wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

A
Fe
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Mn
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Cr
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D Cr
Let us look at the ground state electronic configuration of each element.
Vanadium has a [Ar] 3d3 4s2 electronic configuration.
Chromium has a [Ar]3d5 4s1 electronic configuration. This configuration is much more stable than the otherwise expected configuration of [Ar] 3d4 4s2 because in the
[Ar] 3d5 4s1 arrangement, there is maximum stability due to more avenues of exchange energy. As a very crude generalization, exactly half-filled and fully-filled subshells (with respect to d orbitals) are considered more stable.
Manganese has a [Ar]3d5 4s2 electronic configuration.
Iron has a [Ar]3d6 4s2 electronic configuration.
It is known that exactly half-filled and completely filled d orbitals tend to be more stable than partially filled d orbitals. For instance, this is the reason why Cr and Cu have [Ar]3d5 4s1 and [Ar]3d104s1 configurations respectively. As we can see, Cr has the most stable ground state electronic configuration. It also has the most stable M+ electronic configuration of [Ar]3d5.
Hence, removing two electrons from Cr is going to require the maximum energy. So the answer is Cr.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Ionization Energy
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon