Question
Question: Find oxidation number of \[S\] in \[CuS{O_4}.5{H_2}O\]....
Find oxidation number of S in CuSO4.5H2O.
Solution
We know that coordination complex is that the product of a Lewis acid-base reaction during which anions bond or neutral molecules to a central metal atom (or ion) by coordinate covalent bonds. A ligand is an ion or molecule, which donates a pair of electrons to the central metal atom or ion to make a coordination complex
Complete step by step answer:
Copper is transition metals in the d-block of the fourth energy level of the periodic table it will make the electronic configuration for Cu:[Ar] 3d94s2 However, because the 3d orbitals is so much larger than the 4s orbital and the 3d orbital only needs one more electron to be filled, the 3d orbital pulls an electron from the 4s orbital to fill this empty space. This makes the actual electronic configuration Cu:[Ar] 3d10 4s1. The oxidation state of copper depends on its state. When it is metallic, the oxidation state is zero. In its compounds, the most common oxidation number of copper is +2.
In the above coordination cooper complex coordinate with water is neutral ligand hence its oxidation state oxygen is−2 . All compounds add to 0.
Consider oxidation state of sulfur as S
Cu+S+ 4×−2+5×0 =0
⇒2+S+(−8 ) + 0=0
Simplifying the calculations we get,
⇒2+S+(−8)=0
⇒S=0−2+8
⇒S=+6
Hence, oxidation number of S in CuSO4.5H2O is +6
Note:
We have to remember these things,
1.In solid state CuSO4.5H2O copper coordinated with four water molecules.
2.+1,+3 and+4 These are also an oxidation number of copper.
3.The electrons in the outermost shell are called valence electrons.
4.Ligands can be anions, cations, and neutral ligands. Some ligands are neutral ligands such as ammonia, water, carbon monoxide, and nitrogen. Ligands are classified into monodentate, bidentate, and poly dentate.