Paramagnetism is a property due to the presence of unpaired electrons. In case of transition metals, as they contain unpaired electrons in the `(n-1)` d orbitals , most of the transition metal ions and their compounds are paramagnetic. Para magnetism increases with increases in number of unpaired electrons. Magnetic moment is calculated from 'spin only formula' `Vz`
`mu=sqrt(n(n+2)) B.M n="number of unpaired electrons"`
Similarly the colour of the compounds of transition metals may be attributed to the presence of incomplete `(n-1)` d sub-shell. When an electron from a lower energy of d-orbitals is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to complementary colour of the light observed. The frequency of the light absorbed is determined by the nature of the ligand.
Titanium shows magnetic moments of `1.73 BM` in its compound. What is the oxidation state of titanium in the compound?

Options

A. `+2`
B. `+1`
C. `+3`
D. `+4`

Explanation

To determine the oxidation state of titanium in the compound based on its magnetic moment, we can follow these steps:

### Step 1: Write the formula for magnetic moment
The magnetic moment (\( \mu \)) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \text{ B.M.} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Substitute the given magnetic moment
We know that the magnetic moment of titanium in the compound is \( 1.73 \text{ B.M.} \). Therefore, we can set up the equation:
\[ 1.73 = \sqrt{n(n + 2)} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives us:
\[ (1.73)^2 = n(n + 2) \]
Calculating \( (1.73)^2 \):
\[ 2.9929 \approx 3 \quad (\text{for easier calculation}) \]
Thus, we have:
\[ n(n + 2) = 3 \]

### Step 4: Rearrange the equation
Rearranging gives us a quadratic equation:
\[ n^2 + 2n - 3 = 0 \]

### Step 5: Factor the quadratic equation
Factoring the quadratic equation:
\[ (n + 3)(n - 1) = 0 \]
This gives us two possible solutions for \( n \):
\[ n = -3 \quad \text{or} \quad n = 1 \]

### Step 6: Discard the negative solution
Since \( n \) represents the number of unpaired electrons, we discard \( n = -3 \) and take:
\[ n = 1 \]

### Step 7: Determine the oxidation state of titanium
Now, we need to find the oxidation state of titanium. The atomic number of titanium (Ti) is 22, and its electronic configuration is:
\[ \text{Ti: } [\text{Ar}] 3d^2 4s^2 \]
Since we have determined that there is 1 unpaired electron, we can conclude that titanium must have lost 3 electrons to achieve this state. Therefore, the oxidation state of titanium in this compound is:
\[ \text{Oxidation state} = +3 \]

### Final Answer
The oxidation state of titanium in the compound is \( +3 \).

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