Coordination Compounds 2 Question 66
####68. A metal complex having composition $Cr\left(NH _3\right) _4 Cl _2 Br$ has been isolated in two forms $A$ and $B$. The form $A$ reacts with $AgNO _3$ to give a white precipitate readily soluble in dilute aqueous ammonia, whereas $B$ gives a pale yellow precipitate soluble in concentrated ammonia.
Write the formula of $A$ and $B$ and state the hybridisation of chromium in each. Calculate their magnetic moments (spin-only value).
$(2001,5 M)$
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Solution:
- In complexes $A$ and $B$, one halide $\left(Cl^{-}\right.$or $\left.Br^{-}\right)$is outside coordination sphere, i.e. complexes are :
$\left[Cr\left(NH _3\right) _4 Br _2\right] Cl$ and $\left[Cr\left(NH _3\right) _4 BrCl\right] Br$
$A$ gives white precipitate $AgCl$ with excess of $AgNO _3$ which dissolve in excess ammonia. Therefore, $A$ must be $\left[Cr\left(NH _3\right) _4 Br _2\right] Cl$.
$B$ gives a pale yellow precipitate with excess of $AgNO _3$, which dissolve in concentrated ammonia solution. Therefore, precipitate is $AgBr$ and complex $B$ is $\left[Cr\left(NH _3\right) _4 ClBr\right] Br$.
In both $A$ and $B$, hybridisation of chromium is $d^{2} s p^{3}$ and magnetic moment : $\mu=\sqrt{n(n+2)} BM=0$
$\left(3 d^{6}\right.$, strong ligand, no unpaired electron)
