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

$Na > Cu$

Vedclass · Answer

(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.

List-$I$ (Electronic configuration of neutral atom where $n=2$)	List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)
$A. ns^2$	$I. 2080$
$B. ns^2np^1$	$II. 899$
$C. ns^2np^3$	$III. 800$
$D. ns^2np^6$	$IV. 1402$

List-$I$ ($IE_1, IE_2, IE_3$ in $kJ \ mol^{-1}$)	List-$II$ (Element)
$A$. $1510$	$1$. $H$
$B$. $495, 6500, 10200$	$2$. $Li$
$C$. $840, 1630, 13100$	$3$. $Be$
$D$. $600, 2050, 3100$	$4$. $B$

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

Similar Questions

The $3^{rd}$ ionisation energy amongst $Li$,$Be$,$B$ and $C$ is lowest for

Which factors influence the value of ionization enthalpy?

Match List-$I$ with List-$II$:
List-$I$ (Electronic configuration of neutral atom where $n=2$) List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)
$A. ns^2$ $I. 2080$
$B. ns^2np^1$ $II. 899$
$C. ns^2np^3$ $III. 800$
$D. ns^2np^6$ $IV. 1402$

Which ionisation potential $(IP)$ in the following equations involves the greatest amount of energy?

Match List-$I$ with List-$II$ with correct code:

List-$I$ ($IE_1, IE_2, IE_3$ in $kJ \ mol^{-1}$) List-$II$ (Element)

$A$. $1510$ $1$. $H$

$B$. $495, 6500, 10200$ $2$. $Li$

$C$. $840, 1630, 13100$ $3$. $Be$

$D$. $600, 2050, 3100$ $4$. $B$

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