Step 1: Recall the Formula for Magnetic Moment
The spin-only magnetic moment (\( \mu \)) is given by:
$$ \mu = \sqrt{n(n+2)} $$
where \( n \) is the number of unpaired electrons. The magnetic moment depends only on the number of unpaired electrons in the ion.
Step 2: Analyze the Given Ions
- Has 1 unpaired electron.
- Spin-only magnetic moment:
$$ \mu = \sqrt{1(1+2)} = \sqrt{3} $$
- Has 4 unpaired electrons.
- Spin-only magnetic moment:
$$ \mu = \sqrt{4(4+2)} = \sqrt{24} $$
- Has 5 unpaired electrons.
- Spin-only magnetic moment:
$$ \mu = \sqrt{5(5+2)} = \sqrt{35} $$
- Has 4 unpaired electrons.
- Spin-only magnetic moment:
$$ \mu = \sqrt{4(4+2)} = \sqrt{24} $$
- Has no unpaired electrons.
- Spin-only magnetic moment:
$$ \mu = 0 $$
Step 3: Compare the Magnetic Moments
Step 4: Conclusion
The correct answer is: Option (1) - B and D only, since they have the same spin-only magnetic moment.
Given below are two statements:
Statement I: Ferromagnetism is considered as an extreme form of paramagnetism.
Statement II: The number of unpaired electrons in a $Cr^{2+}$ ion (Z = 24) is the same as that of a $Nd^{3+}$ ion (Z = 60).
In the light of the above statements, choose the correct answer from the options given below:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec: