Question

a) Which electrons experience a greater Z_eff (effective nuclear charge): the valence electrons in Be or the valence electrons of N? Why?

Using the simple equation on page 260, calculate the Z_eff of 3p electron of K atom.

a. Using the trends in ionization energy of sodium and electron affinity of oxygen, explain why sodium oxide has the formula Na₂O and not NaO.

Which specific orbital loses an electron in the fourth ionization step of Carbon?

2) Write the atomic radii and ionic radii of the following using the interactive Periodic   
Table on the web (Link: http://www.chemeddl.org/resources/ptl/index.php).

a-    Explain the trend in the radii for two elements in the same group and that for two      
     elements in the same period. Be sure to include units.

                                                Atomic Radius                                   Ionic Radius

            a. Ra             

        b. Cd

    c. Sr

4)Briefly explain why valence electrons are more important than core electrons in determining the reactivity and bonding in atoms.

Write the abbreviated electron configuration for the following pairs:

   a-Ir (Iridium) atom
   b- Ir³⁺      
   c- Se atom
d- Se^2—

How many unpaired electrons are there in the following: ( you should do their condensed electron configurations first)

a-Fe atom

b-Cu¹⁺ ion

Answer 1

(a)

The effective nuclear charge is best computed using Slater's rules.

Zeff = Z - s

Effective nuclear charge is basically the nuclear charge minus the "shielding" provided by the electrons in between. Nitrogen has the greater nuclear charge, +7 vs +4.

For Be, s = 2 (0.85) + 0.25 = 1.95

Zeff = 4 - 1.95 = 2.05

For N, s = 2(0.85) + 4(0.35) = 3.10

Zeff = 7 - 3.10 = 3.90

The outermost electrons in N have a greater attraction to the nucleus. This accounts for why the ionization energy of N is greater than for Be.

Hence, N valence electrons experience a greater Zeff.

(a)

Formula for sodium ion and oxide ion is Na⁺ and O^2- respectively. Thus, 2 atoms of Na⁺ will bond with 1 atom of O^2- to form Na₂O.

2.

(a)

Atomic size gradually decreases from left to right across a period of elements. This is because, within a period, all electrons are added to the same shell. However, at the same time, protons are being added to the nucleus, making it more positively charged. The effect of increasing proton number is greater than that of the increasing electron number; therefore, there is a greater nuclear attraction. This means that the nucleus attracts the electrons more strongly, pulling the atom's shell closer to the nucleus. The valence electrons are held closer towards the nucleus of the atom. As a result, the atomic radius decreases.

Down a group, atomic radius increases. The valence electrons occupy higher levels due to the increasing principal quantum number (n). As a result, the valence electrons are further away from the nucleus as ‘n’ increases. Electron shielding prevents these outer electrons from being attracted to the nucleus; thus, they are loosely held, and the resulting atomic radius is large.

4.

8 valence are stable: Noble gasses
7 valence have a strong affinity to gain an electron to reach the stable 8 hence they are super reactive: halogens
6 valence are similar but they want to gain two
5 are less reactive, will often gain one or three electrons
4 is a half filled valence and is relatively stable. they do form bonds easily (ex. carbon)

on the other end
1 valence is opposite 7. elements with one valence want to get rid of the one so they can have a stable 8 in the shell underneath
2 is similar to 6
3 is similar to 5

so a earth metal like sodium and a halogen like fluorine will react violently because they both benefit greatly from the reaction. Na looses one electron to F and you get sodium flouride.

(a) Iridium (Ir)    Electron configuration [Xe] 4f14 5d7 6s2

(b) Ir³⁺

+3 ion would have 3 less electrons than ground state. [Xe]6s2 4f14 5d4

(c) Selenium(Se)   Electron configuration    [Ar] 3d10 4s2 4p4

(d) Se^2-

Se2- has the same electron configuration (isoelectronic) as krypton.

You could either write in noble gas notation: [Kr]

or normally: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

Iron (Fe) Electron configuration [Ar] 3d6 4s2

4 unpaired 3d electrons.

Copper     Electron configuration [Ar] 3d10 4s1

Cu⁺ Electron configuration [Ar] 3d10

No unpaired electrons.