The geometry of `[Co(CO)_(4)]^(-)" and "[Cd(CN)_(4)]^(2-)` are :

Options

A. both square planar
B. both tetrahedral
C. tetrahedral and square planar, respectively
D. square planar and tetrahedral, respectively

Explanation

To determine the geometry of the coordination compounds \([Co(CO)_4]^-\) and \([Cd(CN)_4]^{2-}\), we will follow these steps:

### Step 1: Determine the oxidation state of the central metal in each complex.

**For \([Co(CO)_4]^-\):**
- Let the oxidation state of cobalt be \(x\).
- The charge of the complex is \(-1\).
- Carbon monoxide (CO) is a neutral ligand, contributing 0 to the charge.
- Therefore, the equation is:
\[ x + 0 \times 4 = -1 \implies x = -1 \]

**For \([Cd(CN)_4]^{2-}\):**
- Let the oxidation state of cadmium be \(y\).
- The charge of the complex is \(-2\).
- The cyanide ion (CN) has a charge of \(-1\), and there are 4 CN ligands.
- Therefore, the equation is:
\[ y + (-1) \times 4 = -2 \implies y - 4 = -2 \implies y = +2 \]

### Step 2: Determine the electronic configuration of the central metal in each oxidation state.

**For Cobalt (\(Co^{+1}\)):**
- The atomic number of cobalt (Co) is 27.
- The electronic configuration of neutral cobalt is:
\[ [Ar] 3d^7 4s^2 \]
- For \(Co^{+1}\), we remove one electron, typically from the 4s orbital:
\[ [Ar] 3d^8 \]

**For Cadmium (\(Cd^{+2}\)):**
- The atomic number of cadmium (Cd) is 48.
- The electronic configuration of neutral cadmium is:
\[ [Kr] 4d^{10} 5s^2 \]
- For \(Cd^{+2}\), we remove two electrons, typically from the 5s orbital:
\[ [Kr] 4d^{10} \]

### Step 3: Determine the hybridization of the central metal in each complex.

**For \([Co(CO)_4]^-\):**
- CO is a strong field ligand and will cause pairing of electrons.
- The \(Co^{+1}\) ion has a configuration of \(3d^8\).
- The hybridization involves the 3d and 4s orbitals:
- Since there are 4 CO ligands, the hybridization is \(sp^3\).
- The geometry is tetrahedral.

**For \([Cd(CN)_4]^{2-}\):**
- CN is also a strong field ligand.
- The \(Cd^{+2}\) ion has a configuration of \(4d^{10}\).
- The hybridization involves the 4d and 5s orbitals:
- Since there are 4 CN ligands, the hybridization is \(sp^3\).
- The geometry is also tetrahedral.

### Final Summary:
- The geometry of \([Co(CO)_4]^-\) is **tetrahedral**.
- The geometry of \([Cd(CN)_4]^{2-}\) is **tetrahedral**.