Question

A metal complex having composition Cr(NH${}_3$)${}_4$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.The magnetic moments (spin- only value) of A and B are respectively:

• A. $\sqrt{8}BM$,$\sqrt{15}BM$
• B. $\sqrt{8}BM$,$\sqrt{24}BM$
• C. $\sqrt{15}BM$,$\sqrt{15}BM$
• D. $\sqrt{8}BM$,$\sqrt{8}BM$

Answer 1

The correct option is C $\sqrt{15}BM$,$\sqrt{15}BM$

Option (C) is correct.
A metal complex having composition.In this case, $Cr(N{H}_3{)}_{4}C{l}_{2}Br$ has been isolated in two forms A and B. The form A reacts with $AgN{O}_3$ to give a white precipitate readily soluble in dilute aqueous ammonia. This is a test for chloride ion. B gives a pale-yellow precipitate soluble in concentrated ammonia. This is a test for bromide ion.
$Ais\left[Cr\right(N{H}_3{)}_{4}ClBr]Cl.$
$\left[Cr\right(N{H}_3{)}_{4}ClBr]Cl+AgN{O}_3\to \underset{White}{AgCl}\downarrow +[Cr(N{H}_3{)}_{4}ClBr{]}^{+}N{O}_3^{-}$
$AgCl+2N{H}_{4}OH\to \left[Ag\right(N{H}_3{)}_{2}Cl]+2H_{2}O$
B is $\left[Cr\right(N{H}_3{)}_{4}C{I}_2]Br$.
$\left[Cr\right(N{H}_3{)}_{4}C{l}_2]Br+AgN{O}_3\to \underset{Yellow}{AgBr}\downarrow +[Cr(N{H}_3{)}_{4}C{l}_2{]}^{+}N{O}_3^{-}$
$AgBr+2N{H}_{4}OH\to \left[Ag\right(N{H}_3{)}_{2}Br]+2H_{2}O$
In both A and B :
The oxidation number of Cr is $+3$.
Atomic number = 24
$Cr=\left[Ar\right]3d^{5}4s^1$
$C{r}^{3+}=\left[Ar\right]3d^{3}4s^0$
$d^{2}s{p}^3$-Hybridization (octahedral)
There are three unpaired electrons, so the complex is paramagnetic.
Spin magnetic moment
$\mu =\sqrt{n(n+2)}BM=\sqrt{3(3+2)}BM=\sqrt{15}BM$