Question

What is the oxidation state of central atom in $K\left[Cr\right(N{H}_3{)}_{2}C{l}_4]$?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

$K\left[Cr{\left({NH}_3\right)}_{2}C{l}_4\right]$

$\Downarrow$

${K^{+}\left[Cr{\left({NH}_3\right)}_{2}C{l}_4\right]}^{-1}$

Central atom $=Cr$

Oxidation state of $N{H}_3$ in general $=0,Cl=-1$

Applying formula, $\text{Sum of total oxidation state of all atoms}=\text{Overall charge on the compound}.$

Let oxidation state of $Cr=xx+2\times \left(0\right)+4\times (-1)=-1,x=+3$

$\therefore$ Oxidation state of $Cr=+3.$