Question

Why is \[{[Cr{(N{H_3})_6}]^{3 + }}\] a high spin complex?

Answer 1

Hint: A high spin or a low spin complex can be decided by the number of unpaired electrons that are present on the central metal ion. Write down the electronic configuration of the central metal ion keeping its oxidation state in mind and then study the impact of the ligands on it.

Complete answer:
Transition metals belonging to the d-block of the modern periodic table tend to form a wide variety of coordination complexes due to their ability to show variable oxidation states.
A coordination complex can have different geometries depending upon the number of ligands associated with the central metal ion. A complex containing a total of six donor ligands along the axes around the central metal ion is said to exist in an octahedral geometry.
The complex \[{[Cr{(N{H_3})_6}]^{3 + }}\] consists of chromium as the central metal ion and six ammonia ligands are attached to it. Since ammonia is a neutral molecule, each ligand contributes two electrons through a coordinate covalent bond. The charge on chromium is the same as that of the charge on the coordination sphere as ammonia molecules are covalent.
The electronic configuration of chromium metal in its stable ground state is: \[4{s^1}3{d^5}\]
The electronic configuration of \[C{r^{3 + }}\] ion in the complex is: \[3{d^3}\]
A high spin complex is the one which contains more number of unpaired electron on the central metal ion. A low spin complex is the one consisting of strong field ligands that forces the pairing of electrons on the central metal ion and reduces the total spin of the complex.
If six ligands need to be accommodated, the hybridization should be \[{d^2}s{p^3}\] or \[s{p^3}{d^2}\]. Since, \[C{r^{3 + }}\] ion has only three unpaired electrons and the d-orbital has two vacant position, the complex can have a hybridization of \[{d^2}s{p^3}\].
\[{\text{total spin }}s = \dfrac{1}{2} \times ({\text{number of electrons)}}\]
\[{\text{total spin }}s = \dfrac{1}{2} \times 3 = \dfrac{3}{2}\]
Hence, the complex \[{[Cr{(N{H_3})_6}]^{3 + }}\] is a high spin complex as there are no paired electrons on the central metal ion.

Note:
Whether a complex is high spin or low spin is mainly decided by the field strength of the ligands involved. But there are only three unpaired electrons present on chromium ions, hence the strength of the ligand is insignificant in making the complex a high spin complex.