Given below are two statements: One is labelled as "Assertion A" and the other is labelled as "Reason R".
Assertion A: In the complex Ni(CO), and Fe(CO)s, the metals have zero oxidation state.
Reason R: Low oxidation states are found when a complex has ligands capable of π-donor character in addition to the o-bonding.
In the light of the above statements, choose the most appropriate answer from the option given below.
Assertion A: In the complex Ni(CO), and Fe(CO)s, the metals have zero oxidation state.
Reason R: Low oxidation states are found when a complex has ligands capable of π-donor character in addition to the o-bonding.
In the light of the above statements, choose the most appropriate answer from the option given below.
- Both A and R are correct and R is the correct explanation of A.
- Both A and R are correct but R is NOT the correct explanation of A
- A is correct but R is not correct
- A is not correct but R is correct
The Correct Option is C
Approach Solution - 1
Coordination Compound Oxidation States Analysis
Let's analyze both Assertion A and Reason R:
Assertion A: In \( \text{Ni(CO)}_4 \) and \( \text{Fe(CO)}_5 \), the carbonyl ligand (\( \text{CO} \)) is neutral. Thus, the oxidation state of Ni and Fe in these complexes is indeed zero. So, Assertion A is correct.
Reason R: Low oxidation states of metals are stabilized by synergistic bonding, also known as backbonding. This involves σ-donation from the ligand to the metal and π-backdonation from the metal to the ligand. Crucially, the ligand must have empty π orbitals to accept electrons from the metal. \( \text{CO} \) is a π-acceptor ligand, not a π-donor ligand. So, Reason R is incorrect.
Conclusion:
The correct answer is (3): A is correct, but R is not correct.
Approach Solution -2
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Classical Idea Of Redox Reactions – Oxidation And Reduction Reactions Questions
- Element not showing variable oxidation state is:
- In which of the following compound central atom has +4 oxidation state?
- Which of the following can not act as an oxidising agent?
- Structure of residue (A) and compound (B) formed respectively is:
- Low oxidation state of metals in their complexes are common when ligands :
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.