analysis questions:\n1. does f⁻ have a larger or smaller radius than al³⁺?
analysis questions:\n1. does f⁻ have a larger or smaller radius than al³⁺?
Answer
Explanation:
Step1: Determine electron configurations
$F^-$ has gained 1 electron, so its electron configuration is the same as Ne: $1s^2 2s^2 2p^6$. $Al^{3+}$ has lost 3 electrons, so its electron configuration is also the same as Ne: $1s^2 2s^2 2p^6$. So they are isoelectronic (same number of electrons, same electron configuration).
Step2: Compare nuclear charges
The nuclear charge (number of protons) of F is 9, and for Al it is 13. So the nuclear charge of $Al^{3+}$ ($Z = 13$) is greater than that of $F^-$ ($Z = 9$).
Step3: Relate nuclear charge to ionic radius
For isoelectronic species, the one with a larger nuclear charge will attract the electrons more strongly, pulling them closer to the nucleus, thus resulting in a smaller ionic radius. Since $Al^{3+}$ has a larger nuclear charge than $F^-$, $F^-$ will have a larger radius than $Al^{3+}$.
Answer:
$F^-$ has a larger radius than $Al^{3+}$.