The atomic numbers of V, Cr, Mn and Fe are 23 | vedclass

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(B) The electronic configurations of the given elements are:$V (Z=23): [Ar] 3d^3 4s^2$$Cr (Z=24): [Ar] 3d^5 4s^1$$Mn (Z=25): [Ar] 3d^5 4s^2$$Fe (Z=26)

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$Cr$

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(B) The electronic configurations of the given elements are:$V (Z=23): [Ar] 3d^3 4s^2$$Cr (Z=24): [Ar] 3d^5 4s^1$$Mn (Z=25): [Ar] 3d^5 4s^2$$Fe (Z=26): [Ar] 3d^6 4s^2$For the second ionization enthalpy,we remove one electron from the $M^+$ ion:$V^+: [Ar] 3d^3 4s^1$$Cr^+: [Ar] 3d^5$$Mn^+: [Ar] 3d^5 4s^1$$Fe^+: [Ar] 3d^6 4s^1$Removing the second electron from $Cr^+$ is difficult because it has a stable half-filled $3d^5$ configuration. Therefore,$Cr$ has the highest second ionization enthalpy.

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

The atomic numbers of $V, Cr, Mn$ and $Fe$ are $23, 24, 25$ and $26$ respectively. Which of these elements will have the highest second ionization enthalpy?

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Triad-$I$ $[N^{3-}, O^{2-}, Na^{+}]$
Triad-$II$ $[N^{+}, C^{+}, O^{+}]$
Choose the species of lowest $IP$ from triad-$I$ and highest $IP$ from triad-$II$ respectively.

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 of the following statements is incorrect?

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