STATEMENT -1 `:` The graph of `Psi^(2)` and r for 4d orbital has two maximas
STATEMENT-2 `: 3d_(z^(2))` has no nodal plane.
STATEMENT-3 `,` 4s is closer to nucleus than 3d

Options

A. T T T
B. T T F
C. F T F
D. T F T

Answer

To analyze the statements regarding the 4d orbital, 3d_z^2 orbital, and the 4s orbital, we will evaluate each statement step by step.

### Step-by-Step Solution:

**Statement 1:** The graph of \( \Psi^2 \) and \( r \) for the 4d orbital has two maxima.

1. **Understanding the 4d Orbital:**
- The principal quantum number \( n \) for the 4d orbital is 4.
- The azimuthal quantum number \( l \) for d orbitals is 2.

2. **Calculating Radial Nodes:**
- The number of radial nodes can be calculated using the formula:
\[ \text{Number of radial nodes} = n - l - 1 \]
- Substituting the values:
\[ \text{Number of radial nodes} = 4 - 2 - 1 = 1 \]

3. **Graph Behavior:**
- The presence of one radial node indicates that the graph of \( \Psi^2 \) (probability density) versus \( r \) will have two maxima, as the radial node divides the graph into two regions where the probability density is non-zero.
- Therefore, **Statement 1 is true**.

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**Statement 2:** 3d_z^2 has no nodal plane.

1. **Understanding the 3d_z^2 Orbital:**
- The 3d_z^2 orbital has a unique shape characterized by a lobular structure along the z-axis and a toroidal (doughnut-shaped) region in the xy-plane.

2. **Identifying Nodal Planes:**
- A nodal plane is a region in space where the probability density is zero, meaning no electrons are likely to be found.
- The 3d_z^2 orbital does not have any planes where the probability density is zero; it has a nodal surface (the toroidal shape) but no nodal planes.

3. **Conclusion:**
- Therefore, **Statement 2 is true**.

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**Statement 3:** 4s is closer to the nucleus than 3d.

1. **Using the n + l Rule:**
- For the 4s orbital:
- \( n = 4 \), \( l = 0 \) (s orbital)
- \( n + l = 4 + 0 = 4 \)
- For the 3d orbital:
- \( n = 3 \), \( l = 2 \) (d orbital)
- \( n + l = 3 + 2 = 5 \)

2. **Comparing n + l Values:**
- According to the n + l rule, the orbital with the lower \( n + l \) value is closer to the nucleus.
- Since \( 4 < 5 \), the 4s orbital is indeed closer to the nucleus than the 3d orbital.

3. **Conclusion:**
- Therefore, **Statement 3 is true**.

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### Final Evaluation:
- All three statements are true. Thus, the correct option is **Option 1: true, true, true**.

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