`{:(,"Column I (Complex)",,"Column II (Magnetic moment)"),(("A"),K[Cr(H_(2)O)_(2)(C_(2)O_(4))_(2)].3H_(2)O,(p),1.73 BM),(("B"),K_(4)[Mn(CN)_(6)],(q),5.92 BM),(("C"),[Co(NH_(3))_(5)Cl]Cl_(2),(r),3.87BM),(("D"),Cs[FeCl_(4)],(s),"Zero"):}`

Options

A. A-r, B-p, C-s, D-q
B. A-p, B-r, C-q, D-s
C. A-q, B-r, C-p, D-s
D. A-s, B-r, C-q, D-p

Answer

To solve the problem of matching the coordination complexes from Column I with their corresponding magnetic moments from Column II, we will follow these steps for each complex:

### Step 1: Analyze Complex A - K[Cr(H2O)2(C2O4)2].3H2O
1. **Determine the oxidation state of chromium (Cr)**:
- Water (H2O) is a neutral ligand (0 charge).
- Oxalate (C2O4) has a charge of -2. Since there are two oxalate ligands, the total contribution is -4.
- Potassium (K) contributes +1.
- Let the oxidation state of Cr be x.
- The equation becomes: \( x + 0 + (-4) + 1 = 0 \) → \( x - 3 = 0 \) → \( x = +3 \).

2. **Determine the electronic configuration of Cr in +3 state**:
- The atomic number of Cr is 24. In the +3 state, the configuration is: \( [Ar] 3d^3 \).

3. **Count the number of unpaired electrons**:
- In the 3d subshell, there are 3 unpaired electrons.

4. **Calculate the magnetic moment**:
- Use the formula: \( \mu = \sqrt{n(n + 2)} \), where n is the number of unpaired electrons.
- Here, \( n = 3 \):
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \text{ BM}. \]

### Step 2: Analyze Complex B - K4[Mn(CN)6]
1. **Determine the oxidation state of manganese (Mn)**:
- Potassium (K) contributes +4 (4 x +1).
- Cyanide (CN) has a charge of -1, and there are 6 of them, contributing -6.
- Let the oxidation state of Mn be y.
- The equation becomes: \( y + 4 - 6 = 0 \) → \( y - 2 = 0 \) → \( y = +2 \).

2. **Determine the electronic configuration of Mn in +2 state**:
- The atomic number of Mn is 25. In the +2 state, the configuration is: \( [Ar] 3d^5 \).

3. **Count the number of unpaired electrons**:
- In the 3d subshell, there are 5 unpaired electrons (since CN is a strong field ligand and causes pairing).

4. **Calculate the magnetic moment**:
- Here, \( n = 5 \):
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \text{ BM}. \]

### Step 3: Analyze Complex C - [Co(NH3)5Cl]Cl2
1. **Determine the oxidation state of cobalt (Co)**:
- Ammonia (NH3) is neutral (0 charge).
- Chlorine (Cl) contributes -2 (2 x -1).
- Let the oxidation state of Co be z.
- The equation becomes: \( z + 0 - 2 = 0 \) → \( z - 2 = 0 \) → \( z = +3 \).

2. **Determine the electronic configuration of Co in +3 state**:
- The atomic number of Co is 27. In the +3 state, the configuration is: \( [Ar] 3d^6 \).

3. **Count the number of unpaired electrons**:
- NH3 is a strong field ligand, so it causes pairing. Thus, there are 0 unpaired electrons.

4. **Calculate the magnetic moment**:
- Here, \( n = 0 \):
\[ \mu = \sqrt{0(0 + 2)} = 0 \text{ BM}. \]

### Step 4: Analyze Complex D - Cs[FeCl4]
1. **Determine the oxidation state of iron (Fe)**:
- Cesium (Cs) contributes +1.
- Chlorine (Cl) contributes -4 (4 x -1).
- Let the oxidation state of Fe be w.
- The equation becomes: \( w + 1 - 4 = 0 \) → \( w - 3 = 0 \) → \( w = +3 \).

2. **Determine the electronic configuration of Fe in +3 state**:
- The atomic number of Fe is 26. In the +3 state, the configuration is: \( [Ar] 3d^5 \).

3. **Count the number of unpaired electrons**:
- Cl is a weak field ligand, so there are 5 unpaired electrons.

4. **Calculate the magnetic moment**:
- Here, \( n = 5 \):
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \text{ BM}. \]

### Final Matching
- A: 3.87 BM (matches with r)
- B: 5.92 BM (matches with q)
- C: 0 BM (matches with s)
- D: 5.92 BM (matches with p)

### Summary of Matches
- A → r (3.87 BM)
- B → q (5.92 BM)
- C → s (0 BM)
- D → p (5.92 BM)