How many ions are produced from the complex $[Co(NH_3)_6]Cl_2$ in solution?
$i. 6 \quad ii. 4 \quad iii. 3 \quad iv. 2$
$i. 6 \quad ii. 4 \quad iii. 3 \quad iv. 2$
(C) $iii.$ The given complex can be written as $[Co(NH_3)_6]Cl_2$.
In an aqueous solution,it dissociates as follows:
$[Co(NH_3)_6]Cl_2 \rightarrow [Co(NH_3)_6]^{2+} + 2Cl^-$
Thus,one $[Co(NH_3)_6]^{2+}$ ion and two $Cl^-$ ions are produced,resulting in a total of $3$ ions.
In an aqueous solution,it dissociates as follows:
$[Co(NH_3)_6]Cl_2 \rightarrow [Co(NH_3)_6]^{2+} + 2Cl^-$
Thus,one $[Co(NH_3)_6]^{2+}$ ion and two $Cl^-$ ions are produced,resulting in a total of $3$ ions.
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