Name: Reaction Between Ferrous Ion and Potassium Dichromate
Uploaded: 2019-07-19
Duration: 4 min 21 s
Description: This video explains the excellent oxidising agent, potassium dichromate. One mole of potassium dichromate can oxidise six moles of ferrous sulphate molecules...

Question

Predict the nature of the ligands in the following complexes:

(a)

{[Fe {(H_{2} O)}_{6}]}^{2 +},

(b)

{[Fe (CN)_{6}]}^{4 -}

and

(c)

{[Fe (CN)_{6}]}^{3 -} .

How many unpaired electrons are there in each complex, and what would you expect their magnetic moments to be?

Accepted Answer

$(a)$ , here iron metal is in $+ 2$ oxidation state. The ${Fe}^{2 +}$ has the configuration of ${[Fe (CN)_{6}]}^{4 -}$ . $H_{2} O$ is a weak field ligand with ${Fe}^{2 +}$ ; So, it forms a high spin complex. ${[Fe {(H_{2} O)}_{6}]}^{2 +}$ is a high spin complex. If a complex is low spin, then there is pairing occur first in $t_{2 g}$ and then electron field after pairing in $e_{g}$ . If the complex is high spin, then we fill the one electron in each orbitals after that pairing will do.

$Magnetic moment = \sqrt{n (n + 2)} BM Where n = number of unpaired electrons So, n = 4 in case of {[Fe {(H_{2} O)}_{6}]}^{2 +} Magnetic moment = \sqrt{4 (4 + 2)} = \sqrt{6 \times 4} = \sqrt{24} = 4.89 BM$

${[Fe (CN)_{6}]}^{4 -}$ , here iron metal is in $+ 2$ oxidation state. The ${Fe}^{2 +}$ has the configuration of $d^{6}$ . $CN$ is a strong field ligand with ${Fe}^{2 +}$ ; so, it forms a low spin complex. ${[Fe (CN)_{6}]}^{4 -}$ is a low spin complex. If a complex is low spin then there is pairing occur first in $t_{2 g}$ and then electron field after pairing in $e_{g}$ . If the complex is high spin, then we fill the one electron in each orbitals after that pairing will do.

$Magnetic moment = \sqrt{n (n + 2)} BM Where n = number of unpaired electrons So, n = 0 in case of {[Fe {(CN)}_{6}]}^{4 -} Magnetic moment = \sqrt{0} = \sqrt{0 \times 2} = \sqrt{0} = 0 BM$

${[Fe (CN)_{6}]}^{3 -}$ , here iron metal is in $+ 3$ oxidation state. The ${Fe}^{3 +}$ has the configuration of $d^{5}$ . $CN$ is a strong field ligand with ${Fe}^{3 +}$ ; so, it forms a low spin complex. ${[Fe (CN)_{6}]}^{3 -}$ is a low spin complex. If a complex is low spin, then there is pairing occur first in $t_{2 g}$ and then electron field after pairing in $e_{g}$ . If the complex is high spin, then we fill the one electron in each orbitals after that pairing will do.

Splitting in ${[Fe (CN)_{6}]}^{3 -}$ :

We can see there is $1$ unpaired electron present.

$Magnetic moment = \sqrt{n (n + 2)} BM Where n = number of unpaired electrons So, n = 1 in case of {[Fe {(CN)}_{6}]}^{3 -} Magnetic moment = \sqrt{1 (1 + 2)} = \sqrt{1 \times 3} = \sqrt{3} = 1.73 BM$

Predict the nature of the ligands in the following complexes:
$(a)$ ${[Fe {(H_{2} O)}_{6}]}^{2 +},$ $(b)$ ${[Fe (CN)_{6}]}^{4 -}$ and $(c)$ ${[Fe (CN)_{6}]}^{3 -} .$ How many unpaired electrons are there in each complex, and what would you expect their magnetic moments to be?

Important Questions on Group 8 - the Iron Group

Learn and Prepare for any exam you want !

Predict the nature of the ligands in the following complexes: (a) FeH2O62+, (b) Fe(CN)64- and (c) Fe(CN)63-. How many unpaired electrons are there in each complex, and what would you expect their magnetic moments to be?

Important Questions on Group 8 - the Iron Group

Learn and Prepare for any exam you want !

Predict the nature of the ligands in the following complexes:
$(a)$ ${[Fe {(H_{2} O)}_{6}]}^{2 +},$ $(b)$ ${[Fe (CN)_{6}]}^{4 -}$ and $(c)$ ${[Fe (CN)_{6}]}^{3 -} .$ How many unpaired electrons are there in each complex, and what would you expect their magnetic moments to be?