# How many unpaired electrons would you expect on aluminum in aluminum oxide.?

How many unpaired electrons do you expect in aluminum in aluminum oxide?

There are 3 unpaired electrons in aluminum in aluminum oxide. Explanation: The chemical formula of aluminum oxide is . The atomic number of aluminum is 13 and its electronic configuration is . To complete its octet, aluminum will lose or donate 3 electrons to the oxygen atom, since it has 3 excess electrons. Thus, there are three unpaired electrons in an aluminum atom. Since there are two aluminum atoms in , there will be a total of 6 unpaired electrons in two aluminum atoms. Since each oxygen atom is deficient in 2 electrons, each aluminum atom will donate its excess electrons to each oxygen atom.

Answer: The number of unpaired electrons present in iron in iron(II) sulfide is 4 Explanation: The compound given is FeS In this compound, iron is in the oxidation state (+2) and sulfur is in the (-two) oxidation state. As we know iron is a block element and the atomic number of iron is 26. That means there are 26 electrons. The electron configuration of iron is: The electron configuration of Fe²⁺ will be: Thus, there are 4 unpaired electrons present in iron in iron(II) sulfide.

Five unpaired electrons Explanation: This should be the actual equation: [Fe(H20)6]3+ First, ligands are neutral ions or molecules, which bind to a central metal atom or ions. In this case, the ligand here is the aqua(H2O)6 molecule. According to Hund’s rule, electrons always enter an empty orbital before pairing up. This rule is valid because the orbitals have the same energy. In a situation like this where we have a ligand coordinating to a metal ion, the orbital energies are no longer the same. What really affects Hund’s rule is the splitting of the crystal field. In this case, the ligand molecule (H2O) has weak crystal division and is called weak field ligands. The weak field ligand has less field separation energy than the pairing energy. Thus, the electron filling pattern in the d-shell follows Hund’s rule. Therefore, iron(Fe) = 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Yielding the 2 electrons from the 4s shell and one electron from the single paired d electron (Fe3+). The remaining five d-shell electrons become unpaired.

Complex ions will have 4 unpaired electrons.
Explanation It is known that in the composite coordinate complex [Fe(NH3)6]2+, the NH3 ligand exhibits a neutral charge, so the oxidation state of the transition metal Fe is the +2 state in this complex compound.
Now consider the electronic configuration of iron (Fe) in the ground state, that is [Ar]3d64s2. So since Fe is in the form of Fe2+ ions, that means 2 electrons will be removed from the valence shell, so the resulting electron configuration after reaching Fe2+ will be [Ar] 3d6. It is now also known that d orbitals can occupy a maximum number of 10 electrons in their 5 lobes which are represented by dxy, dxz, dyz, dx2-y2 and dz2. As in this case it has only 6 electrons, so according to Pauli’s exclusion principle and Aufbau’s principle, the five lobes of the d-subshell will first be filled with five unpaired electrons and then the 6th extra electron will be filled in the dxy lobe. So the remaining four lobes will have unpaired electrons leading to the formation of four unpaired electrons.
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Iron(III) has 5 valence electrons and the complex is in an octahedral geometry. Since water as a ligand has a small separation energy, the complex will have a high spin and the five electrons will not be paired.

ON of NH3 forms a coordinate
covalent bond with metallic iron as in this case. The iron itself is at the “center” of the bound complex ion. Since the net ionic charge is at 2+, it can be said that there is an unpaired pair of electrons for [fe(nh3)6]2+.

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