A first row transition metal in its +2 oxidation state has a spin-only magnetic moment value of $3.86 \mathrm{~BM}$. The atomic number of the metal is

<p>To determine the atomic number of the metal based on the provided spin-only magnetic moment value, we use the formula for calculating the spin-only magnetic moment:</p> <p>$ \mu = \sqrt{n(n+2)} \mu_{B} $</p> <p>where $ \mu $ is the magnetic moment in Bohr magnetons ($ \mu_{B} $), and $ n $ is the number of unpaired electrons.</p> <p>Given that the spin-only magnetic moment value is $ 3.86 \, \mathrm{BM} $, we can set this value equal to the formula to solve for $ n $:</p> <p>$ 3.86 = \sqrt{n(n+2)} $</p> <p>Squaring both sides gives:</p> <p>$ 14.8996 = n(n+2) $</p> <p>Since $ n $ must be an integer and this equation doesn't solve neatly for an integer $ n $, we look for a value of $ n $ that would give a product close to $ 14.8996 $ when plugged into $ \sqrt{n(n+2)} $.</p> <p>Practically, we can test the square root values near $ 3.86 $ for different $ n $ to see which one gives a close match. Considering the usual spin-only magnetic moments for common numbers of unpaired electrons:</p> <ul> <li>$ n = 1 $, $ \mu = \sqrt{3} \approx 1.73 \mu_B $</li><br> <li>$ n = 2 $, $ \mu = \sqrt{8} \approx 2.83 \mu_B $</li><br> <li>$ n = 3 $, $ \mu = \sqrt{15} \approx 3.87 \mu_B $</li><br> <li>$ n = 4 $, $ \mu = \sqrt{24} \approx 4.90 \mu_B $</li> </ul> <p>The value $ n = 3 $ gives a magnetic moment of approximately $ 3.87 \mu_B $, which is very close to the given value, $ 3.86 \mu_B $. Therefore, the metal has 3 unpaired electrons in its d-orbital configuration.</p> <p>In the +2 oxidation state, metals lose electrons from the s-orbital before the d-orbital. Given that the element is a first row transition metal, starting with scandium (Sc) at atomic number 21 which has an electronic configuration of $[Ar] 3d^1 4s^2$ in its neutral state, we can deduce the configurations:</p> <ul> <li>$ Z = 22 $ for Titanium (Ti), which has a neutral configuration of $[Ar] 3d^2 4s^2$. In the +2 oxidation state, it would have an electronic configuration of $[Ar] 3d^2$, resulting in 2 unpaired electrons.</li><br> <li>$ Z = 23 $ for Vanadium (V), with a neutral configuration of $[Ar] 3d^3 4s^2$. In the +2 state, its configuration would be $[Ar] 3d^3$, presenting 3 unpaired electrons, which matches our calculation.</li><br> <li>$ Z = 26 $ for Iron (Fe), indicating a neutral configuration of $[Ar] 3d^6 4s^2$. In the +2 state, $[Ar] 3d^6$, which would have 4 unpaired electrons, not matching the calculation.</li><br> <li>$ Z = 25 $ for Manganese (Mn), with a neutral configuration of $[Ar] 3d^5 4s^2$. In the +2 state, $[Ar] 3d^5$, indicating 5 unpaired electrons, which also does not match our calculation.</li> </ul> <p>So, the atomic number of the metal with a +2 oxidation state and a spin-only magnetic moment value of $3.86 \mathrm{~BM}$ (corresponding to 3 unpaired electrons) is 23, which identifies the metal as Vanadium (V). Therefore,</p> <p>Option B (23) is the correct answer.</p>