The correct option is
B An ion with
d5 configuration has one unpaired electron both in weak and strong fields
(A)
d5 configuration is
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1548998/original_180A.png)
With strong field ligand, pairing will take place and number of unpaired electrons will become
1
With weak field ligand, electrons does not get paired and number of unpaired electrons will be
5
(B)
d8 configuration
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1549002/original_180S1.png)
So in case of
d8 configuration, number of unpaired electrons in presence of both strong field and weak field ligand is same only, which is
2
(C)
d6 configuration is
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1549004/original_180S2.png)
In case of
d6 configuration in presence of strong field ligand pairing will occur and the number of unpaired electrons will be
0
Therefore, it show diamagnetic nature.
(D) In
d4 electronic configuration, the number of unpaired electrons in presence of weak field ligand is
4 and in presence of strong field ligand it is
2
In
d5 electronic configuration, the number of unpaired electrons in presence of weak field ligand is
5 and in presence of strong field ligand it is
1
In
d6 electronic configuration, the number of unpaired electrons in presence of weak field ligand is
4 and in presence of strong field ligand it is
0
In
d7 electronic configuration, the number of unpaired electrons in presence of weak field ligand is
3 and in presence of strong field ligand it is
1