To solve the problem of identifying which statement is not true regarding the coordination compounds mentioned, we will analyze each statement step by step.
### Step 1: Analyze MnCl4^2-
1. **Identify the oxidation state of Mn**:
- Let the oxidation state of Mn be \( x \).
- The overall charge of the complex is -2, and there are 4 Cl atoms, each with a charge of -1.
- Therefore, the equation is:
\[
x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2
\]
2. **Electron configuration**:
- Mn has an atomic number of 25, so its electron configuration is \( [Ar] 3d^5 4s^2 \).
- In the +2 oxidation state, it loses 2 electrons from the 4s orbital, resulting in \( 3d^5 \).
3. **Determine the geometry and magnetic properties**:
- Cl is a weak field ligand, leading to no pairing of electrons in the \( 3d \) orbitals.
- The geometry is tetrahedral, and since there are 5 unpaired electrons, MnCl4^2- is paramagnetic.
### Step 2: Analyze [Mn(CN)6]^{3-}
1. **Identify the oxidation state of Mn**:
- Let the oxidation state of Mn be \( x \).
- The overall charge is -3, and there are 6 CN^- ligands (each with a charge of -1).
- Therefore, the equation is:
\[
x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3
\]
2. **Electron configuration**:
- In the +3 oxidation state, Mn has the electron configuration \( 3d^4 \).
3. **Determine the geometry and magnetic properties**:
- CN is a strong field ligand, which causes pairing of electrons.
- The geometry is octahedral, and since there are 2 unpaired electrons, [Mn(CN)6]^{3-} is paramagnetic.
### Step 3: Analyze [Cu(CN)4]^{2-}
1. **Identify the oxidation state of Cu**:
- Let the oxidation state of Cu be \( x \).
- The overall charge is -2, and there are 4 CN^- ligands.
- Therefore, the equation is:
\[
x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2
\]
2. **Electron configuration**:
- Cu has an atomic number of 29, so its electron configuration is \( [Ar] 3d^{10} 4s^1 \).
- In the +2 oxidation state, it loses 1 electron from the 4s and 1 from the 3d, resulting in \( 3d^9 \).
3. **Determine the geometry and magnetic properties**:
- CN is a strong field ligand, leading to pairing of electrons.
- The geometry is square planar, and since there is 1 unpaired electron, [Cu(CN)4]^{2-} is paramagnetic.
### Step 4: Analyze Ni(PPh3)3
1. **Identify the oxidation state of Ni**:
- Let the oxidation state of Ni be \( x \).
- The overall charge is 0, and PPh3 is a neutral ligand.
- Therefore, the equation is:
\[
x = 0 \implies x = 0
\]
2. **Electron configuration**:
- Ni has an atomic number of 28, so its electron configuration is \( [Ar] 3d^8 4s^2 \).
3. **Determine the geometry and magnetic properties**:
- The geometry is trigonal bipyramidal, and since there are no unpaired electrons, Ni(PPh3)3 is diamagnetic.
### Conclusion:
The statement regarding [Cu(CN)4]^{2-} being diamagnetic is **not true** because it is actually paramagnetic due to the presence of one unpaired electron.
### Final Answer:
The statement that is not true is: **[Cu(CN)4]^{2-} is diamagnetic.**
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