Unit 1b Periodicity and Stoichiometry Review SheetName:______

  1. The following are the ionization energies apply to magnesium.

Ionization Energies for Element X (kJ mol-1)

First / 735 kJ/mol
Second / 1445 kJ/mol
Third / 7730 kJ/mol

Explain why the third ionization is so high compared to the second ionization energy.

The outer two most electrons come from the valence shell. The third outermost electron would be removed from a new shell that is located closer to the nucleus. As you move closer to the nucleus, the Coulombic attraction increases, requiring more energy to remove those inner electrons.

  1. As atomic number increases from 37 to 54 what happens to the atomic radius? Why?

Atomic radius decreases as you move from 37 to 54. As you move from left to right within the same period there is an increase in effective nuclear charge (Zeff). As Zeff increases (for atoms in the same period) Coulombic attraction increases. The electrons in the outermost shell become increasingly attracted to the nucleus and are pulled in closer, this makes the size of the atoms decrease.

  1. Which has a higher first ionization energy, Br or Kr? Why?

Kr has a higher first ionization energy. Kr has a higher effective nuclear charge. This makes the Coulombic attraction between its outmost electron and nucleus stronger. The stronger the attractive forces, the more energy required to remove an electron.

  1. In general, first ionization energy increases going across a period from left to right. However, the first ionization energy of S is lower than P. What accounts for this deviation?

The outermost electron in S would be placed in a p orbital that is previously occupied. This is known as the “first pairing”. Due to the first pairing, repulsive forces are increased. Since repulsive forces come into play, this lowers the amount of energy required to remove that electron.

  1. Explain why Ca has a lower first ionization energy than Be.

The valence electrons in Ca are located in the fourth principle energy level, while those in Be are located in the second principle energy level. As the distance between the outermost electron and the nucleus increases, shielding increases. This lowers the Columbic attraction and makes it easier to remove an electron.

  1. Explain why the ionic radius of P-3 is larger than P.

P-3has three more electrons than P. The more electrons, the higher the repulsive forces. This causes the anions to be larger in size than their parent atoms.

  1. Explain why aluminum has a lower first ionization energy than magnesium.

Aluminum’s outermost electron is in a p orbital while magnesium’s is in an s orbital. The outer p1 electron is at times, slightly further away from the nucleus and experiences slightly more shielding from the full s orbital. The higher the shielding and further the distance, the lower the attractive forces, and the lower the first ionization energy.

  1. C2H4 reacts with O2 to form carbon dioxide and water.
  2. Write and balance this equation.

C2H4 + 3O2  2CO2 + 2H2O

  1. If 120 grams of C2H4 react with excess oxygen, how many moles of carbon dioxide are produced?

C= 12.01 X 2 = 24.02 120g C2H41 mole C2H42mole CO2 = 8.56 mole CO2

H= 1.008 X 4 = + 4.03228.052 g C2H41 moleC2H4

28.052 g/mol

  1. A 4.00- g sample of an alloy (containing only Pb and Sn) was dissolved in nitric acid (HNO3). Sulfuric acid was added to this solution, which precipitated 1.98 g of PbSO4. Assuming that all of the lead was precipitated, what is the percentage of Sn in the sample?

Pb= 207.21.98g PbSO41 mole PbSO41 mole Pb 207.2 g Pb

S = 32.07 303.27 g PbSO41 mole PbSO4 1 mole Pb

O = + 64.00

303.27 g/mol= 1.35 g Pb

1.35 / 4.00 x 100 = 33.75% Pb

100 – 33.75 = 66.25 % Sn

  1. What is the concentration of sodium hydroxide solution if 19.1 mL of sodium hydroxide solution react with 20.0 mL of 0.103 mol L–1 hydrochloric acid solution?

Mb = ?

Vb = 19.1 mL

Ma = 0.103 M

Va = 20.0 mL

MaVa = MbVb

(0.103 M) (20.0 mL) = (Mb) (19.1 mL)

Mb = 0.108M