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$2.33 \mathrm{~g}$ of compound $\mathrm{X}$ (empirical formula $\left.\mathrm{CoH}_{12} \mathrm{~N}_{4} \mathrm{Cl}_{3}\right)$ upon treatment with excess $\mathrm{AgNO}_{3}$ solution produces $1.435 \mathrm{~g}$ of a white precipitate. The primary and secondary valences of cobalt in compound $\mathrm{X}$, respectively, are
[Given : Atomic mass : $\mathrm{Co}=59, \mathrm{Cl}=35.5, \mathrm{Ag}=108$ ]
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[Given : Atomic mass : $\mathrm{Co}=59, \mathrm{Cl}=35.5, \mathrm{Ag}=108$ ]
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Verified Answer
The correct answer is:
3,6
$$
\begin{array}{l}
\text { M.wt }=59+12+14 \times 4+35.5 \times 3=233.5 \\
\mathrm{C}_{0} \mathrm{H}_{12} \mathrm{~N}_{4} \mathrm{Cl}_{3} \stackrel{\mathrm{AgNO}_{3}}{\longrightarrow} 1.435 \mathrm{~g} \mathrm{AgCl} \\
\frac{2.33}{233.5} \mathrm{~g}=0.01 \text { mole } \quad(0.01 \mathrm{~mole})
\end{array}
$$
(i) $0.01$ mole molecule produce $0.01$ mole $\mathrm{AgCl}$ $\therefore$ one replaceable $\mathrm{Cl}^{\underline{\Theta}}$ ion so formula of complex is
$$
\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}
$$
(ii) Oxidation no. of Co is $+3$ so primary valency is 3 .
(iii) Coordination no. is 6 so sec. valency is 6 so ans. is 3,6
\begin{array}{l}
\text { M.wt }=59+12+14 \times 4+35.5 \times 3=233.5 \\
\mathrm{C}_{0} \mathrm{H}_{12} \mathrm{~N}_{4} \mathrm{Cl}_{3} \stackrel{\mathrm{AgNO}_{3}}{\longrightarrow} 1.435 \mathrm{~g} \mathrm{AgCl} \\
\frac{2.33}{233.5} \mathrm{~g}=0.01 \text { mole } \quad(0.01 \mathrm{~mole})
\end{array}
$$
(i) $0.01$ mole molecule produce $0.01$ mole $\mathrm{AgCl}$ $\therefore$ one replaceable $\mathrm{Cl}^{\underline{\Theta}}$ ion so formula of complex is
$$
\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}
$$
(ii) Oxidation no. of Co is $+3$ so primary valency is 3 .
(iii) Coordination no. is 6 so sec. valency is 6 so ans. is 3,6
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