Considering periods 1, 2 & 3, explain the trends of ionisation energy

Question

Considering periods 1, 2 & 3, explain the trends of ionisation energy.

also For the s block, first five periods (i.e. Li – Cs & Be – Ba) explain the trends in ionisation energies down each block.

Answer 1

I know this looks like a long answer, but if you read through it and look at the two links I provided it will become much easier to understand.

 

Ionization energy is the amount of energy it takes to detach one electron from a neutral atom. The trend  for ionization energy in the periodic table is that ionization energy increases from bottom to top and from left to right. This means that atoms in period 1 will have a lower ionization energy than atoms in period 2 (I.e., ionization energy of Li=520 KJ/mol and Be=900 KJ/mol.) Period 3 will have higher ionization energy than period 2 and this trend continues until you reach period 8 which has the highest ionization energy.

 

*Notice there is a slight change in this trend when crossing from the s-block to the p-block, but the trend continues throughout the rest of the p-block.

 

*There is also a slight change in the trend from period 5 to period 6. Period 5 has slightly higher ionization energy than period 6.

 

Here is a simplified explanation using column 2 of the periodic table:

Period 1 s-block (Li=520 KJ/mol) < Period 2 s-block (Be=900 KJ/mol) >* Period 3 p-block (B=800 KJ/mol < Period 4 p-block (C=1090 KJ/mol) < Period 5 p-block (N=1400 KJ/mol) >* Period 6 p-block (O=1310 KJ/mol) < Period 7 p-block (F= 1680 KJ/mol) < Period 8 p-block (Ne=2080 KJ/mol)

 

Even further simplified:

Period 1 < Period 2 >* Period 3 < Period 4 < Period 5 >* Period 6 < Period 7 < Period 8

 

 

http://www.shodor.org/chemviz/ionization/students/background.html

 

http://www.emsb.qc.ca/laurenhill/science/Trends2/ionization%20energy.jpg