Which among the following orders of second IP | vedclass

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Question

(C) The second ionization potential $(IP_2)$ corresponds to the energy required to remove an electron from a unipositive ion $(M^+ \rightarrow M^{2+}

Vedclass · Accepted Answer

$Na > Cu$

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(C) The second ionization potential $(IP_2)$ corresponds to the energy required to remove an electron from a unipositive ion $(M^+ \rightarrow M^{2+} + e^-)$.$1$. For $Mn$ $(3d^5 4s^2)$,$Mn^+$ is $3d^5 4s^1$. For $Cr$ $(3d^5 4s^1)$,$Cr^+$ is $3d^5$. Since $Cr^+$ has a stable half-filled $d$-subshell,removing the second electron from $Cr^+$ is harder than from $Mn^+$. Thus,$Cr > Mn$.$2$. For $Zn$ $(3d^{10} 4s^2)$,$Zn^+$ is $3d^{10} 4s^1$. For $Cu$ $(3d^{10} 4s^1)$,$Cu^+$ is $3d^{10}$. $Cu^+$ has a stable fully-filled $d$-subshell,making $IP_2$ of $Cu$ higher than $Zn$. Thus,$Cu > Zn$.$3$. For $Na$ $([Ne] 3s^1)$,$Na^+$ is $[Ne]$ (stable noble gas configuration). For $Cu$ $([Ar] 3d^{10} 4s^1)$,$Cu^+$ is $[Ar] 3d^{10}$. The removal of an electron from the stable noble gas core of $Na^+$ requires significantly more energy than from $Cu^+$. Thus,$Na > Cu$.$4$. For $Cr$ $([Ar] 3d^5 4s^1)$,$Cr^+$ is $3d^5$. For $K$ $([Ar] 4s^1)$,$K^+$ is $[Ar]$. $K^+$ has a stable noble gas configuration,so $IP_2$ of $K$ is very high. Thus,$K > Cr$.Comparing the options,$Na > Cu$ is the correct order.

Which among the following orders of second $IP$ (Ionization Potential) is correct?

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