Question #27e03

1 Answer
Dec 16, 2014

The fact that the element has 5p electrons indicates that it is in the fifth period on the periodic table. According to your information, there are five electrons in the 5p sublevel, in which two of the three 5p orbitals contain paired electrons, with the third 5p orbital containing one unpaired electron, for a total of five, represented as #"5p"^5#. The group 17/VIIA elements have five electrons in their outermost (highest energy) p sublevel.

This element would be iodine, atomic number 53, which is the number of protons in an atom of iodine. It is in the fifth period, in group 17/VIIA. A neutral iodine atom would also have 53 electrons. Its ground state electron configuration would be:


There is a noble gas shorthand method for writing long electron configurations. Each element's electron configuration starting in the second period has the electron configuration of the noble gas in the previous period, plus that element's additional electrons. In order to write an electron configuration in noble gas shorthand, you place the symbol for the previous noble gas in brackets, then continue writing the electron configuration of the element after the noble gas.

The noble gas shorthand electron configuration for iodine in its ground state is #"[Kr]5s"^2"4"^10"5p"^5#.

Iodine and fluorine form the covalent compound #"IF"#, named iodine monofluoride.

#"I"_2# + #"F"_2# #rarr# #"2IF"#

The following link will send you to a site where you can download a printable periodic table. It includes the electron configurations of the elements and lots of other important information.