The 'spin only' magnetic moment value of $\mathrm{MO}_4{ }^{2-}$ is ________ BM. (Where M is a metal having least metallic radii. among $\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$ and $\mathrm{Zn}$ ).

(Given atomic number: $\mathrm{Sc}=21, \mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Cr}=24, \mathrm{Mn}=25$ and $\mathrm{Zn}=30$)

<p>To determine the 'spin only' magnetic moment of the complex $\mathrm{MO}_4^{2-}$, we first need to identify the metal $\mathrm{M}$.</p> <p><strong>Step 1: Identify the metal M based on metallic radii.</strong></p> <p>The problem states that $\mathrm{M}$ has the least metallic radii among $\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$, and $\mathrm{Zn}$.</p> <p>The trend for metallic radii across the first transition series (3d series) generally decreases from $\mathrm{Sc}$ to $\mathrm{Cr/Mn}$, stays relatively constant for $\mathrm{Fe, Co, Ni}$, and then increases for $\mathrm{Cu}$ and $\mathrm{Zn}$. However, $\mathrm{Mn}$ has a complex crystal structure leading to a metallic radius that is anomalously high compared to its neighbors. The actual values (in pm) are approximately:</p> <ul> <li><p>$\mathrm{Sc}$: 164</p></li> <li><p>$\mathrm{Ti}$: 147</p></li> <li><p>$\mathrm{V}$: 135</p></li> <li><p>$\mathrm{Cr}$: 129</p></li> <li><p>$\mathrm{Mn}$: 137 (anomalously high)</p></li> <li><p>$\mathrm{Zn}$: 137 (larger due to full d-subshell shielding)</p></li> </ul> <p>Among the options provided ($\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}, \mathrm{Zn}$), <strong>Chromium ($\mathrm{Cr}$)</strong> has the smallest metallic radius.</p> <p>Therefore, $\mathrm{M} = \mathrm{Cr}$.</p> <p><strong>Step 2: Determine the oxidation state of Cr in the complex.</strong></p> <p>The complex is $\mathrm{CrO}_4^{2-}$ (Chromate ion).</p> <p>Let the oxidation state of $\mathrm{Cr}$ be $x$.</p> <p>The oxidation state of Oxygen ($\mathrm{O}$) is generally $-2$.</p> <p>The sum of oxidation states equals the charge on the ion:</p> <p>$x + 4(-2) = -2$</p> <p>$x - 8 = -2$</p> <p>$x = +6$</p> <p>So, Chromium is in the $+6$ oxidation state ($\mathrm{Cr}^{6+}$).</p> <p><strong>Step 3: Determine the electronic configuration.</strong></p> <p>The atomic number of Chromium ($\mathrm{Cr}$) is 24.</p> <p>Ground state configuration of $\mathrm{Cr}$: $[\mathrm{Ar}] 3d^5 4s^1$</p> <p>To form $\mathrm{Cr}^{6+}$, we remove the outer 6 electrons (one from $4s$ and five from $3d$).</p> <p>Configuration of $\mathrm{Cr}^{6+}$: $[\mathrm{Ar}] 3d^0 4s^0$</p> <p><strong>Step 4: Calculate the number of unpaired electrons.</strong></p> <p>In the $3d^0$ configuration, there are no electrons in the d-orbitals.</p> <p>Number of unpaired electrons ($n$) = $0$.</p> <p><strong>Step 5: Calculate the spin-only magnetic moment.</strong></p> <p>The formula for spin-only magnetic moment ($\mu$) is:</p> <p>$\mu = \sqrt{n(n+2)} \text{ BM}$</p> <p>Substituting $n = 0$:</p> <p>$\mu = \sqrt{0(0+2)} = \sqrt{0} = 0 \text{ BM}$</p> <p><strong>Conclusion:</strong></p> <p>The 'spin only' magnetic moment value is 0.</p> <p><strong>Answer:</strong> 0</p>