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The correct order of decreasing second ionization enthalpy
Ti(22), V(23), Cr(24), Mn(25)

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V > Mn > Cr > Ti

Mn > Cr > Ti > V

Cr> Mn > V > Ti

Ti > V > Cr > Mn

answer is D.

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Detailed Solution

$C r^{+} (3 d^{5}), M n^{+} (3 d^{5} 4 s^{1}), V^{+} (3 d^{3} 4 s^{1}), T i^{+} (3 d^{2} 4 s^{1})$

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