Question

The geometry of $Ni{(CO)_4}$ and$\left[ {Ni{{\left( {PP{h_3}} \right)}_2}C{l_2}} \right]$ are:
a.) Both square planar
b.) Tetrahedral and square planar
c.) Both tetrahedral
d.) Square planar and tetrahedral

Answer 1

Hint: To solve this question, we must focus on the connection between coordination number and geometry of a coordination complex. It is important to note that every geometry has a specific coordination number associated with it, but every complex molecule with the specific coordination number can have a choice of several possible geometries.

Complete step by step answer:
Let us take a look at the different geometries associated with coordination numbers.
Complexes having coordination number 2 always have a linear geometry.
Coordination complex molecules having coordination number 3 either have a trigonal planar geometry or a trigonal pyramidal geometry. If there are certain distortions, these molecules become T-shaped.
For coordination number 4, two different geometries are possible. A molecule may have a tetrahedral geometry or a square planar one.
We will now find the coordination number of the molecules assigned to us:
$Ni{(CO)_4}$:
We find that the oxidation number of Nickel here is 0 since CO has 0 charge as well.
If we write the electronic configuration,
Ground state Ni= $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^8}4{s^2}$
Upon excitation, $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}$
We can place 8 electrons in 3d orbital and 2 electrons in 4s.
But we know that CO is a strong ligand, hence electrons are paired up in 3d orbital and 4s remains empty giving the molecule $s{p^3}$hybridization and a tetrahedral geometry.
$\left[ {Ni{{\left( {PP{h_3}} \right)}_2}C{l_2}} \right]$:
The oxidation number of Ni here is +2 since $PP{h_3}$ is neutral and chlorine is -1.
$N{i^{ + 2}} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^8}$
We will place 8 electrons in 3d orbital whereas 4s and 4p remains empty.
Without even pairing up, the 4 subshell remains empty giving it a $s{p^3}$hybridization and a tetrahedral geometry.

Hence, the correct answer is Option (C) Both Tetrahedral.

Note:
The coordination compounds have distinct geometric structures. Only on altering the relative position of atoms we find different geometries in space with the same molecular formula.