w.o.f have \Delta\_o > P.E? | Chemistry - Crystal Field Theory | SolveeIt

Question

w.o.f have $\Delta_o > P.E$ ?

$Co(H_2O)_6^{+3}$

$Fe(EDTA)^{-2}$

$Pt(H_2O)_6^{+4}$

$Co(NO_3)_6^{-3}$

Answer

b) $Fe(EDTA)^{-2}$ and c) $Pt(H_2O)_6^{+4}$

Explanation

Solution

The condition $\Delta_o > P.E$ signifies a low-spin complex. This occurs when the crystal field splitting energy ( $\Delta_o$ ) is greater than the electron pairing energy (P.E.). In such complexes, electrons fill the lower energy $t_{2g}$ orbitals before occupying the higher energy $e_g$ orbitals, leading to a reduced number of unpaired electrons.

Let's analyze each option:

$Co(H_2O)_6^{+3}$ : Cobalt is in the +3 oxidation state with a $3d^6$ electronic configuration. Water ( $H_2O$ ) is a weak to moderate field ligand. For $Co^{+3}$ ( $3d^6$ ), $\Delta_o$ is typically around 18,000 $cm^{-1}$ , while P.E. is approximately 22,000 $cm^{-1}$ . Thus, $\Delta_o < P.E$ , making it a high-spin complex.
$Fe(EDTA)^{-2}$ : Iron is in the +2 oxidation state with a $3d^6$ electronic configuration. EDTA is a strong field ligand. With a strong ligand, $\Delta_o$ is significantly larger than P.E. for $Fe^{+2}$ ( $3d^6$ ), resulting in a low-spin complex. Therefore, $\Delta_o > P.E$ .
$Pt(H_2O)_6^{+4}$ : Platinum is in the +4 oxidation state with a $5d^6$ electronic configuration. Although water ( $H_2O$ ) is a weak field ligand, platinum is a 3rd-row transition metal in a high oxidation state (+4). These factors lead to a very large $\Delta_o$ , which is much greater than P.E. This complex is low-spin. Therefore, $\Delta_o > P.E$ .
$Co(NO_3)_6^{-3}$ : Cobalt is in the +3 oxidation state with a $3d^6$ electronic configuration. Nitrate ( $NO_3^-$ ) is a weak field ligand. For $Co^{+3}$ ( $3d^6$ ) with a weak ligand, $\Delta_o$ is smaller than P.E., resulting in a high-spin complex. Thus, $\Delta_o < P.E$ .

Therefore, the complexes satisfying $\Delta_o > P.E$ are $Fe(EDTA)^{-2}$ and $Pt(H_2O)_6^{+4}$ .

Question: w.o.f have $\Delta_o > P.E$?...

Solution