Trends on the Periodic Table Chapter 7. Written by JoAnne L. Swanson University of Central Florida. Effective Nuclear Charge. Z eff = Z – S = (The protons in the nucleus) – (the number of electrons between the nucleus and electron of interest)
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Trends on the Periodic TableChapter 7
JoAnne L. Swanson
University of Central Florida
Zeff = Z – S
= (The protons in the nucleus) – (the number of electrons between the nucleus and electron of interest)
Zeff allows us to estimate trends in properties
atomic and ionic size
Electrons in an atom are attracted to the positive pull of the nucleus while at the same time they are __________________ ________________________________________________-.
Outer electrons are also ________________________________ ________________________________________________
These attractions and repulsions are taken into consideration when interpreting trends on the periodic table.
Ionization energy is the energy required to remove the most loosely held electron from an atom in its gaseous state.
Example to demonstrate:
Al(g) Al+1(g) + e-1st Ei = 580 KJ / mol
Al+1(g) Al+2(g) + e-2nd Ei = 1815 KJ / mol
Al+2(g) Al+3(g) + e-3rd Ei = 2740 KJ / mol
Al+3(g) Al+4(g) + e-4th Ei = 11600 KJ / mol
The outermost electron is always ____________________.
This is because it is ______________________________________ ______________________. In this case the 3p electron is removed.
The second ionization energy is considerably higher because now the electron (the 3s) is being removed from ___________________. ______________________________________________.
The fourth ionization energy is the highest for Al+3 because this would be removing a “____” electron. That is, _______________ ______________________________________________________ ______________________________________________________.
In the graph of ionization energies, there is a general trend of _________________________________________________.
There are deviations from this trend due to shielding, Zeff, and repulsions.
Generally, ionization energy increases going to the right of a period because the electrons are in the same PEL, and are not shielded from the nuclear charge. That is, the positive protons are increasing across the period along with the negative electrons, in the same energy level. There is a stronger positive to negative attraction as both protons and electrons increase.
Therefore it would take more energy to remove an electron.
The ionization energy _________________________. Here the electron is in a higher energy level, farther from the nucleus, and thus held less tightly.
Deviations from the trend –
Be has a higher ionization energy than B
the 2p electron in B is shielded from the nuclear charge by the 2s electrons and is thus easier to remove.
N has a higher ionization energy than O
N has 3 unpaired electrons. O has a paired electron in the 2px orbital and thus an increase in repulsion. The electron is easier to remove due to the repulsion.
Electron affinity (Ea) is _______________________________ ____________________________________. The more negative the number, the more ____________________. Therefore, atoms with a more negative electron affinity tend to _______________.
Repulsions and nuclear charge affect electron affinity –
N- (g) is unstable and so has a positive Ea (endothermic). Here an electron would be added to an already occupied ‘p’ orbital in which the repulsion is greaterthan the Zeff (the positive attraction of the nucleus).
O(g) + e- O-(g) is exothermic. Here the difference in Zeff (it has more protons therefore more attraction to electrons), overcomes the repulsion of adding a 2nd electron to an orbital.
Oddly enough, O2- is not stable and a 2nd electron cannot be added to O-. The reason we see so many compounds of O2- has to do with the O2- being stabilized by the large attraction of the positive charge of the ion it is bonding to.
So the basic trend for Electron affinity with some deviations is –
a more negative Ea going to the right in a period.
Going down a group, Ea becomes more positive with some deviations. The electrons are being added at increasing distances from the nucleus and also to increasingly larger orbitals. Zeff and repulsion play a smaller role.
It is helpful to look at the basic trends, deviations aside, as depending on metallic or non metallic character. Metallic character increases down and left. Non metallic character increases up and right.
Metals tend to lose electrons therefore, have low ionization energies and more positive electron affinities.
Non metals tend to gain electrons therefore have high ionization energies and more negative, exothermic, electron affinities.
Ionic Size –
Cations are ____________than their respective neutral atoms due to less repulsions and the emptying of the largest, outer orbital.
Anions on the other hand are larger than their neutral atoms due to repulsions between the added electrons while no additional protons are added. This will cause the electrons to spread out more.
The representative elements (s and p filling elements) exhibit trends that change in a regular way
1. The number and type of valence electrons primarily determine an atoms chemistry.
2. The electron configuration is one of the most valuable pieces of information for being able to predict behavior of elements.
3. The most basic division of elements is that of the metals and nonmetals.
Metals with more metallic character (lower left), are the most reactive metals. They have low ionization energies and so lose electrons more easily.
The non metals with more nonmetallic character (upper right, excluding inert gases), are the most reactive non metals and tend to gain electrons. They have high ionization energies and more exothermic Ea (more negative).
Metals tend to form ionic compounds with non metals, that is,_____________________________________________. These are held together by strong ________________ (+ and – attractions).
Metals in water form H2 gas and base.(Since metals have low Ei they form bases that will ionize in water.
Na + H2O NaOH + H2(g)
Metal oxides in water are basic.
Na2O + H2O 2NaOH
NiO + H2O Ni(OH)2
THEREFORE SINCE METAL OXIDES ARE BASIC, WHEN PUT IN ACID THEY FORM SALT AND WATER:
Na2O + 2 HCl 2NaCl + H2O
THE Lithium REACTS WITH OXYGEN TO FORM A METAL OXIDE AND ALSO REACTS WITH HYDROGEN TO FORM A METAL HYDRIDE:
4Li + O2 2Li2O
2Li + H2 2LiH
But other alkali metals form peroxides (O22-) with O2:
Na + O2 Na2O2
What is a hydride???
The alkaline earth metals get more reactive going down a group:
Be + H2O(l) NR
Be + H2O(g) NR
Mg + H2O(l) NR
Mg + H2O(g) Mg(OH)2
Ca + H2O(l) Ca(OH)2
Sr + H2O(l) Sr(OH)2
THEY ALSO FORM METAL OXIDES: Mg + O2 MgO
NON METAL OXIDES ARE _________, THAT IS, WHEN REACTED WITH WATER PRODUCE ______. IT FOLLOWS THAT NON METAL OXIDES WITH BASE SHOULD THEN PRODUCE _____________________.
P4O6 (s) + 6 H2O 4 H3PO3
P4O6 (s)+ 12 KOH(aq) 4 K3PO3 + 6 HOH
PHYSICAL CHARACTERISTICS OF METALS:
LUSTEROUS AND SOFT
HIGH MELTING POINTS
GOOD CONDUCTORS OF ELECTRICITY (WHY?) AND HEAT
MALLEABLE AND DUCTILE
THEY ARE SOLIDS LIQUIDS AND GASES AT ROOM TEMP.
SOLIDS ARE BRITTLE AND DULL
LOWER MELTING POINTS THAN METALS
POOR CONDUCTORS OF ELECTRICITY
WHEN NON METALS BOND TO OTHER NONMETALS THEY GENERALLY FORM __________________ AND THEREFORE ______________ SUBSTANCES.
IN THESE COVALENT BONDS, ELECTRONS ARE _________ BETWEEN ATOMS BY OVERLAP OF ORBITALS.
HALOGENS (X) REACT WITH HYDROGEN TO FORM HX(g)
HALOGENS REACT WITH METALS TO FORM MX(S). F AND Cl ARE THE MOST REACTIVE HALOGENS.
FLUORINE IS THE MOST REACTIVE HALOGEN AND WILL TAKE ELECTRONS FROM ALMOST ANY SUBSTANCE.
F2 + H2O HF + O2 EXOTHERMIC
F2 + SiO2 SiF4 + O2 EXOTHERMIC
THUS, FLUORINE GAS IS VERY DANGEROUS TO WORK WITH.
SULFUR AND OXYGEN HAVE SIMILAR PROPERTIES AND FORM SULFIDES AND OXIDES WITH OTHER NONMETALS.
OXYGEN BEING MORE REACTIVE DOES SO MORE EXOTHERMICALLY.