Answers

1. (a) The Pauli Exclusion Principle states that no two electrons in an atom can have the set of four_______________ .

b). Each atomic orbital, eg "2p(x)", can be occupied by no more than _____ (give number) electrons and these electrons must have opposite ___________ .

c) For each pair of atomic orbitals given below indicate which one has the lower relative energy. Think "n+l"

d) When an anion is formed from an atom, is the size of the anion larger than, smaller than, or the same size as the atom which was ionized ? Briefly explain:

e) Circle the following ion/s which are not likely to be found in chemical reactions under usual conditions.

f) Arrange the following ions in order of increasing ionic radius:O (2-), F (1-), Na (1+) and N (3-) Briefly expain your reasoning.

2 . Identify each of the following elements based on the information provided:

a) the electronic configuration is 1s^2, 2s^2, 2p^6, 3s^2, 3p^4 ___________.

b) the Group 1 A (alkali metal) with the lowest, first ionization energy ______.

c) the element whose +2 ion has the same electonic configuration as argon ______.

d) the element in the second row that has three unpaired electrons in its 2p orbitals ______.

3.i) Using spectroscopic notation, write the electron configurations for the following species:

ii) What do you note about the configurations of these four ions?

iii)Which one would have the largest ionic radius? Explain.

4.i) Write the "Lewis Dot" symbols for the elements, hydrogen, carbon, nitrogen and oxygen.

ii) In each case, how is the number of elecrons in the valence shell (outershell) related to the Group number of the element in the periodic table?

iii) Write the spectroscopic notation electron configurations for H, C, N, and O.What is the relation between the Lewis Dot Symbol and the number of elecrons in the highest "n" level for each element?

iv) How do the Lewis Dot Symbols support the idea that one definition of chemistry could be "what happens to the valence shell or outer shell electrons" ? Comments

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1 a) quantum numbers: n, l, m(l), and m(s)

b) 2 electrons, opposite spins, +1/2 and -1/2

c) 4s is n=4 and l=0 or "n+l"=4 while 3d is n=3 and l=2 or "n+l"=5 therefore "4s" is energetically slightly lower than "3d"

The second example: 4p is n=4 and l=1 or "n+l"=5 while 5s is n=5 and l=0 or "n+l"=5 so the n+l is the same for these two but 4 is a lower than 5 so "4p" fills before "5s"

The last case: 4f is n=4 and l=3 or "n+l"=7 while 5d is n=5 and l=2 or "n+l"= 7 again the same n+l for both but n=4 is a lower relattve "n" value so 4f will fill before 5d.

d) When a anion is formed, one or more electrons are added to the valence shell of the atom resulting in more electrons inthis outer shell but no change in the effective nuclear charge experienced by each one of the valence shell electrons.Since the effective nucelar charge remsins constant, the overall radius increases and the ionic radius of anions are larger than their "parent" neutral atoms. Thus O (2-) has a larger radius than an O atom. Compare Fig. 8.10, page 377 with Fig.8.15, page 384.

e) Li (2+) , Ne (1+) and S (3-)

f) smallest = Na(1+) < F(1-) < O(2-) < N(3-) All of these ions have the same number of valence shell electrons, 8. However, the effective nuclear charge increases from +7 in nitrogen to +11 in sodium resulting in the trend in ionic radii.

2 a) sulphur, S element number 16

b) either francium or cesium. Generally, the elements in the lower left corner of the periodic table have are easiest to ionize.

c) Ca (2+) 18 electrons the same as argon, Ar

d) nitrogen 1s^2, 2s^2, 2p^3.

3. Cl = 17 electrons or 1s^2, 2s^2, 2p^6, 3s^2, 3p^5

K = 19 electrons or 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1

S = 16 electrons of 1s^2, 2s^2, 2p^6, 3s^2, 3p^4

Ca = 20 electrons or 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2

The ions, Cl(1-), K(1+),S(2-) and Ca(2+) all have the same number of elecrons = 18 and the same electron configuration= 1s^2, 2s^2, 2p^6, 3s^2, 3p^6 a "Argon configuration" What can you say about the relative size of these ions? The largest radius would be S(2-) and the smallest is Ca(2+). Same nymber of electrons but different effective nuclear charges. See 1f above.

4.