Solveeit Logo

Question

Question: The hybridization of \( Ni \) in \( Ni{(CO)_4} \) is: A. \( s{p^2} \) B. \( ds{p^2} \) C. \(...

The hybridization of NiNi in Ni(CO)4Ni{(CO)_4} is:
A. sp2s{p^2}
B. dsp2ds{p^2}
C. sp3s{p^3}
D. sp3ds{p^3}d

Explanation

Solution

Hint : It is a coordination number 44 complex and the complexes having coordination number 44 have two geometry that is tetrahedral and square planar. Tetrahedral geometry is shown when the hybridization is sp3,d3ss{p^3},{d^3}s and square planar geometry is shown when the hybridization is dsp2,sp2dds{p^2},s{p^2}d .

Complete Step By Step Answer:
Coordination number is the total number of atoms or molecules bonded to the central atom. Coordination number is also called ligancy which means number of ligands bonded to the central metal atom.
In Ni(CO)4Ni{(CO)_4} , NiNi is attached to 44 molecules of COCO that’s why its coordination number is 44 .
In coordination chemistry, hybridization depends on the number of d-electrons (of the central metal atom). In case of Ni(CO)4Ni{(CO)_4} the central metal atom is nickel so we have to count the number of d-electrons in nickel:
Configuration of nickel is as follows:
NiNi = 1s22s22p63s23p63d84s21{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^8}4{s^2}
Valence shell configuration: 3d84s23{d^8}4{s^2}

As nickel has zero charge on because carbonyl is a neutral ligand now, as we can see that ss is not vacant so 44 ligands can't get bonded to nickel.
For bonding in such cases, electrons of ss orbital get shifted to dd orbital and new configuration becomes 3d104s03{d^{10}}4{s^0} .
It is called dsd - s pairing and the d10{d^{10}} is called a converted dsystemd'system .
3d104s04p03{d^{10}}4{s^0}4{p^0} :
Hybridization will take place with ss and pp orbital as sp3s{p^3} . Hence, nickel is sp3s{p^3} hybridized with geometry tetrahedral and spin only moment zero. Spin only moment is zero because the number of unpaired electrons are zero. As one of its properties is also derived by the number of unpaired electrons that is magnetism. There are no unpaired electrons so it is a diamagnetic compound.
So, option C is correct.

Note :
Ni(CO)4Ni{(CO)_4} was first synthesised in 18901890 by Ludwig Mond by the direct reaction of nickel metal with COCO . This pioneering work foreshadowed the existence of many other metal carbonyl compounds, including those of V,Cr,Mn,FeV,Cr,Mn,Fe and CoCo . It was also applied industrially to the purification of nickel by the end of the nineteenth century.