# Practice essay sample answers/periodic trends

 Date 20.02.2019 Size 82 Kb. #81331
Use the principals of atomic structure to answer each of the following. . .

1. The radius of the O atom is 0.730 Angstroms; the radius of the O2- ion is 1.40 Angstroms. Account for this difference.

Outline:

• Describe charged particles

• Electron configuration for location of electrons

• Coulomb’s law

• Summary: repulsion represents a larger ionic size (refer to data)

ANSWER: Oxygen has 8 p+ and 6 valence e’. The e’ are arranged as [He]2s22p4. When becoming an oxide ion, oxygen acquires an additional 2 e’ and places them in its 2nd n shell. This location is the same valence shell that neutral oxygen uses. Upon accepting 2 additional electrons, O2- is now [He]2s22p6 or [Ne].

Coulomb’s Law is a mathematical relationship that describes the attractive/repulsive forces between two charged particles and the distance between them:

Fel = kQ1Q2

d2

Q1 represents a (-) charged electron. Q2 represents another (-) repelling electron. Oxygen is a very small atom so the distances between the e’s are quite small. When more (-) charged particles are placed in proximity to one another, greater repulsive forces will result. This repulsive force pushes the electrons apart, causing them to occupy more volume as demonstrated by the data. [It could also be noted that the nuclear attraction (+) is not increasing with the addition of the two e’.]

1. The lattice energy of MgO(s) is-3,795 kJ per mole; the lattice energy for SrO(s) is

-3,217 kJ per mole. Account for this difference.

Outline:

• Define lattice energy (relative to ionic bond strength)

• Electron configuration (to establish n shells)

• Coulomb’s Law (focus on cation only)

• Summary: relate Fel to LE.

Lattice energy is a relative measure of the strength of an ionic bond, which is proportional to the electrostatic force Fel. Formally, Lattice Energy represents the amount of energy released when a mole of gaseous cations and anions come together, forming ionic bonds. The larger the negative (LE) value, the greater the force of attraction in the bond.

Mg is [Ne]3s2. Sr is [Kr]5s2. If nd, then Sr with an n of 5 is much larger than Mg.
Coulomb’s Law explains the attractive forces acting upon these charged particles relative to the distance between them:
Fel = kQ1Q2

d2

Coulomb’s shows that as d   Fel . The cation charges of alkaline earth metals and the anion charge of oxide are the same. Both MgO and SrO are essentially +2/-2 ionic pairs. Therefore, if Q1 and Q2 represent the +2/-2 charges, then the numerator of the equation is the same for both ionic salts. It becomes a difference in the distance that determines attractive force.
The smaller cation, Mg, has a higher Fel, and therefore a higher lattice energy due to the fact that it can get closer to the oxide ion.

1. When comparing the electron affinities of Chlorine and Fluorine, it becomes apparent that this is a violation of the typical trend. Account for this difference.
 F -328 kJ/mol Cl -349 kJ/mol Br -325 kJ/mol I -295 kJ/mol

Outline:

• Define e’ affinity (stability)

• Halogens are small (size – shielding, protons, etc)

• Describe why halogens have high e’ affinities

• Describe trend, violation of trend

• Relate that e’/e’ repulsion is high in small atoms (F is smallest)

Electron affinity is a measure of how stable/unstable an atom becomes after accepting an electron. The (-) values reveal a loss of energy upon acquiring an electron. This means that the halogen was higher E and more unstable prior to gaining an e’. Resultantly, halogens have a high affinity for electrons as they become more stable halides.

ENERGY DIAGRAM:

 [Reactants] H = Hprod - Hreact Halogen + e’ ____________ I \ H is (-) exothermic  \ H \ I \___________ halide [products]

E

RXN PROGRESS 
The halogens are the smallest atoms in their period that can accept electrons. They are small because as you move from left to right in a period the # of p+ and e’ increase while the core electrons that shield nuclear effective attractions remains the same. Because the atoms are small the strong nuclear attractive forces can pull an additional electron into the atom’s valence shell.

One would expect the smallest halogens to have the greatest negative affinities due to this increased nuclear force. This trend is true from I to Cl. However, Cl/F values appear to violate this trend because F is the smallest halogen, [He]2s22p5. Cl is [Ne]3s23p5.

If nd, then F’s nuclear attractive force is strongest, relative to its valence shell. This should imply that it would pull electrons in, in a more energetically favorable way. Chlorine is actually more favorable (higher (-) number). This has to do with the extreme level of e’/e’ repulsion that occurs when an extra e’ is added to such a small F atom. Due to Coulomb’s Law:
Fel = kQ1Q2

d2

Q1 and Q2 can be viewed as any of the eight (-) charged electrons. F is a very small atom so the distances between the e’s are quite small. As one can see, when more (-) charged particles are placed in proximity to each other, the more repulsive force there is between them. Chlorine, [Ne]3s23p5, is just large enough that it holds the additional electron in a more favorable way. As we move down the group towards Iodine, the atoms grow larger as n shells are added. This distance between the shells makes for more predictable electron affinities. As valence shells increase, this extra distance creates a less powerful attractive force.