Question

Calculate the second ionization potential of \[{\text{L}}{{\text{i}}^{\text{ + }}}\]
a.\[{\text{112}}{\text{.4eV }}\]
b.\[{\text{122}}{\text{.4eV }}\]
c.\[{\text{244}}{\text{.8eV }}\]
d.\[{\text{61}}{\text{.2eV}}\]

Answer 1

Hint: As we know that in chemistry, periodic tables play a vital role. In the periodic table totally \[118\] elements. In the periodic table there are totally \[18\] columns and \[7\] rows. The columns are called groups. Hence, \[18\] groups in the periodic table. The rows are called as period. Hence, totally \[7\] period in the table. Each and every element having a specific symbol in the periodic table. Each element having a specific position in the periodic table, that position is dependent on the physical and chemical properties of the element.
Formula used:
The energy of the nth of the atom \[{\text{ = }}\dfrac{{{\text{( - 13}}{\text{.6)}}{{\text{z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}{\text{eVato}}{{\text{m}}^{{\text{ - 1}}}}\]
Here n is represented as the orbital number of the atom.
The atomic number of the atom is represent as Z

Complete answer:
Lithium has three electrons. \[{\text{L}}{{\text{i}}^{\text{ + }}}\] is having one orbital species.
The energy of the nth of the atom \[{\text{ = }}\dfrac{{{\text{( - 13}}{\text{.6)}}{{\text{z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}{\text{eVato}}{{\text{m}}^{{\text{ - 1}}}}\]
The value of n is one
The value of z if three
The first ionization potential of \[{\text{L}}{{\text{i}}^{\text{ + }}}\] is calculated as \[{{\text{E}}_1}\],
\[{\text{ = }}\dfrac{{{\text{( - 13}}{\text{.6)}}{{\text{z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}{\text{eVato}}{{\text{m}}^{{\text{ - 1}}}}\]
Now we can substitute the known values we get,
\[{\text{ = }}\dfrac{{{\text{( - 13}}{\text{.6)(3}}{{\text{)}}^{\text{2}}}}}{{{1^{\text{2}}}}}{\text{eVato}}{{\text{m}}^{{\text{ - 1}}}}\]
On simplification we get,
\[ = - 122.4{\text{eVato}}{{\text{m}}^{{\text{ - 1}}}}\]
Energy at infinity =\[{{\text{E}}_\infty }{\text{ = 0}}\]
The second ionization energy = \[{{\text{E}}_\infty }{\text{ = 0}}\]-\[{{\text{E}}_1}\]
\[{\text{ = 0 - ( - 122}}{\text{.4)eV}}\]
On simplification we get,
\[{\text{ = 122}}{\text{.4eV}}\]
The second ionization potential of \[{\text{L}}{{\text{i}}^{\text{ + }}}\] is \[{\text{122}}{\text{.4 eV }}\]

Hence, Option b is correct because the second ionization value is \[{\text{122}}{\text{.4eV}}\]

Note:
We have to remember that lithium is one of the elements in the periodic table. The atomic number of lithium is \[3\]. The symbol of lithium is \[{\text{Li}}\]. Lithium is present in the second period and first group in the periodic table. In \[{\text{L}}{{\text{i}}^{{\text{2 + }}}}\] is a one electron species. In general, first ionisation energy is always less than second ionisation energy of the atom. Because the electron removal in the neutral atom is very easy compared to the positive ion of the atom.