Download MARKS App - Trusted by 15,00,000+ IIT JEE & NEET aspirants! Download Now
Search any question & find its solution
Question: Answered & Verified by Expert
$2.33 \mathrm{~g}$ 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 : $\mathrm{Co}=59, \mathrm{Cl}=35.5, \mathrm{Ag}=108$ ]
ChemistrySome Basic Concepts of ChemistryKVPYKVPY 2016 (SB/SX)
Options:
  • A 3,6
  • B 3,4
  • C 2,4
  • D 4,3
Solution:
1303 Upvotes 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

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.