium even though they both have two electrons in it.

Don't get confused. As you move across a Period (from left to right) the atoms ALWAYS get smaller.
Helium, has two protons and two electrons so it pulls them in more tightly. That is why helium is smaller than hydrogen and (thus) to the right of hydrogen in the Periodic Table. BUT when you have extra or fewer electrons (thus an ion) its size depends more on the effects of electrostatic attraction than on the shell.

Back to the question.

A 15.

You must use the Table of first ionization energies to figure out how much energy (electron volts, eV) is needed to ionize Na to Na+.

Back to the question.

A 16.

Ionization energies decrease as you go down a Group because the larger shells and more electrons make it easier to remove an electron.

You can see from the Table that you need 3.89 eV to ionize cesium, so you need no more (actually you need less) than 3.89 eV to ionize francium, because it is below cesium.

Similarly, you need (less than) 5.21 eV to ionize radium, because 5.21 will ionize barium (right above it).

Back to the question.

A 17.

You can't tell! Not from this table.
This is a table of FIRST ionization energies.
This is the energy to remove the first (and easiest) electron.
It cannot be used to calculate the energy needed to ionize the second electron.

To do that you need to also consult a Table of SECOND ionization energies.
Certainly 15 eV is enough to ionize the first electron from two calcium atoms
(2Ca X 6.11 eV = 12.22 eV, so you even have a little energy left over.)

But that is not what the question asks.

Back to the question.

A 18.

The second ionization energy decreases down the Group, just like the first ionization energy. The reason is the same as for the first ionization. Larger atoms have larger shells with more electrons. The larger shells mean the electrons in the outer shell are easier to remove than they would be if they were in the Period above it. The same trend is seen in second ionization energy as seen in the first ionization energy.

BUT notice that the second ionization energy decreases as you move from the Group I to the Group II elements. This is not the same trend as seen in the first ionization energy.
You might have expected the ionization energy to increase as you move to the right, because the shells are drawn more closely to the nucleus. That is what happens to the first ionization energy. It increases as you go across the Table. (Remember?)

The second ionization removes the next electron. The next electron from a Group II single cation (say for example, Mg+) is the remaining electron in that shell, which makes the whole new shell (of Mg+2) look exactly like a noble shell. (Think about that a bit so you understand it.)
BUT, the next electron from a Group I single cation (say for example, Na+) must be an electron from the next inner shell. Think about that. A Group I ion (A+) has a complete outer shell so it looks like a noble element. Removing another electron will cause the outer shell to become incomplete. The atom "doesn't want to do that"!

Therefore, it is easier to remove the first electron from Group I than Group II because that first ionization brings the Group I element to the "noble state". But, it is easier to remove the second electron from Group II than Group I because that second ionization brings the Group II element to the "noble state". (Of course, you can't remove the second electron until you've removed the first!)

Read through that again to be sure you got it.

Back to the question.

A 19.

The Table doesn't show you the second ionization energies of these two elements and you are expected to use the "trends" noted in the previous question.

Francium will need less than 25.1 eV to remove its second electron because cesium needs 25.1 eV to do the same. Because ionization energies decrease as you move down a Group, francium needs less energy than cesium to do the same job. BUT how much less is unclear. Maybe francium's second ionization energy is 15 eV. That's less than cesium's, so the Table would be "correct" as far as the trends are concerned. BUT, 10 eV would not be enough! So 10 eV might remove Fr+'s second electron or it might not! You can't tell (from this Table).

However, 10 eV will definitely remove the second electron from the radium cation (Ra+). You can see that 10 eV will remove the second electron from the barium ion (Ba+) which is just above radium. That means radium will have a second ionization energy less than 10 eV. So 10 eV will remove the second electron from Ra+ to make Ra+2 (and there will be a wee bit of energy left over).

Back to the question.

A 20.

Calcium's first ionization energy is 6.11 eV. That removes the first electron, changing the neutral Ca atom into Ca+.

Calcium's second ionization energy is 11.9 eV. That removes the second electron, changing Ca+ to Ca+2.

Therefore you need 18.01 eV (6.11 eV + 11.9 eV = 18.01 eV) to remove both electrons, converting Ca to Ca+2

You can express that as two equations linked together like this

Ca ----6.11 eV----> Ca+ ----11.9 eV------> Ca+2

or simply

Ca ----18.01 eV------> Ca+2

Back to the question.

A 21.

With 15 eV, two calcium atoms would lose one electron each.
(2Ca X 6.11 eV = 12.22 eV).
You would have a little energy left over (2.78 eV) but that is not enough to remove the first electron from the third calcium atom.

So you would have two ions (2Ca+) and one neutral atom (Ca).

Note; 15 eV is NOT enough to turn even one calcium atom (Ca) into Ca+2. (That requires 18.01 eV, like you figured out in Q 20.)

Back to the question.

A 22.

This is NOT an easy question! There are TWO possibilities.

1) All three calcium ions could lose their first electron to become 3Ca+. (Because, 3Ca X 6.11 eV = 18.33 eV.) That leaves 6.67 eV left over, enough to remove the first electron from a forth calcium atom, if you had one, but since you don't the extra energy is just wasted as heat.
So, you could get 3Ca+.
3Ca ----3 X 6.11 eV----> 3Ca+

OR

2) One atom of calcium could lose two electrons. That uses up 18.01 eV (you worked that out in Q 20). The 6.99 eV remaining (25 eV - 18.01 eV = 6.99 eV) is enough to remove one electron from either of the two remaining calcium atoms (and you would have 0.88 eV left over as heat). The third atom would not be touched because the remaining 0.88 eV isn't enough to do anything.
So, you could get Ca+2, Ca+ and Ca (all three types)..
Ca ---------------->Ca
Ca ----6.11 eV----> Ca+
Ca ----6.11 eV----> Ca+ ----11.9 eV------> Ca+2
-------------------------------------------------------------
3Ca ----24.12 eV-------> Ca, Ca+, Ca+2

That was a VERY difficult question and you should be very proud of yourself if you got either answer. However, you should read through this answer and make sure you know what I have done to get the two possible answers.

Oh, which is the right answer? Well, that is hard to say. A great deal depends on how the energy is delivered to each atom. Even an advanced Alchemist would have trouble proving which way these three atoms would end up!

Back to the question.

A 23.

Electron affinity is a measure of how easily an atom accepts an extra electron to become an anion. It is NOT the opposite of ionization energy. By that I mean that an atom with low or even negative electron affinity is not necessarily good at becoming a cation.

Back to the question.

A 24.

Their high ionization energy means the that noble elements require a great deal of energy to remove an electron. So they are not likely to form cations. The low (actually, negative) electron affinities of the noble elements mean they are very bad at picking up extra electrons. So they are not likely to form anions either.

These two values, a high ionization energy and low electron affinity, simply put numbers to the ideas you already know about forming stable outer shells. All the noble elements have 8 electrons in their outer shell, making it a full (or complete) shell. Eight is the "magic number" of electrons that all atoms "want" in their outer shell. It is the basis of Lewis' "octet rule", Newlands' "law of octaves", and explains the shape of Mendeleev's Periodic Table by defining where the Period ends.

Back to the question.

A 25.

Electronegativity is a measure of how tightly an atom holds onto electrons - its own electrons or spare electrons it finds.
It is a value specifically designed to show how electrons behave in molecular bonds.
Alchemists use electronegativity to predict what kinds of bonds are formed between two atoms.

Back to the question.

A 26.

Fluorine is the most electronegative of all elements and that makes it the most reactive of all elemental molecules.

Because fluorine has the greatest electronegativity of any element (3.98, just to remind you) it can pull electrons away from just about any molecule it approaches. The only atom safe around a fluorine atom is another fluorine atom. Two fluorine atoms (F2) are covalently linked together because they both have the exact same electronegativity, so they can't steal an electron away from the partner. All they can do is share a pair of electrons as they try to satisfy their "desire" for a complete outer shell (8 electrons). But when an atom of any other element comes along, fluorine tries to grab an electron away from it. Most of the time the other atom doesn't stand a chance! If the other element has a low electronegativity, fluorine will steal an electron. That causes fluorine to become an anion and the other atom to become a cation.
The same reactions happen when F2 contacts skin (or eyes, mouth, etc.). The very aggressive fluorines steal electrons from the "weaker" atoms. The "stronger" atoms (those with pretty high electronegativities, themselves) will end up sharing an electron with fluorine in a new covalent bond.

So to summarize, fluorine gas (F2) is only stable with itself, because only then can the tug-of-war between the two powerful atoms cancel each other out. But when that gas meets any other kind of atom it either steals an electron from it (if the "victim" has a very low electronegativity of its own), or forms new covalent bonds to it (if the "victim" has a reasonable electronegativity of its own).

Back to the question.

A 27.

The hydrogen gas (H2) is a covalently linked molecule. The two hydrogen atoms share a pair of electrons because they have the exact same electronegativity (2.20).

But when mixed with fluorine gas (which you described in the previous question) the electrons in the two hydrogen atoms cannot resist the pull of the fluorine. This causes the hydrogen molecule (H2) to break up as the fluorine steals away the electrons.

The difference in electronegativity between fluorine and hydrogen (3.98 - 2.20 = 1.78) proves that hydrogen cannot put up a fight with those strong fluorines.
So the two elemental molecules (H2 and F2) become compound molecules. Like this:
H2 + F2 ------> 2HF

Back to the question.

A 28.

The huge difference in electronegativity (1.78) is more than the 1.50 that I like to use as my "cut off". But that only goes to illustrate that there is no clear cut off between covalent and ionic bonds.
Perhaps you said that MOST of the time the H-F bond is ionic (electrovalent), but it must have some covalent "character" SOME of the time. That would have been a reasonable answer, but unfortunately it is wrong. It just so happens that MOST of the time the H-F bond is covalent, but it has some ionic (electrovalent) "character" SOME of the time. You would not have known that from the information given. As a matter of fact, many Alchemists were surprised to discover that HF is mostly covalent in spite of that large difference in electronegativity! (It took lots of experiments to convince them.)

Recall the O-H bonds in water are mostly covalent but have some ionic bonding some of the time. And the O-H bonds have a difference in electronegativity of only 1.24. Water's slight ionic character is helped by oxygen's lone pair orbitals and other details involving its structure. HF has certain properties that affect its character too. H-F bonds, like O-H bonds, are covalent for the most part but have some electrovalent bonding some of the time. Surprise!

As you work through the next questions you will see how this affects the covalent bond

Back to the question.

A 29.

It sure is and it sure can!
The H-F bond is a very polarized bond. The difference in electronegativity (1.78) is very high and tells you that most of the time the electrons that form the covalent bond will be near the fluorine atom. The fluorine atom will have a slight negative charge and the hydrogen will have a slight positive charge. Or, as we Alchemists like to say, fluorine is delta minus and hydrogen delta plus.
That's a polarized bond.

You will also have noticed that a very polarized bond like H-F will allow the hydrogen to form hydrogen bonds. HF's extremely polarized bond causes the hydrogen to be extremely delta plus, so it would easily form a hydrogen bond with any atoms that is even slightly negative.

Back to the question.

A 30.

Hydrogen halides are hydrogen bonded to any halogen. So the hydrogen halides include hydrogen fluoride (HF, but you knew that one), hydrogen chloride (HCl), hydrogen bromide (HBr), and hydrogen iodide (HI) . [Let's ignore hydrogen astatinide because it is very rare.]

Because the electronegativities of the halogens (like most elements) decrease down the Group, you can expect decreasing polarization down the Group. That would mean the H-F bond is more polarized than the H-I bond. It also means that hydrogen fluoride should make better hydrogen bonds than hydrogen iodide, and it does!

You may also have guessed that as you move down the Group the bonds in the hydrogen halides become more covalent and less electrovalent.

Back to the question.

A 31.

Hydrogen fluoride has the highest electronegativity difference (1.78) among the hydrogen halides. So, it will be the last molecule to give up its hydrogen.

Before leaving the subject of hydrohalides and the acids they form, it is worth repeating that HF does NOT make as strong an acid as the other hydrogen halides (HCl for example). This confuses some people.
Hydrofluoric acid is used to "etch" glass - that is to "burn" designs onto glass. Many people (even some Alchemists!) make the mistake of assuming that hydrofluoric acid MUST be a very strong acid to do that. No other hydrohalic acid etches glass as well as hydrofluoric acid! BUT, the ability of hydrofluoric acid to etch glass has to do with fluorine's ability to react with glass molecules (SiO2) in ways that have nothing to do with acids.

Back to the question.

A 32.

Yes, freon (CCL2F2) has 4 polarized bonds; two C-Cl bonds and two C-F bonds. The C-Cl bond has an electronegativity difference of 0.61 and the C-F bonds have an electronegativity difference of 1.43! So all four bonds are very polarized (especially the C-F bonds).

Recall that carbon has a covalency of four, meaning it forms 4 covalent bonds (when it can). Each halide has a covalency of one. So, freon is a carbon surrounded by the four halide atoms. (It's shaped like methane but with chlorines and fluorines instead of hydrogen.)
Carbon is the least electronegative of the atoms in the freon molecule, so it does not hold its shared electrons as tightly as the chlorine and fluorine atoms. The carbon atom at the center of freon will have a partial positive charge (delta plus) and the halides will all have a partial negative charge (delta minus).

Back to the question.

A 33.

Freon has four atoms sticking out from it which have slightly negative charges. These negative, partial charges are caused by the delta minus on the four halide atoms. The over all effect is to create a "shield" of partial negative charges around each freon molecule. This causes freon molecules to repel each other! Remember, molecules of the same charge will fly away from each other because of electrostatic repulsion (the exact opposite of electrostatic attraction). Therefore, you have to push very hard to make two freon molecules get close enough together to form any kind of intermolecular bonds (bonds between molecules). Actually, it is hard to imagine ANY intermolecular forces could form with all those partial negative charges on their surface. But it CAN be done if you put enough pressure on them. As soon as the pressure is released the molecules quickly fly away from each other, returning to their gaseous state.

This property, along with some others that are beyond the scope of this class, made freon the molecule of choice in refrigerators. Freon is a very stable molecule, most of the time, and many folks came to think of it as inert. But ultraviolet light can break up freon and create ions of chlorine and fluorine that will attack and destroy ozone (O3). It is for this reason that nations decided late in the 20th century to stop releasing freon into the atmosphere. (There are lots of other things to be said about freon's chemistry and physics, but they are too advanced a subject for this course.)

Back to the question.

A 34.

NaCl is an ionic compound, held together by electrovalent bonds. These bonds are formed because there is an electrostatic attraction between the sodium cation (Na+) and the chlorine anion (Cl-).
The O-H bonds in water have an electronegativity difference of 1.24. It is this difference in electronegativity that causes the O-H bonds to be polar. (Remember? Of course you do.)
Water's "polarity" helps it to pull ionic molecules apart and keep them apart.
But hexane is a non-polar liquid so it does not have the power to pull ionic molecules apart or keep them apart.

Polar and ionic molecules dissolve best in a polar liquid (water) but they do not dissolve in a non-polar liquid (hexane).

Back to the question.

A 35.

Fingernail polish and most paints are non-polar, so the polar nature of water is no help at all. As a matter of fact, water's polar nature causes the water molecules to stick to each other and hardly touch any of the non-polar molecules. That is, water doesn't "interact" with non-polar molecules very well (hardly at all!).

But when you wipe hexane on fingernail polish it sweeps away the layers of polish because...
"Like dissolves like."
The non-polar fingernail polish mixes in with the hexane by "non-polar forces". Actually, these forces are simply the lack of "polar forces". In a mixture of hexane and fingernail polish, the two types of molecules interact so well that they intermingle (get between each other). That causes the polish to come off the surface of the nail.
Polar forces are "real". They are caused by the big differences in electronegativity of polar substances (salt in water).
Non-polar forces are not "real" in the sense that they are not due to a real action. Instead these non-polar forces are caused by the lack of polar forces.

Note and Warning! Hexane is a carcinogen (causes cancer) and is flammable (burns easily). It should NOT be used on skin, nails or any other part of your body.
Many years ago Alchemists often washed their hands in hexane to remove nasty stains. Years later many of them developed cancer. (At least those that didn't set themselves on fire first!)
Fingernail polish remover contains acetone, another flammable non-polar liquid, but at least it isn't a carcinogen (in small amounts).

Back to the question.

A 36.

Hydrogen has an electronegativity of 2.20 which is pretty low compared to the elements on the right side of the Table. However, the elements on the left side of the Table have much lower electronegativities. The Na-H bond has an electronegativity difference of 1.27, enough to make a strong bond. As a matter of fact, it's an ionic one.

(Remember the "cut-off" of 1.5 is not a true cut-off.)

The NaH molecule is formed because sodium is not able to hold onto its outer electron when hydrogen pulls on it. In this tug-of-war the sodium gives up its outer electron to become a cation. (Nothing unusual about that.) The hydrogen gains that electron and becomes an anion (H-)! Then the two ions are drawn together by electrostatic attraction to form the ionic molecule NaH.
Hydrogen as an anion (H-) has two electrons in its outer (only) shell. When naming the ionic molecule formed by an anion of hydrogen (or any other anion), the anion goes to the back of the naming and takes on the "ide" ending.

Back to the question.

A 37.

The electronegativity of ALL the elements (not just Group I) decreases down the Group, because the extra shells added with each Period make it easier to remove an electron and harder to hold onto an extra electron.

Back to the question.

A 38.

The reactivity of the Group I elements with water increases down the Group because with decreasing electronegativity, the water molecules can more easily grab an electron away.
(You also know that electron affinity and ionization energy decrease down the Group. That is another way of looking at the same problem. But electronegativity gives a better picture.)

Back to the question.

A 39.

The Group I elements are less electronegative than their Group II neighbors to the right because as you move from left to right across a Period the atoms draw their outer shells closer. That makes it harder to remove an electron, so their electronegativity increases as you go from Group I to Group II.

Therefore any Group II element will be less reactive with water than the Group I element to its left because it is more electronegative so it is less willing to give up its electron(s) to water.

Back to the question.

A 40.

The alkaline metals are alkaline metals!
By alkaline, we mean that they form hydroxides (OH-) when added to water, causing an excess of hydroxide ions to be released into the water. Alkaline means to have an excess of OH-.
They are metals because they have all the properties of a metal; they conduct electricity and heat, are malleable and ductile, and have a luster (are shiny).

Back to the question.

A 41.

No, it's the other way around.
The alkali metals (Group I) are more alkaline than the alkaline metals (Group II).

Back to the question.

A 42.

No, it's the other way around.
Metallic properties increase down any Group because the shells get larger. Atoms with larger shells conduct electricity and heat better, are more malleable and ductile, and are easier to cut, than atoms with smaller shells.

Back to the question.

A 43.

Electronegativity of all the elements decreases down the Groups (because the extra shells make the atoms larger).

Back to the question.

A 44.

The reactivity of the halides decreases down the Group because the electronegativity also decreases as you go down (any) Group.

Back to the question.

A 45.

The reactivity of Groups I and VII in water change in opposite directions because the way they react is opposite.

Group I elements react in water by donating an electron, to become an cation. Because electronegativity decreases as you go down any Group, the elements can release their electron more easily as you go down a Group.

Group VII elements react by stealing an electron from water, to become an anion. The more electronegative any element, the easier it is for it to steal an electron. Because electronegativity decreases as you go down any Group, the Group VII elements become less able to steal away an electron as you go down Group VII.

Back to the question.

A 46.

No, there's no such thing as a semimetal bond.

Semimetals are elements that sometimes have some of the properties of a metal (depending on other factors like molecular arrangement and "contaminants").
Semimetals make metal bonds but only sometimes.

Back to the question.

A 47.

Metals have metallic properties. Semimetals have some metal properties under certain conditions.

All metal properties increase down a Group
(because the outer shells get larger)
but decrease across a Period
(because the outer shells get smaller).

These two opposing effects produce the staircase.

Back to the question.

A 48.

Some folks might call this a "trick question" but it's really meant to make you think about the shells.
Perhaps you used the Period Table to counted the number of elements in each Period then assumed that would be the maximum number of electrons that the shell could hold. If you did you would have decided that the M-shell can hold 8 electrons, the N and O-shells can both hold 18 electrons and the P shell can hold 32 electrons.
This should be clear to you from a look at the Table and knowing what it means. Each element in the Table has one more proton than the one before it. Therefore each element must also have one more electron (to keep its overall charge neutral).
Each Period represents a shell. By counting how many elements are in each Period, you can determine how many electrons can fit in the shell that the Period represents. Is that right?
Well, no. Remember, most big atoms are the transition and inner transition elements.
They hold their extra electrons in shells below their outer shell. The position of an atom in the Table (its Group) reflects its outermost shell but the number of atoms in the Period does not correctly reflect the number of atoms that shell is capable of holding.

The M-shell is capable of holding 18 electrons but it only does that when it is forced to do so by the larger outer shell. In the third Period the M-shell is the outer shell and it will only hold eight electrons (2 in an s orbital and 6 in the three p orbitals). That seems OK. But in the fourth Period the M-shell can be forced to store electrons in its d orbitals. That opens up a "basement" for storing an additional 10 electrons! So, in the fourth Period elements the outer shell (a N-shell) can force the M-shell to hold up to 18 electrons (2 in an s orbital, 6 in the three p orbitals and 10 in the five d orbitals). That's exactly what the first set of transition elements do. They (the transition elements in the fourth Period ) force the "basement" shell (the M-shell) to use its d orbitals.
So the M-shell can be forced to hold up to 18 electrons.

The N-shell is larger than the M-shell and it can be forced to hold extra electrons in its "basement" by the shells above it. An element in the fifth Period will use its O-shell as an outer shell forcing the N-shell to use its d orbitals. That's exactly what the second set of transition elements do. They (the transition elements in the fifth Period) force the "basement" shell (the N-shell) to use its d orbitals. But that's not the end of the story. An element in the sixth Period can use its P-shell to force the N-shell to use its f orbitals! That's exactly what the first set of inner transition elements do. They (the inner transition elements in the sixth Period ) force the "deep-basement" shell (the N-shell) to use its f orbitals. You will recall that there are 7 f orbitals so all together they are capable of holding 14 more electrons.
So the N-shell can be forced to hold up to 32 electrons (2 in an s orbital, 6 in the three p orbitals, 10 in the five d orbitals and 14 in the f orbitals).

What of the O-shell?
Well, it will be able to hold at least as many electrons as the N-shell because higher shells will forced it to do so. But which higher shells? The P-shell will force it to hold d orbitals and the Q-shell will force it to hold f orbitals. That means the O-shell can hold as many as 32 electrons (just like the N-shell). If the Periodic Table went on forever (it doesn't, but if it did) you would expect the "R-shell" to force more electrons into its "super-deep-basement" and they would be in strange orbitals that don't even exist!

You should be proud if you got that question right. Indeed, you should be proud if you understood that explanation the first time you read it. So read it again and focus on those "basement" problems.

Back to the question.

A 49.

12C has 6 protons (that's why it's carbon, C), 6 neutrons (to give it a total mass of 12) and has its electrons arranged as 2 (in the K-shell) and 4 (in the L-shell).
The L-shell electrons are the outer electrons and they will be arranged in orbitals as s2 and p2.

Carbon has 4 electrons in its outermost shell and it will try to get 4 more electrons to complete its outer shell, so it is a Group IV element. From other Tables we know that carbon strives to complete its outer shell by sharing electrons (thus forming covalent bonds) so carbon has a covalency of 4.

Back to the question.

A 50.

40Ar has 18 protons, 22 neutrons and the electrons are arranged as 2 (K), 8 (L), and 8 (a complete M-shell).
The M-shell electrons are the outer electrons and they will be arranged in orbitals as s2 and p6.

Because argon has a complete outer shell it is a noble element and belongs in Group VIII. Also, because its outer shell is complete, argon will be inert (not form bonds). It is unwilling to lose or gain electrons.

Back to the question.

A 51.

Calcium's electron structure is 2 (K), 8(L), 8(M), 2(N).
Scandium's electronic structure should be 2 (K), 8(L), 8(M), 3(N). But there is good evidence that it has only 2 electrons in its N-shell. Where does the other electron go?

If you guessed that it was squeezed into the M shell, you are right!
So scandium's electronic structure is really 2 (K), 8(L), 9(M), 2(N). Scandium places that extra electron in a d orbital in the M-shell!
Notice that scandium didn't use its M-shell d orbital until its N-shell had a complete s orbital. The "basement" was used only AFTER the top (outer) shell had a complete s orbital.
Perhaps you recall this element from the discussion I had with Arthur.

Back to the question.

A 52.

This isn't as easy as it might first appear, but I hope you got the outer electrons right. By looking at the table you should have guessed that the outer shell is the P-shell and it has two electrons (in an s orbital).
[Don't get confused! I said those two electrons are in the P-shell. I did not say they were in the p orbitals. Many students get mixed up by not reading the sentence carefully. Atoms have shells (the first quantum number) and orbitals (the next two quantum numbers). They are related to each other but they are NOT the same! P-shells and p orbitals are different things so pay close attention to which one I am talking about. Always keep track of whether I am talking about shells or orbitals. If you don't pay attention here you will become hopelessly confused.]

137Ba has 56 protons (you read that from the Table), 81 neutrons (you calculated by subtracting 56 from 137) and its 56 electrons are arranged as 2 (K), 8 (L), 18 (M), 18 (N), 8 (O) and 2 (in the P-shell).
That arrangement may have surprised you.
Remember, the M shell can have d orbitals but it doesn't like to use them. Here it has to. So the third shell, the M-shell has an s orbital, three p orbitals and five d orbitals for a total of 9 orbitals. All orbitals can hold two electrons each so the M-shell holds 18 electrons.
The N-shell can also have d orbitals. In fact, it can have f orbitals too (if it must). The N-shell of barium has full s, p and d orbitals for a total of 18 electrons. That leaves only ten more electrons to assign somewhere. At this point it would have been an easy mistake to stuff the 10 electrons into the N-shell's f orbitals. But maybe you remembered that f orbitals aren't used until the NEXT shell (two shells up) is full. So move up to the next shell, the O-shell. Two electrons go into the O-shell's s orbital and six electrons go into the O-shell's three p orbitals.
That leaves you with two electrons to assign. Where do they go? Perhaps you thought they would now go into the "basement" of the N-shell's f orbitals. After all, there's plenty of room there. Or maybe you wanted them to go into O-shell's d orbitals. Neither of these possibilities are right. Working with these "basement" orbitals is always difficult, because you never really known when to use them.
That's where a look at the Table helps. If you found barium (Ba) on the Table, you would see that it is in the same Group as beryllium (Be), magnesium (Mg) and calcium (Ca). You know that they all have two electrons in their outer shell. And you can see from the Table that the P-shell is used. That's the clue you need to help you. The last two electrons are not in the "basement" orbitals of lower shells, they are in the s orbital of the P-shell! That's why barium is in Group II.

The P-shell electrons are barium's outer electrons and they will be arranged in orbitals as s2. Maybe you figured that out from a look at the Table. That's great! That's what the Table is for and it would have saved you all that work I showed you (above).

Barium could take on the electron configuration of a noble element (xenon) by losing two electrons. Therefore, it is in Group II and will try to lose its two (outer) electrons to become a cation (Ba+2). Barium will form ionic bonds to anions (to two anions at once if they each have a charge of -1).

That was a very difficult puzzle to solve. You should be proud if you got it right. I didn't get that one right the first time I saw it. Even the explanation is hard to understand! So read through this answer again and be sure you understand it before going onto the next problem.

Back to the question.

A 53.

187Au has 79 protons, 108 neutrons and you might expect its electrons to be distributed as 2 (K), 8 (L), 18 (M), 32 (N), 19 (O) but that would have made the O-shell the last shell and you know that's not right because the table shows gold is in the 6th Period so it must have the outer electrons in its P-shell. Looking at the other elements in the same column as gold is of little help.
Don't feel bad. I would have been surprised if you got it right! I hope you gave it a good try and will now carefully read the correct answer (below). Read it as an "explanation" and a review of how complex these atoms can be.

The K-shell and L-shell are easy. Two electrons go into the K-shell and 8 into the L-shell. That's 10 down and 69 electrons to go (!).
The M-shell can have d orbitals so it is capable of holding 18 electrons, but doesn't "want to". Now it must! The gold atom is so large and has so many electrons that it must use its "basement" to store them. So 18 electrons go into the M-shell. That's 28 down and 51 electrons to go.
The next shell, the N-shell, can have d and f orbitals as well as the s and p orbitals. Like the M-shell, it doesn't like to use its weird orbitals unless it has to. But here it must! When you count up all the different types of each orbital and place two electrons in each, you'll find that there is room for 32 electrons in the N-shell! (Recall that there is one s orbital, three p orbitals, five d orbitals and seven f orbitals for a total of 16 orbitals in the N-shell and each orbital is capable of holding two electrons. That's 32 electrons!)
So far we have ...
K-shell with 2 electrons (easy).
L-shell with 8 electrons (also easy).
M-shell with 18 electrons (because it is forced to use its d orbitals).
N-shell with 32 electrons (because it is forced to use its d and f orbitals).
That's 60 electrons with 19 electrons to go into the higher shells, the O-shell and P-shell. Both of these shells can have d and f orbitals, but now we are getting near the limit of the electrons needed to "force" each weird orbital to be used. At this point we can look at the O-shell and see how many electrons we can put in it without using weird shells.

The O-shell takes 2 electrons into its s orbital and 6 more electrons into its three p orbitals. Easy! That leaves us with only 11 electrons to go. But where do they go?
You may be surprised to learn that 10 of them are stuffed into the O-shell's d orbitals leaving one electron to go into the P-shell.

So the electronic configuration for gold is
K-shell with 2 electrons (s orbital).
L-shell with 8 electrons (s orbital and three p orbitals).
M-shell with 18 electrons (s orbital, three p orbitals and five d orbitals).
N-shell with 32 electrons (s orbital, three p orbitals, five d orbitals and seven f orbitals).
O-shell with 18 electrons (s orbital, three p orbitals and five d orbitals).
P-shell with 1 electrons (s orbital).

Phew! That was hard. Please don't be disappointed if you could not figure out the outer shell of gold. Unlike the previous question about barium, the Table offers very few clues to guide you. Besides, all that hard work describing how the electrons fill each shell and fight over positions in the various orbitals is really something that very few Alchemists worry about. Any Alchemist working with gold would simply look up the electronic structure of a gold atom in a special reference book.

The transition (and inner transition) elements fill their shells in a more complex manner than the "non-transition" elements. This gives the transition (and inner transition) elements their unusual properties.
You should remember that gold (like all transition metals) can form metal bonds.

Oh, by the way, the transition elements don't go into real Groups. Actually, some 20th century Alchemists have labeled the transition "columns" with Group names (like "1B" for the column containing gold, silver and copper). Unfortunately there is no global standard for naming the transition elements' "Groups" (in the 20th century, but in a few more years..?).

Back to the question.

A 54.

At first the number of protons equals the number of neutrons. But as the elements get larger they take on more than an equal number of neutrons.
Carbon had 6 protons and 6 neutrons.
Oxygen had 8 protons and 8 neutrons.
Argon had 18 protons but 22 neutrons.
Barium had 56 protons and 81 neutrons.
Gold had 79 protons and 108 neutrons.

The green line on this graph represents all the stable isotopes of the elements. The yellow line represents the situation where protons always equal neutrons. (That is, for each proton added, one neutron is added too.)

Notice that the two lines agree for the smaller elements but the real situation (the green line) shows that as the atomic number increases, the atomic mass increases even more. That is because of the extra neutrons needed to make stable nuclei for the heavier elements.

Note: unstable isotopes (radioisotopes) will be "off" the green line, having too few or too many neutrons to make the nucleus stable. So they release or convert nucleons to reach a stable point somewhere on the green line. Radioactivity is caused by this reshuffling of nucleons as the atom tries to reach the stable position (on the green line).

Back to the question.

A 55.

I hope you got these symbols for those elements and their isotopes
168O, 126C, 4018Ar, 13756Ba, 19779Au.

Back to the question.

A 56.


The first quantum number for the K-shell is 1.
The first quantum number for the L-shell is 2.
The first quantum number for the M-shell is 3.
The first quantum number for the N-shell is 4.
The first quantum number for the O-shell is 5.
The first quantum number for the P-shell is 6.
The first quantum number for the Q-shell is 7.
(I told you it was easy.)

Let's start thinking about the shells by their first quantum number.
You may recall that the first quantum number is often called the "principle" quantum number.

Back to the question.

A 57.

Perhaps this colored diagram will show you what I was looking for.
By following along the diagonal from the upper right to the lower left, you will see that there is a pattern that lists of all the orbitals in order of increasing energy!

You now have a simple way to follow the orbital filling order. Once you have written the pattern (you don't need colors) you can create the electronic configuration of any atom!. (The next few questions will give you a chance to see how.)

Starting with 1s put a line through it from right to left as you write down "1s". To start the second series you move back to the upper right and cross through the 2s while writing down "2s". Now move back to the upper right again. This time you will cross through 2p and then 3s so write them down in that order ("2p, 3s"). Back up to the upper right to start again and you will now pass through the 3p and 4s as you sweep diagonally downward from right to left.

Each time you reach the end of the diagonal line (by going all the way left) you start over again at the next set on the upper right. So now you sweep through 3d, 4p and 5s (the red ones) in that order and you write them down in that order. You can continue to do this until you have all the orbitals of every shell written down. (You can stop at 7p because atoms don't get any bigger.)

The sequence of orbitals produced in this way will look a wee bit out of order (some principal quantum numbers won't be in order) but that is good because this series represents the orbitals in the order in which they are filled.
You will recall that orbitals are filled from the lowest energy to the highest energy. So this diagram allows you to create a list of all the orbitals in order of increasing energy!

This is a very useful technique and you should try it in the next few questions.
(And like most of Merlin's Qs and As you should add this new idea to your notebook.)

Back to the question.

A 58.

Here is the pattern I got.
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
That should be enough to complete this atom.
It represents the order of orbitals in increasing energy. Notice that the numbers do not follow a direct pattern. That's because the energies of the orbitals do not follow a direct pattern either. (If they did, life would be so simple!)

Now place the electrons into their proper places. Start at the lowest energy orbital (1s) and work your way up as you fill them.
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6, 7s2, 5f14, 6d10, 7p6
That's a total of 118 electrons!

Now look carefully in order to determine the outermost shell. What's the highest principal quantum number in this list? Seven! That means this monster atom has 7 shells and you can count that out, starting with K =1, to figure out that we are talking about the Q-shell! (Perhaps it doesn't surprise you that such a large atom would need such a large shell to hold all those electrons.)
But that's not the end of our problem. How are the electrons arranged in that last shell. Look carefully at the list and note all those with principal number 7. Those include 7s2 and 7p6 and no others. That means this atom has eight electrons in its outer shell. Indeed, it has a complete set of eight in its outer shell. You would have been right to think that this imaginary atom is a noble gas! It would have all the properties you would expect of a Group VIII element.

Back to the question.

A 59.

Here's a long string of the orbital series that you worked out earlier. (I've only included the part of the series you need for these problems.)
Now all you have to do is fill in electrons from left to right until you run out.

1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10

Oxygen's 8 electrons go 1s2, 2s2, 2p4.
Notice that the 2p orbital gets the last electrons and there is only 4 electrons to place there.
You can write that with shells as
K-shell(1s2) = 2,
L-shell(2s2, 2p4) = 6.
Notice that the L-shell needs two more electrons to fill its outer shell so oxygen is a Group VI element.
(But you knew that.)

Carbon's 6 electrons go 1s2, 2s2, 2p2.
You can write that with shells as
K-shell(1s2) = 2,
L-shell(2s2, 2p2) = 4.
Notice that the L-shell needs four more electrons to fill its outer shell so oxygen is a Group IV element.
(But you knew that too!)

Argon's 18 electrons go 1s2, 2s2, 2p6, 3s2, 3p6.
That's
K-shell(1s2) = 2,
L-shell(2s2, 2p6) = 8,
M-shell(3s2, 3p6) = 8.
Argon's outer shell (M-shell) is full (has eight electrons) so it is inert, a Group VIII element. (Right?)

Barium's 56 electrons took some work but hopefully you decided they are arranged as
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2.
Arranging that into shells takes a bit of work because the numbers aren't following each other very conveniently, but I hope you found they are
K-shell(1s2) = 2,
L-shell(2s2, 2p6) = 8,
M-shell(3s2, 3p6, 3d10) = 18,
N-shell(4s2, 4p6, 4d10) = 18
O-shell(5s2, 5p6) = 8,
P-shell(6s2) = 2.
With only 2 electrons in its outermost shell we have proven that barium is a Group II element (like calcium or magnesium).

And finally there was gold! You had to extend the series a bit further but I hope you found
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d9
There's a lot to keep track here but if you did it slowly and carefully you would get
K-shell(1s2) = 2,
L-shell(2s2, 2p6) = 8,
M-shell(3s2, 3p6, 3d10) = 18,
N-shell(4s2, 4p6, 4d10, 4f14) = 32,
O-shell(5s2, 5p6, 5d9) = 17,
P-shell(6s2) = 2.
Hmm, that isn't right! You may recall that the last time (when we were using the Table) I looked it up in a book and the book told me that gold had one electron in its s orbital. What gives?
Most elements, even most transition elements, will follow the series but there are some exceptions. It isn't fair and it certainly doesn't make Alchemy any easier but there are a few transition elements that do not fit this logical pattern (or any other logical pattern). Rather than ask you to memorize the exceptions I think it is better that you simply be aware that these things happen. When asked to determine the electronic structure of any element you can either use the Table to guide you or use this series you made from the chart. Regardless, you will get an answer that is usually right. But if you are working with a transition element you might get a surprise.

Alchemists make up excuses for the behavior of the transition elements that misbehave. Actually, these aren't excuses, they are probable causes. In the case of gold the excuse (I mean cause) is like this. For a brief period of time the two outer shells are exactly as you have figured out above
O-shell(5s2, 5p6, 5d9) = 17,
P-shell(6s2) = 2.
But the O-shell only needs to take one electron from the s orbital in the shell above in order to complete its d orbitals and shells. In these large atoms the energy levels are pretty close anyway so the d orbitals in the O-shell swipe an electron from above. (Atoms like to complete shells by pairing up their electrons. Sometimes.) The final (true) outer shells of gold are
O-shell(5s2, 5p6, 5d10) = 18,
P-shell(6s2) = 1 and that is a more "comfortable" position for the electrons.
(Hey, I didn't create the universe. I just teach how it works!)

You have to pay attention when adding up these orbitals. Notice that barium and gold both end up with a P-shell(6s), but gold has many more electrons stored in its "basement" shells. It is very important to follow the series when assigning the electrons to the orbitals and it is equally important to add the shells up correctly.

Back to the question.

A 60.

Here's a string of the orbital energies we will need and I’ve listed them as all filled.
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10.

Nickel has 28 electrons so I can fill them as
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d8.
That's
K-shell(1s2) = 2,
L-shell(2s2, 2p6) = 8,
M-shell(3s2, 3p6, 3d8) = 16.
N-shell(4s2) = 2. So Nickel has two s orbital electrons in its outer (N) shell.

Copper has 29 electrons so I can fill them as
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d9.
That's
K-shell(1s2) = 2,
L-shell(2s2, 2p6) = 8,
M-shell(3s2, 3p6, 3d9) = 17.
N-shell(4s2) = 2. So Copper has two s orbital electrons in its outer (N) shell.

Zinc has 30 electrons so I can fill them as
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10.
That's
K-shell(1s2) = 2,
L-shell(2s2, 2p6) = 8,
M-shell(3s2, 3p6, 3d10) = 18.
N-shell(4s2) = 2. So Zinc has two s orbital electrons in its outer (N) shell.

There's something special happening here. I checked the book and see that we have found the correct electron configuration for nickel and zinc but copper's M-shell and N-shell are wrong! The book says they should be
M-shell (3s2, 3p6, 3d10) = 18.
N-shell (4s2) = 1.

Just like with gold, the s orbital electron in the outer shell has been dragged into a lower shell!

Back to the question.

A 61.

Argon (Ar) is the last noble element before reaching the fourth Period where nickel, cooper and zinc reside.
Argon's electron configuration is
K-shell(1s2) = 2,
L-shell(2s2, 2p6) = 8,
M-shell(3s2, 3p6) = 8.
So we can write these past three elements to have an electron configuration like that off argon but with some extra orbitals and shells to be considered.

Nickel is argon with the addition of a 3d8 and a 4s2 so it is shorter to write it as [Ar]3d8, 4s2. Notice that I am adding those extra electrons to Argon's M-shell (the 8 electrons in the 3d orbitals) and two electrons into the new shell that argon does not have, the N-shell (s orbital electrons).

Copper's correct configuration, using argon as a platform to build on, is [Ar]3d10, 4s1.
(Remember to use copper's corrected electron configuration.)

Zinc is [Ar]3d10, 4s2.

Notice that you still have to start with the same old long series of electron orbitals but this method makes it easier to write the final answer and makes it more obvious how each atom differs from the others.

Be sure to write this shorthand method into your notes.

Back to the question.

A 62.

The transuranic elements are elements bigger than uranium (atomic number 92). They are hidden among the "inner transitional elements".

The transuranic elements are VERY unstable (radioactive) elements produced by man since the 20th century.

Back to the question.

A 63.

There is only one type of s orbital and it is the orbital in the outer shell of elements in Groups I and II. It can hold only two electrons (because that is the maximum that any single orbital can hold).

P orbitals help shape the atoms of the "post-transition" elements (Groups III to VIII) and there are three types (x, y and z) which all together can hold a total of 6 electrons. They are part of the outer shell of the post-transition elements (which they share with the two s orbital electrons).

Indeed, s and p orbitals are the make up the outer shell of all atoms! the next two orbitals are “basement” orbitals.

D orbitals are found in the "almost" outer shell of transition metals. That is, the transition elements place electrons into the d orbitals of their second outermost shell so d orbital electrons are never in the outer shell. (Not and warning: there is one element that breaks this "rule" so it isn't a rule it is a trend.)
There are 5 types of d orbitals and all together they can hold up to 10 electrons.

The inner transition metals have f orbitals in their third outermost shell but never as their outer shell.
There are 7 types of f orbitals and all together they can hold up to 14 electrons.

Note: All the transition elements (with one exception) have s orbitals in their outermost shell and d orbitals in the shell beneath. All the inner transition elements have s orbitals as their outer shell and (with one exception) f orbitals buried in their third outermost shell with some inner transition elements holding d orbital electrons in the second outermost shell. (Yes, they are very confusing atoms and they are a real headache to most Alchemists trying to make sense of these exceptions.)

Back to the question.


This work was created by Dr Jamie Love and licensed under a Creative Commons Attribution-ShareAlike 4.0 International License Creative Commons Licence.