Question

What will be the value of x in Fe⁺⁺, if the magnetic moment \(\mu = \sqrt{24}\) BM

• A. 0
• B. 2
• C. 1
• D. 3

Answer 1

The Correct Option is B

The value of x in Fe⁺⁺ can be determined by considering the magnetic moment (µ) and the number of unpaired electrons in the Fe⁺⁺ ion.
In an Fe⁺⁺ ion, the iron atom loses two electrons, resulting in a 2⁺ charge. 
To determine the number of unpaired electrons, we can use the fact that the magnetic moment (µ) is given by the formula: 
\(\mu = \sqrt{n(n+2)}\) BM Where n represents the number of unpaired electrons. 

Given that \(µ = \sqrt{24}\) BM, 
we can solve for n: \(\sqrt{n(n+2)} = \sqrt{24}\)
Squaring both sides: \(n(n+2) = 24\)
Expanding the equation: \(n^2 + 2n = 24\)
Rearranging and simplifying: \(n^2 + 2n - 24 = 0\)
Factorizing the quadratic equation: \((n - 4)(n + 6) = 0\)
Solving for n: n = 4 or n = -6 

Since the number of unpaired electrons cannot be negative, we discard the n = -6 solution.
 Therefore, the Fe⁺⁺ ion has 4 unpaired electrons. 
In Fe⁺⁺, x represents the oxidation state of iron. 
Since Fe⁺⁺ has a 2⁺ charge, the value of x is 2. 
Therefore, the correct value of x in Fe⁺⁺ is (2) 2.

Answer 2

Given: Magnetic moment \( \mu = \sqrt{24} \) BM 

Formula for magnetic moment (spin-only): $$ \mu = \sqrt{n(n+2)} \, \text{BM} $$ where \( n \) = number of unpaired electrons.

Compare: $$ \sqrt{n(n+2)} = \sqrt{24} \Rightarrow n(n+2) = 24 $$ Try \( n = 4 \):
\( 4(4+2) = 4 \times 6 = 24 \Rightarrow n = 4 \)

So, Fe⁺⁺ has 4 unpaired electrons.

Fe atomic number = 26 ⇒ electronic configuration = [Ar] 3d⁶ 4s²
Fe⁺⁺ ⇒ 2 electrons removed ⇒ 3d⁶

In 3d⁶, if there are 4 unpaired electrons, the ion is in the high-spin state (common for weak field ligands).

Correct value of x = number of unpaired electrons = 4
But options given are 0, 2, 1, 3. So if they ask for:

• x = number of paired electrons ⇒ total d-electrons = 6, unpaired = 4 ⇒ paired = 2

Correct answer: 2

Answer 3

We are given that the magnetic moment, \(\mu = \sqrt{24}\) BM for Fe^x+. We need to find the value of x.

The spin-only magnetic moment formula is:

\(\mu = \sqrt{n(n+2)}\) BM

Where n is the number of unpaired electrons.

We have \(\mu = \sqrt{24}\), so

\(\sqrt{24} = \sqrt{n(n+2)}\)

\(24 = n(n+2)\)

\(n^2 + 2n - 24 = 0\)

\((n+6)(n-4) = 0\)

Since n must be a positive integer, \(n = 4\). This means Fe^x+ has 4 unpaired electrons.

The electronic configuration of neutral Fe is [Ar] 3d⁶4s².

If Fe^x+ has 4 unpaired electrons in the 3d orbitals, we can analyze the possible oxidation states:

• If x = 1 (Fe⁺): The configuration would be [Ar] 3d⁷. Hund's rule gives us 3 unpaired electrons. Incorrect.
• If x = 2 (Fe²⁺): The configuration would be [Ar] 3d⁶. Hund's rule gives us 4 unpaired electrons. Correct.
• If x = 3 (Fe³⁺): The configuration would be [Ar] 3d⁵. Hund's rule gives us 5 unpaired electrons. Incorrect.

Therefore, the value of x is 2.

Answer:

2