In how many of the following complex ions, the central metal ions use (n-1)d, ns and np orbitals for hybridisation ?
`[Mn(CN)_6]^(4-), [Ni(NH_3)_6]^(3-), [Co( o x)_3]^(3-), [Cu(NO_2)_6]^(4-), [AgF_4]^(-)`

Explanation

To determine how many of the given complex ions have central metal ions that use (n-1)d, ns, and np orbitals for hybridization, we will analyze each complex ion one by one.

### Step 1: Analyze [Mn(CN)6]^(4-)
- **Central Metal Ion**: Manganese (Mn)
- **Oxidation State**: In this complex, CN^- is a strong field ligand and does not affect the oxidation state. The oxidation state of Mn can be calculated as follows:
- Let the oxidation state of Mn be x.
- x + 6(-1) = -4 → x - 6 = -4 → x = +2
- **Electron Configuration**: Mn (Atomic number 25) has the electron configuration [Ar] 3d^5 4s^2.
- **Hybridization**: The hybridization involves 3d, 4s, and 4p orbitals.
- **Conclusion**: This complex uses (n-1)d, ns, and np orbitals.

### Step 2: Analyze [Ni(NH3)6]^(3-)
- **Central Metal Ion**: Nickel (Ni)
- **Oxidation State**: For NH3, the oxidation state of Ni can be calculated similarly:
- Let the oxidation state of Ni be y.
- y + 6(0) = -3 → y = -3
- **Electron Configuration**: Ni (Atomic number 28) has the electron configuration [Ar] 3d^8 4s^2.
- **Hybridization**: The hybridization involves 3d, 4s, and 4p orbitals.
- **Conclusion**: This complex also uses (n-1)d, ns, and np orbitals.

### Step 3: Analyze [Co(ox)3]^(3-)
- **Central Metal Ion**: Cobalt (Co)
- **Oxidation State**: For oxalate (ox), which is a bidentate ligand:
- Let the oxidation state of Co be z.
- z + 3(-2) = -3 → z - 6 = -3 → z = +3
- **Electron Configuration**: Co (Atomic number 27) has the electron configuration [Ar] 3d^7 4s^2.
- **Hybridization**: The hybridization involves 3d, 4s, and 4p orbitals.
- **Conclusion**: This complex uses (n-1)d, ns, and np orbitals.

### Step 4: Analyze [Cu(NO2)6]^(4-)
- **Central Metal Ion**: Copper (Cu)
- **Oxidation State**: For NO2, which is a bidentate ligand:
- Let the oxidation state of Cu be w.
- w + 6(-1) = -4 → w - 6 = -4 → w = +2
- **Electron Configuration**: Cu (Atomic number 29) has the electron configuration [Ar] 3d^10 4s^1.
- **Hybridization**: The hybridization involves 3d, 4s, and 4p orbitals.
- **Conclusion**: This complex also uses (n-1)d, ns, and np orbitals.

### Step 5: Analyze [AgF4]^(-)
- **Central Metal Ion**: Silver (Ag)
- **Oxidation State**: For F, which is a monodentate ligand:
- Let the oxidation state of Ag be v.
- v + 4(-1) = -1 → v - 4 = -1 → v = +3
- **Electron Configuration**: Ag (Atomic number 47) has the electron configuration [Kr] 4d^10 5s^1.
- **Hybridization**: The hybridization involves 4d, 5s, and 5p orbitals.
- **Conclusion**: This complex does not use (n-1)d, ns, and np orbitals.

### Final Count:
- The complexes that use (n-1)d, ns, and np orbitals are:
- [Mn(CN)6]^(4-)
- [Ni(NH3)6]^(3-)
- [Co(ox)3]^(3-)
- [Cu(NO2)6]^(4-)

Thus, the total number of complexes that use (n-1)d, ns, and np orbitals is **4**.

### Summary of the Answer:
The answer is **4**.