Among the following which one is not paramagnetic ? [ Atomic numbers , `Be =4, Ne =10`,As =33 ,Cl=17]`

Options

A. `Ne^(2+)`
B. `Be^(+)`
C. `Cl^(-)`
D. `As^(+)`

Answer

To determine which of the given elements is not paramagnetic, we need to analyze the electronic configurations of each option and check for the presence of unpaired electrons. Paramagnetic substances have unpaired electrons, while non-paramagnetic substances have all electrons paired.

### Step-by-Step Solution:

1. **Identify the Elements and Their Atomic Numbers:**
- Be (Beryllium) = 4
- Ne (Neon) = 10
- Cl (Chlorine) = 17
- As (Arsenic) = 33

2. **Write the Electronic Configurations:**
- **Neon (Ne)**: Atomic number 10
- Configuration: \(1s^2 2s^2 2p^6\)
- **Ne2+**: Removing 2 electrons from Ne
- Configuration: \(1s^2 2s^2 2p^4\) (2 unpaired electrons in 2p)
- **Beryllium (Be)**: Atomic number 4
- Configuration: \(1s^2 2s^2\)
- **Be+**: Removing 1 electron from Be
- Configuration: \(1s^2 2s^1\) (1 unpaired electron in 2s)
- **Chlorine (Cl)**: Atomic number 17
- Configuration: \(1s^2 2s^2 2p^6 3s^2 3p^5\)
- **Cl-**: Adding 1 electron to Cl
- Configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6\) (all paired)
- **Arsenic (As)**: Atomic number 33
- Configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3\)
- **As+**: Removing 1 electron from As
- Configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^2\) (2 unpaired electrons in 4p)

3. **Determine the Presence of Unpaired Electrons:**
- **Ne2+**: 2 unpaired electrons (paramagnetic)
- **Be+**: 1 unpaired electron (paramagnetic)
- **Cl-**: 0 unpaired electrons (not paramagnetic)
- **As+**: 2 unpaired electrons (paramagnetic)

4. **Conclusion:**
- The element that is **not paramagnetic** is **Cl-** because it has no unpaired electrons.

### Final Answer:
**Cl- is not paramagnetic.**