Question:
\([\mathrm{Fe(H_2O)_6}]^{2+}\) is paramagnetic whereas \([\mathrm{Fe(CO)_5}]\) is diamagnetic. Justify the statement.
[Atomic number of Fe = 26]
\([\mathrm{Fe(H_2O)_6}]^{2+}\) is paramagnetic whereas \([\mathrm{Fe(CO)_5}]\) is diamagnetic. Justify the statement.
[Atomic number of Fe = 26]
Show Hint
Always determine the oxidation state of the metal ion first. Then identify whether the ligand is a weak-field ligand (\(\mathrm{H_2O}, \mathrm{F^-}, \mathrm{Cl^-}\)) or a strong-field ligand (\(\mathrm{CO}, \mathrm{CN^-}, \mathrm{NH_3}\)). Weak-field ligands generally produce high-spin (paramagnetic) complexes, whereas strong-field ligands produce low-spin (often diamagnetic) complexes.
Updated On: Jul 22, 2026
Show Solution
Verified By Collegedunia
Solution and Explanation
Concept:
The magnetic nature of a coordination compound depends upon the number of unpaired electrons present in the central metal ion.
• A compound containing one or more unpaired electrons is paramagnetic.
• A compound having all electrons paired is diamagnetic.
• Strong field ligands cause pairing of electrons, whereas weak field ligands generally do not.
Step 1: Analysis of \([\mathrm{Fe(H_2O)_6}]^{2+}\).
• Oxidation state of Fe = \(+2\).
• Electronic configuration of Fe: \[ [Ar]\,3d^6\,4s^2 \]
• Electronic configuration of \(Fe^{2+}\): \[ [Ar]\,3d^6 \]
• Water (\(\mathrm{H_2O}\)) is a weak field ligand.
• It cannot force pairing of the \(3d\) electrons.
• Therefore, the complex is a high-spin complex having four unpaired electrons.
• Hence, \[ [\mathrm{Fe(H_2O)_6}]^{2+} \] is paramagnetic.
Step 2: Analysis of \([\mathrm{Fe(CO)_5}]\).
• Carbon monoxide (\(\mathrm{CO}\)) is a neutral ligand.
• Therefore, the oxidation state of Fe is \(0\).
• Electronic configuration of Fe: \[ [Ar]\,3d^6\,4s^2 \]
• Carbon monoxide is a strong field ligand.
• It causes pairing of the electrons in the \(3d\) orbitals.
• The complex undergoes \(dsp^3\) hybridisation with all electrons paired.
• Therefore, \[ [\mathrm{Fe(CO)_5}] \] contains no unpaired electrons and is diamagnetic.
Final Answer: \[ [\mathrm{Fe(H_2O)_6}]^{2+} \] contains four unpaired electrons because \(\mathrm{H_2O}\) is a weak field ligand. Hence, it is paramagnetic. \[ [\mathrm{Fe(CO)_5}] \] contains no unpaired electrons because \(\mathrm{CO}\) is a strong field ligand that causes electron pairing. Hence, it is diamagnetic.
• A compound containing one or more unpaired electrons is paramagnetic.
• A compound having all electrons paired is diamagnetic.
• Strong field ligands cause pairing of electrons, whereas weak field ligands generally do not.
Step 1: Analysis of \([\mathrm{Fe(H_2O)_6}]^{2+}\).
• Oxidation state of Fe = \(+2\).
• Electronic configuration of Fe: \[ [Ar]\,3d^6\,4s^2 \]
• Electronic configuration of \(Fe^{2+}\): \[ [Ar]\,3d^6 \]
• Water (\(\mathrm{H_2O}\)) is a weak field ligand.
• It cannot force pairing of the \(3d\) electrons.
• Therefore, the complex is a high-spin complex having four unpaired electrons.
• Hence, \[ [\mathrm{Fe(H_2O)_6}]^{2+} \] is paramagnetic.
Step 2: Analysis of \([\mathrm{Fe(CO)_5}]\).
• Carbon monoxide (\(\mathrm{CO}\)) is a neutral ligand.
• Therefore, the oxidation state of Fe is \(0\).
• Electronic configuration of Fe: \[ [Ar]\,3d^6\,4s^2 \]
• Carbon monoxide is a strong field ligand.
• It causes pairing of the electrons in the \(3d\) orbitals.
• The complex undergoes \(dsp^3\) hybridisation with all electrons paired.
• Therefore, \[ [\mathrm{Fe(CO)_5}] \] contains no unpaired electrons and is diamagnetic.
Final Answer: \[ [\mathrm{Fe(H_2O)_6}]^{2+} \] contains four unpaired electrons because \(\mathrm{H_2O}\) is a weak field ligand. Hence, it is paramagnetic. \[ [\mathrm{Fe(CO)_5}] \] contains no unpaired electrons because \(\mathrm{CO}\) is a strong field ligand that causes electron pairing. Hence, it is diamagnetic.
Was this answer helpful?
0
0
Top CBSE CLASS XII Chemistry Questions
- ${AgNO_3}$ does not decompose where :
- Azeotropic mixture of water and HCl boils at 381.5 K. By distilling the mixture it is possible to obtain
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylanilineAccount for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
View More Questions
Top CBSE CLASS XII Bonding in Coordination Compounds Questions
- Answer the following questions about the complexes $\mathrm{[NiCl_4]^{2-}}$ and $\mathrm{[Ni(CN)_4]^{2-}}$ :
(i) Write the hybridization involved in each case.
(ii) Which of them is the inner orbital complex and which one is the outer orbital complex ?
(iii) Compare their magnetic behaviour. [Atomic number : Ni = 28] - Why are low spin tetrahedral complexes rarely formed ?
- Answer the following questions about the complexes \[ [NiCl_4]^{2-} \quad \text{and} \quad [Ni(CN)_4]^{2-} \] (i) Write the hybridization involved in each case. (ii) Which of them is the inner orbital complex and which one is the outer orbital complex ? (iii) Compare their magnetic behaviour. (Atomic Number of Ni = 28)
- Write the oxidation state and hybridisation of the central metal in the following complex : \[ [Fe(H_2O)_6]^{3+} \] [Atomic number of Fe = 26]
- Answer the following questions about the complexes \([NiCl_4]^{2-}\) and \([Ni(CN)_4]^{2-}\): \[ (i)\;\text{Write the hybridization involved in each case.} \] \[ (ii)\;\text{Which of them is the inner orbital complex and which one is the outer orbital complex?} \] \[ (iii)\;\text{Compare their magnetic behaviour.} \] \[ \text{Atomic number: }Ni=28 \]
View More Questions
Top CBSE CLASS XII Questions
- 2, b, c are in A.P. and the range of determinant $\begin{vmatrix}1&1&1\\ 2&b&c\\ 4&b^{2}&c^{2}\end{vmatrix}$ is [2, 16]. Then find range of c is
- A boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. He runs to the other, end. The distance through which the centre of mass of the boat boy system moves is
- A capillary tube of radius r is dipped inside a large vessel of water. The mass of water raised above water level is M. If the radius of capillary is doubled, the mass of water inside capillary will be
- A circular disc is rotating about its own axis. An external opposing torque 0.02 Nm is applied on the disc by which it comes rest in 5 seconds. The initial angular momentum of disc is
- A circular disc is rotating about its own axis at uniform angular velocity \(\omega.\) The disc is subjected to uniform angular retardation by which its angular velocity is decreased to \(\frac {\omega}{2}\) during 120 rotations. The number of rotations further made by it before coming to rest is
View More Questions