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

Among Ni(CO)4,[Ni(CN)4]2− and [NiCl2−4]
[IIT-JEE, 1991]

A
Ni(CO)4 and NiCl24 are diamagnetic and [Ni(CN)4]2 is paramagnetic.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
NiCl24 and [Ni(CN)4]2 are diamagnetic and Ni(CO)4 is paramagnetic.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Ni(CO)4 and [Ni(CN)4]2 are diamagnetic and NiCl24 is paramagnetic.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Ni(CO)4 is diamagnetic and NiCl42 and [Ni(CN)4]2 are paramagnetic.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C Ni(CO)4 and [Ni(CN)4]2 are diamagnetic and NiCl24 is paramagnetic.
The concept you need to apply is simple. Figure out which ligand is strong and weak, and hence the electronic configuration of the metal in question.
Let's do this:
(i) [Ni(CO)4] OS Ni is 0
Ni = (Z=28) 3d8 4s2

CO is a strong field ligand, it causes pairing.

sp3-Hybridisation (tetrahedral)
There are no unpaired electrons, so the complex is diamagnetic.
Spin magnetic moment = zero.

(ii) [Ni(CN4)2, Ni2+=3d8

Cyano is a strong field ligand, it causes pairing.

dsp2-hybridisation (square planar)
There are no unpaired electrons so, the complex is diamagnetic.
Spin magnetic moment = zero.

(iii) [NiCl4]2, Ni2+=3d8

Chlorido is a weak field ligand, no pairing
sp3-hybridisation (tetrahedral)
There are two unpaired electrons, so the complex is paramagnetic.

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