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

Answer: (B)

2

Answer 2

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.