Which of the following statements is correct for complex `[Cr(NH_(3))(CN)_(4)(NO)]^(2-)`(given that n=1)?

Options

A. It is `d^(2)sp^(3)` hybridised.
B. The chromium is in +1 oxidation state
C. It is heteroleptic complex and its aqueous solution is coloured
D. All of these

Explanation

To solve the question regarding the complex \([Cr(NH_3)(CN)_4(NO)]^{2-}\), we will follow a systematic approach to determine the correct statements about the complex.

### Step-by-Step Solution:

**Step 1: Determine the oxidation state of chromium (Cr).**
- Let the oxidation state of chromium be \(X\).
- The ligands are:
- \(NH_3\) (neutral, oxidation state = 0)
- \(CN^-\) (negative, oxidation state = -1, and there are 4 of them, contributing -4)
- \(NO\) (neutral, but typically considered as a cationic ligand, contributing +1)

The overall charge of the complex is -2. Therefore, we can set up the equation:
\[ X + 0 + (4 \times -1) + 1 = -2 \]
\[ X - 4 + 1 = -2 \]
\[ X - 3 = -2 \]
\[ X = +1 \]
Thus, the oxidation state of chromium is +1.

**Step 2: Determine the electronic configuration of chromium in the +1 state.**
- The atomic number of chromium (Cr) is 24. Its ground state electronic configuration is \([Ar] 3d^5 4s^1\).
- In the +1 oxidation state, it loses one electron from the 4s orbital:
\[ \text{Configuration of } Cr^{+1} = [Ar] 3d^5 \]

**Step 3: Determine the geometry and hybridization of the complex.**
- The complex has six coordination sites (one \(NH_3\), four \(CN^-\), and one \(NO\)), indicating an octahedral geometry.
- In an octahedral field, the \(d\) orbitals split into two sets: \(T_{2g}\) and \(E_g\).
- With 5 electrons in \(3d\), they will fill the \(T_{2g}\) orbitals first:
\[ \text{Filling: } \uparrow \uparrow \uparrow \uparrow \uparrow \quad (5 \text{ electrons}) \]
- There is one unpaired electron in the \(T_{2g}\) level.

**Step 4: Determine the hybridization.**
- The hybridization for an octahedral complex with 2 \(d\) orbitals, 1 \(s\) orbital, and 3 \(p\) orbitals is \(d^2sp^3\).

**Step 5: Analyze the nature of the complex.**
- A heteroleptic complex contains different types of ligands. Here, we have \(NH_3\), \(CN^-\), and \(NO\), which are different types of ligands. Thus, it is indeed a heteroleptic complex.

**Step 6: Determine the color of the complex.**
- The presence of one unpaired electron in the \(d\) orbitals allows for d-d transitions, which typically results in the complex being colored.

### Conclusion:
All statements about the complex \([Cr(NH_3)(CN)_4(NO)]^{2-}\) are correct:
1. It is \(d^2sp^3\) hybridized.
2. Chromium is in the +1 oxidation state.
3. It is a heteroleptic complex.
4. Its aqueous solution is colored.

### Final Answer:
All statements given are correct, and the correct option is option four.