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In the brown ring complex [Fe(H$_2$O)$_5$(NO)]SO$_4$ nitric oxide behaves as:
A) NO$^{+}$
B) neutral NO molecule
C) NO$^{-}$
D) NO$_2$$^{+}$

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Hint: The behaviour of nitric oxide can be determined by calculating the oxidation state. The electron transfer can be observed, but the whole complex is neutral with the charge 0. So, it will help to observe the behaviour in the coordination compound.

Complete step by step answer:
Now, we are given the complex [Fe(H$_2$O)$_5$(NO)]SO$_4$. As mentioned, the complex is neutral. So, we can calculate the oxidation state of Fe in the ring. The ligands present in the complex are aqua, nitrosyl, and sulphate ion.
Thus, the oxidation state of Fe can be written as:
x + 5 (0) + (+1) + (-2) = 0, as the complex is neutral, so it is equal to zero.
Thus, x = +1, so, we can say that the oxidation state of Fe is +1.
Now, we can say that the charge on anion i.e. sulphate ion is -2, and cationic complexes must have +2 charge due to the neutral ring.
Therefore, the charge on Fe will be +2. In other terms, we can say that the NO to Fe, the electrons will be transferred.
Thus, both the atoms will attain a charge of +1, and it leads to the +1 oxidation state of NO.
In the last we can conclude that the NO will exist in +1 oxidation state, as the charge is transferred. So, in the brown ring complex [Fe(H$_2$O)$_5$(NO)]SO$_4$, nitric oxide behaves as NO$^{+}$.

Hence, the correct option is (A).

Note: Don’t get confused about how electrons are transferred from NO to Fe. We can also say that the water ligand, or aqua is replaced by the NO, and it gets oxidized to the ligand named nitrosyl. So, the oxidation state of Fe is +1.