Question

Among the following which is the smallest anion according to ionic radius?
(A)$F^{-}$
(B)$C{l}^{-}$
(C)$B{r}^{-}$
(D)$H^{-}$

Answer 1

Hint: Atomic and ionic radii are the distances which are away from the nucleus or central atom having different periodic trends. The distance away from the nucleus is called atomic radius. Atomic radius increases while going from top to bottom in the periodic table and decreases while going across the periodic table. The distance away from the central atom is known as ionic radius.

Complete answer:
Radius depends on the effective nuclear charge. The net positive charge of an electron in an atom having multiple electrons is called effective nuclear charge. Here, the term effective is being used because the shielding effect of negatively charged electrons will prevent higher orbital electrons from experiencing the full nuclear charge. Repulsion is more dominant than nuclear attraction and hence the size increases. The ${\displaystyle \frac{e}{p}}$ ratio for hydrogen is $2$. The ${\displaystyle \frac{e}{p}}$ ratio of $F^{-}$is $1.11$. As we go down the group the atomic radius increases. $F^{-}$has the smallest atomic radius. The order is: $F^{-}<C{l}^{-}<B{r}^{-}<H^{-}$

Therefore, option A is correct.

Note:
Whenever an atom loses an electron to form a cation, other electrons will be attracted more towards the nucleus and the radius of the ion becomes smaller. Ionic radius decreases with an increase in the positive charge and it increases with an increase in the negative charge. Nuclear charge is defined as the total charge of a nucleus whereas an effective nuclear charge is the net charge that a valence shell experiences.