Q80 / 200 · NEET 2008 (Screening) · Chemistry

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

  • Correct answer
    A
    $\mathrm{Cr}>\mathrm{Mn}>\mathrm{V}>$ Ti
  • B
    V $>$ Mn $>$ Cr $>$ Ti
  • C
    Mn $>$ Cr $>$ Ti $>$ V
  • D
    Ti $>$ V $>$ Cr $>$ Mn

Solution

Key Idea : The amount of energy required to remove an electron from a unipositive ion is called second IP which generally increases in a period from left to right. The second ionisation energies of the first transition series increase almost regularly with increase in atomic number. However, the value for $\mathrm{Cr}$ is sufficiently higher than those of its neighbour, ie, (Mn). This is due to stable configuration of $\mathrm{Cr}^{+}$( $3 d^5$ exactly half filled). Note : The halffilled and completely filled configurations are more stable.

Hence, the correct answer is option: 'A'.

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