Organometallic compounds Questions in English

(B) $\pi$-complex is an organometallic compound where the metal atom is bonded to an organic ligand through $\pi$-electrons of the ligand.
Ferrocene,with the formula $Fe(\eta^5-C_5H_5)_2$,is a classic example of a sandwich compound where the iron atom is bonded to two cyclopentadienyl rings via $\pi$-electron clouds.
Trimethylaluminium and diethylzinc are primarily $\sigma$-bonded organometallic compounds.
Ethylenediamine is a chelating ligand,not an organometallic $\pi$-complex.
Therefore,the correct answer is Ferrocene.

(C) In metal carbonyls,the metal-carbon bond possesses both $\sigma$ and $\pi$ character.
The $M-C$ $\sigma$-bond is formed by the donation of a lone pair of electrons from the carbonyl carbon into a vacant orbital of the metal.
The $M-C$ $\pi$-bond is formed by the back-donation of a pair of electrons from a filled $d$-orbital of the metal into the vacant antibonding $\pi^*$ orbital of carbon monoxide.

(D) The sandwich structure of ferrocene,$Fe(C_5H_5)_2$,was established through experimental evidence.
$X$-ray diffraction studies provided the structural geometry,showing the iron atom is sandwiched between two parallel cyclopentadienyl rings.
Infrared spectroscopy further confirmed the equivalence of all $C-H$ bonds and the presence of the metal-ring interaction.
Therefore,both techniques were instrumental in confirming the sandwich structure.

(C) In $K[Pt(C_2H_4)Cl_3]$,the $Pt$ is bonded to the $C_2H_4$ ligand through the carbon atoms (Zeise's salt).
In $[Ni(CO)_4]$,the $Ni$ is bonded to the $C$ atom of the $CO$ ligand.
In $C_2H_5MgBr$,the $Mg$ is directly bonded to the $C$ atom of the ethyl group (Grignard reagent).
In $Al(OC_2H_5)_3$,the $Al$ atom is bonded to the oxygen atom of the ethoxide group,not the carbon atom. Therefore,it does not contain a metal-carbon bond.

(B) The complex $Fe(C_5H_5)_2$ is a well-known organometallic compound called ferrocene.
According to $IUPAC$ nomenclature for coordination compounds,the ligand $C_5H_5^-$ is named 'cyclopentadienyl'.
Since there are two such ligands,the prefix 'bis' is used.
The central metal atom is iron,and its oxidation state is calculated as $x + 2(-1) = 0$,so $x = +2$.
Thus,the correct $IUPAC$ name is bis(cyclopentadienyl)iron $(II)$.

(A) The correct answer is $A$.
Grignard reagent is represented as $R-Mg-X$,which is an organometallic compound containing a polar $C-Mg$ $\sigma$-bond.
It does not contain any $\pi$-bonding.
In contrast,Dibenzene chromium,Zeise’s salt,and Ferrocene are all organometallic compounds that involve $\pi$-bonding between the metal center and the organic ligands (such as $\eta^6-C_6H_6$,$\eta^2-C_2H_4$,and $\eta^5-C_5H_5$ respectively).

(C) The organometallic compound of $Mg$ is known as Grignard reagent $(R-Mg-X)$,where $R$ is an alkyl or aryl group and $X$ is a halogen.

(A) Grignard's reagent,represented as $RMgX$,contains a direct $\sigma$-bond between the carbon atom of the alkyl group and the magnesium metal atom.
Therefore,it is classified as a $\sigma$-bonded organometallic compound.
In contrast,ferrocene,cobaltocene,and ruthenocene are examples of $\pi$-bonded organometallic compounds (metallocenes).

(A) In metal carbonyl complexes,the metal-carbon bond is formed by the donation of lone pairs from $CO$ to the metal,and back-bonding occurs from the metal $d$-orbitals to the empty $\pi^*$ antibonding orbitals of $CO$.
As the negative charge on the metal carbonyl complex increases,the electron density on the metal increases,which enhances the back-donation of electrons into the $\pi^*$ antibonding orbitals of $CO$.
This increases the population of the antibonding orbitals,which weakens the $C-O$ bond and increases its bond length.
Comparing the complexes:
$1. [Mn(CO)_6]^+$: Metal has a positive charge,least back-bonding.
$2. [Ni(CO)_4]$: Neutral complex.
$3. [Co(CO)_4]^-$: Negative charge,more back-bonding.
$4. [Fe(CO)_4]^{2-}$: Highest negative charge,maximum back-bonding.
Therefore,$[Fe(CO)_4]^{2-}$ has the longest $C-O$ bond length. The order is $[Mn(CO)_6]^+ < [Ni(CO)_4] < [Co(CO)_4]^- < [Fe(CO)_4]^{2-}$.

(A) The strength of the $C-O$ bond in metal carbonyls depends on the extent of back-bonding from the metal $d$-orbitals to the $\pi^*$-antibonding orbitals of the $CO$ ligand.
As the negative charge on the metal complex increases,the metal atom becomes more electron-rich and donates more electron density into the $\pi^*$-orbitals of $CO$,which weakens the $C-O$ bond.
Conversely,as the positive charge on the central metal atom increases,the metal's ability to donate electron density decreases.
Therefore,the $C-O$ bond is strongest in the species with the highest positive charge or lowest negative charge,as there is minimal back-bonding.
Among the given options,$Mn(CO)_6^+$ has the highest positive charge,resulting in the least back-bonding and the strongest $C-O$ bond.

(D) In metal carbonyls,the metal-carbon bond is formed by the donation of lone pair electrons from $CO$ to the metal (sigma donation) and the back-donation of electrons from the metal $d$-orbitals to the empty antibonding $\pi^*$ orbitals of $CO$ ($\pi$-backbonding).
This $\pi$-backbonding increases the electron density in the antibonding $\pi^*$ orbital of $CO$,which decreases the $CO$ bond order.
Simultaneously,the $\pi$-backbonding introduces partial double bond character to the $M-C$ bond,thereby increasing the $M-C$ bond order.

(D) Synergic bonding occurs in metal carbonyls and metal cyanides where there is a donation of electron density from the ligand to the metal ($L \rightarrow M$ $\sigma$-bond) and a back-donation of electron density from the metal to the empty antibonding $\pi^*$ orbitals of the ligand ($M \rightarrow L$ $\pi$-bond).
$1$. $[Mo(CO)_6]$ is a metal carbonyl complex that exhibits this synergic bonding.
$2$. $[Mn(CO)_6]^-$ is an anionic metal carbonyl complex that also exhibits this synergic bonding.
$3$. $[Ni(CN)_4]^{4-}$ is a metal cyanide complex where $CN^-$ acts as a $\pi$-acceptor ligand,exhibiting synergic bonding similar to $CO$.
Therefore,all the given species show synergic bonding.

(B) The statement "All the transition metals form monometallic carbonyls" is incorrect.
There are many metal carbonyls that contain two or more metal atoms per molecule,known as polynuclear metal carbonyls.
For example,$Mn_2(CO)_{10}$ and $Fe_2(CO)_9$ are polynuclear carbonyls.

(D) Ferrocene is an organometallic compound with the formula $[(C_5H_5)_2Fe]$.
It is a sandwich complex consisting of two cyclopentadienyl rings $(C_5H_5^-)$ bound on opposite sides of a central iron atom $(Fe^{2+})$.
It is an orange solid that sublimes above room temperature and is soluble in most organic solvents.

(D) Ziese's salt is $[PtCl_3(C_2H_4)]^-$.
In the ethylene ligand $(C_2H_4)$,there are $4$ $C-H$ sigma bonds and $1$ $C-C$ sigma bond.
There are $3$ $Pt-Cl$ sigma bonds.
The interaction between $Pt$ and the ethylene ligand involves a coordinate bond (sigma donation from the $\pi$-bond of ethylene to $Pt$),which counts as $1$ sigma bond.
Total sigma bonds = $4$ $(C-H)$ + $1$ $(C-C)$ + $3$ $(Pt-Cl)$ + $1$ $(Pt-ethylene)$ = $9$ sigma bonds.
Since $9$ is not among the options,the correct answer is $D$ (none of these).

(C) In Zeise's salt,the ethylene molecule acts as an $\eta^2$-ligand.
Due to the synergic bonding (back-donation from $Pt$ $d$-orbitals to the $\pi^*$ orbital of ethylene),the $C=C$ bond length increases and the hybridization of carbon atoms changes from $sp^2$ towards $sp^3$.
Consequently,the ethylene molecule becomes distorted,and the $HCH$ bond angle decreases compared to that in a free ethylene molecule.
Therefore,statement $C$ is correct.

(D) The $C-O$ stretching frequency in metal carbonyls depends on the extent of back-bonding from the metal to the $CO$ ligand.
Greater back-bonding increases the electron density in the $CO$ antibonding $\pi^*$ orbital,which weakens the $C-O$ bond and decreases the stretching frequency.
Back-bonding is inversely proportional to the negative charge on the complex and directly proportional to the positive charge.
Comparing the complexes:
$1. [Fe(CO)_4]^{2-}$ (Charge: $-2$)
$2. [Co(CO)_4]^-$ (Charge: $-1$)
$3. [Ni(CO)_4]$ (Charge: $0$)
$4. [Mn(CO)_6]^+$ (Charge: $+1$)
As the positive charge increases,the metal becomes more electron-deficient,reducing the back-donation to the $CO$ ligands.
Therefore,$[Mn(CO)_6]^+$ has the least back-bonding,resulting in the strongest $C-O$ bond and the highest $IR$ stretching frequency.

(D) In a redox reaction,an oxidising agent accepts electrons and undergoes reduction.
For a metal carbonyl complex to act as an oxidising agent,it should ideally have an $EAN$ (Effective Atomic Number) less than the atomic number of the nearest noble gas to achieve stability by gaining an electron.
For $[V(CO)_6]$,the $EAN$ is calculated as: $V$ $(Z=23)$ + $6 \times 2$ (from $CO$) = $23 + 12 = 35$.
The nearest noble gas is Krypton $(Kr)$ with $Z=36$.
Since $35 < 36$,the complex $[V(CO)_6]$ readily accepts an electron to form $[V(CO)_6]^-$,which has an $EAN$ of $36$,thereby acting as an oxidising agent.

(C) In $V(CO)_6$, the vanadium atom is in a neutral oxidation state. In $[V(CO)_6]^-$, the vanadium atom is in a $-1$ oxidation state.
Due to the negative charge on the complex $[V(CO)_6]^-$, there is an increased back-bonding from the metal $d$-orbitals to the $\pi^*$ antibonding orbitals of the $CO$ ligands.
This increased back-bonding strengthens the $V-C$ bond (increasing its double bond character) and weakens the $C-O$ bond.
Consequently, the $V-C$ bond length in $V(CO)_6$ is approximately $200 \ pm$, whereas in $[V(CO)_6]^-$, the increased electron density on the metal facilitates stronger back-donation, resulting in a shorter $V-C$ bond length of approximately $193 \ pm$.

(B) Organometallic compounds containing carbonyl $(CO)$ ligands exhibit both $\sigma$ and $\pi$ bonding.
In $[Co(CO)_5(NH_3)]^{2+}$,the $CO$ ligand forms a $\sigma$ bond by donating a lone pair to the metal and a $\pi$ bond by accepting electron density from the metal $d$-orbitals into its empty $\pi^*$ antibonding orbitals (synergic bonding).
Therefore,the correct option is $B$.

(D) In $Fe(CO)_5$,the $Fe-C$ bond is formed by the donation of a lone pair of electrons from the $CO$ ligand to the $Fe$ atom,which forms a $\sigma$ bond.
Additionally,there is $d\pi-p\pi$ back-bonding where electrons from the filled $d$-orbitals of $Fe$ are donated into the empty antibonding $\pi^*$ orbitals of the $CO$ ligand,which forms a $\pi$ bond.
Therefore,the $Fe-C$ bond possesses both $\sigma$ and $\pi$ character.

(A) In metal carbonyls like $Fe(CO)_5$,the metal-carbon bond involves synergic bonding.
This involves the donation of electron density from the $CO$ lone pair to the metal $d$-orbitals ($\sigma$-donation) and the back-donation of electron density from the metal $d$-orbitals to the empty antibonding $\pi^*$ orbitals of $CO$ ($\pi$-backbonding).
This back-donation increases the electron density in the antibonding $\pi^*$ orbital of $CO$,which decreases the bond order of the $C-O$ bond.
$A$ decrease in bond order leads to an increase in the $C-O$ bond length.
Therefore,the $C-O$ bond length in $Fe(CO)_5$ $(1.158 \ \mathring{A})$ is greater than that in free $CO$ $(1.128 \ \mathring{A})$.

(C) The Ziegler-Natta catalyst is a combination of an organometallic compound of a Group $1-3$ metal (typically an alkyl aluminum compound like $R_3Al$ or $R_2AlCl$) and a transition metal compound (typically a titanium halide like $TiCl_4$ or $TiCl_3$).
Therefore,the correct composition is $R_3Al + TiCl_4$.

(A) The structure of the metal carbonyl complex $[Co_2(CO)_8]$ in the solid state consists of two $Co(CO)_4$ units linked by a metal-metal bond.
It contains one $Co-Co$ bond.
There are six terminal $CO$ ligands (three on each $Co$ atom) and two bridging $CO$ ligands.
Therefore,the correct option is $A$.

(C) The Wilkinson catalyst is a well-known organometallic complex used as a homogeneous catalyst for the hydrogenation of alkenes.
Its chemical formula is $RhCl(PPh_3)_3$,which is represented as $[(Ph_3P)_3RhCl]$.
It consists of a rhodium center coordinated to three triphenylphosphine ligands and one chloride ion.

(A) The structure of $Co_2(CO)_8$ contains two bridging $CO$ ligands that connect the two cobalt atoms.
Additionally,there is a direct metal-metal bond between the two cobalt atoms ($Co-Co$ bond).
Therefore,the number of bridging $CO$ ligands is $2$ and the number of $Co-Co$ bonds is $1$.

(A) An organometallic compound is defined as a compound containing at least one direct chemical bond between a metal atom and a carbon atom of an organic group or molecule.
In $Mn_2(CO)_{10}$,the manganese $(Mn)$ atoms are directly bonded to the carbon $(C)$ atoms of the carbonyl $(CO)$ ligands.
Therefore,the presence of the $Mn-C$ bond makes it an organometallic compound.

(D) In $PR_3$,the phosphorus atom has vacant $d$-orbitals that can participate in synergic bonding.
Specifically,the vacant non-axial $d$-orbitals of the phosphorus atom accept electron density from the filled metal $d$-orbitals,facilitating $M \xrightarrow{\pi} PR_3$ back-bonding in its complexes.

(B) Zeise's salt is $K[PtCl_3(\eta^2-C_2H_4)]$.
In this organometallic complex,the ethylene ligand $(C_2H_4)$ acts as a $\sigma$-donor by donating electron density from its $\pi$-orbital to an empty $d$-orbital of the $Pt^{2+}$ metal center.
Simultaneously,the metal center acts as a $\pi$-donor by back-donating electron density from its filled $d$-orbital into the empty $\pi^*$-antibonding orbital of the ethylene ligand,making the ethylene ligand a $\pi$-acceptor.
Therefore,the metal center acts as a $\sigma$-acceptor (receiving $\sigma$-donation) and a $\pi$-donor (providing $\pi$-back-donation).
However,in the context of the ligand-metal interaction description,the metal center is characterized by its ability to accept $\sigma$-electrons and donate $\pi$-electrons.

(A) $Mn(CO)_5$ is an oxidizing agent because it has an odd number of electrons ($17$ valence electrons).
It gains an electron to achieve the stable $18$-electron configuration (Effective Atomic Number of $Kr = 36$).
The reaction is: $Mn(CO)_5 + e^- \rightarrow [Mn(CO)_5]^-$.
Since it accepts an electron,it acts as an oxidizing agent.

(D) $(D)$. $CO$ is called a $\pi$-acid ligand.
In metal carbonyl complexes,there is a donation of an electron pair from the carbon atom to the empty orbital of the metal.
Simultaneously,a back $\pi$-bonding is formed by the sideways overlap of a filled $d$-orbital on the metal with the empty antibonding $\pi^*_{2py}$ orbital of $CO$.
This is known as $d \pi - p \pi$ back bonding,not $p \pi - p \pi$ back bonding.

(D) The $CO$ ligand is a $2e^-$ donor,while the $NO$ ligand acts as a $3e^-$ donor in neutral metal carbonyls.
To maintain the $18e^-$ rule,the total number of electrons must remain constant.
Substituting $CO$ with $NO$ requires that the total electron count remains unchanged.
For $Cr(CO)_6$ $(18e^-)$,replacing three $CO$ $(3 \times 2e^- = 6e^-)$ with two $NO$ $(2 \times 3e^- = 6e^-)$ gives $Cr(CO)_3(NO)_2$,which is valid.
For $Fe(CO)_5$ $(18e^-)$,replacing three $CO$ with two $NO$ gives $Fe(CO)_2(NO)_2$,which is valid.
For $Ni(CO)_4$ $(18e^-)$,replacing two $CO$ with one $NO$ would give $Ni(CO)_2(NO)$,but $Ni(CO)_2(NO)_2$ would result in $20e^-$,which is incorrect.
Thus,$Ni(CO)_2(NO)_2$ is not a stable neutral complex following the $18e^-$ rule.

(B) The $CO$ bond order is inversely proportional to the extent of back-bonding from the metal to the $CO$ ligand $(M \to CO)$.
$1$. In $H_3B \leftarrow CO$,there is no back-bonding; instead,$B$ acts as a Lewis acid,which can increase the $CO$ bond order slightly above $3.0$.
$2$. In free $CO$,the bond order is $3.0$.
$3$. In $Fe(CO)_5$,back-bonding occurs from $Fe$ to $CO$,reducing the $CO$ bond order to less than $3.0$.
$4$. In $[Mn(CO)_5]^-$,the negative charge on the metal center increases the electron density,leading to stronger back-bonding compared to $Fe(CO)_5$,thus further reducing the $CO$ bond order.
Therefore,the correct sequence of $CO$ bond order is $R > Q > P > S$.

(A) Wilkinson's catalyst is $[RhCl(PPh_3)_3]$,which is red-violet in color and has a square planar structure.
It is used for the selective hydrogenation of organic molecules at room temperature and pressure.
$TiCl_4 + (C_2H_5)_3Al$ is the Ziegler-Natta catalyst.
$(C_2H_5)_4Pb$ is an anti-knocking agent.
$[PtCl_2(NH_3)_2]$ is cisplatin,which is used as an anti-cancer agent.

(C) Tartar Emetic is potassium antimonyl tartrate. Its chemical formula is $K[C_4H_2O_6(SbO)] \cdot 0.5H_2O$ or written as a structural representation: $\begin{array}{*{20}{c}} CH(OH)COO(SbO) \\ | \\ CH(OH)COOK \end{array}$.

(C) Zeise's salt is a well-known organometallic compound with the chemical formula $K[Pt(\eta^2-C_2H_4)Cl_3] \cdot H_2O$.
It is historically significant as one of the first organometallic compounds discovered,containing a platinum atom coordinated to an ethylene molecule through its $\pi$-electrons ($\eta^2$-bonding).

(D) Organometallic compounds are classified based on the nature of the metal-carbon bond.
$A$. $Al_2(CH_3)_6$ (Trimethylaluminium dimer) contains $\sigma-$bonded methyl groups.
$B$. $Pb(CH_3)_4$ (Tetramethyllead) contains $\sigma-$bonded methyl groups.
$C$. $Zn(C_2H_5)_2$ (Diethylzinc) contains $\sigma-$bonded ethyl groups.
$D$. Ferrocene,$[Fe(\eta^5-C_5H_5)_2]$,is a $\pi-$bonded organometallic compound (sandwich complex) where the metal is bonded to the $\pi-$electron cloud of the cyclopentadienyl rings.
Therefore,Ferrocene is not a $\sigma-$bonded organometallic compound.

(D) Organometallic compounds are classified based on the nature of the metal-carbon bond.
$\pi-$bonded organometallic complexes involve ligands like alkenes,alkynes,or arenes that donate $\pi-$electron density to the metal center.
$1$. Ferrocene $[Fe(\eta^5-C_5H_5)_2]$ is a sandwich complex with $\pi-$bonding.
$2$. Dibenzene chromium $[Cr(\eta^6-C_6H_6)_2]$ is a sandwich complex with $\pi-$bonding.
$3$. Zeise's salt $[K[PtCl_3(\eta^2-C_2H_4)]]$ contains an ethylene ligand bonded via $\pi-$electrons.
$4$. Tetraethyl lead $[Pb(C_2H_5)_4]$ contains $Pb-C$ $\sigma-$bonds only,where the ethyl group is bonded to the metal through a single carbon atom. Therefore,it is a $\sigma-$bonded organometallic compound.

(C) In the neutral complex $[V(CO)_6]$, the vanadium atom is in a lower oxidation state compared to the anionic complex $[V(CO)_6]^-$.
However, the extent of back-bonding from the metal $d$-orbitals to the $\pi^*$ orbitals of the $CO$ ligand is significant in both.
Experimental data shows that the $V-C$ bond distance in $[V(CO)_6]$ is approximately $200 \ pm$.
In the anionic complex $[V(CO)_6]^-$, the increased electron density on the metal enhances the back-bonding, which strengthens the $V-C$ bond and shortens the bond length to approximately $193 \ pm$.
Therefore, the distances are $200 \ pm$ and $193 \ pm$ respectively.

(A) In $Fe(CO)_5$,there exists a synergic bond between $Fe$ and $CO$.
Due to the back-donation of electrons from the metal $d$-orbitals into the antibonding $\pi^*$ orbitals of $CO$,the bond order of the $C-O$ bond decreases.
As the bond order decreases,the bond length increases compared to free $CO$ $(1.128 \ \mathring{A})$.
Among the given options,the only value greater than $1.128 \ \mathring{A}$ is $1.158 \ \mathring{A}$.
Therefore,option $A$ is the correct answer.

(A) In metal carbonyls,the metal-carbon bond possesses both $ \sigma $ and $ \pi $ character.
$1$. The $ \sigma $ bond is formed by the donation of an electron pair from the $ \text{CO} $ molecule into a vacant $ d $-orbital of the metal.
$2$. The $ \pi $ bond (synergic bonding) is formed by the back-donation of electrons from a filled metal $ d $-orbital into the empty antibonding $ \pi ^* $ molecular orbital of the $ \text{CO} $ ligand.
Therefore,the $ \pi ^* $ molecular orbital of the ligand is used in the synergic bond.

(A) $1$. Identify the components of the complex: The central metal is platinum $(Pt)$ in oxidation state $(II)$.
$2$. The ligands are three chloro groups $(Cl^-)$ and one ethylene molecule $(C_2H_4)$.
$3$. The coordination sphere is $[PtCl_3(C_2H_4)]^n$.
$4$. Calculate the charge $(n)$: $n = (+2) + 3(-1) + 0 = -1$.
$5$. To balance the charge of $-1$,one potassium ion $(K^+)$ is required.
$6$. Thus,the formula is $K[PtCl_3(C_2H_4)]$,which is known as Zeise's salt.

(D) Organometallic compounds are classified based on the nature of the metal-carbon bond.
$1$. Zeise's salt $(K[PtCl_3(\eta^2-C_2H_4)])$ contains an ethylene ligand bonded via $\pi -$ electrons to the metal.
$2$. Ferrocene $([Fe(\eta^5-C_5H_5)_2])$ is a sandwich complex with $\pi -$ bonded cyclopentadienyl rings.
$3$. Dibenzene chromium $([Cr(\eta^6-C_6H_6)_2])$ is a sandwich complex with $\pi -$ bonded benzene rings.
$4$. Tetraethyl lead $((C_2H_5)_4Pb)$ contains only $\sigma -$ bonds between the lead atom and the ethyl groups.
Therefore,tetraethyl lead is not a $\pi -$ bonded complex.

(A) Ferrocene is a classic example of an organometallic sandwich compound.
It consists of an iron atom $(Fe)$ sandwiched between two cyclopentadienyl rings $(C_5H_5^-)$.
The iron atom is bonded to the rings through the $\pi$-electron cloud of the cyclopentadienyl ligands,which is denoted by the hapticity symbol $\eta ^5$.
Therefore,the correct chemical formula is $Fe(\eta ^5 - C_5H_5)_2$.

(B) Mononuclear carbonyls contain only one metal atom per molecule.
$Fe$ forms $Fe(CO)_5$,$Ni$ forms $Ni(CO)_4$,and $W$ forms $W(CO)_6$.
$Mn$ forms a binuclear carbonyl,$Mn_2(CO)_{10}$,because it has an odd number of valence electrons ($7$ valence electrons),and it dimerizes to satisfy the $18$-electron rule.
Therefore,$Mn$ does not form a stable mononuclear carbonyl.

(A) Metal carbonyls are coordination complexes where carbon monoxide $(CO)$ acts as a ligand.
$CO$ is a neutral ligand with an oxidation state of $0$.
In metal carbonyls,the metal is bonded to neutral $CO$ ligands,and there is no overall charge on the complex.
Therefore,the oxidation state of the metal in metal carbonyls is $0$.

(A) An organometallic compound is defined as a compound containing at least one $M-C$ bond (metal-carbon bond).
$1$. $Cisplatin$ is $[Pt(NH_3)_2Cl_2]$. It does not contain any $Pt-C$ bond,so it is not an organometallic compound.
$2$. $Ferrocene$ is $[Fe(\eta^5-C_5H_5)_2]$,which contains $Fe-C$ bonds.
$3$. $Zeise's$ $salt$ is $K[PtCl_3(\eta^2-C_2H_4)]$,which contains a $Pt-C$ bond.
$4$. $Grignard$ $reagent$ is $R-Mg-X$,which contains a $Mg-C$ bond.
Therefore,$Cisplatin$ is the correct answer.

(B) Organometallic compounds that are both $\sigma$ and $\pi$ bonded typically involve ligands like $CO$,$C_2H_4$ (ethene),or $C_5H_5^-$ (cyclopentadienyl) where the metal-ligand bond involves both $\sigma$-donation and $\pi$-backbonding.
In $[PtCl_3(\eta^2-C_2H_4)]^-$,known as Zeise's salt,the ethene ligand acts as a $\pi$-donor to the $Pt$ atom and receives electron density from the $Pt$ $d$-orbitals into its $\pi^*$ antibonding orbital ($\pi$-backbonding).
While ferrocene $[Fe(\eta^5-C_5H_5)_2]$ also involves $\pi$-bonding,Zeise's salt is the classic example of a $\sigma$-$\pi$ bonded organometallic complex in coordination chemistry textbooks.

(A) Tetraethyl lead,with the formula $(C_2H_5)_4Pb$,is historically used as an antiknock agent in gasoline to prevent engine knocking and improve fuel efficiency.