`Na_(2)[Cr(NO)(NH_(3))(C_(2)O_(4))_(2)], u=sqrt(3)BM`, Then total no. of electron in `d_(x^(2)-y^(2)) and d_(z^(2))` orbitals of metals:

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the oxidation state of chromium in the complex
The complex is given as \( Na_2[Cr(NO)(NH_3)(C_2O_4)_2] \).

1. Sodium (Na) has a +1 oxidation state, and there are 2 sodium ions, contributing +2.
2. The nitrosyl (NO) ligand is a cationic ligand with a +1 oxidation state.
3. Ammonia (NH3) is a neutral ligand with a 0 oxidation state.
4. Oxalate (C2O4) is a bidentate ligand, and since there are 2 oxalate ligands, they contribute -4 (each oxalate has a -2 charge).

Setting up the equation for the oxidation state \( x \) of chromium:
\[ x + 2 + 1 + 0 - 4 = 0 \]
\[ x - 1 = 0 \implies x = +1 \]

### Step 2: Determine the electronic configuration of chromium in the +1 oxidation state
The atomic number of chromium (Cr) is 24. The ground state electronic configuration is:
\[ [Ar] 3d^5 4s^1 \]
In the +1 oxidation state, one electron is removed from the 4s orbital:
\[ Cr^{+1}: [Ar] 3d^5 4s^0 \]

### Step 3: Identify the coordination number and geometry
The coordination number is determined by the number of donor atoms:
- 1 from NO
- 1 from NH3
- 4 from 2 oxalate ligands (each oxalate has 2 donor sites)

Total donor sites = 1 + 1 + 4 = 6. Therefore, the coordination number is 6, indicating that the complex is octahedral.

### Step 4: Analyze the d-orbital splitting in an octahedral field
In an octahedral field, the d-orbitals split into two sets:
- \( t_{2g} \) (which includes \( d_{xy}, d_{yz}, d_{zx} \))
- \( e_g \) (which includes \( d_{x^2-y^2}, d_{z^2} \))

### Step 5: Distribute the electrons in the d-orbitals
With 5 electrons in the \( 3d \) orbitals, they will fill the \( t_{2g} \) orbitals first according to Hund's rule:
- The first three electrons will occupy the \( d_{xy}, d_{yz}, \) and \( d_{zx} \) orbitals.
- The remaining two electrons will pair up in the \( d_{xy} \) orbital.

Thus, the distribution is:
- \( t_{2g} \): 5 electrons in \( d_{xy}, d_{yz}, d_{zx} \)
- \( e_g \): 0 electrons in \( d_{x^2-y^2} \) and \( d_{z^2} \)

### Step 6: Conclusion
The total number of electrons in the \( d_{x^2-y^2} \) and \( d_{z^2} \) orbitals is:
\[ 0 + 0 = 0 \]

### Final Answer
The total number of electrons in \( d_{x^2-y^2} \) and \( d_{z^2} \) orbitals of the metal is **0**.

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