Question

$2^{nd}$ Ionization potential of Al would not be numerically the same as

• A. $2^{nd}$ electron affinity of Al
• B. $1^{st}$ ionization potential of $A{l}^{+}$
• C. $2^{nd}$ electron affinity of $A{l}^{3+}$
• D. $3^{rd}$ ionization potential of $A{l}^{-}$

Answer 1

The correct option is A $2^{nd}$ electron affinity of Al

$2^{nd}$ ionization potential of Al is the energy required to remove an electron from $A{l}^{+}$ i.e. to remove an electron from $3s^2$ orbital.
a)$2^{nd}$ electron affinity of $Al$ is the energy required to add two electron on its valence shell i.e. on $3p^1$ orbital. So the energy for $2^{nd}$ ionization potential of $Al$ and $2^{nd}$ electron affinity of $Al$ will be different.
b)$1^{st}$ ionization potential of $A{l}^{+}$ is same as the $2^{nd}$ ionization potential of $Al$ i.e. to remove an electron from $3s^2$ orbital.
c)$2^{nd}$ electron affinity of $A{l}^{3+}$ is the energy required to add an electron in $3s^1$ orbital. So it also same as the $2^{nd}$ ionization potential of $Al$.
d)$3^{rd}$ ionization potential of $A{l}^{-}$ is also energy to required to remove an electron from the $3s^2$ orbital.