Search any question & find its solution
Question:
Answered & Verified by Expert
An element ' $X$ ' with the atomic number 13 forms a complex of the type $\left[\mathrm{XCl}\left(\mathrm{H}_2 \mathrm{O}\right)_5\right]^{2+}$. The covalency and oxidation state of $X$ in it are respectively
Options:
Solution:
1397 Upvotes
Verified Answer
The correct answer is:
$6,+3$
Element ' $X$ ' is forming a complex of the type
$\left[X \mathrm{Cl}\left(\mathrm{H}_2 \mathrm{O}\right)_5\right]^{2+}$.
Total number of ligands attached to $X$ is 6 (one $\mathrm{Cl}$ and five $\mathrm{H}_2 \mathrm{O}$ ).
$\mathrm{Cl}$ is a unidentate ligand with charge -1 .
$\mathrm{H}_2 \mathrm{O}$ is a unidentate ligand with charge 0.
$\therefore \quad x+(-1)+0=+2$
$x-1=+2$
$x=+2+1=+3$
$\therefore$ Oxidation state of $x$ is +3 .
$\left[X \mathrm{Cl}\left(\mathrm{H}_2 \mathrm{O}\right)_5\right]^{2+}$.
Total number of ligands attached to $X$ is 6 (one $\mathrm{Cl}$ and five $\mathrm{H}_2 \mathrm{O}$ ).
$\mathrm{Cl}$ is a unidentate ligand with charge -1 .
$\mathrm{H}_2 \mathrm{O}$ is a unidentate ligand with charge 0.
$\therefore \quad x+(-1)+0=+2$
$x-1=+2$
$x=+2+1=+3$
$\therefore$ Oxidation state of $x$ is +3 .
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.