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Ti (Z = 22) is 1s2 2s2 2p6 3s2 3p6 3d2 4s2. V (Z = 23) is 1s2 2s2 2p6 3s2 3p6 4s2 3d3. Cr (Z = 24) is 1s2 2s2 2p6 3s2 3p6 3d5 4s1. Mn (Z = 25) is 1s2 2s2 2p6 3s2 3p6 4s2 3d5. The second electron in all the cases (except Cr) is taken out from 4s-orbital and for Cr it is taken from half-filled 3d-orbital. The force required for removal of second electron will be more for Mn than others (except for Cr) due to having more positive charge. Based on this, we find the correct order Mn > V > Ti.

1. Ti (Z = 22) is 1s2 2s2 2p6 3s2 3p6 3d2 4s2.
2. V (Z = 23) is 1s2 2s2 2p6 3s2 3p6 4s2 3d3.
3. Cr (Z = 24) is 1s2 2s2 2p6 3s2 3p6 3d5 4s1.
4. Mn (Z = 25) is 1s2 2s2 2p6 3s2 3p6 4s2 3d5.

Correct answer: Mn (Z = 25) is 1s2 2s2 2p6 3s2 3p6 4s2 3d5.

Solution

The question discusses the second ionization energy trend, which depends on the electronic configuration and effective nuclear charge. Mn has a higher nuclear charge and stable half-filled 3d5 configuration, making its second ionization energy higher than others (except Cr).

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