Question

Which of the facts about the complex $\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]C{{l}_{3}}$ is wrong?
[A] The complex involves ${{d}^{2}}s{{p}^{3}}$ hybridisation and it’s octahedral in shape.
[B] The complex is paramagnetic.
[C] The complex is an outer orbital complex.
[D] The complex gives white precipitate with silver nitrate solution.

Answer 1

To solve this, firstly find out the hybridisation of the given complex using the fact that the coordination number of the central atom is 6. Remember the shape assigned to coordination number 6 according to VBT. Remember that the presence of unpaired electrons will make the complex paramagnetic.

Complete step by step solution:
To solve this question we will proceed option wise.
Here the complex given to us is $\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]C{{l}_{3}}$.
The first option is the complex involving ${{d}^{2}}s{{p}^{3}}$ hybridisation and it’s octahedral in shape. We can see we have 6 ligands here so the shape is octahedral. For octahedral shape the possible hybridisation according to VBT is ${{d}^{2}}s{{p}^{3}}$ or $s{{p}^{3}}{{d}^{2}}$ .
If the complex is inner orbital then the hybridisation will be ${{d}^{2}}s{{p}^{3}}$ and if it is an outer orbital complex then the hybridisation will be $s{{p}^{3}}{{d}^{2}}$.
We know chromium belongs to the 3d series and its electronic configuration is $\left[ Ar \right]3{{d}^{4}}4{{s}^{2}}$.
For $C{{r}^{3+}}$ the electronic configuration will be $\left[ Ar \right]3{{d}^{3}}$. Therefore, two 3d, one 4s and one 4p orbital will be involved in hybridisation. It will give us a ${{d}^{2}}s{{p}^{3}}$ hybridisation. So, this option is correct.
Then we have that the complex is paramagnetic. For a complex to be paramagnetic it needs to have unpaired electrons. Here, due to the three chlorine atoms we will have 3 unpaired electrons. Therefore, it will be paramagnetic. So, this option is also correct.
Then we have that the complex is an outer orbital complex. We have already discussed above that it is an inner orbital complex. Therefore, this option is incorrect.
And then we have the complex that gives white precipitate with a silver nitrate solution.
With silver nitrate solution, the chlorine atoms present outer the complex sphere will give a white precipitate of silver chloride which is white in colour.
$\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]C{{l}_{3}}+3AgN{{O}_{3}}\to {{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{3-}}+3N{{O}_{3}}^{+}+AgCl\downarrow$
Therefore, this option is correct.
We can understand from the above discussion that the fact that it is an outer orbital complex is incorrect.

Therefore, the correct answer is option [C] the complex is an outer orbital complex.

Note: Here, the ligand given to us is ammonia. Ammonia is known as the ‘notorious ligand’ as it forms both high spin as well as low spin complexes. A substance which is not attracted by a magnetic field due to the absence of unpaired electrons is known as diamagnetic substances. When two electrons are paired with each other in an orbital, their total spin is zero and they repel magnetic fields. For an element / atom / substance to own diamagnetic property, it should not have unpaired electrons. A substance which is attracted by a magnetic field due to the presence of unpaired electrons are known as paramagnetic substances. When electrons are unpaired in an orbital, they have a net spin which is not zero and thus they attract magnetic fields.