Question

How many unpaired electrons are present in $[Cr(NH{)}_3]B{r}_3$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

$[Cr(N{H}_3{)}_5]B{r}_3$
Suppose oxidation state of $Cr$ is $x,$
then. $x+(5\times 0)+(-1\times 3)=0$
$x-3=0$
$x=+3$
${}_{24}Cr=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5,4s^1$
$C{r}^{3+}=1s^{2}2s^{2}2p^6,3s^{2}3p^{6}3d^3$
So, three unpaired electrons are present in
$[Cr(N{H}_3)]B{r}_3.$