Question

The number of unpaired electrons in nickel carbonyl is :
(A). Zero
(B). one
(C). four
(D). five

Answer 1

Hint: The number of unpaired electrons in nickel carbonyl depends on the complex formation, which further depends on multiple factors such as the electronic configuration of Ni and the influence of Co (whether it is a weak field or a strong field ligand).

Complete step by step answer:
 We know that nickel carbonyl is also known as nickel tetracarbonyl $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$ . It is a colorless liquid . So let us understand the structure and safety of nickel carbonyl . In nickel tetracarbonyl , nickel $\left( {Ni} \right)$ is $ + ve$ central atom with atomic number 28 and ground state electronic configuration $\left[ {{}_{18}Ar} \right]3{d^8}4{s^2}$ .
i.e
Ni :- Atomic number 28 :-
Ground state electronic configuration of valence shell .
         $3{d^8}$ $4s$ $4p$

As carbon monoxide is the strong field legend and oxidation state 0 . When it approaches near nickel atom being strong field legend , it causes pairing of electron such that electrons of 4s orbital gets pair up with electrons to give valence shell configuration $\left[ {{}_{18}Ar} \right]3{d^{10}}4{s^0}$ .
So we gate 3 empty subshell in nickel atom i.e.
Orbital of 4s subshell and 3 subshell of 4p orbital . Now these for empty orbital undergo hybridization to form bonds with 4 C O atoms by accepting then subshell electrons.
Electronic configuration of $Ni{\left( {CO} \right)_4}$ .
Ni :- 28 $3{d^{10}}$ $4{s^2}$ $4{p^6}$

Since 1 orbital of subshell and 3 orbitals of p subshell undergo hybridization . So the hybridization of $Ni{\left( {CO} \right)_4}$ will be $s{p^3}$ its shape is tetrahedral . Moreover we see that there are no unpaired electrons present in $Ni{\left( {CO} \right)_4}$ , so it is a diamagnetic compound .
So the number of unpaired electrons in $Ni{\left( {CO} \right)_4}$ is zero .

So, the correct answer is Option A..

Note:
The legends which do not cause pairing of electrons of central atom are known has weak field legends and legend which cause pairing of ${e^{ - s}}$ are known as strong field legends the arrangement of the legend in increasing order of causing pairing strength are known has spectrochemical series .
${F^ - } < O{H^ - } < {C_2}O_4^{2 - } < {H_2}O < NC{S^ - } < edta < N{H_3} < en < C{N^ - } < CO$