How many unpaired electrons would you expect for t...

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How many unpaired electrons would you expect for the following complex ions?1. [Cr(CN)6]4−2. [MnF6]4−3. [Ru(en)3]2+

Jules Bruno · Accepted Answer

We have to determine the number of unpaired electrons in each of the given complexes.The number of unpaired electrons in a complex can be determined through:Calculating the oxidation number of metal. This gives the number of electrons in the d orbital.Finding out whether the ligand attached to the metal is a strong-field or a weak-field ligand.Filling the d electrons of the metal into the split d orbitals to find out how many are unpaired.[readmore][Cr(CN)6]4−The total charge on the complex is -4. CN has a -1 charge.Cr + (6x-1) = -4 Cr - 6 = -4 Cr = -4 + 6Cr = +2The electronic configuration of Cr (Z = 24) is: 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d54s is the outermost orbital therefore it will be the first to lose an electron.There are 2 electrons lost so the 2nd electrons will be from 3d. The electron configuration of Cr2+ ions is:Cr2+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d4The spectrochemical series can be used to determine the field strength of the ligand.O22−< I− < Br− < S2− < SCN− (S–bonded) < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < NCS− (N–bonded) < CH3CN < gly (glycine) < py (pyridine) < NH3 < en (ethylenediamine) < bipy (2,2'-bipyridine) < phen (1,10-phenanthroline) < NO2− < PPh3 < CN− < CO < CH2When a complex is formed, the d orbital of metal is split into two sets:t2g set containing dxy, dyz and dzxeg set containing dx2-y2 and dz2CN- is a strong-field ligand. A strong-field ligand causes the electrons to stay in the lower lying t2g set of orbitals.Cr2+ has 4 d electrons so, we will fill 4 electrons in the split d orbitals. All 4 electrons will stay in the low-lying t2g set due to the large energy gap produced by the strong field from CN-.No. of unpaired electrons = 2[MnF6]4−The total charge on the complex is -4. F has a -1 charge.Mn + (6x-1) = -4 Mn - 6 = -4 Mn = -4 + 6Mn = +2The electronic configuration of Mn (Z = 25) is: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d54s is the outermost orbital therefore it will lose 2 electrons.The electron configuration of Mn2+ ions is:Mn2+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d5F- is a weak-field ligand. A weak-field ligand does not cause the electrons to stay in the lower lying t2g set of orbitals. They are distributed equally.Mn2+ has 5 d electrons so, we will fill 5 electrons in the split d orbitals. All 5 electrons will be filled unpaired in all the split d orbitals. This is because F- is a weak-filed ligand and the energy gap produced is small.No. of unpaired electrons = 5[Ru(en)3]2+The total charge on the complex is +2. en has a 0 charge because it is a neutral ligand.Ru + 0 = +2Ru = +2The electronic configuration of Ru (Z = 44) is: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s1, 4d75s is the outermost orbital therefore it will lose the electron first and the 2nd electrons will be lost from the 4d orbital.Ru2+ = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 4d6en is a strong-field ligand. A strong-field ligand causes the electrons to stay in the lower lying t2g set of orbitals.Ru2+ has 6 d electrons so, we will fill 6 electrons in the split d orbitals. All 4 electrons will stay in the low-lying t2g set due to the large energy gap produced by the strong field from en.

Problem: How many unpaired electrons would you expect for the following complex ions?1. [Cr(CN)6]4−2. [MnF6]4−3. [Ru(en)3]2+

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Problem: How many unpaired electrons would you expect for the following complex ions?1. [Cr(CN)6]4−2. [MnF6]4−3. [Ru(en)3]2+

FREE Expert Solution

Problem Details

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