The correct order of decreasing second ionization enthalpy of is:
$(T i = 22, V = 23, C r = 24, M n = 25)$

M n > C r > T i > V

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T i > V > C r > M n

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C r > M n > V > T i

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V > M n > C r > T i

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Solution

The correct option is D $C r > M n > V > T i$
The value is second IE generally increases towards the right in period, however ${C r}^{+}$ has $3 d^{5}$ configuration which is extra-stable.

Hence $2^{n d}$ IE of Cr is more than that of Mn. Thus correct order of $2^{n d}$ IE is $C r > M n > V > T i$

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Similar questions

The correct order of decreasing second ionisation enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is [CBSE AIPMT 2008]

Which one of the following ions is the most stable in aqueous solution?
(Atomic number. Ti = 22, V = 23, Cr = 24, Mn = 25)

T i = 22, V = 23, C r = 24, M n = 25

4 s

T i = 22, V = 23, C r = 24, M n = 25

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