Question

The species in which $\left( n-1 \right){{d}_{{{x}^{2}}-{{y}^{2}}}}$orbital take part in hybridization?
A.$PC{{l}_{5}}$
B.$~{{\left[ NiC{{l}_{4}} \right]}^{2-}}$
C.${{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{3+}}$
D.$\left[ Ni{{\left( CO \right)}_{4}} \right]$
E.B and C

Answer 1

We should recall the concept of hybridisation. We should know that choice of ‘d’ orbital for a particular type of hybridization depends on the spatial orientation of the orbital and the geometry of the molecule or ion which the hybridized orbitals are expected to form.

Complete answer:
To determine the hybridization of any compound, we must know the structure of that compound. Hybridization is defined as a concept of mixing two atomic orbitals having the same energy levels to give new degenerate orbitals. It will be interesting for us to know that the idea that atoms form covalent bonds by sharing pairs of electrons was first proposed by G. N. Lewis in $1902.$
It was in $1927$ that Walter Heitler and Fritz London showed how the sharing of pairs of electrons holds a covalent molecule together. The Heitler-London model of covalent bonds was the basis of the valence-bond theory. The last major step in the evolution of this theory was the suggestion by Linus Pauling that atomic orbitals mix to form hybrid orbitals. Either finds out the hybridization of each species and then reaches the answer or it can also be told by looking at the structure and then number of bond pairs and lone pairs or any charge present. This could lead to the answer very easily.
Here to answer similar types of questions, the hybridization concept should be well known which can be calculated by the formula as well as by looking at the structure and then number of bond pairs and lone pairs. So, any method could be used to find out the hybridization of any atom or species by this. $PC{{l}_{5}}$ is $s{{p}_{3}}d$ since its trigonal bipyramidal geometry, here the d-orbital must be $d{{z}^{2}}.$ ${{\left[ NiC{{l}_{4}} \right]}^{-2}}$ is $s{{p}^{3}}$ and it has no orbitals but ${{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{3+}}$ is $s{{p}^{3}}{{d}^{2}}$ which is square pyramidal, here d-orbitals are ${{d}_{{{x}^{2}}-{{y}^{2}}}}$ , $d{{z}^{2}}.$
Hence, the correct option is C.

Note:
Remember that when a lone pair is present, then the molecular geometry will be the three-dimensional arrangement of the atoms without the lone pair. If there are no lone pairs, only bond pairs then the number of bond pairs can be equal to the hybridization number.