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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Answer:
4.16g of MgCl2
Explanation:
First, let us generate a balanced equation for the reaction:
Mg + 2HCl —> MgCl2 + H2
Molar Mass of Mg = 24g/mol
Molar Mass of MgCl2 = 24 + (2x35.5) = 24 + 71 = 95g
From the equation,
24g of Mg produced 95g of MgCl2.
Therefore, 1.05g of Mg will produce = (1.05x95)/24 = 4.16g of MgCl2
Answer:
6.517.
Explanation:
- It is known that for pure water: [H₃O⁺] = [OH⁻], so water is neutral.
<em>Kw = [H₃O⁺][OH⁻] = 9.25 x 10⁻¹⁴.</em>
∵ [H₃O⁺] = [OH⁻].
∴ [H₃O⁺]² = 9.25 x 10⁻¹⁴.
∴ [H₃O⁺] = √(9.25 x 10⁻¹⁴) = 3.041 x 10⁻⁷ M.
∵ pH = - log[H₃O⁺]
<em>∴ pH</em> = - log(3.041 x 10⁻⁷) = <em>6.517.</em>
