Chemistry 101 In-Class Assignment 1 (Due by Sept. 24) IonsIonic

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Chemistry 101 In-Class Assignment 1 (Due by Sept. 24) IonsIonic Powered By Docstoc
					                                          Chemistry 101
                             In-Class Assignment 1 (Due by Sept. 24)
                                     Ions & Ionic Compounds
                           and an Introduction to Lewis Dot Structures

1. Draw lewis dot structures for an atom of each of following elements:

   (a) K                                                 (c) Ar

   (b) Si                                               (d) As

2. Identify the ion likely to be formed by the following elements:

   (a) Ca                                                (e) Cl
   (b) P                                                 (f) Al
    (c) Rb                                              (g) Cs
   (d) Ba                                               (h) Se

3. Predict the formulas of the ionic compounds that would be formed by the combination of each pair of

   (a) Ca and Br                                         (c) Al and O

   (b) Li and F                                         (d) Mg and N

4. Name the following compounds:
   (a) RbCl

   (b) Na2 O

    (c) AlCl3

   (d) BaO
5. What are the chemical formulas for:
   (a) potassium fluoride
   (b) magnesium oxide
    (c) calcium chloride
   (d) aluminum bromide

  Naming binary ionic compounds.
  Binary ionic compounds (binary = composed of 2 elements) are named in the following matter:
   (a) The name of the positive metal ion comes first. This is followed by the name of the negative ion.
   (b) For positive ions, we typically use the name of the element.
    (c) For negative ions, we use the element name with the ending changed to -ide.
     • NaCl = sodium chloride
     • Mg3 N2 = magnesium nitride
Chemistry 101, Fall 2003                                                             In-Class Assignment 4, page 2

Metals that form more than one type of ion (text section: 4.4)
   Many transition metals commonly form more than one type of ion. Iron, for instance, forms a +2 ion as
well as a +3 ion.

                                                      Fe2+ and Fe3+

    Just a few main group metals do this as well - such as tin and lead. To identify compounds containing
such elements we must specify the charge on the metal ion. This is done in two different ways. The first
is through common names that have been used historically for the different ions. The other is through a
formal naming system which uses a roman numeral after the element name, indicating the charge:

                        Element      Ions       common name (historical)      systematic name
                          Iron       Fe2+              ferrous                     iron (II)
                                     Fe3+                ferric                   iron (III)
                        Copper       Cu+               cuprous                   copper (I)
                                     Cu2+               cupric                   copper (II)
                          Gold       Au+                aurous                      gold (I)
                                     Au3+                auric                    gold (III)
                          Tin        Sn2+             stannous                      tin (II)
                                     Sn4+              stannic                     tin (IV)
                          Lead       Pb2+             plumbous                     lead (II)
                                     Pb4+              plumbic                    lead (IV)

   The cation name is used in combination with the anion name to name the compound. By balancing the
charges of the ions, you can also determine the formula of the compound. For example:

   •       – ferrous chloride = iron (II) chloride = FeCl2
             ferrous = Fe2+ , need 2 Cl− ’s to balance charge
           – ferric chloride = iron (III) chloride = FeCl3
             ferric = Fe3+ , need 3 Cl− ’s to balance charge

  6. Try this by completing the following table:
       .          common name               .     .       formal name            .   .       formula         .

                  ferrous chloride                       iron (II) chloride                     FeCl2

                   cupric chloride


                                                          tin (IV) sulfide

                   plumbic oxide


                                                          tin (IV) oxide
Chemistry 101, Fall 2003                                                       In-Class Assignment 4, page 3

Covalent Bonds (Text section: 4.5)
   An atom can achieve a noble gas configuration by sharing electrons with other atoms. We have seen
that metals and nonmetals combine to form ionic compounds. However, when two nonmetals combine, they
typically share electrons in covalent bonds and form what are know as covalent compounds.

For example, consider two hydrogen atoms. Each one has one valence electron. However, both hyrogens
would like to have two electrons, to obtain the same configuration as [He]. To do this, each H atom shares

an electron with the other, forming a covalent bond. In H2 , each H atom has 2 electrons.

   H   .       +       .H              H    .                      H   H                 H2
                                                                                     a hydrogen
                                     H’s share e−’s                             (diatomic = 2 atoms)
                                  forming a covalent
     each H has 1 e−                      bond
                                                          shared pairs can
    and wants 1 more                                     be represented as
                                                       lines between atoms

Here’s another example, this time for chlorine. A Cl atom has 7 valence electrons, so it needs 1 more to
achieve an octet.

           .           . Cl
. .
. .

                       . .
                       . .


   .                          .       . . .               .
                                                          .    .           .
   .Cl         +              .       . . .
                                       Cl    Cl                .   Cl Cl   .             Cl

                                                                                      a chlorine
                                  each Cl shares 1 e−’s
                                   forming a covalent
        each Cl has 7 e−            bond consisting of
                                      2 electrons              the shared electron pair
       and wants 1 more                                        is represented as a line

Each chlorine now has 8 electrons, because you can count the shared ones:

    this Cl now has an
                                                                                    this Cl atom also has an octet

   octet of 8 electrons                          . . .
                                                 . . .
                                                  Cl      Cl

Sometimes you need to share more than one pair of electrons to achieve an octet:

           .           .O     .
. .

                       . .

   .   O           +          .       . .. .
                                      . .. .
                                        O     O
                                                               .   O   O
                                                                           .             O2




                                                                                      an oxygen
                                   each O shares 2e−’s
                                   forming a double
        each O has 6 e−              covalent bond
                                                          the four shared electrons
       and wants 2 more
                                                           are represented as two lines between the atoms
Chemistry 101, Fall 2003                                                        In-Class Assignment 4, page 4

We can check to see that each oxygen has the appropriate number of electrons:

                           each oxygen atom                          total # valence
     . .. .
     . .. .
       O       O           now has an octet                          electrons = 12
                                                                     (note the 12 dots)

For problems 7-12, calculate the total number of valence electrons in each of the following molecules. Then,
draw lewis dot structures for each:

  7. Br2                                                                       # valence e−

  8. HCl                                                                       # valence e−

  9. HBr                                                                       # valence e−

 10. N2                                                                        # valence e−

 11. NO                                                                        # valence e−
     This one is tricky, there is no way to achieve an octet. So, get each atom as close to an octet as possible
     without going over.

 12. SiH4                                                                      # valence e−
     Another tricky one. Take Si to be the central atom and arrange the H atoms around it. Try to get an
     octet around silicon and two electrons around each of the hydrogens.

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