Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

The geometry of Ni(CO)4 and Ni(PPh3)2Cl2 are:
(A) Both square planar
(B) Tetrahedral and square planar, respectively
(C) Both tetrahedral
(D) Square planar and tetrahedral, respectively

Answer
VerifiedVerified
438.6k+ views
like imagedislike image
Hint: First find the electronic configuration of Ni in Ni(CO)4 and check whether it can be excited or not. Due to the presence of a strong ligand, there will be forced pairing of electrons. Hence, again determine the electronic configuration of Ni and use it to find the hybridisation which will tell the shape of Ni(CO)4 . The same steps will be applied to find the hybridisation and shape of Ni(PPh3)2Cl2 .

Complete step by step answer:
Here we will use Valence Band Theory (VBT) to determine the geometry of both Ni(CO)4 and Ni(PPh3)2Cl2 .
First let’s determine the geometry of Ni(CO)4 .
Atomic number of Ni=28
Hence the configuration of Ni is 3d84s2 .
The configuration can be calculated as follows:
For an atomic number of 28 , the electronic configuration is:
 1s2,2s22p6,3s23p63d8,4s2
seo images

Now if we adjust it for noble-gas configuration, we get,
 Ni=[Ar]3d84s2
We chose Argon because it is the closest element to Ni and it is a noble-gas. Also, this configuration is in ground state.
But we cannot excite Ni as the oxidation number of Ni in Ni(CO)4 is 0 . However, the electrons from s orbital moved to d orbital because of the forced pairing of electrons due to the presence of a strong ligand, CO , and it will change the configuration to:
 1s2,2s22p6,3s23p63d10
When we draw hybridisation state of Ni ,
seo images

We see that hybridisation is sp3 , and we know that for sp3 hybridisation, the geometrical shape of that molecule is tetrahedral.
Hence, the geometrical shape of Ni(CO)4 will be tetrahedral.
Now, let’s determine the geometry of Ni(PPh3)2Cl2 .
The configuration of Ni in ground state will remain same as above,
 1s2,2s22p6,3s23p63d8,4s2
seo images

Here, the oxidation number of Ni in Ni(PPh3)2Cl2 is 2 which means Ni can be excited to Ni2+ :
 1s2,2s22p6,3s23p63d10
There won’t be any forced pairing here because Cl2 is a weak ligand.
Again, let’s draw the hybridisation state of Ni , which is:
seo images

Again, the hybridisation of Ni is sp3 which results in a tetrahedral geometric shape.
Hence, the geometrical shape of Ni(PPh3)2Cl2 will be tetrahedral.
Hence option (C) is correct.

Note:
Pay attention while writing the electronic configuration of Ni in ground state as well as in hybridised state. Make the arrow-head representation to write the electronic configuration as it will help you in verifying the hybridization and electronic configuration. Also, try to remember the shape for each hybridisation since it will save you a lot of time.
Latest Vedantu courses for you
Grade 10 | CBSE | SCHOOL | English
Vedantu 10 CBSE Pro Course - (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for CBSE students
PhysicsPhysics
Social scienceSocial science
ChemistryChemistry
MathsMaths
BiologyBiology
EnglishEnglish
₹41,000 (9% Off)
₹37,300 per year
EMI starts from ₹3,108.34 per month
Select and buy