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Arrange the elements of second and third period in increasing order of ionisation...

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prak10 | Student, Grade 10 | eNotes Newbie

Posted May 11, 2013 at 11:28 AM via web

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Arrange the elements of second and third period in increasing order of ionisation energy. 

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mvcdc | Student, Undergraduate | (Level 1) Associate Educator

Posted May 11, 2013 at 12:55 PM (Answer #1)

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The following are the ionization energies for the elements in the second period:

  • Lithium - 520
  • Beryllium - 900
  • Boron - 801
  • Carbon - 1087
  • Nitrogen - 1402
  • Oxygen - 1314
  • Fluorine - 1681
  • Neon - 2081

Hence, arranging them according to increasing ionization energy, we get:

Li < B < Be < C < N < O < F < Ne.

The following are the ionization energies for elements in the third period:

  • Sodium - 496
  • Magnesium - 738
  • Aluminum 578
  • Silicon - 787
  • Phosphorus - 1012
  • Sulfur - 1000
  • Chlorine - 1251
  • Argon - 1521

Hence, the order is as follows:

Na < Al < Mg < Si < S < P < Cl < Ar

Notice that the general trend for ionization energy is that it increases from left to right within a period; and decreases from top to bottom within a group.  

Ionization energy is the energy required to remove an electron from an atom. (Note that the values listed here are the first ionization energies - the energy required to remove the first electron.)

Hence, the more attracted the electrons are to the nucleus, the harder it is to remove the electron. A higher atomic number, then, which corresponds to higher number of protons and electrons, would mean a higher ionization energy - and hence the trend within a period. The further the electrons are from the nucleus, the easier they are to remove from an atom. This explains the trend within a group. 

Note that there are only a few exceptions to this - such as Boron and Beryllium, and Magnesium and Aluminum. These special cases can be explained by looking at the electron configuration. The ionization energy of B is less than that of Be because Boron has a complete 2s orbital. The shielding effect of the electrons in the 2s orbitals make it easier for the electron in the 2p orbital to be more easily removed. In general, however, the trend for the ionization energy holds.


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