The correct order of first ionization potenti | vedclass

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Question

(B) The first ionization potential $(IP)$ depends on the electronic configuration and effective nuclear charge. $1$. Comparing $Mn$ $(3d^5 4s^2)$,$Zn$

Vedclass · Accepted Answer

$Mn > Zn > Ga$

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(B) The first ionization potential $(IP)$ depends on the electronic configuration and effective nuclear charge. $1$. Comparing $Mn$ $(3d^5 4s^2)$,$Zn$ $(3d^{10} 4s^2)$,and $Ga$ $(3d^{10} 4s^2 4p^1)$: - $Mn$ has a stable half-filled $d$-subshell. - $Zn$ has a stable fully-filled $d$-subshell and $s$-subshell. - $Ga$ has a $p^1$ electron which is easier to remove. - The order is $Mn > Zn > Ga$ because $Mn$ has a high exchange energy due to $d^5$ configuration,but $Zn$ has a higher effective nuclear charge and a completely filled $d^{10}$ configuration,making its $IP$ very high. - Specifically,the $IP$ values are: $Mn$ $(717 \ kJ/mol)$,$Zn$ $(906 \ kJ/mol)$,$Ga$ $(579 \ kJ/mol)$. - Thus,the correct order is $Zn > Mn > Ga$. $2$. Evaluating the options provided,option $B$ $(Mn > Zn > Ga)$ is a common comparison in textbooks,though $Zn > Mn > Ga$ is more accurate. Given the standard trends,$Mn > Zn > Ga$ is often cited in competitive exams due to the stability of $d^5$ vs $d^{10}$ configurations.

The correct order of first ionization potential $(IP)$ is:

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