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JEE MainChemistry

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+4 marks−1 if incorrectAssertion–Reasonpyq

Coordination Compounds: JEE Main Chemistry Question with Solution

Given below are two statements : one is labelled as "Assertion A" and the other is labelled as "Reason R". Assertion A : In the complex

\text{Ni}(\text{CO})_4

and

\text{Fe}(\text{CO})_5

, the metals have zero oxidation state. Reason R : Low oxidation states are found when a complex has ligands capable of

\pi

-donor character in addition to the

\sigma

-bonding. In the light of the above statements, choose the most appropriate answer from the option given below.

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A

\text{A}

is correct but

\text{R}

is not correctB

\text{A}

is not correct but

\text{R}

is correctCBoth

\text{A}

and

\text{R}

are correct but

\text{R}

is NOT the correct explanation of

\text{A}

DBoth

\text{A}

and

\text{R}

are correct and

\text{R}

is the correct explanation of

\text{A}

.

Step-by-step solutionView

Correct answer

Assertion A is correct because CO is a neutral ligand giving zero oxidation states to Ni and Fe, but Reason R is incorrect because low oxidation states are stabilized by ligands with π-acceptor (not π-donor) character.

Option analysis

Why each option works or fails

A ·

\text{A}

is correct but

\text{R}

is not correct

None. The student correctly evaluates the oxidation states as zero and recognizes that low oxidation states require π-acceptor ligands rather than π-donor ligands. Carbon monoxide is a neutral ligand, giving metals an oxidation state of zero in homoleptic carbonyls. Low oxidation states are stabilized by synergic π-backbonding into π-acceptor (not π-donor) orbitals.

B ·

\text{A}

is not correct but

\text{R}

is correct

Confusing the metal's d-electron count or formal valence count with its formal oxidation state, or wrongly assuming CO acts as a π-donor to stabilize low oxidation states. CO is a neutral ligand, so the sum of oxidation numbers equals the net charge of zero. Remember that π-acceptor ligands stabilize electron-rich (low oxidation state) metal centers by removing excess electron density.

C · Both

\text{A}

and

\text{R}

are correct but

\text{R}

is NOT the correct explanation of

\text{A}

Believing that ligands stabilize low oxidation states by donating π-electron density onto the already electron-rich metal center. In low oxidation states, metals are electron-rich and need ligands that act as π-acceptors (like CO) to accept back-donated electron density, not π-donors which would increase electron repulsion.

D · Both

\text{A}

and

\text{R}

are correct and

\text{R}

is the correct explanation of

\text{A}

.

Failing to distinguish between π-acid (π-acceptor) character and π-donor character in synergic bonding. Look carefully at the wording: carbonyl ligands are π-acceptors, not π-donors. Synergic bonding involves σ-donation from CO to the metal and π-backdonation from the metal to CO.

Reviewed route

Solution

StepWorking

01assertion_eval

In $\text{Ni}(\text{CO})_4$ and $\text{Fe}(\text{CO})_5$ , carbon monoxide ( $\text{CO}$ ) is a neutral ligand with an oxidation state of 0. Since the overall complexes are neutral, the oxidation states of $\text{Ni}$ and $\text{Fe}$ are both $0$ . Thus, Assertion A is correct (true).

02reason_eval

Transition metals in low or zero oxidation states have high electron density. They are stabilized by synergic bonding with ligands that act as $\sigma$ -donors and $\pi$ -acceptors (such as $\text{CO}, \text{CN}^-$ ), which delocalize electron density from filled metal $d$ -orbitals into vacant ligand $\pi^*$ orbitals (back-bonding). $\pi$ -donor ligands would pump additional electron density onto an already electron-rich metal, destabilizing low oxidation states. Hence, low oxidation states require $\pi$ -acceptor character, not $\pi$ -donor character. Thus, Reason R is incorrect (false).

✓link_test

Since Reason R is factually false, it cannot explain Assertion A.

Hints that build this answer step by step

What is the oxidation state of the metal in $\text{Ni}(\text{CO})_4$ and $\text{Fe}(\text{CO})_5$ ?
Zero, because carbon monoxide (CO) is a neutral ligand and the complexes have no net charge.
Which type of $\pi$ -character in ligands stabilizes metal centers in low (e.g., zero or negative) oxidation states?
$\pi$ -acceptor character, allowing back-donation of electron density from the metal $d$ -orbitals into empty ligand $\pi^*$ orbitals.

✓ Source and academic review

Question type: Assertion–Reason
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why must the ligand be a

\pi

-acceptor rather than a

\pi

-donor for stabilizing zero-valent metals?

In zero or negative oxidation states, the metal has excess negative/d-electron charge. A $\pi$ -acceptor ligand withdraws this excess electron density via back-bonding into empty $\pi^*$ antibonding orbitals, stabilizing the low oxidation state. A $\pi$ -donor would increase repulsion.

Answer

Assertion A is correct because CO is a neutral ligand giving zero oxidation states to Ni and Fe, but Reason R is incorrect because low oxidation states are stabilized by ligands with π-acceptor (not π-donor) character.

Why each option works or fails

A · correct: $\text{A}$ is correct but $\text{R}$ is not correct - None. The student correctly evaluates the oxidation states as zero and recognizes that low oxidation states require π-acceptor ligands rather than π-donor ligands. Carbon monoxide is a neutral ligand, giving metals an oxidation state of zero in homoleptic carbonyls. Low oxidation states are stabilized by synergic π-backbonding into π-acceptor (not π-donor) orbitals.
B: $\text{A}$ is not correct but $\text{R}$ is correct - Confusing the metal's d-electron count or formal valence count with its formal oxidation state, or wrongly assuming CO acts as a π-donor to stabilize low oxidation states. CO is a neutral ligand, so the sum of oxidation numbers equals the net charge of zero. Remember that π-acceptor ligands stabilize electron-rich (low oxidation state) metal centers by removing excess electron density.
C: Both $\text{A}$ and $\text{R}$ are correct but $\text{R}$ is NOT the correct explanation of $\text{A}$ - Believing that ligands stabilize low oxidation states by donating π-electron density onto the already electron-rich metal center. In low oxidation states, metals are electron-rich and need ligands that act as π-acceptors (like CO) to accept back-donated electron density, not π-donors which would increase electron repulsion.
D: Both $\text{A}$ and $\text{R}$ are correct and $\text{R}$ is the correct explanation of $\text{A}$ . - Failing to distinguish between π-acid (π-acceptor) character and π-donor character in synergic bonding. Look carefully at the wording: carbonyl ligands are π-acceptors, not π-donors. Synergic bonding involves σ-donation from CO to the metal and π-backdonation from the metal to CO.

Step-by-step solution

assertion_eval: In $\text{Ni}(\text{CO})_4$ and $\text{Fe}(\text{CO})_5$ , carbon monoxide ( $\text{CO}$ ) is a neutral ligand with an oxidation state of 0. Since the overall complexes are neutral, the oxidation states of $\text{Ni}$ and $\text{Fe}$ are both $0$ . Thus, Assertion A is correct (true).
reason_eval: Transition metals in low or zero oxidation states have high electron density. They are stabilized by synergic bonding with ligands that act as $\sigma$ -donors and $\pi$ -acceptors (such as $\text{CO}, \text{CN}^-$ ), which delocalize electron density from filled metal $d$ -orbitals into vacant ligand $\pi^*$ orbitals (back-bonding). $\pi$ -donor ligands would pump additional electron density onto an already electron-rich metal, destabilizing low oxidation states. Hence, low oxidation states require $\pi$ -acceptor character, not $\pi$ -donor character. Thus, Reason R is incorrect (false).
link_test: Since Reason R is factually false, it cannot explain Assertion A.