Question

The oxidation state of Cr in $[Cr(N{H}_3{)}_{4}C{l}_2{]}^{+}$ is :

• A. 0
• B. +1
• C. +2
• D. +3

Answer 1

Answer: (D)

$[Cr(N{H}_3{)}_{4}C{l}_2{]}^{+}$
Let oxidation state of $Cr=x$
$N{H}_3=0$
$Cl=-1$
Net charge $=+1$
$\therefore$ $[Cr(N{H}_3{)}_{4}C{l}_2{]}^{+}$
$x+4\times 0+2(-1)=+1$
$\therefore x=3$