How many of the following complexes are correctly matched with given properties ?
(a)`[Ni(NH_3)_6]^(2+)`-Octahedral outer orbital complex and two unpaired electrons.
(b)`[Cr (NH_3)_6]^(3+)`-Octahedral inner orbital complex and one unpaired electron.
( c)`[Zn(NH_3)_6]^(2+)`-Octahedral outer orbital complex and diamagnetic.
(d)`[Pt (NH_(3))_4]^(2+)`-Square planar and diamagnetic
(e)`[Co (SCN)_4]^(2-)`-Tetrahedral and three unpaired electrons.
(f)`[NiBr_2(PPh_3)_2]`-Square planar and diamagnetic.

Explanation

To determine how many of the given complexes are correctly matched with their properties, we will analyze each complex one by one based on the oxidation state, electronic configuration, geometry, and magnetic properties.

### Step-by-Step Solution:

**(a)** **Complex: `[Ni(NH_3)_6]^(2+)`**
- **Oxidation State of Ni:** +2
- **Electronic Configuration:** Ni (atomic number 28) → [Ar] 3d^8 4s^2. For Ni^2+, it is 3d^8.
- **Geometry:** Since Ni^2+ has a 3d^8 configuration, it forms an outer orbital complex (octahedral).
- **Magnetic Properties:** The 3d^8 configuration can have 2 unpaired electrons.
- **Conclusion:** The statement is **correct**.

**(b)** **Complex: `[Cr(NH_3)_6]^(3+)`**
- **Oxidation State of Cr:** +3
- **Electronic Configuration:** Cr (atomic number 24) → [Ar] 3d^5 4s^1. For Cr^3+, it is 3d^3.
- **Geometry:** Cr^3+ with a 3d^3 configuration forms an inner orbital complex (octahedral).
- **Magnetic Properties:** The 3d^3 configuration has 3 unpaired electrons.
- **Conclusion:** The statement is **incorrect** (it states one unpaired electron).

**(c)** **Complex: `[Zn(NH_3)_6]^(2+)`**
- **Oxidation State of Zn:** +2
- **Electronic Configuration:** Zn (atomic number 30) → [Ar] 3d^10 4s^2. For Zn^2+, it is 3d^10.
- **Geometry:** Zn^2+ forms an outer orbital complex (octahedral).
- **Magnetic Properties:** The 3d^10 configuration is diamagnetic (all electrons are paired).
- **Conclusion:** The statement is **correct**.

**(d)** **Complex: `[Pt(NH_3)_4]^(2+)`**
- **Oxidation State of Pt:** +2
- **Electronic Configuration:** Pt (atomic number 78) → [Xe] 4f^14 5d^9 6s^1. For Pt^2+, it is 5d^8.
- **Geometry:** Pt^2+ typically forms a square planar geometry.
- **Magnetic Properties:** The 5d^8 configuration is diamagnetic (all electrons are paired).
- **Conclusion:** The statement is **correct**.

**(e)** **Complex: `[Co(SCN)_4]^(2-)`**
- **Oxidation State of Co:** +2
- **Electronic Configuration:** Co (atomic number 27) → [Ar] 3d^7 4s^2. For Co^2+, it is 3d^7.
- **Geometry:** Co^2+ with SCN^- (a weak field ligand) forms a tetrahedral complex.
- **Magnetic Properties:** The 3d^7 configuration has 3 unpaired electrons.
- **Conclusion:** The statement is **correct**.

**(f)** **Complex: `[NiBr_2(PPh_3)_2]`**
- **Oxidation State of Ni:** +2
- **Electronic Configuration:** Ni^2+ is 3d^8.
- **Geometry:** With PPh3 (a strong field ligand), Ni^2+ typically forms a square planar complex. However, due to steric effects, it can also be tetrahedral.
- **Magnetic Properties:** The 3d^8 configuration can have unpaired electrons, making it paramagnetic.
- **Conclusion:** The statement is **incorrect** (it states diamagnetic).

### Summary of Correctness:
- (a) Correct
- (b) Incorrect
- (c) Correct
- (d) Correct
- (e) Correct
- (f) Incorrect

### Total Correct Statements: **4**