Hybridisation and oxidation state of sulphur in caro's acid
Answers
Caro's acid, Marshall's acid also has a peroxide linkage so that in which S shows +6 oxidation state. or 2 + 2x – 12 – 2 = 0 or x = + 6. This cannot be true as maximum O. N. of Cr cannot be more than + 6. ... This exceptional value is due to the fact that four oxygen atoms in CrO5 are in peroxide linkage.
Answer:
Hybridisation and oxidation state of sulphur in caro's acid is sp³ hybridisation and +6 oxidation state.
Explanation:
For hybridization → Redistribution of the energy of orbitals of individual atoms to give orbitals of equivalent energy happens when two atomic orbitals combine to form a hybrid orbital in a molecule. This process is called hybridization.
Hybridization of the central atom can be determined by the formula →
Steric Number (SN) = ↓ [V+M-C+A] 2
V = Number of valence electrons in central atom
M = No of monovalent atoms
C = Algebraic charge of cation
A = Algebraic charge of anion
For H₂S04 →
SN = 1 / 2 [V+ M- C+ A]
V = 6 , M= 2
SN = 1 / 2 [ 6 + 2]
= 8 /2 = 4
Hybridisation → sp³
oxidation state
The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state.
∴ The oxidation state of sulphur in Caro's acid is +6.
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