Coordination compounds plays many important roles in animals and plants. The are essential in the storage and transport of oxygen as electrons transfer agents as catalysts and in photosynthesis Wide range of application in daily life takes place through formation of complexes Photographic fixing qualitative and quantitative analysis purification of water metallurgical extraction are some specific worth mentioning
Arrange of the following in order of decreasing number of unpaired electrons
(I) `[Fe(H_(2)O))_(6)]^(2+)`
(II) `[Fe(CN)_(6)]^(3-)`
(III) `[Fe(CN)_(6)]^(4-)`
(IV) `[fe(H_(2)O)_(6)]^(3+)`
(a) IV,I,II,III
(b) `I,II,III,IV`
(c) `III,II,IIV`
(d) II,III,I,IV` .

Options

A. IV,I,II,III
B. I,II,III,IV
C. III,II,I,IV
D. II,III,I,IV

Answer

To solve the problem of arranging the given coordination compounds in order of decreasing number of unpaired electrons, we need to analyze each complex and determine the oxidation state of iron in each case, followed by the electronic configuration and the number of unpaired electrons.

### Step-by-Step Solution:

1. **Identify the complexes and their oxidation states:**
- (I) \([Fe(H_2O)_6]^{2+}\)
- (II) \([Fe(CN)_6]^{3-}\)
- (III) \([Fe(CN)_6]^{4-}\)
- (IV) \([Fe(H_2O)_6]^{3+}\)

2. **Calculate the oxidation state of iron in each complex:**
- For (I): \[Fe^{2+}\] → Oxidation state = +2
- For (II): \[Fe^{3+}\] → Oxidation state = +3
- For (III): \[Fe^{4+}\] → Oxidation state = +2
- For (IV): \[Fe^{3+}\] → Oxidation state = +3

3. **Determine the electronic configuration of iron for each oxidation state:**
- Iron has an atomic number of 26, with the ground state configuration: \[ [Ar] 3d^6 4s^2 \]
- For (I) \([Fe(H_2O)_6]^{2+}\):
- Configuration: \[3d^6\] (2 electrons removed from 4s)
- For (II) \([Fe(CN)_6]^{3-}\):
- Configuration: \[3d^5\] (3 electrons removed)
- For (III) \([Fe(CN)_6]^{4-}\):
- Configuration: \[3d^6\] (2 electrons removed)
- For (IV) \([Fe(H_2O)_6]^{3+}\):
- Configuration: \[3d^5\] (3 electrons removed)

4. **Analyze the ligand field strength:**
- Water (\(H_2O\)) is a weak field ligand, while cyanide (\(CN^-\)) is a strong field ligand.
- This affects the pairing of electrons in the d-orbitals.

5. **Determine the number of unpaired electrons:**
- For (I) \([Fe(H_2O)_6]^{2+}\) (3d^6):
- Weak field ligand → 4 unpaired electrons (configuration: ↑↓ ↑ ↑ ↑ ↑)
- For (II) \([Fe(CN)_6]^{3-}\) (3d^5):
- Strong field ligand → 1 unpaired electron (configuration: ↑ ↑ ↑ ↑ ↑)
- For (III) \([Fe(CN)_6]^{4-}\) (3d^6):
- Strong field ligand → 0 unpaired electrons (configuration: ↑↓ ↑↓ ↑ ↑)
- For (IV) \([Fe(H_2O)_6]^{3+}\) (3d^5):
- Weak field ligand → 5 unpaired electrons (configuration: ↑ ↑ ↑ ↑ ↑)

6. **Arrange the complexes in order of decreasing number of unpaired electrons:**
- (IV) → 5 unpaired electrons
- (I) → 4 unpaired electrons
- (II) → 1 unpaired electron
- (III) → 0 unpaired electrons

### Final Arrangement:
The order of decreasing number of unpaired electrons is:
**(IV), (I), (II), (III)**

### Answer:
(a) IV, I, II, III