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The correct order of decreasing second ionization enthalpy of Ti (22), V (23), Cr (24) and Mn (25) is -
A
Mn > Cr > Ti > V
B
Ti > V > Cr > Mn
C
Cr > Mn > V > Ti
D
V > Mn > Cr > Ti
Detailed Solution
Ionization enthalpy (both first and second) generally increases from left to right across the period because the nuclear charge increases.
After losing one electron the configurations are: $Ti^+ = 3d^2 4s^1$, $V^+ = 3d^3 4s^1$, $Cr^+ = 3d^5$, $Mn^+ = 3d^5 4s^1$.
For Ti, V and Mn the second electron is removed from the 4s orbital.
For chromium, $Cr^+$ has the stable half-filled $3d^5$ configuration, so the second electron has to be removed from this stable arrangement and needs extra energy.
Only chromium is exceptional due to this stable configuration, so its second ionization enthalpy is higher than that of Mn.
Approximate values: Ti = 1310, V = 1414, Cr = 1592, Mn = 1509 kJ mol$^{-1}$.
So the correct order is: Cr > Mn > V > Ti
After losing one electron the configurations are: $Ti^+ = 3d^2 4s^1$, $V^+ = 3d^3 4s^1$, $Cr^+ = 3d^5$, $Mn^+ = 3d^5 4s^1$.
For Ti, V and Mn the second electron is removed from the 4s orbital.
For chromium, $Cr^+$ has the stable half-filled $3d^5$ configuration, so the second electron has to be removed from this stable arrangement and needs extra energy.
Only chromium is exceptional due to this stable configuration, so its second ionization enthalpy is higher than that of Mn.
Approximate values: Ti = 1310, V = 1414, Cr = 1592, Mn = 1509 kJ mol$^{-1}$.
So the correct order is: Cr > Mn > V > Ti