Among, mathrmSc$\mathrm{Sc}$, mathrmMn$\mathrm{Mn}$, mathrmCo$\mathrm{Co}$ and mathrmCu$\mathrm{Cu}$, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is ________ BM (in nearest integer).

### Related Formula Spin-only magnetic moment (mu$\mu$) is given by: mu = sqrtn(n+2)mathrm~BM$$\mu = \sqrt{n(n+2)}\mathrm{~BM}$$ where n$n$ is the number of unpaired d-electrons. ### Core Logic Enthalpies of atomization of the given 3d transition elements (in mathrmkJ/mol$\mathrm{kJ/mol}$): - Scandium (mathrmSc$\mathrm{Sc}$): 326$326$ - Manganese (mathrmMn$\mathrm{Mn}$): 281$281$ - Cobalt (mathrmCo$\mathrm{Co}$): 425$425$ - Copper (mathrmCu$\mathrm{Cu}$): 339$339$ Thus, Cobalt (mathrmCo$\mathrm{Co}$) has the highest enthalpy of atomization. ### Step 1: Determine unpaired electrons in mathrmCo^2+$\mathrm{Co}^{2+}$ Electronic configuration of Cobalt (Z=27$Z=27$): mathrmCo: [mathrmAr] 3d^7 4s^2$$\mathrm{Co}: [\mathrm{Ar}] 3d^7 4s^2$$ For divalent Cobalt ion (mathrmCo^2+$\mathrm{Co}^{2+}$): mathrmCo^2+: [mathrmAr] 3d^7$$\mathrm{Co}^{2+}: [\mathrm{Ar}] 3d^7$$ In the d-subshell (five orbitals): - Three orbitals are paired, and three are unpaired (n=3$n=3$). ### Step 2: Calculate spin-only magnetic moment mu = sqrt3(3+2) = sqrt15 approx 3.87mathrm~BM$$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\mathrm{~BM}$$ Rounding to the nearest integer gives 4$4$. ### Pattern Recognition Enthalpy of atomization generally peaks near the middle of transition series due to maximum metallic bonding. However, mathrmMn$\mathrm{Mn}$ (3d^5 4s^2$3d^5 4s^2$) is an anomaly with an exceptionally low value (281mathrm~kJ/mol$281\mathrm{~kJ/mol}$) due to its highly stable half-filled d^5$d^5$ subshell configuration which reduces electron delocalization in metallic bonding. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements