ANSWER SOLUTION for PROBLEM SET #2

Questions 3.52, 3.60 (a,b,c) and 4.30 (a,b,c) were graded carefully for 6 points each.  Other questions given one point for completion.  Total possible = 25

 (Specified page numbers are from the text Introduction to General, Organic, and Biological Chemistry. Matta, Wilbraham, and Stanley.)

 3.18    Elements are defined by the number of protons that constitute their atoms.  Therefore, an atom of carbon is the only atom that has 6 protons and 6 electrons.

 3.44    In generating electron configurations we need to now the orbitals available, their energy level ordering, and the rules for filling them.

 The orbitals available are summarized in a table from lecture 6.

 For multielectron elements, the energy filling is summarized by the following mnemonic also from lecture 6

 The rules for filling orbitals are to fill the lowest energy first (Aufbau), orbitals can hold two electrons (Pauli exclusion principle), and fill when there are multiple orbitals of the same energy, fill them singly first then pair them up (Hund’s rule). 

 (a) Using these rules, we see that the electron configuration can be written as 1s²2s²2p_x ²2p_y²2p_z¹.  Normally we do not distinguish the different types of p-orbitals (or d-orbitals) in writing electron configurations, so the correct answer is  1s²2s²2p⁵.  This corresponds to the configuration of the atom flourine.

 (b) The 1s²2s²2p⁵ configuration does not plainly tell us that the number of unpaired electrons.  If we remember that there are three p-orbitals and consider Hund’s rule, we then see that there is one unpaired electron as shown more clearly by the configuration where we spell out the filling of the various p-orbitals 1s²2s²2p_x ²2p_y²2p_z¹.

 3.52    Following problem 3.44 we see:

(a) Selenium has 34 electrons; 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴

(b) Vanadium has 23electrons; 1s²2s²2p⁶3s²3p⁶4s²3d³

(c)    Nickel has 28 electrons; 1s²2s²2p⁶3s²3p⁶4s²3d⁸

 3.60    The symbol of the isotope is written with the mass number as a superscript and the atomic number as a subscript.

The atomic number is equal to the number of protons.

The mass number is the sum of the number of protons and the numbers of neutrons and represents the approximate mass of the atom in the atomic mass unit (amu).

 4.10    Follow the group (column) and the period (row) to where they intersect at an element.

(a)    sodium

(b)   boron

(c)    bromine

(d)   oxygen

 4.14    Nonmetals are found to the right of the diagonal zigzag line on the right of the periodic table.  Metalloids are directly to the right of the diagonal zigzag line and display some properties of metals and nonmetals.  The noble gases are group 0.  So, the nonmetals are (listed by period) 

     Period 1:  hydrogen, H

Period 2:  carbon, C; nitrogen, N; oxygen, O; fluorine, F

Period 3:  phosphorous, P; sulfur, S; chlorine, Cl

Period 4:  selenium, Se; bromine, Br

Period 5: iodine, I

4.22    Refer to figure 4.6 and to the figure below from Lecture 7s notes..  Groups 1A and 2A, on the left of the periodic table, correspond to the s sublevel.  Groups 3A through 7A and group 0, on the right side of the periodic table, correspond to the p sublevel.  The transitions metals, the group B metals in the middle of the periodic table, correspond to the d sublevel.  The inner transition metals, at the bottom of the periodic table, correspond to the f sublevel.

4.30    The atomic radius:

• decreases moving left to right across a period because the effective nuclear charge (positive) is increasing while electrons are filling orbitals of similar “size” (for the main group elements, same principle energy level).  As an electron’s attraction to the nucleus increases as it becomes more positive, the size of the atom shrinks.

• increases while moving down a group because electrons are occupying progressively higher principle energy levels.

The pairs of elements in the question are always members of the same group.  Hence the one further down the group is the large one.

(a)    sodium; because sodium is below lithium in group 1A

(b)    strontium

(c)    germanium

(d)    selenium

4.32  The ionization energy:

• increases moving left to right across a period because the effective nuclear charge (positive) is increasing resulting in the electrons being more tightly held, and hence more difficult to remove.

• decreases moving down a group because outer electrons are occupying progressively higher primary energy levels which are further from the nucleus and thus less tightly held and more easily removed.

As the non-metals are on the far right of the periodic table they are expected to have higher ionization energies than the metals on the left of the periodic table.

  4.48  

(a)    Kr; the noble gases are in group 0 (Also known as Group 8A)

(b)   Ca, He, Cl;  the representative elements (also known as the main group elements) are the group A elements and group 0 (noble gases).  These are also the s- and p-block elements.

(c)    Rb; the alkali metals are in group 1A.

(d)   F; the halogens are found in group 7A.

(e)    Sr, Ca; the alkaline earth metals are in group 2A.

(f)     Pu; the inner transition metals (also known as the Lanthanides and Actinides) are in the two rows found at the bottom of the periodic table.

(g) Cu, Co, Cr; the transition metals are the group B elements.