The total number of ions produced from the complex $[Cr(NH_3)_6]Cl_3$ in aqueous solution will be____________.
Dissociation of $[Cr(NH_3)_6]Cl_3$ Complex
When the coordination complex $[Cr(NH_3)_6]Cl_3$ is dissolved in an aqueous solution, it dissociates into ions based on its ionic structure.
The complex consists of a complex cation, $[Cr(NH_3)_6]^{3+}$, and three chloride anions, $Cl^-$.
The dissociation reaction in water can be represented as:
$ [Cr(NH_3)_6]Cl_3 \xrightarrow{H_2O} [Cr(NH_3)_6]^{3+} + 3Cl^- $
Calculating Total Ions Produced
By examining the dissociation equation, we can count the total number of ions formed:
- One complex cation: $[Cr(NH_3)_6]^{3+}$
- Three chloride anions: $3 \times Cl^-$
The total number of ions is the sum of the cation and the anions:
Total ions = (Number of complex cations) + (Number of chloride anions)
Total ions = 1 + 3 = 4
Therefore, a total of 4 ions are produced from the complex $[Cr(NH_3)_6]Cl_3$ in an aqueous solution.
Important Questions from Coordination Compounds
The chemical formula of sodium nitroprusside is
The allowed transition in an atomic system is
The correct set of information is
The products A and B for the given reaction [Co(NH3)5CI]2+ + [Cr(OH2)6]2+ + 5H3O+ → A + B are, respectively
The rate of hydration of [CrIII(H2O)5X]n+, (X = \(\rm N_3^-\), F-, CN- and NH3) in neutral aqueous medium remains unaffected in acidic medium at room temperature, if 'X' is
