JEE MAIN - Chemistry (2023 - 10th April Evening Shift - No. 6)
The correct order of the number of unpaired electrons in the given complexes is
A. $$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$$
B. $$\left[\mathrm{Fe} \mathrm{F}_{6}\right]^{3-}$$
C. $$\left[\mathrm{CoF}_{6}\right]^{3-}$$
D. $$\left.[\mathrm{Cr} \text { (oxalate})_{3}\right]^{3-}$$
E. $$\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$$
Choose the correct answer from the options given below:
Explanation
The number of unpaired electrons in a complex can be determined by the oxidation state of the central metal atom and the ligand field splitting energy. In general, a complex with a high-spin configuration will have more unpaired electrons than a complex with a low-spin configuration.
In the given complexes, the oxidation states of the central metal atoms are:
- Fe(II) in [Fe(CN)6]3-
- Fe(III) in [FeF6]3-
- Co(III) in [CoF6]3-
- Cr(III) in [Cr(ox)3]3-
- Ni(0) in [Ni(CO)4]
The ligand field splitting energy of a complex depends on the type of ligand. In general, ligands that are strong field ligands will split the d orbitals more than ligands that are weak field ligands.
The following table shows the number of unpaired electrons in each complex :
| Complex | Oxidation state | Ligand field splitting energy | Number of unpaired electrons |
|---|---|---|---|
| [Fe(CN)6]3- | Fe(II) | Low-spin | 1 |
| [FeF6]3- | Fe(III) | High-spin | 5 |
| [CoF6]3- | Co(III) | High-spin | 4 |
| [Cr(ox)3]3- | Cr(III) | High-spin | 3 |
| [Ni(CO)4] | Ni(0) | None | 0 |
Therefore, the correct order of the number of unpaired electrons in the given complexes is :
E < A < D < C < B
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