To calculate the spin magnetic moment of the compound [X(CN)2(CO)4]+, we need to follow these steps: 1. Determine the oxidation state of the metal X(Z=26). The metal X is iron ( Fe ) since it has an atomic number of 26 . Let the oxidation state of Fe in the complex be x . Given the ligands, CN (cyanide) and CO (carbon monoxide), both are considered to be strong field ligands which typically cause pairing of electrons. The overall charge of the complex is +1 : ‌x+2(−1)+4(0)=+1 ‌x−2=+1 ‌x=+3 So, the oxidation state of Fe in the compound is +3 . 2. Find the electronic configuration of Fe3+ : The electronic configuration of neutral Fe is: [Ar]3d64s2 When Fe loses 3 electrons to form Fe3+, the configuration becomes: [Ar]3d5
3. Determine the number of unpaired electrons: Since CN−and CO are strong field ligands, they will pair up the electrons in the 3 d orbitals: 3d orbitals with strong field ligands: ↑↓↑↓↑↓↑↑ This configuration shows that there is only one unpaired electron in Fe3+. 4. Calculate the spin magnetic moment: The spin-only formula for the magnetic moment µ‌spin ‌ is given by: µ‌spin ‌=√n(n+2)‌BM Where n is the number of unpaired electrons. For Fe3+ in this compound, n=1.
‌µ‌spin ‌=√1(1+2)‌BM ‌µ‌spin ‌=√3‌BM ‌µ‌spin ‌≈1.73‌BM Therefore, the spin magnetic moment value for the compound [X(CN)2(CO)4]+is 1.73 BM . Option B: 1.73 BM