Chemistry 1111: Test 2 - Ch 8 - 11

Front 1

What is isoelectronic?

Back 1

if they have the same number of electrons[1] or a similar electron configuration[2] and the same structure

Front 2

Cation

Back 2

has positive charge.
> loses e-

Front 3

Anion

Back 3

has negative charge
> gains e-

Front 4

Which elements are noble gases?

Back 4

Column 18, or 8
> He, Ne, Ar, Kr, Xe, and Rn
> all noble gases are monatomic species. We use their symbols.

Front 5

Electron Config
- ions derived from Representative elements

Back 5

Na: [Ne]3s1 Na+:[Ne]
Ca: [Ar]4s2 Ca2+:[Ar]
Al: [Ne]3s2, 3p1, Al3+: [Ne]

Front 6

Effective Nuclear Charge

Back 6

> presence of other e- in an atom reduces the electrostatic attraction between a given e- and the positively charged protons in the nucleus

> Effective nuclear charge (Zeff) is the nuclear charge felt by and e- when both actual nuclear charge (Z) and repulsive effects (shielding) of other e- are taken into acct.

Zeff = Z - sigma(shielding constant)

> core e- are closer to nucleus than Val e-
> core e- shield val e- much more than val e- shield one another.
> 2nd period elements (Li to Ne). Left to right, number of core e- 91s2) remains constant, while nuclear charge increases.

> effective nuclear charge increases from left to right. The added Val e- at each step is more strongly attached by the nucleus than the one before.

Front 7

Atomic Radius

Back 7

> size of an atom in terms of its Atomic Radius, which is half the distance between the two nuclei in two adjacent metal atoms or in a diatomic molecule.

> Going left to right, size goes down
> going top to bottom, size goes up.

> Atomic Radius increases from Li to Ne.
> within a group, atomic radius increases w/ atomic number.
> w/in a group (column) size of atomic radius increases even though effective nuclear charge also increases from Li to Cs.

Front 8

Example 8.2 (p.338)

Referring to a periodic table, arrange the following atoms in order of increase atomic radius: P, Si, N

Back 8

Solution: N and P are in same group. N is above P. (atomic radius increases as we go down a group).
- Both Si and P are in the third period (row), Si is left of P.
- Radius of P is smaller than Si (atomic radius decreases as we go left to right)
- Order of increasing radius: N < P < Si

Front 9

Ionic Radius

Back 9

> Radius of a Cation of an Anion.
> When a neutral atom is converted to an Ion, we expect a change in size.
> If atom forms an ANION, size (radius) increases - because nuclear charge remains the same, but repulsion from additional e- enlarges e- cloud

> Removing one or more e- from an atom reduces electron-electron repulsion, but nuclear charge remains same
> e- cloud shrinks, and Cation is smaller than Atom.
> Radius increases as we go from ions w/ uninegative charge (-) to those w/ dinegative charge (2-) and so on.

eg) oxide ion is larger than flouride ion because oxygen has one few proton than flourine; e- cloud is spread out more in 02-

Front 10

Example 8.3 (p.340)

For each of the following pairs indicate which one of the two species is larger:
a) N3- or F- b) Mg2+ or Ca2+ c) Fe2+ or Fe3+

Back 10

Strategy: comparing ionic radii, useful to classify the ions into 3 categories: 1) isoelectronic ions, 2) ions that carry the same charges and are generated from atoms of the same periodic group 3) ions that carry different charges but are generated from the same atom.

Case 1) ions carrying a greater negative charge are always larger
Case 2) ions from atoms having a greater atomic number, are always larger
Case 3) ions having smaller positive charge are always larger

Solution:
a) N3- and F- are isoelectronic anions, both w/ 10 e-. Because N3- has only 7 protons and F- has 9, smaller attraction exerted by the nucleus on the e- results in a larger N3- ion.

b) Both Mg and Ca belong to group 2A (alkaline earth metals). Ca2+ ion is larger than Mg2+ because Ca's val e- are in a larger shell (n=4) than are Mg's (n=3)

c) Both ions have the same nuclear charge, but Fe2+ has one more e- (24 e- compared to 23 e- for Fe3+) and hence greater electron-electron repulsion. Radius of Fe2+ is >

Front 11

Practice Exercise (p. 340)

Select the smaller ion in each of the following pairs:
a) K+, Li+ b) Au+, Au3+, C) P3-, N3-

Back 11

a) Li+
b) Au3+
c) N3-

Front 12

Ionization Energy

Back 12

Minimum E (kJ/mol) req'd to remove and e- from a gaseous atom in its ground state.
> amt of E in kj, needed to strip 1 mole of e- from 1 mole mole of gaseous atoms.
> Ionization E is a measure of how tightly the e- is held in the atom.
> higher ionization E, harder to remove the e-
> When an e- is removed from an atom, repulsion among the remaining e- decreases.
> nuclear charge remains constant, more E needed to remove another e- from the positively charged ion.
> Ionization E increases: IE1<IE2<IE3...

Front 13

Ionization E (part 2)

Back 13

> Ionization is always an endothermic process.
> E absorbd by atoms (or ions) in ionization process has a positive value
> IE are all positive quantities.
> We associate full valence shell e- configurations w/ chemical stability. High IE of the noble gases (large effective nuclear charge) is one reason for this stability.
> Group 1A (alkali metals) has lowest first IE. Each of these metals has one Val e-, which is shielded by the completely filled inner shells.

Front 14

Ionization E (part 3)

Back 14

It is easy to remove an e- from the atom of an Alkali metal (1A) to form a unipositive ion (Li+, Na+...)
> metals have low IE compared to nonmetals
> suggests why metals form cations (+), and non metals form Anions (-) in ionic compounds

> IE decreases w; increasing atomic number (as we move down the group)
> principle quantum number increases, so does avg distance from val e- from the nucleus.
> Greater separation between e- and nucleus, means weaker attraction, easier to remove.

Front 15

Ionization E

Irregularities

Back 15

>general trend: first IE increase from left to right

Exceptions:
> group 3A elements have lower 1st IE than 2A elements, because they have a single e- in the outermost p subshell (ns2, np1)
> which is sheilded by inner e- and the ns2 e-
> less E needed to remove single p e- than to remove an s e- from same E level.

Front 16

Example 8.4 (p346)

a) Which atom should have a smaller first IE: oxygen or sulfur?

b) which atom should have a higher 2nd IE: lithium or beryllium?

Back 16

Strategy: a) 1st IE decreases as we go down a group, because outermost e- is farther away from the nucleus and feels less attraction.
b) removal of outermost e- req's less E if its shielded by a filled inner subshell

Solution:
a) oxygen and sulfur are members of Group 6A. Have same Val e- config, but 3p e- in sulfur is farther from the nucleus and experiences less nuclear attraction than the 2p e- in oxygen.
- Sulfur should have a smaller first IE

b) e- config of Li and Be are 1s2,2s1 and 1s2,2s2.
- Because 1s e- shield 2s e- more effectively than they shield each other, should be easier to remove a 2s e- from Be+ than to remove a 1s e- from Li+

Front 17

Example 8.4 (p.346)

Practice Exercise:
a) Which of the following atoms should have a larger first IE: N or P?

b) WHich of the following atoms should have a smaller 2nd IE: Na or Mg?

Back 17

a) N

b) Mg

Front 18

8.5

Electron Affinity

Back 18

> Is a measure of the tendency of an atom to gain an e-
> ability of an atom o accept one or more e-
> more positive the e- affinity of an element, greater is the affinity of an atom of the element to accept an e-

> The E that must be supplied to remove an e- from the Anion.
> a large positive e- affinity means that the negative ion is very stable.
> atom has a great tendency to accept an e-

> Overall trend: an increase in tendency to accept e- (e- affinity becomes more positive) from Left to Right across a period.
> E.Aff of metals are lower than nonmetals.
> Values vary little w/ in a group.
> halogens (7A) have highest e- aff. values.

Front 19

Electron Affinity (part 2)

Back 19

> there is a general correlation between e- affinity and effective nuclear charge - also increases from left to right in a given period.

Exceptions:
> Group 2A e- Aff is lower than Group 1A
> Group 5A is lower than Group 4A
> due to Val e- config of elements

Front 20

Example 8.5 (p. 349)

Why are the electron affinities of the Alkaline earth metals, shown in Table 8.3, either negative or small positive values?

Back 20

Strategy: What are the e- confis of alkaline earth metals? Would the added e- to such an atom be held strongly by the nucleus?

Solution: The Val e- config of the alkaline earth metals is ns2, where n is the highest principle quantum number.

> The extra e- must enter the np subshell, which is effectively shielded by the two ns e- (the ns e- are more penetrating than the np e-) and the inner e-. Consequently, the alkaline earth metals have little tendency to pick up an extra e-.