The dand fblock elements - Result Question 21
####21. The correct order of decreasing second ionisation enthalpy of $Ti(22), V(23), Cr(24)$ and $Mn(25)$ is :
[2008]
(a) $Cr>Mn>V>Ti$
(b) $V>Mn>Cr>Ti$
(c) $Mn>Cr>Ti>V$
(d) $Ti>V>Cr>Mn$
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Solution:
(a) $Ti$; Z (22) is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{2}$
$V ; \mathbf{Z}(23)$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{3}$
$Cr ; Z(24)$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{1}$
$Mn ; Z(25)$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 d^{5} 4 s^{2}$
The second electron in all the cases (except $Cr$ ) is taken out from $4 s$-orbital and for $Cr$ it is taken from half-filled $3 d$-orbital. The force required for removal of second electron will be more for $Mn$ than others (except for $Cr$ ) due to having more positive charge. Based on this, we find the
\begin{tabular}{|c|c|} \hline $7^{3+}=$ & \\ \hline$i^{3+}=$ & \\ \hline 111 & \begin{tabular}{|l|l|} 1 & 1 \\ \end{tabular} \\ \hline$=$\begin{tabular}{|l|l|} 1 & 1 \end{tabular} & \\ \hline \end{tabular} correct order $Mn>V>Ti$.
i.e. $Cr>Mn>V>$ Ti.
