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# In the complex ion${{[Fe{{({{H}_{2}}O)}_{5}}NO]}^{2+}}$A. Fe is in the +1 oxidation state, and NO coordinates as $N{{O}^{+}}$ (nitrosonium ion)B. Fe is in the +2 oxidation state, and NO coordinates as neutral NO (nitrosyl) radicalC. Fe is in the +3 oxidation state and NO coordinates as $N{{O}^{-}}$D. Fe is in the +2 oxidation state and NO coordinates as $N{{O}^{+}}$

Last updated date: 09th Aug 2024
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Hint: In order to find the oxidation number of iron, the oxidation numbers for ${{H}_{2}}O$ and NO need to be known. Then an equation can be formed, which on solving would give the required oxidation number for iron.

In order to answer our question, we need to learn about the coordination compounds. Now, when a salt is dissolved in water, it dissociates into ions, for example, NaCl dissociates into $N{{a}^{+}}$and $C{{l}^{-}}$ions. A double salt is the complex form of the salt which has more entities in it, for example, Mohr’s salt. Coordination compounds also have many entities attached to the central metal atom, but they do not dissociate into ions, dissolving in water. Coordination may be defined as the number of coordinate bonds formed with central atom/ion the ligands. For example in coordination entity ${{[Ag{{(CN)}_{2}}]}^{-}},{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}},{{[Cr{{({{H}_{2}}O)}_{6}}]}^{3+}}$, the coordination number of Ag, Cu and Cr are 2, 4 and 6 respectively. Similarly in ${{[Fe{{({{C}_{2}}{{O}_{4}})}_{3}}]}^{3-}}\,and\,{{[Co{{(en)}_{3}}]}^{3+}}$ the coordination number of both Fe and Co is 6 because ${{C}_{2}}{{O}_{4}}^{2-}$(oxalate) and en (ethylene-1, 2-diamine) forms two coordinate bonds each with central metal atom/ion.
Oxidation number: Oxidation number of the central metal atom ion in a complex is the charge present on it if all the ligands are removed along with the electron pairs that are shared with the central atom. It is represented by Roman numerals in parentheses after the name of central atom For example, oxidation number of Co Fe and Ni in ${{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}},{{[Fe{{(CN)}_{6}}]}^{4-}}\,and\,[Ni{{(CO)}_{4}}]$ is +3, +2 and 0, and written as Co(i), Fe(ii) and Ni(0) respectively. The compound looks like:
Let the oxidation number of iron be ‘x’. Now, we have:$x+(5\times 0)+0=2$,as NO and ${{H}_{2}}O$have oxidation number 0. Also the compound is in+2 state. So, x comes out to be +2 and NO is present as nitrosyl radical.