Give step-by-step solution with explanation and final answer:
What is the correct order of C-O stretching frequency of the following compounds:
[V(CO),], Cr(CO),, [Co(CO),], [Mn(CO),]
13-[Mn(CO)][

Question

Give step-by-step solution with explanation and final answer: 
What is the correct order of C-O stretching frequency of the following compounds: 
[V(CO),], Cr(CO),, [Co(CO),], [Mn(CO),] 
13-[Mn(CO)]>[Co(CO)] > Cr(CO), >[V(CO)。] 
[Mn(CO)]>[Co(CO)] > Cr(CO), > [V(CO),] 
[V(CO),]>Cr(CO), > [Co(CO),] > [Mn(CO),]

VIPSolutions · Accepted Answer

Let's solve the question step by step:

### **Question Recap**
You are given four metal carbonyl complexes:
- [V(CO)$_6$]$^{-}$
- Cr(CO)$_6$
- [Co(CO)$_4$]$^{-}$
- [Mn(CO)$_5$]$^{-}$

You are to **order them according to their C–O stretching frequency (ν$_{CO}$) in IR**, from highest to lowest.

### **Step 1: Understanding the Trend**

**The C–O stretching frequency in metal carbonyls depends on the electron density at the metal:**
- **More electron density on metal → more back-donation to CO π*\* orbitals → weaker C–O bond → lower ν$_{CO}$**
- **Less electron density → less back-donation → stronger C–O bond → higher ν$_{CO}$**

**The order is usually:**
- **More negative charge on the complex → more electron density on metal → lower ν$_{CO}$.**
- **Higher oxidation state of metal (more positive) → less electron density on metal → higher ν$_{CO}$.**

### **Step 2: Assigning Electron Count and Charge**

1. **[V(CO)$_6$]$^{-}$:**
   - Charge: -1
   - V is in -I oxidation state (since CO is neutral, V must be -1).
   - More electron-rich.

2. **Cr(CO)$_6$:**
   - Charge: 0
   - Cr is in 0 oxidation state.
   - Neutral.

3. **[Co(CO)$_4$]$^{-}$:**
   - Charge: -1
   - Co is in -I oxidation state.
   - Electron-rich.

4. **[Mn(CO)$_5$]$^{-}$:**
   - Charge: -1
   - Mn is in -I oxidation state.
   - Electron-rich.

### **Step 3: Comparing Electron Richness**

- **[V(CO)$_6$]$^{-}$**, **[Mn(CO)$_5$]$^{-}$**, **[Co(CO)$_4$]$^{-}$** all have a negative charge.
- But, **the more d-electrons on the metal center, the greater the back-donation**:
    - V$^-$: d$^{6}$
    - Cr: d$^{6}$
    - Mn$^-$: d$^{8}$
    - Co$^-$: d$^{10}$

#### **Let's check electron counts:**
- [V(CO)$_6$]$^{-}$: V is group 5, +1e (from - charge): 6e$^-$
- Cr(CO)$_6$: Cr is group 6: 6e$^-$
- [Mn(CO)$_5$]$^{-}$: Mn is group 7, +1e: 8e$^-$
- [Co(CO)$_4$]$^{-}$: Co is group 9, +1e: 10e$^-$

But the important thing is **how much back-donation each metal can do**. The more electrons on the metal, the more back-donation.

### **Step 4: Predict the Trend**

- **[Co(CO)$_4$]$^{-}$: d$^{10}$, most electron-rich → most back-donation → lowest ν$_{CO}$**
- **[Mn(CO)$_5$]$^{-}$: d$^{8}$, next most electron-rich**
- **[V(CO)$_6$]$^{-}$: d$^{6}$, less electron-rich**
- **Cr(CO)$_6$: d$^{6}$, neutral, least electron-rich → least back-donation → highest ν$_{CO}$**

But, actually, **[V(CO)$_6$]$^{-}$** and **Cr(CO)$_6$** both have d$^{6}$ configuration, but [V(CO)$_6$]$^{-}$ has a negative charge, so slightly more back-donation (slightly lower ν$_{CO}$).

So the order from highest to lowest C–O stretching frequency is:

1. **Cr(CO)$_6$** (least electron-rich, highest ν$_{CO}$)
2. **[V(CO)$_6$]$^{-}$**
3. **[Mn(CO)$_5$]$^{-}$**
4. **[Co(CO)$_4$]$^{-}$** (most electron-rich, lowest ν$_{CO}$)

### **Step 5: Matching to Options**

But the options use the symbol [Mn(CO)] for [Mn(CO)$_5$]$^{-}$, etc. Let's match:

#### The option:  
**[V(CO)$_6$]$^{-}$ > Cr(CO)$_6$ > [Co(CO)$_4$]$^{-}$ > [Mn(CO)$_5$]$^{-}$**
- This is not correct, because [Co(CO)$_4$]$^{-}$ should be lowest.

#### The option:  
**[Mn(CO)$_5$]$^{-}$ > [Co(CO)$_4$]$^{-}$ > Cr(CO)$_6$ > [V(CO)$_6$]$^{-}$**
- This is not correct, as [Mn(CO)$_5$]$^{-}$ should not be the highest.

#### The option:  
**[V(CO)$_6$]$^{-}$ > Cr(CO)$_6$ > [Co(CO)$_4$]$^{-}$ > [Mn(CO)$_5$]$^{-}$**
- Again, order is not matching with our conclusion.

#### The typical correct order (from literature and textbooks):
**Cr(CO)$_6$ > [V(CO)$_6$]$^{-}$ > [Mn(CO)$_5$]$^{-}$ > [Co(CO)$_4$]$^{-}$**

But none of the options matches this order exactly as written.

### **Step 6: Final Answer**

**The correct order of C–O stretching frequency from highest to lowest is:**

$$
\boxed{\text{Cr(CO)}_6 > [\text{V(CO)}_6]^− > [\text{Mn(CO)}_5]^− > [\text{Co(CO)}_4]^−}
$$

**Explanation:**  
- Higher electron density at the metal increases π-back-donation to CO, which weakens the C–O bond and lowers the C–O stretching frequency.
- [Co(CO)$_4$]$^{-}$ is most electron-rich, so lowest ν$_{CO}$.
- Cr(CO)$_6$ is neutral, least electron-rich, so highest ν$_{CO}$.

**If you have to choose from the given options** (as you wrote them), **none is exactly correct**. If you want to match as closely as possible, the order should be:

$$
\text{Cr(CO)}_6 > [\text{V(CO)}_6]^− > [\text{Mn(CO)}_5]^− > [\text{Co(CO)}_4]^−
$$

**If you want to memorize: More negative charge → more back-donation → lower C–O stretching frequency.**

---

### **Summary Table**

| Complex                 | Electron Count | ν$_{CO}$ (cm$^{-1}$) |
|-------------------------|---------------|--------------------------|
| Cr(CO)$_6$            | d$^{6}$     | Highest                  |
| [V(CO)$_6$]$^{-}$   | d$^{6}$     | High                     |
| [Mn(CO)$_5$]

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