What is the sum of radial and angular nodes in the following orbitals of H-atom?
(I) `Psi_(2px)`, (ii) `Psi_(2s)`, (iii) `Psi_(3d_(x))`, (iv) `Psi_(3d_(x^(2)-y^(2)))`

Explanation

To find the sum of radial and angular nodes for the given orbitals of the hydrogen atom, we will use the following formulas:

1. **Radial Nodes**: The number of radial nodes is given by the formula:
\[ \text{Radial Nodes} = n - l - 1 \]
where \( n \) is the principal quantum number and \( l \) is the azimuthal quantum number.

2. **Angular Nodes**: The number of angular nodes is equal to \( l \).

Now, let's calculate the sum of radial and angular nodes for each of the given orbitals:

### (I) For \( \Psi_{2px} \):
- **Principal Quantum Number (n)**: 2
- **Azimuthal Quantum Number (l)**: 1 (since \( p \) orbitals have \( l = 1 \))

**Calculating Radial Nodes**:
\[ \text{Radial Nodes} = n - l - 1 = 2 - 1 - 1 = 0 \]

**Calculating Angular Nodes**:
\[ \text{Angular Nodes} = l = 1 \]

**Total Nodes**:
\[ \text{Total Nodes} = \text{Radial Nodes} + \text{Angular Nodes} = 0 + 1 = 1 \]

### (II) For \( \Psi_{2s} \):
- **Principal Quantum Number (n)**: 2
- **Azimuthal Quantum Number (l)**: 0 (since \( s \) orbitals have \( l = 0 \))

**Calculating Radial Nodes**:
\[ \text{Radial Nodes} = n - l - 1 = 2 - 0 - 1 = 1 \]

**Calculating Angular Nodes**:
\[ \text{Angular Nodes} = l = 0 \]

**Total Nodes**:
\[ \text{Total Nodes} = \text{Radial Nodes} + \text{Angular Nodes} = 1 + 0 = 1 \]

### (III) For \( \Psi_{3d_{x}} \):
- **Principal Quantum Number (n)**: 3
- **Azimuthal Quantum Number (l)**: 2 (since \( d \) orbitals have \( l = 2 \))

**Calculating Radial Nodes**:
\[ \text{Radial Nodes} = n - l - 1 = 3 - 2 - 1 = 0 \]

**Calculating Angular Nodes**:
\[ \text{Angular Nodes} = l = 2 \]

**Total Nodes**:
\[ \text{Total Nodes} = \text{Radial Nodes} + \text{Angular Nodes} = 0 + 2 = 2 \]

### (IV) For \( \Psi_{3d_{x^2-y^2}} \):
- **Principal Quantum Number (n)**: 3
- **Azimuthal Quantum Number (l)**: 2 (since \( d \) orbitals have \( l = 2 \))

**Calculating Radial Nodes**:
\[ \text{Radial Nodes} = n - l - 1 = 3 - 2 - 1 = 0 \]

**Calculating Angular Nodes**:
\[ \text{Angular Nodes} = l = 2 \]

**Total Nodes**:
\[ \text{Total Nodes} = \text{Radial Nodes} + \text{Angular Nodes} = 0 + 2 = 2 \]

### Final Calculation of the Sum of All Nodes:
Now, we sum the total nodes from all orbitals:
\[ \text{Total Sum} = 1 (\Psi_{2px}) + 1 (\Psi_{2s}) + 2 (\Psi_{3d_{x}}) + 2 (\Psi_{3d_{x^2-y^2}}) = 1 + 1 + 2 + 2 = 6 \]

### Answer:
The sum of radial and angular nodes in the given orbitals is **6**.