Magnetic moment (spin only) of octahedral complex having CFSE=−0.8Δo and surrounded by weak field ligands can be : Q
To answer this, the Crystal Field Stabilization Energy has to be calculated for a (d3 metal in both configurations. The geometry with the greater stabilization will be the preferred geometry. So for tetrahedral d3, the Crystal Field Stabilization Energy is: CFSE = -0.8 x 4/9 Δo = -0.355 Δo.
[Co(CN)64-] is also an octahedral d7 complex but it contains CN-, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron. In this case the CFSE is −(6)(25)ΔO+(1)(35)ΔO+P=−95ΔO+P.
The crystal field stabilization energy (CFSE) (in kJ/mol) for complex, [Ti(H2O)6]3+. According to CFT, the first absorption maximum is obtained at 20,3000cm−1 for the transition.
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The right answer is ( D ) Movement Of Tectonic Plates
Answer:
mass Na2SO4 = 14.3816 g
Explanation:
sln Na2SO4:
∴ V = 450 mL
∴ <em>C </em>=<em> </em>0.2250 mol/L
∴ Mw ≡ 142.04 g/mol..... from literature
⇒ mol Na2SO4 = (0.2250 mol/L)(0.450 L) = 0.10125 mol
⇒ mass Na2SO4 = (0.10125 mol)(142.04 g/mol) = 14.3816 g
Answer:
4 Al +3 02-------> 2 AL2O3
Explanation:
