Question

If the second I.P. of calcium is ${\text{275}}{\text{.12kcalsmo}}{{\text{l}}^{{\text{ - 1}}}}$ , then, the equation that represents this ionization can be written as:
A.${\text{C}}{{\text{a}}^{\text{ + }}}\left( {\text{s}} \right) \to {\text{C}}{{\text{a}}^{{\text{3 + }}}}\left( {\text{s}} \right) + 2{{\text{e}}^{\text{ - }}} - 275.12{\text{kcals}}$
B.${\text{C}}{{\text{a}}^{\text{ + }}}\left( {\text{s}} \right) \to {\text{C}}{{\text{a}}^{{\text{3 + }}}}\left( {\text{s}} \right) + 2{{\text{e}}^{\text{ - }}} + 275.12{\text{kcals}}$
C.${\text{C}}{{\text{a}}^{\text{ + }}}\left( {\text{s}} \right) \to {\text{C}}{{\text{a}}^{{\text{2 + }}}}\left( {\text{s}} \right) + {{\text{e}}^{\text{ - }}} + 275.12{\text{kcals}}$
D.${\text{C}}{{\text{a}}^{\text{ + }}}\left( {\text{s}} \right) \to {\text{C}}{{\text{a}}^{{\text{2 + }}}}\left( {\text{s}} \right) + {{\text{e}}^{\text{ - }}} - 275.12{\text{kcals}}$

Answer 1

Hint: In chemistry, the term ionization energy refers to the amount of energy which is required to remove an electron from an isolated atom which is in gaseous state in its ground state. When expressed in electron volts, it is called ionization potential. It is also represented by ‘I.P.’. Ionization energy is always positive.

Complete step by step answer:
The ionization energy which is required to remove the first electron from the isolated gaseous atom in the ground state or unexcited state is termed as the first ionization energy or the first ionization potential of that atom.
The ionization energy which is required to remove the ${{\text{2}}^{{\text{nd}}}}$ electron from the isolated gaseous atom is termed as the second ionization energy or the second ionization potential of that atom and that which is required to remove the ${{\text{3}}^{{\text{rd}}}}$ electron from the isolated gaseous atom is called the third ionization energy or the third ionization potential of that atom and so on.
For calcium, the first ionization energy ${\left( {{\text{I}}{\text{.P}}{\text{.}}} \right)_{\text{1}}}$ is the energy needed for removal of the first electron from calcium to form the ${\text{C}}{{\text{a}}^{\text{ + }}}$ ion. The equation that represents this ionization is:
${\text{Ca}} \to {\text{C}}{{\text{a}}^{\text{ + }}} + {{\text{e}}^{\text{ - }}}$
Now, the second ionization energy ${\left( {{\text{I}}{\text{.P}}{\text{.}}} \right)_2}$ is the energy needed for removal of the second electron from ${\text{C}}{{\text{a}}^{\text{ + }}}$ ion to form the ${\text{C}}{{\text{a}}^{{\text{2 + }}}}$ ion. The equation that represents this ionization is:
${\text{C}}{{\text{a}}^ + } \to {\text{C}}{{\text{a}}^{{\text{2 + }}}} + {{\text{e}}^{\text{ - }}}$

Moreover, the ionization energy is always positive, and so, the correct answer is C.

Note: In general, the ionization energy tends to decrease down the group from top to bottom and tends to increase along a period from left to right. But there are some exceptions. Whenever a stable electronic configuration is present, the ionization energy for further ionization increases exceptionally. This means that the ionization energy for an element with a half-filled or a fully-filled configuration will be very high.