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In [Fe(CN) 6 ] 4− , the strong field of six cyanide ligands produces a large Δ oct . Under these conditions, the electrons require less energy to pair than they require to be excited to the e g orbitals (Δ oct >P). The six 3 d electrons of the Fe 2+ ion pair in the three t 2 g orbitals ( [link] ). Complexes in which the electrons are paired because of the large crystal field splitting are called low-spin complexes because the number of unpaired electrons (spins) is minimized.
In [Fe(H 2 O) 6 ] 2+ , on the other hand, the weak field of the water molecules produces only a small crystal field splitting (Δ oct <P). Because it requires less energy for the electrons to occupy the e g orbitals than to pair together, there will be an electron in each of the five 3 d orbitals before pairing occurs. For the six d electrons on the iron(II) center in [Fe(H 2 O) 6 ] 2+ , there will be one pair of electrons and four unpaired electrons ( [link] ). Complexes such as the [Fe(H 2 O) 6 ] 2+ ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called high-spin complexes because the number of unpaired electrons (spins) is maximized.
A similar line of reasoning shows why the [Fe(CN) 6 ] 3− ion is a low-spin complex with only one unpaired electron, whereas both the [Fe(H 2 O) 6 ] 3+ and [FeF 6 ] 3− ions are high-spin complexes with five unpaired electrons.
(a) K 3 [CrI 6 ]
(b) [Cu(en) 2 (H 2 O) 2 ]Cl 2
(c) Na 3 [Co(NO 2 ) 6 ]
(a) Cr 3+ has a d 3 configuration. These electrons will all be unpaired.
(b) Cu 2+ is d 9 , so there will be one unpaired electron.
(c) Co 3+ has d 6 valence electrons, so the crystal field splitting will determine how many are paired. Nitrite is a strong-field ligand, so the complex will be low spin. Six electrons will go in the t 2 g orbitals, leaving 0 unpaired.
d 4 , d 5 , d 6 , and d 7
4; because Δ tet is small, all tetrahedral complexes are high spin and the electrons go into the t 2 orbitals before pairing
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