Question

Give the number of core electrons for $ {F^ - } $
$A.\;\;\;\;\;22 \\
  B.\;\;\;\;\;75 \\
  C.\;\;\;\;\;9 \\
  D.\;\;\;\;\;4 \\
  E.\;\;\;\;\;\;2 $

Answer 1

Hint: The core electrons are the inner shell orbitals. They are present in the lower energy shells. They are the electrons that do not take part in any chemical bond formation. The number of core electrons can be found out by subtracting valence electrons from atomic numbers.

Complete answer:
Core electrons are the electrons in an atom that are not valence electrons and do not participate in chemical bond formation as these electrons are not exposed much. They are the inner electrons of an atom. The energy required to remove the core electrons is very high as these electrons are present in the inner shells.
Now let us find out the number of core electrons in the $ {F^ - } $ element:
The electronic configuration of $ {F^ - } $ is $ 1{s^2}2{s^2}2{p^6} $ . It has a total of $ 10 $ electrons. These $ 10 $ electrons are core electrons as well as valence electrons. In the electronic configuration of $ {F^ - } $ we can see that the inner shell is $ 1s $ and the outer shell are $ 2s{\text{ }}and{\text{ }}2p $ and so we can say that the electrons in the $ 1s $ shell are the core electrons. Also as a neutral atom, the number of protons it has must be equal to the number of electrons, as they are particles of opposite charge and a neutral atom has a total charge of zero. Therefore, $ F $ atom has $ 9 $ electrons in total and $ {F^ - } $ has $ 10 $ electrons in total. And thus there are only $ 2 $ electrons in the core and the rest $ 8 $ electrons are considered as the valence electrons.
Therefore the correct option is $ E.\;2 $ .

Note:
In elements that do not have any charge, one can find out the core electrons by just subtracting the number of valence electrons from atomic numbers. For example: in the case of aluminum the electronic configuration is $ [Ne]3{s^2}3{p^1} $ . The valence electrons in aluminum are $ 3 $ . Thus core electrons equals atomic number minus valence electrons $ = 13 - 3 = 10 $ . Therefore $ 10 $ electrons are core electrons in aluminum.