2.33 g of compound X (empirical formula = Co | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reaction is: $CoH_{12}N_{4}Cl_{3} + AgNO_{3} \rightarrow AgCl$ (white ppt).Molar mass of $CoH_{12}N_{4}Cl_{3} = 59 + 12(1) + 4(14) + 3(35.5) =

Vedclass · Accepted Answer

$3, 6$

Vedclass · Answer

(A) The reaction is: $CoH_{12}N_{4}Cl_{3} + AgNO_{3} \rightarrow AgCl$ (white ppt).Molar mass of $CoH_{12}N_{4}Cl_{3} = 59 + 12(1) + 4(14) + 3(35.5) = 59 + 12 + 56 + 106.5 = 233.5 \ g/mol$.Moles of $X = \frac{2.33 \ g}{233.5 \ g/mol} = 0.01 \ mol$.Moles of $AgCl = \frac{1.435 \ g}{143.5 \ g/mol} = 0.01 \ mol$.Since $0.01 \ mol$ of $X$ produces $0.01 \ mol$ of $AgCl$,there is $1 \ Cl^-$ ion outside the coordination sphere.The complex is $[Co(NH_{3})_{4}Cl_{2}]Cl$.Primary valency (oxidation state of $Co$): $x + 4(0) + 2(-1) = +1 \Rightarrow x = +3$.Secondary valency (coordination number of $Co$): $4(NH_{3}) + 2(Cl) = 6$.

$2.33 \ g$ of compound $X$ (empirical formula $= CoH_{12}N_{4}Cl_{3}$) upon treatment with excess $AgNO_{3}$ solution produces $1.435 \ g$ of a white precipitate. The primary and secondary valences of cobalt in compound $X$,respectively are.
[Given,atomic mass : $Co = 59, Cl = 35.5, Ag = 108$]

Similar Questions

How can the coordination number of a metal atom in a complex be determined?

In a coordination compound,the coordination number of the metal is equal to:

Which of the following statements is incorrect?

Addition of excess of $AgNO_3$ to an aqueous solution of $1$ mole of $PdCl_2 \cdot 4 NH_3$ gives $2$ moles of $AgCl$. The conductivity of this solution corresponds to

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$2.33 \ g$ of compound $X$ (empirical formula $= CoH_{12}N_{4}Cl_{3}$) upon treatment with excess $AgNO_{3}$ solution produces $1.435 \ g$ of a white precipitate. The primary and secondary valences of cobalt in compound $X$,respectively are.[Given,atomic mass : $Co = 59, Cl = 35.5, Ag = 108$]