Question

analysis questions:

1. does f⁻ have a larger or smaller radius than al³⁺?

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 have the same number of electron shells and the same electron configuration.

Step2: Compare nuclear charges

The nuclear charge of $F^-$ comes from F nucleus, which has 9 protons. The nuclear charge of $Al^{3+}$ comes from Al nucleus, which has 13 protons. Since $Al^{3+}$ has a greater nuclear charge (more protons), it will pull the electron cloud more strongly towards the nucleus.

Step3: Relate nuclear charge to ionic radius

With the same electron cloud (same number of electrons and same electron configuration), a greater nuclear charge will result in a smaller ionic radius. So $F^-$ has a larger radius than $Al^{3+}$ because it has a smaller nuclear charge, so the electron cloud is less pulled in.

Answer:

$F^-$ has a larger radius than $Al^{3+}$