Question

A metal complex having composition $Cr(N{H}_3{)}_{4}C{l}_{2}Br$ has been isolated in two forms $\left(A\right)$ and $\left(B\right)$. The form $\left(A\right)$ reacts with $AgN{O}_3$ to give a white precipitate readily soluble in dilute aqueous ammonia, whereas $\left(B\right)$ gives a pale yellow precipitate soluble in concentrated ammonia. Write the formula of $\left(A\right)$ and $\left(B\right)$ and state the hybridization of chromium in each. Calculate their magnetic moments (spin- only value).

• A. $\sqrt{8}BM$
• B. $\sqrt{15}BM$
• C. $\sqrt{24}BM$
• D. None of these

Answer 1

Answer: (B)

$\sqrt{15}BM$

From reactions it can be seen that,

$A:\left[Cr\right(N{H}_3{)}_{4}ClBr]Cl$ gives white ppt of AgCl

$B:\left[Cr\right(N{H}_3{)}_{4}C{l}_2]Br$ gives a pale yellow precipitate of AgBr.

Here, as
$C{r}^{+3}$ is $d^{2}s{p}^3$-hybridized

$C{r}^{+3}\Rightarrow \left[Ar\right]4s^{0}3d^3$

and unpaired electrons $N=3$

so, magnetic moment$=\sqrt{N(N+2)}BM=\sqrt{15}BM$