Which of the following species is/are coloured ?

Options

A. `K_(2)Cr_(2)O_(7)`
B. `CrO_(2)Cl_(2)`
C. `Na[Ag(CN)_(2)]`
D. `Cl_(2)(g)`

Explanation

To determine which of the following species are colored, we need to analyze the oxidation states and electronic configurations of the central metal atoms in each compound. The color of coordination compounds is typically due to d-d transitions or charge transfer processes. Let's evaluate each species step by step.

### Step 1: Analyze K2Cr2O7 (Potassium dichromate)
1. **Identify the oxidation state of chromium**:
- The formula is K2Cr2O7.
- Potassium (K) has a +1 charge, and there are 2 potassium ions, contributing +2.
- Oxygen (O) has a -2 charge, and there are 7 oxygen atoms, contributing -14.
- Let the oxidation state of chromium be \( x \). The equation becomes:
\[ 2x + 2 - 14 = 0 \implies 2x - 12 = 0 \implies x = +6 \]
- Thus, the oxidation state of chromium in K2Cr2O7 is +6.

2. **Determine the electronic configuration**:
- Chromium in +6 oxidation state has the electronic configuration of 3d0 (since it loses all d electrons).

3. **Check for color**:
- Since there are no d-electrons, there cannot be d-d transitions. However, the presence of O2- (a pi donor ligand) allows for ligand-to-metal charge transfer (LMCT), which can impart color. Thus, K2Cr2O7 is colored.

### Step 2: Analyze CrO2Cl2 (Chromium oxychloride)
1. **Identify the oxidation state of chromium**:
- The formula is CrO2Cl2.
- Oxygen contributes -4 (2 oxygen atoms at -2 each).
- Chlorine contributes -2 (2 chlorine atoms at -1 each).
- Let the oxidation state of chromium be \( x \). The equation becomes:
\[ x - 4 - 2 = 0 \implies x = +6 \]
- Thus, the oxidation state of chromium in CrO2Cl2 is +6.

2. **Determine the electronic configuration**:
- Similar to K2Cr2O7, chromium in +6 oxidation state has the electronic configuration of 3d0.

3. **Check for color**:
- Again, there are no d-d transitions, but LMCT can occur due to the presence of pi donor ligands (O2-). Therefore, CrO2Cl2 is also colored.

### Step 3: Analyze Na[Ag(CN)2] (Sodium silver cyanide)
1. **Identify the oxidation state of silver**:
- The formula is Na[Ag(CN)2].
- Sodium (Na) has a +1 charge.
- Cyanide (CN) has a -1 charge, and there are 2 cyanide ions, contributing -2.
- Let the oxidation state of silver be \( x \). The equation becomes:
\[ x - 2 + 1 = 0 \implies x = +1 \]
- Thus, the oxidation state of silver in Na[Ag(CN)2] is +1.

2. **Determine the electronic configuration**:
- Silver in +1 oxidation state has the electronic configuration of 4d10 (fully filled d-orbitals).

3. **Check for color**:
- Since the d-orbitals are fully filled, there are no d-d transitions possible. Therefore, Na[Ag(CN)2] is colorless.

### Step 4: Analyze Cl2 (Chlorine gas)
1. **Identify the electronic transitions**:
- Chlorine gas (Cl2) is a diatomic molecule.
- It has electronic transitions between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO).

2. **Check for color**:
- The electronic transitions in Cl2 lead to a yellow-green color due to the energy gap between HOMO and LUMO. Therefore, Cl2 is colored.

### Conclusion:
- **Colored species**: K2Cr2O7, CrO2Cl2, Cl2
- **Colorless species**: Na[Ag(CN)2]