Question

The hybridization involved in complex $ {[Ni(CN)4]^{2-}}$ is (At.No. Ni = 28)

• A. $d^2sp^2$
• B. $d^2sp^3$
• C. $dsp^2$
• D. $sp^3$

Answer 1

The Correct Option is C

$ {[Ni(CN)_4]^{2-}}$ oxidation state of Ni is +2 x - 4 = 2 x = +2

Answer 2

The atomic number of Ni is given as 28. So, as we can see, there will be 28 electrons located in the Ni atom.

Here it’s given that the charge on the complex [Ni(CN)₄]^2- is (-2). Each cyanide ion has a charge of (-1). So, we can say that Ni atoms are in a +2 oxidation state. Now, we will first write the ground state electronic configuration of Ni²⁺.

Ground state electronic configuration of Ni²⁺ : [Ar]3d⁸

   Now, the excited state electronic configuration of Ni2+ can be given as[Ar]3d8 but the arrangement of electronics is as below.

Here, we’ll notice  1 d-orbital, 1 s-orbital, and 3 p-orbitals become vacant. The 8 electrons present in the d-orbitals are coupled together.

- So, the cyanide ligands produce the pair of electrons to the vacant orbitals of the nickel atom.

- We can see that there are 4 cyanide ligands present in that complex. So, four produce two electrons each. So, a total of four orbitals will get the electron pairs by each.

- First the orbital with the lower energy level gets] the electrons. So, electron pairs are produced in one d-orbital, 1 s-orbital, and 2 p-orbitals and so on, 4 cyanide ligands bind to the metal.

- So, we can say that the hybridization of the complex element is dsp2

 

So, the correct answer is (C).