The magnetic moment of two dioxygen species x and y are 2.83 & 1.73 BM Which of the following option is always correct about the above species.

Options

A. Bon order of `x gt y `
B. Bond order of `y gtx `
C. x is`O_(2)` and y is `O_(2)^(-)`
D. x is`O_(2)`

Explanation

To solve the problem regarding the magnetic moments of two dioxygen species (x and y) with values of 2.83 and 1.73 Bohr Magnetons (BM), we will follow these steps:

### Step 1: Understand the formula for magnetic moment
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Calculate the number of unpaired electrons for species y (1.73 BM)
Given that the magnetic moment for species y is 1.73 BM, we can set up the equation:
\[ \sqrt{n(n + 2)} = 1.73 \]
Squaring both sides gives:
\[ n(n + 2) = (1.73)^2 \]
Calculating \( (1.73)^2 \):
\[ (1.73)^2 = 2.9929 \approx 3 \]
So, we have:
\[ n(n + 2) = 3 \]

### Step 3: Solve the quadratic equation
This can be rearranged to form a quadratic equation:
\[ n^2 + 2n - 3 = 0 \]
Factoring the quadratic:
\[ (n - 1)(n + 3) = 0 \]
Thus, \( n = 1 \) (since \( n \) cannot be negative).

### Step 4: Identify species y
Species y has 1 unpaired electron. The dioxygen species with 1 unpaired electron is \( O_2^- \) (superoxide ion), which has 17 electrons.

### Step 5: Calculate the number of unpaired electrons for species x (2.83 BM)
Now, for species x with a magnetic moment of 2.83 BM:
\[ \sqrt{n(n + 2)} = 2.83 \]
Squaring both sides gives:
\[ n(n + 2) = (2.83)^2 \]
Calculating \( (2.83)^2 \):
\[ (2.83)^2 = 8.0089 \approx 8 \]
So, we have:
\[ n(n + 2) = 8 \]

### Step 6: Solve the quadratic equation for species x
This can be rearranged to form another quadratic equation:
\[ n^2 + 2n - 8 = 0 \]
Factoring the quadratic:
\[ (n - 2)(n + 4) = 0 \]
Thus, \( n = 2 \) (since \( n \) cannot be negative).

### Step 7: Identify species x
Species x has 2 unpaired electrons. The dioxygen species with 2 unpaired electrons is \( O_2 \) (molecular oxygen), which has 16 electrons.

### Conclusion
- Species x is \( O_2 \) with 2 unpaired electrons.
- Species y is \( O_2^- \) with 1 unpaired electron.

### Final Answer
The correct option regarding the above species is that species x is \( O_2 \) and species y is \( O_2^- \).

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