The correct order of decreasing second ionisation enthalpy of Ti (22), V (23), Cr (24) and $\mathrm{Mn}(25)$…

The correct order of decreasing second ionisation enthalpy of Ti (22), V (23), Cr (24) and $\mathrm{Mn}(25)$ is
  1. $\mathrm{Ti} > \mathrm{V} > \mathrm{Cr} > \mathrm{Mn}$
  2. $\mathrm{Cr} > \mathrm{Mn} > \mathrm{V} > \mathrm{Ti}$
  3. $\mathrm{V} > \mathrm{Mn} > \mathrm{Cr} > \mathrm{Ti}$
  4. $\mathrm{Mn} > \mathrm{Cr} > \mathrm{Ti} > \mathrm{V}$

Solution

$\mathrm{Cr}(24) \rightarrow[\mathrm{Ar}] 3 d^5 4 s^1$ After removing one electron from chromium, the resulting structure becomes more stable. Hence $\mathrm{Cr}$ has higher second ionisation enthalpy. Thus, the correct order is $\mathrm{Cr} > \mathrm{Mn} > \mathrm{V} > \mathrm{Ti}$

Asked in: NEET 2008 (Mains)

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