Question

The 'spin only' magnetic moment value of \(MO_4^{2-}\) is ____ \(\text{BM}\). (Where \(M\) is a metal having least metallic radii among \(\text{Sc, Ti, V, Cr, Mn, Zn}\)). \([ \text{Given atomic number: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25, Zn = 30} ]\)

Answer 1

Correct Answer: 0

To find the 'spin only' magnetic moment of \(MO_4^{2-}\), where \(M\) is a metal with the least metallic radii among \(\text{Sc, Ti, V, Cr, Mn, Zn}\), follow these steps:

Step 1: Determine the Metal \(M\)

• The atomic number determines the metallic radius: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25, Zn = 30.
• Typically, a higher atomic number indicates a smaller atomic radius within a period (consider the transition metals' trend).
• Zinc (\(\text{Zn}\)) has the highest atomic number (30) in this set, suggesting the smallest radius among the given metals.
• Therefore, \(M = \text{Zn}\).

Step 2: Determine D-electrons for \(\text{Zn}^{2+}\)

• Zinc starts with an electron configuration of \([Ar]3d^{10}4s^2\).
• As \(\text{Zn}^{2+}\), it loses 2 electrons from the 4s orbital, resulting in a configuration of \([Ar]3d^{10}\).
• This implies 10 d-electrons and no unpaired electrons.

Step 3: Calculate the Magnetic Moment

• The 'spin only' magnetic moment \(\mu\) is calculated using \(\mu = \sqrt{n(n+2)}\) where \(n\) equals the number of unpaired electrons.
• For \(\text{Zn}^{2+}\), \(n = 0\) since there are no unpaired electrons.
• Thus, \(\mu = \sqrt{0(0+2)} = \sqrt{0} = 0\) Bohr Magneton (BM).

Step 4: Confirm the Solution

• The calculated magnetic moment is 0 BM, which matches the expected range (0,0).

Conclusion: The 'spin only' magnetic moment value of \(MO_4^{2-}\) where \(M = \text{Zn}\) is 0 BM.

Answer 2

Among the given elements, Cr has the least metallic radii. The CrO$_4^{2-}$ ion has Cr$^{6+}$, which has a $d^0$ configuration (diamagnetic).
The spin-only magnetic moment:
\[\mu = \sqrt{n(n+2)} \, \text{BM},\]
where $n = 0$.
Final Answer:
\[0 \, \text{BM}.\]