Question

Which of the following complexes is inner orbital as well as low spin complex?
A) ${\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$
B) ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$
C) ${\left[ {Cu{{\left( {CN} \right)}_4}} \right]^{3 - }}$
D) ${\left[ {Mn{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }}$

Answer 1

We know that a spectrochemical arrangement is a rundown of ligands in the request for ligand strength and a rundown of metal particles dependent on oxidation number. In precious stone field hypothesis ligands adjust the distinction in energy between the d orbitals called the ligand field parting boundary for ligands which is essentially reflected in contrasts in shade of comparative metal ligand edifices.
Series of ligands from weak field to strong field is,
${{I^ - } < B{r^ - } < {S^{2 - }} < SC{N^ - } < C{l^ - } < {N_3}^ - < {F^ - } < NC{O^ - } < O{H^ - } < {C_2}{O_4}^ - < {O^{2 - }} < {H_2}O < NC{S^ - }} { < C{H_3}CN < gh < py < N{H_3} < bipy < NO_2^ - < PP{h_3} < C{N^ - } < CO}$

Complete step by step answer:
A complex can be delegated high spin or low spin. When discussing all the atomic calculations, we analyze the gem field parting energy $\left( \Delta \right)$ and the matching energy ( P ). Ordinarily, these two amounts decide if a specific field is low spin or high spin.
The hybridization of the complex ${\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$ is ${d^2}s{p^3}$ and it is inner orbital complex. Therefore, the option A is incorrect.
The hybridization of the complex ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$ ${d^2}s{p^3}$ and it is a low spin and spin paired complex. Therefore, option B is the correct answer.
The hybridization of the complex ${\left[ {Cu{{\left( {CN} \right)}_4}} \right]^{3 - }}$ is $s{p^3}$ and ${\left[ {Mn{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }}$ is $s{p^3}{d^2}$ and the outer orbital and it is high spin, spin free complex. Hence option C and D is incorrect.
At the point when the precious stone field parting energy is more noteworthy than the blending energy, electrons will top off all the lower energy orbitals solitary at that point pair with electrons in these orbitals prior to moving to the higher energy orbitals. Electrons will in general fall in the most minimal conceivable energy state, and since the matching energy is lower than the gem field parting energy, it is all the more vivaciously good for the electrons to combine up and totally top off the low energy orbitals until there is no room left by any means, and really at that time start to fill the high energy orbitals. Then again, when the blending energy is more prominent than the precious stone field energy, the electrons will possess all the orbitals first and afterward pair up, regardless of the energy of the orbitals. In the event that each orbital of a lower energy had one electron, and the orbitals of the next higher energy had none, an electron for this situation would involve the higher energy orbital. This adheres to Hund's standard that says all orbitals must be involved prior to blending starts. Keep in mind, this circumstance possibly happens when the blending energy is more prominent than the gem field energy. These wonders happen in view of the electron's propensity to fall into the most reduced accessible energy state.

So, the correct answer is Option B.

Note: We have to remember that another technique to decide the spin of a complex is to see its field strength and the frequency of shading it assimilates. In the event that the field is solid, it will have not many unpaired electrons and in this manner low spin. On the off chance that the field is feeble, it will have more unpaired electrons and hence high spin. Regarding frequency, a field that retains high energy photons (at the end of the day, low frequency light) has low spin and a field that assimilates low energy photons (high frequency light) has high spin.
Indeed, regardless of whether a complex is high spin or low spin relies upon two primary factors: the gem field parting energy and the blending energy. The electrons will take the easy way out - the way that requires minimal measure of energy. On the off chance that the pairing energy is more prominent than $\delta$ at that point electrons will move to a higher energy orbital on the grounds that it takes less energy. In the event that the blending energy is not exactly $\delta$ at that point the electrons will match up instead of moving separately to a higher energy orbital. Underneath, tips and models are given to help sort out whether a specific particle is high spin or low spin.