Download Periodic Properties Worksheet Key and more Summaries Chemistry in PDF only on Docsity!
Periodic Properties Worksheet Key
- State which atom is larger according to periodic trends. a) Br, F Br
b) Ca, Mg Ca
c) C, Si Si
d) He, Ar Ar
- Explain the difference in size for each of the pairs in Problem 1.
The outer electrons of the larger atom are in a subshell further away from the nucleus than the outer electrons of the smaller atom.
- State which atom is smaller according to periodic trends. a) P, S S
b) Be, Li Be
c) I, Xe Xe
d) As, Ge As
- Explain the difference in size for each of the pairs in Problem 3.
The outer electrons of both atoms are in the same subshell; however, the smaller atom has a greater nuclear charge that attracts the electrons closer to the nucleus.
- State which atom has a smaller first ionization energy according to periodic trends. a) O, S S
b) K, Na K
c) Cd, Zn Cd
d) Al, Ga Ga
- Explain the difference in ionization energy for each of the pairs in Problem 5.
The outer electrons of the atom with the smaller ionization are in a subshell further away from the nucleus than the outer electrons of the atom with the larger ionization energy.
- State which atom has a larger ionization energy according to periodic trends. a) Ba, Cs Ba
b) Cl, S Cl
c) Sc, Ti Ti
d) Sb, Sn Sb
- Explain the difference in size for each of the pairs in Problem 8.
The outer electrons of both atoms are in the same subshell; however, the atom with the larger ionization has a greater nuclear charge that attracts the electrons closer to the nucleus making the electrons more difficult to remove.
- Explain why the second ionization energy of rubidium is higher than the second ionization energy of strontium.
The second electron removed from the rubidium atom is from the 4p subshell; whereas, the second electron removed from strontium is from the 5s subshell. Since the 4p subshell has a lower energy (a more negative energy) than the 5s subshell, the second ionization energy of rubidium is much greater than the second ionization energy of strontium.
- Which of the following is smaller? a) Mg, Mg2+ Mg2+
b) Al, Al 3+ Al3+
c) Cs, Cs + Cs +
- Which of the following is larger? a) O, O2- O 2-
b) P, P3- P3-
c) I, I - I -
- Arrange in order of increasing size a) Fe, Fe 2+, Fe 3+ Fe3+^ , Fe2+^ , Fe
b) Ti, Ti 3+, Ti 4+ Ti4+^ , Ti3+^ , Ti
- Arrange in order of increasing size: Br - , Ca 2+, Cl - , F-, Li +, Mg 2+, N3-, Na +^ , O2-, Rb+, S 2-
First determine the number of electrons in each ion since each of these ions will have the same number of electrons as a noble gas.
Br - - 36 e-, Ca2+^ - 18 e-^ , Cl-^ - 18 e-, F-^ - 10 e-, Li+^ - 2 e-, Mg 2+^ - 10 e-, N3-^ - 10 e-, Na+^ - 10 e-, O2-^ - 10 e-, Rb+^ - 36 e-, S2-^ - 18 e-
Group the ions according to number of electrons. Size will increase according to number of electrons (in the case of ions with noble gas electron configurations).
Li+^ - 2 e-^ F-^ , Mg2+^ , N3-^ , Na+^ , O2-^ - 10 e-^ Ca2+^ , Cl-, S2-^ - 18 e-^ Br - , Rb+^ - 36 e-
Arrange ions in groups in order of decreasing nuclear charge which causes decreased attraction and increasing size. Then put groups together.
Li+^ , Mg2+^ , Na+^ , F-, N3-, O 2-, S2-, Cl-, Ca2+^ , Rb+^ , Br -
Cations are smaller than the neutral atom because they have fewer electrons. Removing electrons either empties a larger subshell or reduces the repulsion with a subshell.
Anions are larger than the neutral atom because they have more electrons in the same subshell which causing increased repulsion. The increased repulsion causes the electrons to move further from each other making the anion larger.
Increasing number of electrons causes increasing repulsion which causes increasing size.
