Question

Red precipitate is obtained when ethanol solution of dimethylglyoxime is added to ammonical ${\text{Ni(II)}}$. Which of the following statements is not true?
A) Red complex has a tetrahedral geometry
B) Complex has symmetrical H-bonding
C) Red complex has a square planar geometry
D) Dimethylglyoxime functions as bidentate ligand

Answer 1

Recall the structure of dimethylglyoxime (DMG). It contains two nitrogen atoms and both can donate its electron density to the metal ion. Also, DMG is a strong field ligand therefore, pairing of electrons will occur in the nickel-DMG complex. After pairing, find the hybridisation of nickel-DMG complex and then its geometry.

Complete step by step solution:
We are given that a red precipitate is obtained when ethanol solution of dimethylglyoxime (DMG) is added to ammonical ${\text{Ni(II)}}$. This chemical reaction can be represented as shown below:

Now, structure of DMG is:

DMG contains two nitrogen atoms and both N will donate electron pairs to the central metal ion, hence, DMG is a bidentate ligand. Thus, the statement given in option D is true.

-Now, as you can see in the above nickel-DMG complex that there is symmetrical hydrogen bonding present in the complex. Thus, statement B is also true.
-As you know, Ni is ${d^8}$ system, therefore:
-Valence electronic configuration of $Ni = 3{d^8}4{s^2}$
-Valence electronic configuration of $N{i^{2 + }} = 3{d^8}$
-Now, we know that dimethylglyoxime (DMG) is a strong field ligand, therefore when nickel is present in complex i.e., $[Ni{(DMG)_2}]$, pairing of d-electrons of $N{i^{2 + }}$ will occur. Thus, we can show this as:

Hence, hybridisation of the $[Ni{(DMG)_2}]$ complex (red complex) will be $ds{p^2}$ and geometry will be square planar. Thus, statement C is also true. On concluding, we find that the statement given in option A i.e., red complex has a tetrahedral geometry, is not true.

Thus, option A is the answer.

Note: When coordination number of a complex is 4 i.e., four ligands are linked to central metal ions, then the geometry will be either tetrahedral or square planar. For tetrahedral geometry, hybridisation is $s{p^3}$ whereas for square planar geometry, hybridisation is $ds{p^2}$.