Correct sequence of CO bond order in given compounds is:
(P) `Fe(CO)_(5)`
(Q) `CO`
(R) `H_(3) B larr CO`
(S) `[Mn(CO)_(5)]^(-)`

Options

A. `P gt R gt S gt Q`
B. `S gt P gt R gt Q`
C. `Q gt S gt P gt R`
D. `R gt Q gt P gt S`

Explanation

To determine the correct sequence of CO bond order in the given compounds (P) `Fe(CO)_(5)`, (Q) `CO`, (R) `H_(3)B ↔ CO`, and (S) `[Mn(CO)_(5)]^(-)`, we need to analyze the bond order of CO in each compound.

### Step 1: Analyze the bond order in CO
The bond order of carbon monoxide (CO) is 3. This is because there are three bonds formed between carbon and oxygen: one sigma bond and two pi bonds.

**Hint:** Remember that the bond order is calculated as the number of bonding electrons minus the number of antibonding electrons divided by two.

### Step 2: Analyze the bond order in `Fe(CO)_(5)`
In `Fe(CO)_(5)`, there is back bonding due to the presence of a transition metal (Fe). The CO acts as a good π-acceptor, allowing for back donation from the metal to the CO ligand. This back bonding reduces the bond order of CO in this complex. The bond order in `Fe(CO)_(5)` is less than 3, but greater than that in `[Mn(CO)_(5)]^(-)`.

**Hint:** Consider how back bonding affects the bond order; it typically decreases the bond order in coordination complexes.

### Step 3: Analyze the bond order in `H_(3)B ↔ CO`
In `H_(3)B ↔ CO`, the carbon in CO donates a lone pair to the boron atom, which has an empty p-orbital. This interaction leads to a slight increase in the bond order of CO, making it slightly greater than 3.

**Hint:** Look for interactions that can increase the bond order, such as donation of lone pairs to electron-deficient atoms.

### Step 4: Analyze the bond order in `[Mn(CO)_(5)]^(-)`
In `[Mn(CO)_(5)]^(-)`, the presence of a negative charge on the manganese increases the back donation to CO. This results in a significant decrease in the bond order of CO, making it the lowest among the given compounds.

**Hint:** The more negative the charge on the central metal ion, the greater the back donation, which lowers the bond order.

### Step 5: Summarize the bond orders
- (Q) `CO`: Bond order = 3
- (R) `H_(3)B ↔ CO`: Bond order > 3
- (P) `Fe(CO)_(5)`: Bond order < 3
- (S) `[Mn(CO)_(5)]^(-)`: Lowest bond order due to significant back donation

### Final Sequence of Bond Order
Based on the analysis:
1. (R) `H_(3)B ↔ CO` (Bond order > 3)
2. (Q) `CO` (Bond order = 3)
3. (P) `Fe(CO)_(5)` (Bond order < 3)
4. (S) `[Mn(CO)_(5)]^(-)` (Lowest bond order)

Thus, the correct sequence of CO bond order in the given compounds is:
**(R) > (Q) > (P) > (S)**