The IUPAC definition of a transition element is that it is an element that has an incomplete d-subshell in either the neutral atom or its ion. Thus the group 12 elements are member of the d-block but are not transition elements.
Chemically solft members of the d-block occurs as sulphide minerals and are partially oxidised to obtain the metal, the more electropositive 'hard' metals occurs as oxides and are extracted by reduction.
Opposite to p-block elements, the higher oxidation states are favoured by the heavier elements of d-block Metals on the right of the d-block tend to exist in low oxidation states and form complexes with the ligands. Square-planar complexes are common for the platinum metals and gold in oxidation states that yield `d^8` electronic configuration, which include RH(I),Ir(I),Pd(II),Pt(II) and Au(III).
The most distinctive features/properties of transition metal complex is their wide range of colours.The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron.It is important to note that (a) in absence of ligand, crystal field spilling does not occur and hence the substances is colourless, (b) the type of ligand also influences the colour of the complexes.
Which of the following has `dsp^2` hybridisation and is diamagnetic in nature ?
(i)`Na_4[Cr(CO)_4]` , (ii)`[Ni(DMGH)_2]` , (iii)`[PtHBr(PEt_3)_2]`
(iv)`[As(SCN)_4]^(3-)` , (v)`[AuBr_4]^(-)`

Options

A. I,II,IV only
B. I,III,IV,V only
C. II,III,V only
D. III,V only

Explanation

To solve the problem, we need to determine which of the given complexes has `dsp^2` hybridization and is diamagnetic in nature. We will analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding `dsp^2` Hybridization:**
- `dsp^2` hybridization occurs in square planar complexes, typically involving a central metal atom with a coordination number of 4. This hybridization is common for transition metals in a +2 oxidation state, especially those with a `d^8` electron configuration.

2. **Analyzing Each Complex:**
- **(i) `Na_4[Cr(CO)_4]`:**
- Chromium (Cr) in this complex is in the +4 oxidation state.
- The electron configuration of Cr is `[Ar] 3d^5 4s^1`. In +4 state, it becomes `[Ar] 3d^4`.
- CO is a strong field ligand, promoting pairing of electrons. The configuration becomes `3d^4` (all paired).
- The hybridization is `sp^3` (due to 4 ligands), not `dsp^2`.
- **Conclusion:** Not diamagnetic and not `dsp^2`.

- **(ii) `[Ni(DMGH)_2]`:**
- Nickel (Ni) is in the +2 oxidation state.
- The electron configuration of Ni is `[Ar] 3d^8 4s^2`. In +2 state, it becomes `[Ar] 3d^8`.
- DMGH is a strong field ligand, leading to pairing of electrons.
- The hybridization is `dsp^2` (due to 4 ligands).
- **Conclusion:** Diamagnetic and `dsp^2`.

- **(iii) `[PtHBr(PEt_3)_2]`:**
- Platinum (Pt) is in the +2 oxidation state.
- The electron configuration of Pt is `[Xe] 4f^{14} 5d^8 6s^2`. In +2 state, it becomes `[Xe] 4f^{14} 5d^8`.
- The ligands (HBr and PEt3) are strong field ligands, leading to pairing of electrons.
- The hybridization is `dsp^2` (due to 4 ligands).
- **Conclusion:** Diamagnetic and `dsp^2`.

- **(iv) `[As(SCN)_4]^{3-}`:**
- Arsenic (As) is in the -3 oxidation state.
- The electron configuration of As is `[Ar] 4s^2 3d^{10} 4p^3`. In -3 state, it becomes `[Ar] 4s^2 3d^{10} 4p^6`.
- The hybridization is `sp^3` (due to 4 ligands).
- **Conclusion:** Not diamagnetic and not `dsp^2`.

- **(v) `[AuBr_4]^{-}`:**
- Gold (Au) is in the +3 oxidation state.
- The electron configuration of Au is `[Xe] 4f^{14} 5d^{10} 6s^1`. In +3 state, it becomes `[Xe] 4f^{14} 5d^8`.
- The ligands (Br) are strong field ligands, leading to pairing of electrons.
- The hybridization is `dsp^2` (due to 4 ligands).
- **Conclusion:** Diamagnetic and `dsp^2`.

### Final Answer:
The complexes that have `dsp^2` hybridization and are diamagnetic are:
- **(ii) `[Ni(DMGH)_2]`**
- **(iii) `[PtHBr(PEt_3)_2]`**
- **(v) `[AuBr_4]^{-}`**