CHM 218 (Dr. Tremain) QUIZ 2 February 5th, 2003
NAME__KEY_____________________________
SHOW ALL WORK IN ORDER TO RECEIVE FULL CREDIT.
1. (1 point) Which one of the following neutral elements has the largest atomic radius?
N O S F Cl Br
2. (1 points) The graph to the right shows the ionization
energies of an unknown element. Deduce to which group
in the periodic table it belongs.
Five valence electrons. Ionization of the 6th and 7th
electrons is incredibly high since electrons are being
removed from the noble gas (inner) core. The group in the
periodic table is Group 5 (transition metals) or Group 15
(main group elements).
3. (2 points). Which element, nitrogen or oxygen, has the higher first ionization energy? Which element,
nitrogen or oxygen, has the higher second ionization energy? Explain your reasoning.
Nitrogen (1s2 2s2 2p3) has the higher 1st IE due to stability of the ½ filled 2p subshell (thus more difficult
to remove an electron). Oxygen (1s2 2s2 2p4) has a lower 1st IE due to interelectron repulsions between
the paired electrons in the 2p subshell (thus easier to remove an electron).
Oxygen has the higher 2nd IE because of the difficulty in removing an electron from a ½ filled 2p
subshell. Moreover, Zeff is larger for oxygen than nitrogen, so removal of a second electron is more
difficult for oxygen.
4. (2 points) Explain why fluorine has a lower electron affinity (EA = –328 kJ/mol) than chlorine (EA = –398
kJ/mol).
The main reason why fluorine has a lower electron affinity than chlorine is interelectron repulsions.
Fluorine is much smaller in size than chlorine, so the electrons are crowded close together near the
nucleus in fluorine. The addition of an electron to this region of space already crowded with electron
density decreases fluorine’s electron affinity. Chlorine is larger in size (more diffuse electron density),
so interelectron repulsions are not as dominant. Keep in mind that halogens in general have the highest
electron affinity of all the elements because of the stability of the completely filled 2p subshell (nobel gas
configuration).
5. (2 point) BrF3 has a T-shaped molecular geometry (Br is the central atom). What are the possible
F–Br–F bond angles assuming the ideal T-shaped geometry? Describe how the bond angles distort from
the ideal T-shaped geometry due to the presence of lone pairs.
The F–Br–F bond angles in a T-shaped molecule are 90° and 180°. In BrF3, F atoms occupy the axial
positions. One F atom and 2 lone pairs of electrons occupy the equatorial positions. These lone pairs
distort the T-shaped geometry slightly, decreasing the Faxial–Br–Fequatorial bond angles to less than 90°
(86° to be exact).
