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The atomic numbers of vanadium (V), Chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23, 24, 25 and 26 Which one of these may be expected to have the highest second ionization enthalpy?

The atomic numbers of vanadium (V), Chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23, 24, 25 and 26 Which one of these may be expected to have the highest second ionization enthalpy?

Grade:12

2 Answers

Suraj Prasad IIT Patna
askIITians Faculty 286 Points
6 years ago
See the half filled and full filled orbitals are more stable.
if you look at the electronic configuration of chromium Cr, its 3d5 4s1 . So, to remove the 2nd electron from this atom will be difficult than the other atoms.
Hence need to provide more energy. So, Cr will be the correct answer.
SIDDHI AMRALE
28 Points
2 years ago
V:(Ar) 3d3 4s2
Cr:(Ar) 3d5 4s1 ( in order to gain extra stability 3d4 takes in e- from raw and becomes 3d5 4s1)
Mn:(Ar) 3d5 4s2
Fe:(Ar) 3 d6 4s2
First electron from 4s1  ( 1st I.E.)will go then 3d5 remains
It is half filled, hence symmetrical, so hardest to remove
So2end I.E. odd Cr greatest

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