The total number of electrons present in all `s` orbitals, all the `p` orbitals, and all the `d` orbitals of cesium ion are, respectively,

Options

A. 8,26,10
B. 10, 24, 20
C. 8, 22, 24
D. 12, 20, 22

Explanation

To find the total number of electrons present in all `s`, `p`, and `d` orbitals of the cesium ion (Cs⁺), we will follow these steps:

### Step 1: Determine the Atomic Number of Cesium
The atomic number of cesium (Cs) is 55. This means that a neutral cesium atom has 55 electrons.

### Step 2: Determine the Electron Configuration of Neutral Cesium
The electron configuration of neutral cesium is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s¹

### Step 3: Adjust for the Cesium Ion (Cs⁺)
Since we are dealing with the cesium ion (Cs⁺), we need to remove one electron from the neutral cesium configuration. The electron removed will be from the outermost shell, which is the 6s orbital.

Thus, the electron configuration for Cs⁺ becomes:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶

### Step 4: Count the Electrons in the `s` Orbitals
From the configuration:
- 1s² → 2 electrons
- 2s² → 2 electrons
- 3s² → 2 electrons
- 4s² → 2 electrons
- 5s² → 2 electrons

Total in `s` orbitals = 2 + 2 + 2 + 2 + 2 = 10 electrons.

### Step 5: Count the Electrons in the `p` Orbitals
From the configuration:
- 2p⁶ → 6 electrons
- 3p⁶ → 6 electrons
- 4p⁶ → 6 electrons
- 5p⁶ → 6 electrons

Total in `p` orbitals = 6 + 6 + 6 + 6 = 24 electrons.

### Step 6: Count the Electrons in the `d` Orbitals
From the configuration:
- 3d¹⁰ → 10 electrons
- 4d¹⁰ → 10 electrons

Total in `d` orbitals = 10 + 10 = 20 electrons.

### Final Answer
The total number of electrons present in all `s` orbitals, all `p` orbitals, and all `d` orbitals of the cesium ion (Cs⁺) are:
- `s` orbitals: 10 electrons
- `p` orbitals: 24 electrons
- `d` orbitals: 20 electrons

Thus, the answer is: **10, 24, 20**.

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