Atomic Properties and the Periodic Table

Question 1

Mendeleev (1869)

Answer 1

Endeavoured to order elements in a logical sequence using chemical properties. ordered the elements using increasing relative atomic masses.

Question 2

Benefits of Mendeleev’s periodic table

Answer 2

Allowed predictions for elements such as gallium before they were even invented.

Question 3

Moseley (1913)

Answer 3

Found that the proper sequence criterion was not determined by relative atomic mass but by atomic number (no of electrons). Also found that hroups in the periodic table were not only chemically similar but also electronically. Ordering the elements based on electronic configuration creates Mendeleev’s arrangement.

Question 4

What can be found from electronic configuration?

Answer 4

The element’s chemistry can be predicted.

Question 5

How were the groups in the periodic table arranged?

Answer 5

The noble gases were placed at the far right of the table, as they have full ns2np6 configurations. Alkaline metals (ns1) and alkaline earth metals (ns2) became groups 1 and 2 on the far left. Halogens (ns2np5), calcogens (ns2np4), and pnictogens (ns2np3) became groups 17, 16, and 15, respectively.

Question 6

Main group elements

Answer 6

Outer shells consisting of only s and p electrons.

Question 7

Transition elements

Answer 7

An element with an incomplete d orbital in either a metal compound or complex—zinc is an exception as it has a complete d orbital

Question 8

Lanthanides

Answer 8

Strip at the bottom of the periodic table contain an incomplete f orbital

Question 9

Multi-electron system assumptions

Answer 9

Assume the orbitals in ME systems look exactly as they do in hydrogen. Quantum numbers and angular functions (shape) are the same. Radial functions are similar but are contracted due to a higher nuclear charge in ME systems.

Question 10

Valence electrons

Answer 10

Outermost electrons, highest energy, partake in chemical reactions.

Question 11

Core electrons

Answer 11

Low-energy electrons in filled shells. Shield the nucleus

Question 12

Shielding

Answer 12

In ME systems, electrons are attracted to the positive nucleus and repelled by negative electrons; repulsion between electrons causes shielding from nuclear charge. The lessened nuclear charge is known as effective nuclear charge. Extent of shielding is dependent on n

Question 13

Penetration

Answer 13

The relative electron density of an electron to the nucleus. Electrons in different orbitals have different wavefunctions and therefore different RDFs. E.g., a 2s orbital has more electron density near the nucleus in comparison to a 2p orbital, therefore is said to be more penetrating. The extent of penetration is dependent on n and l.

Question 14

Trend of electron penetration in an orbital

Answer 14

Provided the n value is the same: s > p > d > f. S orbitals are spherically symmetrical, therefore protect the nucleus in all directions, and have high electron density near the nucleus. P, d and f orbitals, however, have more complex shapes and therefore do not protect in all directions. They also have more nodes, resulting in lower electron density near the nuclear.

Question 15

Why is the 2s orbital filled before the 2p?

Answer 15

Through observing the RDF graphs, the 2s orbital has a maximum peak closer to the nucleus than the 2p, therefore has lower energy/more stability and is filled first.

Question 16

Why is the 4s orbital filled before the 3d?

Answer 16

Although on the RDF graph the 3d orbital has a maximum closer to the nucleus, the 4s orbital penetrates the core electrons. The 4s orbitals experience lower shielding effects than 3d orbitals, making them more stable when filled first.

Question 17

Effective nuclear charge equation

Answer 17

Zeff = Z - S (shielding effects)

Question 18

“Perfect shield”

Answer 18

If an electron were to screen the effects of one positive charge in the nucleus. Would contribute a value of 1 to the shielding constant. However, shielding is not always perfect.

Question 19

Criteria for calculating shielding constant

Answer 19

Electrons with higher values of n contribute 0 to the constant. Those with the same principal quantum number as the observed electron contribute 0.35. Electrons with a n value of one below the observed contribute 0.85. Electrons with low n values contribute 1.

Question 20

Criteria for calculating shielding constant when the observed electron is nf or nd

Answer 20

All electrons with lower n values contribute a value of 1 because d and f orbitals are completely shielded by lower shells.

Question 21

General trend of Zeff

Answer 21

Increases down a group and across a row.

Question 22

Ionisation enthalpy: noble gases

Answer 22

High ionisation energy due to complete shells, therefore very stable. Full outer shells also have high attraction to the nucleus due to opposite charges.

Question 23

Ionisation enthalpy: alkaline metals

Answer 23

Low ionisation energy due to the singular valence electron, which experiences high levels of shielding from core electrons.

Question 24

Ionisation enthalpy: elements in the same group

Answer 24

Analogous elements tend to have similar IEs. With alkaline metals, elements all with 1 s electron outside of a noble gas configuration, it would be expected that the IE would decrease at a large rate with increasing n value, however, extra shielding does not cause a drastic effect to the Zeff, and has a lessened impact with each added shell.

Question 25

Ionisation enthalpy: elements in the same period

Answer 25

General increase in IE across a period. Even though the number of electrons increases, as does the number of protons in the nucleus - therefore Zeff does not increase by a huge amount.

Question 26

Exceptions: B and Be

Answer 26

Boron has a lower IE than beryllium despite being further along across the period. This occurs because the first ionisation energy of boron removes a singular electron from the 2p orbital forming a complete 2s orbital, which requires much less energy than removing a paired electron from a complete 2s orbital

Question 27

Exceptions: O and N

Answer 27

The IE of oxygen is lower than that of nitrogen because N's 2p orbital is made from 3 unpaired electrons, whereas O's is made from that plus one paired electron orbital. It requires less energy to remove electrons from an orbital with a repelling pair of electrons.

Question 28

Electron affinity values

Answer 28

Can be positive or negative. Negative is more common, as when electrons are added, they become attracted to the nucleus and make the element more stable - releasing energy. In the rare event that adding an electron makes an element less stable e.g. noble gases, it will require energy to occur, thus causing a positive value. Same for group 2 elements, which have full 2s orbitals.

Question 29

Electron affinity: G14 v G15

Answer 29

Group 15 has a lower negative EA because each element has 3 half-occupied orbitals, which would require energy to add to due to electron repulsion; therefore, less energy is released. Group 14, however, has only 2 half-occupied orbitals; therefore, less energy is required to add an electron and more energy is released.

Question 30

Electron affinity: first row elements

Answer 30

Tends to be lower than expected due to their small size, leading to stronger electron repulsion.

Question 31

Pauling electronegativity scale

Answer 31

Used bond dissociation energies (energy required to break a bond) to calculate electronegativities. Mathematical formula: electronegativity difference = (bond energy of compound - sum of individual bond energies)^1/2. More difference = more electronegativity. Fluorine was given the highest electronegativity of 4 and EN tends to increase diagonally to the right

Question 32

Mulliken electronegativity

Answer 32

Argued that the arithmetic mean of the IE and EA should be a good measure of the electronegativity.

Question 33

Alfred-Rochow electronegativity

Answer 33

Calculated based off Zeff and covalen radius.

Question 34

What can electronegativity show

Answer 34

Can be used to explain electron distribution in a bond. If two atoms have the same electronegativity, electrons are shared equally, and the bond is non-polar. If electronegativity differs, the polarity is proportional to the difference in electronegativity. Thus, atoms with extreme electronegativities are more likely to form ionic compounds.

Question 35

Differing sizes of atoms

Answer 35

For the same atoms, atomic radii can differdepending upon the nature of the bond e.g. metallic, covalent etc.

Question 36

Why is it impossible to measure the size of an atom?

Answer 36

The probability of finding an electron at any given distance from the nucleus is never zero. Even if you measured the distance at which the probability was 95%, it would still differ between compounds. Only distances between nuclei can be measured.

Question 37

Measuring covalent radii

Answer 37

Experimentally measured bond lengths can be used to find atomic radii if divided by 2. If elements do not form single bonds in the elemental state, the radii can be found from bond lengths in compounds with known bond lengths.

Question 38

Unsaturated compounds covalent atomic radii

Answer 38

Bond lengths in double or triple bonds tend to be shortened, so calculated radii must be adjusted accordingly. e.g., C-C = 154pm C=C = 135pm.

Question 39

Trends in atomic radius

Answer 39

Decreases across a period due to a higher number of electrons, therefore increased Zeff. Increases down a gorup due to electrons being placed into orbitals of a higher n value.

Question 40

Covalent: Group 13 atomic radii

Answer 40

The atomic radii for Al and Ga are the same despite the general increasing trend. Al has valence electrons in the 3p orbitals and Ga has valence electrons in the 4s orbitals; however, because the 3d orbital has quite poor shielding abilities overall, the Zeff of Ga remains quite similar to Al regardless of the extra shell.

Question 41

Metallic atomic radii

Answer 41

Half of the internuclear distance in a metal. Dependent on how closely packed the atoms are and their coordination number

Question 42

Van der Waal's atomic radii

Answer 42

One half of the internuclear distance between non-bonded atoms in a solid or liquid state. This is only known for some elements and can vary considerably. Tend to be larger on average because atoms aren't bonded.

Question 43

Ionic atomic radii

Answer 43

Determined from crystal structures where the sum of the cation and anion radii is equal to the internuclear distance. Half are the atomic radii. The only difficulty comes from knowing where the radius of the cation ends and the anion begins.

Question 44

Trends in ionic radii

Answer 44

The radius decreases with increasing positive charge and with increasing oxidation states.