Question

The complex ${\left[ {{\text{Fe}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{5}}}{\text{NO}}} \right]^{{\text{2 + }}}}$ is formed in the brown ring test for nitrates when freshly prepared ${\text{FeS}}{{\text{O}}_{\text{4}}}$ solution is added to aqueous solution of ${\text{NO}}_{\text{3}}^ - $ followed by addition of concentrated ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$. Select the correct statement about this complex.
A.Colour change is due to charge transfer.
B.It has iron in $ + 1$ oxidation state and nitrosyl as ${\text{N}}{{\text{O}}^{\text{ + }}}$
C.It has magnetic moment of $3.87{\text{BM}}$ confirming three unpaired electrons of ${\text{Fe}}$
D.All of the above statements are correct

Answer 1

Hint: To solve this question, you must recall the electronic arrangement in the ferrous electron in the given compound. Nitrosyl is a strong field ligand and water is a weak field ligand.

Complete step by step answer:
Colour change in a coordination compound happens due to either ${\text{d - d}}$ transfer or charge transfer. We can see that ${\text{d - d}}$transfer is not possible in the given compound due to absence of d orbitals in the ligand, thus, it is clear that the colour change is due to charge transfer.
We know that the oxidation state of the nitrosyl ligand is $ + 1$. Thus, the oxidation state of iron is $ + 1$ and it is present as ${\text{F}}{{\text{e}}^{\text{ + }}}$.
The electronic configuration of ${\text{F}}{{\text{e}}^{\text{ + }}}$is ${\text{F}}{{\text{e}}^{\text{ + }}}:\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^6}{\text{4}}{{\text{s}}^1}$
Since nitrosyl is a strong field ligand, it causes the pairing of an electron. However water is a weak field ligand and does not cause the pairing of electrons. Thus the electron in the $4s$ orbital is paired with an electron in the $3d$ orbital
 The electronic configuration is now, ${\text{F}}{{\text{e}}^{\text{ + }}}:\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^7}{\text{4}}{{\text{s}}^0}$
It is clear that the number of unpaired electrons is $3$.
We can write the magnetic moment of the complex as, $\mu = \sqrt {n\left( {n + 2} \right)} = \sqrt {3 \times 5} $
$\therefore \mu = 3.87{\text{BM}}$
All the above options are correct.

Thus, the correct answer is D.

Note:

${\text{CO,}}{\left( {{\text{CN}}} \right)^ - }{\text{,NO}}_2^ - {\text{,phen,dipy,en,N}}{{\text{H}}_{\text{3}}}$ are strong field ligands. These ligands cause the pairing of unpaired electrons in the atom with which they form a coordination bond.
Generally, the ligands which donate through their $\pi $ orbitals are weak (known as weak field ligands) while those which donate through sigma orbitals are comparatively stronger and those which donate through sigma and accept through $\pi $orbitals are even stronger (known as strong field ligands)