Question

The energy needed for the following process is 1.960 × 104 kJ/mol:

Li(g) → Li3+(g) + 3e−

If the first ionization energy of lithium is 5.20 × 102 kJ/mol, calculate the second ionization energy of lithium—that is, the energy required for the process:

Li+(g) → Li2+(g) + e−

A hydrogen-like ion is an ion containing only one electron. The energies of the electron in a hydrogen-like ion are given by

En = −(2.18 × 10−18 J) (Z2)(1/n2)

where En is the energy of the electron in the hydrogen-like ion, n is the principal quantum number, and Z is the atomic number of the element.

Second ionization energy of lithium:   kJ/mol (Enter your answer in scientific notation)

Answer 1

The energy needed to remove 3 electrons from Li atom given by the following equation is .

Note that we can write the above process as a step wise gradual removal of electrons one by one as follows:

Step 1

This is the equation for first ionization of Li, which requires an energy of

Step 2

This is the equation for the second ionization of lithium which we are trying to find out. Lets denote it by .

Step 3

This is the third ionization where the third electron is removed making the tripositive ion. Let the energy required for this process be denoted by .

Note that is a hydrogen-like species with only 1 electron in its 1s orbital in ground state.

Hence, the energy of the electron of in its shell is given by

Since the electron is in n= 1, and Z = 3 for Li,

Now, for 1 mole of such electrons in 1 mol of atoms, the energy of all these electrons will be

Note that two ionize those electrons with of potential energy, we have to supply the same amount of energy to the electrons.

Hence, the third ionization energy is basically negative of the above value.

Now, the sum of step 1 , step 2 and step 3 can be seen as

------------------------------

hence, we can write

Hence,the second ionization energy of lithium is about .