Question:

Arrange the following \(3d\)-series elements in decreasing order of their second ionisation enthalpy: (A) \(\mathrm{Ti} \ (Z=22)\) (B) \(\mathrm{V} \ (Z=23)\) (C) \(\mathrm{Cr} \ (Z=24)\) (D) \(\mathrm{Mn} \ (Z=25)\)

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Questions involving successive ionisation enthalpies should always be solved by examining the electronic configuration of the ion formed after the previous ionisation step.
Updated On: Jun 11, 2026
  • (C), (B), (D), (A)
  • (D), (C), (B), (A)
  • (D), (B), (C), (A)
  • (C), (D), (B), (A)
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The Correct Option is B

Solution and Explanation

Concept: The second ionisation enthalpy is the energy required to remove an electron from a singly charged gaseous cation. \[ \mathrm{M^+(g)} \rightarrow \mathrm{M^{2+}(g)} + e^- \] The value of second ionisation enthalpy depends strongly on the electronic configuration of the \(M^+\) ion. If the \(M^+\) ion possesses an exceptionally stable configuration such as half-filled or fully-filled subshells, removing an electron from it becomes difficult and hence the second ionisation enthalpy becomes very high.

Step 1:
Write the electronic configurations of the neutral atoms.
\[ \mathrm{Ti}=[Ar]3d^24s^2 \] \[ \mathrm{V}=[Ar]3d^34s^2 \] \[ \mathrm{Cr}=[Ar]3d^54s^1 \] \[ \mathrm{Mn}=[Ar]3d^54s^2 \]

Step 2:
Write the configurations of the singly charged ions.
After removal of one electron: \[ \mathrm{Ti^+}=[Ar]3d^24s^1 \] \[ \mathrm{V^+}=[Ar]3d^34s^1 \] \[ \mathrm{Cr^+}=[Ar]3d^5 \] \[ \mathrm{Mn^+}=[Ar]3d^54s^1 \]

Step 3:
Compare the stability of the \(M^+\) ions.
The ion \(\mathrm{Cr^+}\) possesses an exactly half-filled \(3d^5\) configuration. Similarly, \(\mathrm{Mn^+}\) contains a half-filled \(3d^5\) core with an additional \(4s^1\) electron. Removal of electrons from such stable arrangements requires comparatively higher energy. Among these, the stability considerations lead to the order: \[ \mathrm{Mn} > \mathrm{Cr} > \mathrm{V} > \mathrm{Ti} \] for second ionisation enthalpy.

Step 4:
Write the decreasing order.
\[ (D)>(C)>(B)>(A) \] Hence the correct answer is: \[ \boxed{(D),(C),(B),(A)} \]
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