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# Which of the following statements is/are correct?A. The oxidation state of iron in sodium nitroprusside $N{{a}_{2}}[Fe{{(CN)}_{5}}(NO)]$ is +2.B. ${{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}$ is liner in shapeC. In ${{[Fe{{({{H}_{2}}O)}_{6}}]}^{+3}}$ , Fe is ${{d}^{2}}s{{p}^{3}}$ hybridized.D. In $Ni{{(CO)}_{4}}$ , the oxidation state of Ni is zero.

Last updated date: 12th Jul 2024
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Hint: The oxidation number of the central metal atom in coordination complexes is going to depend on the oxidation number of the ligands attached to the metal. Every ligand has its own oxidation state.

- In the question, it is asked which statement is correct among the given options.
- Coming to the given options, Option A. The oxidation state of iron in sodium nitroprusside $N{{a}_{2}}[Fe{{(CN)}_{5}}(NO)]$ is +2.
- Oxidation of the iron in $N{{a}_{2}}[Fe{{(CN)}_{5}}(NO)]$ is as follows.
X+ 5 (-1) + 2 (1) + 1 = 2.
Here X = oxidation number of the iron.
- So, option A is correct.
- Coming to option B, ${{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}$ is liner in shape.
- ${{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}$ undergoes through sp hybridization with zero lone pair of electrons. So, the shape of the ${{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}$ is linear.
- So, option B is also correct.
- Coming to option C, In ${{[Fe{{({{H}_{2}}O)}_{6}}]}^{+3}}$, Fe is ${{d}^{2}}s{{p}^{3}}$ hybridized. Generally iron won’t form an inner orbital complex with ligands. So, the hybridization of iron (${{d}^{6}}$ ) is $s{{p}^{3}}{{d}^{2}}$ .
- So, option C is incorrect.
- Coming to option D, In $Ni{{(CO)}_{4}}$, the oxidation state of Ni is zero.
- We know that carbonyl (CO) ligands have zero oxidation state. Therefore the oxidation number of the nickel in $Ni{{(CO)}_{4}}$ is zero.
- So, option D is also correct. So, the correct answer is “Option A,B and D”.

Note: The shape of the coordination complex also depends on the presence of the lone pair of electrons on the central metal atom. Due to the bond pair and lone pair repulsions, the shape of the coordination complex is going to change.