Coordination Compounds

There are two ligands in the coordination entity, that is, ammine and aqua, but according to the IUPAC rule, alphabetical order is followed, that is, ammine is named first. Now calculating the oxidation state of Co, the net charge on the coordination entity is +3 as Cl has $-$1 and there are three Cl^$-$ outside the square brackets. So the oxidation state of Co will be III as both ammine and aqua are neutral ligands. So, its correct name will be diamminetetraaquacobalt (III) chloride.

${\left[ {NiC{l_4}} \right]^{2 - }}\,\left( {{d^8}} \right)$



i.e. the number of unpaired electrons in ${\left[ {NiCl{}_4} \right]^{2 - }}\,\,$ is $2.$

$n = \sqrt {n\left( {n + 2} \right)} = \sqrt {2\left( 4 \right)} $

$ = 2\sqrt 2 = 2 \times 1.41$

$ = 2.82\,BM$

$\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}\,\,$ is an inner orbital complex, because in this complex $d-$orbital used is of lower quantum number i.e. $(n-1).$ It results from ${d^2}\,\,\,$ $s{p^3}$ (inner orbital) hybridization.

(i) Nickel in +2 oxidation state $\left(\mathrm{Ni}^{2+}\right)$ prefers to form tetrahedral complexes with chlorine, e.g., $\left[\mathrm{NiCl}_4\right]^{2-}$.

The oxidation state of nickel in $\left[\mathrm{NiCl}_4\right]^{2-}$

$ \begin{gathered} x+4 \times(-1)=-2 \\\\ x=-2+4=+2 \end{gathered} $

Electronic configuration of $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8 4 s^0$

Since, chlorine is a weak ligand, electrons in $\mathrm{Ni}^{2+}$ do not pair up. One $4 s$ and three $4 p$ orbitals undergo hybridisation to form four $s p^3$ hybrid orbitals. These vacant orbitals accept a pair of electron from each of the chloride ion.

The geometry of $s p^3$ hybridised nickel $\left[\mathrm{Ni}^{2+}\right]$ is tetrahedral

(ii) Nickel in +2 oxidation state $\left(\right.$ as $\left.\mathrm{Ni}^{2+}\right)$ forms square planner complex with cyanide ligands, e.g., $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$.

Oxidation state of $\mathrm{Ni}$ in $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

$ \begin{aligned} x+4 \times(-1) & =-2 \\ x & =-2+4=+2 \end{aligned} $

The oxidation state of nickel in $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is $+2$

Electronic configuration of $\mathrm{Ni}^{2+}=[\operatorname{Ar}] 3 d^8 4 s^0$

Since, cyanide is a strong ligand, electrons in $3 d$-orbitals are paired up.



One $3 d$, one $4 s$ and two $4 p$ orbitals undergoes hybridisation to form four $d s p^2$ hybrid orbitals. These vacant orbitals accept a pair of electrons from each of cyanide ligand.



The geometry of $d s p^2$ hybridised nickel complex $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is square planner.



(iii) Nickel in +2 oxidation state as $\mathrm{Ni}^{2+}$ forms octahedral complex with water as ligands, e.g., $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Oxidation state of nickel in $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is:

$ \begin{aligned} x+6 \times 0 & =+2 \\\\ x & =+2 \end{aligned} $

Electronic configuration of $\mathrm{Ni}^{2+}=$



Since, water is a weak ligand, electrons in $3 d$ orbitals of Nickel remains unpaired. One $4 s$, three $4 p$ and two $4 d$ orbitals undergo hybridisation to form $s p^3 d^2$ hybrid orbitals. These vacant orbitals accept lone pair of electron from each of the water molecule.


The geometry of $s p^3 d^2$ hybridised nickel complex $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is octahedral.

The geometries of different $\mathrm{Ni}^{2+}$ complexes are as follows:

$ \begin{aligned} {\left[\mathrm{NiCl}_4\right]^{2-} } & \rightarrow \text { tetrahedral } \\\\ {\left[\mathrm{NiCN}_4\right]^{2-} } & \rightarrow \text { square planner } \\\\ {\left[\mathrm{Ni}_{(}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} } & \rightarrow \text { octahedral } \end{aligned} $

In K₃Fe(CN)₆ :

CN^$-$ being a strong field ligand, it causes pairing of electrons, so only one unpaired electron. The complex is paramagnetic.

In Co[(NH₃)₆]Cl₃ :

NH₃ being a strong field ligand it causes pairing of spins. There is no unpaired electron, hence the complex is diamagnetic.

In Na₃[Co(oxalate)₃] :

Oxalate being a strong field ligand it causes pairing of electron spins. There is no unpaired electron, hence the complex is diamagnetic.

In [Ni(H₂O)₆]Cl₂ :

Since H₂O is a weak field ligand, no pairing takes place. There are two unpaired electrons, so the complex is paramagnetic.

In K₂[Pt(CN)₄] :

CN^$-$ being a strong field ligand, it causes pairing of electrons spins, so no unpaired electron. Hence, the complex is diamagnetic.

In [Zn(H₂O)₆(NO₃)₂] : Zn²⁺ has 3d¹⁰ configuration. Therefore, it is a diamagnetic complex.

$\mathop {CoC{l_3}.6N{H_3}}\limits_{2.675g} \buildrel \, \over \longrightarrow xC{l^ - }$

$xC{l^ - } + AgN{O_3}\buildrel \, \over \longrightarrow \mathop {x\,AgCl \downarrow }\limits_{4.78g} $

Number of moles of the complex

$ = {{2.675} \over {267.5}} = 0.01$ moles

Number of moles of $AgCl$ obtained

$ = {{4.78} \over {143.5}} = 0.03$ moles

$\therefore$ $\,\,\,$ Number of moles of $AgCl$ obtained

$ = 3 \times \,\,$ No. of moles of complex

$\therefore$ $\,\,\,$ $n = {{0.03} \over {0.01}} = 3$

For a substance to be optical isomers following conditions should be fulfilled

$(a)\,\,\,$ A coordination compound which can rotate the plane of polarized light is said to be optically active.

$(b)\,\,\,$ When the coordination compounds have same formula but differ in their abilities to rotate directions of the plane of polarized light are said to exhibit optical isomerism and the molecules are optical isomers. The optical isomers are pair of molecules which are non-super-imposable mirror images of each other.

$(c)\,\,\,$ This is due to the absence of elements of symmetry in the complex.

$(d)\,\,\,$ Optical isomerism is expected in tetrahedral complexes of the type $Mabcd.$

Based on this only option $(2)$ shows optical isomerism ${\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }}$



Complexes of $Z{n^{ + + }}\,\,$ cannot show optical isomerism as they are tetrahedral complexes with plane of symmetry.

${\left[ {Co{{\left( {{H_2}O} \right)}_4}\left( {en} \right)} \right]^{3 + }}\,\,$ have two planes of symmetry hence it is also optically inactive.

Hence the formula of the complex is $\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$

EDTA (ethylenediaminetetracetic acid) is a hexadentate ligand. It is also called the amino poly-carboxylic acid.

It is structure consists of the ethane group which is attached to tertiary amine groups at the terminal positions forming an ethylene diamine. Tertiary amine group is situated by the acetic acid groups $\left(-\mathrm{CH}_2 \mathrm{COOH}\right)$.

The ionisation isomers give different ions in solution. In the complex given in option (B), Cl$^-$ is replaced by NO$_2^ - $ in ionisation sphere.

$\mathrm{[Cr(H_2O)_4Cl(NO_2)]Cl}$ and $[Cr{({H_2}O)_4}C{l_2}](N{O_2})$ have different ions inside and outside the coordinate sphere and they are isomers. Therefore, they are ionisation isomers.

$\begin{aligned} & {\left[\mathrm{NiCl}_4\right]^{2-}} \\\\ & \text { Oxidation state of }\left[\mathrm{NiCl}_4\right]^{2-}=+2 \\\\ & \mathrm{Ni}(28)=[\mathrm{Ar}] 3 d^8 4 s^2 \\\\ & \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8\end{aligned}$

Among the given complexes of Ni , only $\mathrm{Cl}^{-}$is a weak ligand and does not cause pairing of electrons to take place.

Thus, the spin only magnetic moment, with two unpaired electrons $(n=2)$ configuration is

$ \sqrt{n(n+2)}=\sqrt{2(2+2)}=2.82 \mathrm{BM}=2.82 \mathrm{BM} $

The $SC{N^ - }$ ion can coordinate through $S$ or $N$ atom giving rise to linkage isomerism

$M \leftarrow SCN\,\,\,$ thiocyanto

$M \leftarrow NCS\,\,\,$ isothiocyanato.

In [Cr(CO)$_6$]: Cr(24) = [Ar] 3$d^5$ 4$s^1$

Since CO is a strong field ligand, so pairing of electrons will take place and the configuration will become $3d^6$. There will be no unpaired electrons, so, the spin only magnetic moment is zero.

Among the given compounds, the compounds of the type [MA$_2$X$_2$] and [M(AA)$_2$X$_2$] will exhibit geometrical isomerism.

In octahedral complex the magnitude of ${\Delta _o}$ will be highest in a complex having strongest ligand. Of the given ligands $C{N^ - }$ is strongest so ${\Delta _o}$ will be highest for ${\left( {Co{{\left( {CN} \right)}_6}} \right)^{3 - }}.$ Thus option $(a)$ is correct.

In the given complex we have two bidentate ligands (i.e. $en$ and ${C_2}{O_4}$ ), so coordination number of $E$ is $6$ $\left( {2 \times 2 + 1 \times 2 = 6} \right)$

Let the oxidation state of $E$ in complex be $x,$

then $\left[ {x + \left( { - 2} \right) = 1} \right]$ or $x - 2 = 1$

or $\,\,\,\,$ $x=+3,$ so its oxidation state is $+3$

Thus option $(d)$ is correct.

The electronic configuration of Cu$^{2+}$ is [Ar]3d$^9$. It contains one unpaired electron. Hence, CuF$_2$ is coloured compound.

In all other compounds, Cu$^+$ ion is present. It has electronic configuration of Cu$^{2+}$ is [Ar]3d$^{10}$. All electrons are paired.

Hence, they are colourless.

In [Ni(CO)$_4$], Ni is in 0 oxidation state, so there are 8 electrons in 3d subshell and 2 electrons in 4s. CO is the strong ligand, so it cause pairing of electrons in 3d subshell leaving one d subshell vacant. Now, the electron from s shell shifts to d creating vacant space in s subshell and thus the hybridisation is sp$^3$.

Ni(28) = [Ar] 4s$^2$3d$^8$

CO is a strong ligand, causes coupling.

In [Ni(CN)$_4$]$^{2-}$, Ni is in +2 oxidation state so it has 8 electrons in d subshell and CN is also a strong ligand which pairs with d subshell atom and leaves one d orbital empty and thus its hybridisation is dsp$^2$.

In [Ni(CN)$_4$]$^{2-}$, the oxidation state of Ni is +2

Ni$^{2+}$ = [Ar] 3d$^8$4s$^0$

CN$^-$ is strong ligand, causes coupling.

The IUPAC name for [Ni(NH$_3$)$_4$] [NiCl$_4$] is : Tetraamminenickel (II) - tetrachloronickelate (II)

The name of the complex cation is written first followed by the name of the complex anion. Complex cation contains four ammine ligands and complex anion contains four chloride ligands.

The oxidation state of Fe in the complex, [Fe(H$_2$O)$_5$NO]SO$_4$ is +1 [NO has +1 charge]

Fe$^+$: [Ar] 3d$^+$4s$^1$

When the weak ligand H$_2$O and strong ligand NO attacks, the configuration changes.

NO$^+$ causes pairing of 4s electrons inside. Thus the configuration is 3d$^7$ and number of unpaired electrons = 3.

Fe$^+$ : [Ar] 3d$^7$4s$^0$

$\therefore$ Fe$^+$ has 3 unpaired electrons.

For a complex to be optically active, it should not possess either centre of symmetry or axis of symmetry or plane of symmetry. Here, the given complex is an octahedral complex that possesses plane of symmetry as well as axis of symmetry and thus is optically inactive.

The $cis$ and $trans$ both form of complex [M(NH$_3$)$_4$Cl$_2$] are optically inactive.

Complexes with $ds{p^2}$ hybridisation are square planar. So $\,\,{\left[ {PtC{l_4}} \right]^{2 - }}\,\,$ is square planar in shape.

(A) $\mathrm{Mn^+=3}d^5 4s^1$; in presence of CO, effective configuration = $3d^64s^0$

Three lone pair of electrons in metal 3d orbitals for back bonding with vacant ($\pi$) pi* orbital of C in CO.

(B) $\mathrm{Fe^0}=3d^6 4s^2$; in presence of CO, effective configuration = $3d^8$

Four lone pair for back bonding with CO.

(C) $\mathrm{Cr^0}=3d^5 4s^1$; effective configuration = $3d^6$

Three lone pair for back bonding with CO.

(D) $\mathrm{V^-}=3d^4 4s^2$; effective configuration = $3d^6$

Three lone pair for back bonding with CO.

Since, Vanadium has a negative charge, it is present in lower oxidation state and can easily back donate the electrons from metal $d$ orbitals to ($\pi$) pi* orbital of carbonyl ligand. Hence, it forms a much stronger bond with carbon or carbonyl thereby weakening the carbon oxygen bond.

(A) The complex [Co(NH$_3$)$_4$(H$_2$O)$_2$]Cl$_2$ shows geometrical isomers (cis-trans isomers). It is paramagnetic due to presence of unpaired electrons. The central metal (Co) ion has +2 oxidation state.

(B) The complex [Pt(NH$_3$)$_2$Cl$_2$] shows geometrical isomers (cis-trans isomers). The complex is diamagnetic as all the electrons are paired. The Pt metal ion has +2 oxidation state.

(C) The complex [Co(H$_2$O)$_5$Cl]Cl cannot show geometrical isomerism. It is paramagnetic due to presence of unpaired electrons. The cobalt ion has +2 oxidation state.

(D) The complex [Ni(H$_2$O)$_6$]Cl$_2$ is paramagnetic in nature due to presence of unpaired electrons. In [Ni(H$_2$O)$_6$]Cl$_2$, oxidation state of nickel is +2

Ni$^{2+}$ : [Ar] 3d$^8$ 4s$^2$;

2 unpaired electrons.

$EDTA$ has hexadentate four donar $O$ atoms and $2$ donar $N$ atoms and for the formation of octahedral complex one molecule is required

Due to some back-bonding by side-wise overlapping of between $d$-orbitals of metal and $p$-orbital of carbon, the $Fe-C$ bond in $Fe{\left( {CO} \right)_5}$ has both $\sigma $ and $\pi $ character.

${\left[ {Ni{X_4}} \right]^{2 - }},$ the electronic configuration of $N{i^{2 + }}$ is



It contains two unpaired electrons and the hybridisation is $s{p^3}$ (tetrahedral).

The IUPAC name for the complex $[\text{Co(NO}_2)(\text{NH}_3)_5]\text{Cl}_2$ is: Pentaammine(nitrito-$\text{N,O}$)cobalt(III) chloride

Here's how to arrive at the name:

The complex is a cobalt complex, and cobalt has a +3 charge in this compound (since it has one negatively charged NO$_2$ ligand and one negatively charged Cl ion, for a total of -2, to balance the +3 charge on cobalt).

The ligands are NH$_3$ (ammonia) and NO$_2$ (nitrito).

Ammonia is a neutral ligand, so its name is just "ammine" and it is designated by the prefix "penta" because there are five of them.

Nitrito is a negatively charged ligand, so its name is "nitrito-$\text{N,O}$" (the "-$\text{N,O}$" indicates that the NO$_2$ ligand is bound through both the nitrogen and oxygen atoms).

Finally, the compound has a chloride ion, so its name is "chloride" and it is designated by the prefix "di" because there are two of them.

Putting it all together, we get the name "pentaammine(nitrito-$\text{N,O}$)cobalt(III) chloride".

The first step is to find out the chemical formula for complex A and B.

The complete reaction for the formation of A and B are given below:

$\mathrm{NiCl}_{2}+\mathrm{KCN}\left(\right.$ excess) $\rightarrow \mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]+4 \mathrm{KCl}$

The complex A is Tetra cyano nickelate (II) $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]$.

$\mathrm{NiCl}_{2}+$ Conc. $\mathrm{HCl}$ (excess) $\rightarrow$ $\mathop {{{[\mathrm{NiC{l_4}}]}^{2 - }}}\limits_{\mathrm{Chloro\,complex}}$

$\mathrm{K_2[Ni(CN)_4]+4~Conc.~HCl(excess)\to K_2[Ni(Cl)_4]^{2-}+4HCN}$

The complex B is $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$.

The name of the complex A is Potassium tetracyanonickelate (II).

The name of the complex B is Potassium tetrachloronickelate (II).

The complex A is $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$.

The electronic configuration of $\mathrm{Ni}^{2+}$ ion is given below.

Electronic configuration of $\mathrm{Ni}^{2+}$ :

$ 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8 $

The $3 d$ orbital of $\mathrm{Ni}^{2+}$ can be shown as:

Since, cyanide is a strong ligand, electrons in $3 d$ orbitals are paired up. One $3 d$, one $4 s$ and two $4 p$ orbitals undergo hybridization to form four $d s p^2$ hybrid orbitals. Hence, it is diamagnetic in nature. The $d s p^2$ hybridisation occur. The four cyanide ions $\left(\mathrm{CN}^{-}\right)$ions donate electrons to vacant $d s p^2$ hybrid orbitals.

In $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$, potassium is a counter ion and complex ion is $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$. In $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}, \mathrm{Ni}$ exists as $\mathrm{Ni}^{2+}$ ion.

The electronic configuration of $\mathrm{Ni}^{2+}$ ion is given below.

Electronic configuration of $\mathrm{Ni}^{2+}$ :

$ 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8 $

Since, chloride is a weak ligand, electrons in $3 d$ orbitals remains unpaired. One $4 s$ and three $4 p$ orbitals under hybridisation to form four $s p^3$ hybridised orbitals. The $4 \mathrm{Cl}^{-}$ligands will accommodate electrons in empty $s p^3$ hybrid orbitals.

As shown above, there are 2 unpaired electrons present in $\mathrm{Ni}^{2+}$. Hence, it is paramagnetic in nature.

Therefore, A is diamagnetic and B is paramagnetic with two unpaired electrons.

The complex A has $d s p^2$ hybridisation and complex B has $s p^3$ hybridisation as discussed above.

The reaction involving neutron capture by $\mathrm{N}-14$ to form $\mathrm{C}-14$ of ${ }^{14} \mathrm{C}$ to ${ }^{12} \mathrm{C}$ is given as. The proportion of living matter is $1: 10^{12}$.

The structure of metal carbonyl complex, $\mathrm{Fe}(\mathrm{CO})_5$ is shown below:

In $\mathrm{Fe}(\mathrm{CO})_5$ complex, CO groups are ligands attached to central metal ion, Fe. This happens by donation of lone pair of electrons on carbonyl carbon to vacant d orbitals of metal forming a sigma bond and back donation of electron by metal d orbitals to pi ( $\pi$ ) molecular orbitals of carbonyl group forming pi ( $\pi$ ) bond. The back bonding between Fe and CO groups generates a synergic effect which leads to the strengthening of bond between metal ion and ligands and weakening the bond between carbon and oxygen.

This also results in increase in bond length of $\mathrm{C} \equiv \mathrm{O}$ slightly.

On comparing the bond length given in options A to D, option A shows bond length as $1.15 \mathop {\rm{A}}\limits^{\rm{o}}$. This is slightly higher than that of previous bond length that is $1.128 \mathop {\rm{A}}\limits^{\rm{o}}$. The option C shows bond length as $1.72 \mathop {\rm{A}}\limits^{\rm{o}}$ which is much higher than $1.128\mathop {\rm{A}}\limits^{\rm{o}}$.

Hence, Option C is incorrect.

			No. of unpaired electron
a)	Co3+		0
b)	Fe3+		1
c)	Mn3+		2
d)	Cr3+		3

The effective magnetic moment is given by the number of unpaired electrons in a substance, the lesser the number of unpaired electrons lower is its magnetic moment in Bohr $-$ Magneton and lower shall be its paramagnetism

${K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\,\,\,$ is potassium hexacyano ferrate $\left( {{\rm I}{\rm I}{\rm I}} \right).$

Oxidation state of $Cr$ in ${\left[ {Cr{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + }.$

Let it be $x,\,\,1 \times x + 4 \times 0 + 2 \times \left( { - 1} \right) = 1$

Therefore $\,\,\,$ $x=3.$

NOTE : The greater the number of unpaired electrons, greater the magnitude of magnetic moment. Hence the correct order will be $${\left[ {MnC{l_4}} \right]^{ - - }} > {\left[ {CoC{l_4}} \right]^{ - - }} > {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}$$

$C{N^ - }$ ion acts good complexing as well as reducing agent.

Hybridisation

$\mathop {{{\left[ {Fe{{\left( {CN} \right)}_6}} \right]}^{4 - }},}\limits_{{d^2}s{p^3}} \,\,\mathop {{{\left[ {Mn{{\left( {CN} \right)}_6}} \right]}^{4 - }},}\limits_{{d^2}s{p^3}} $

$\mathop {{{\left[ {Co{{\left( {N{H_3}} \right)}_3}} \right]}^{3 + }},}\limits_{{d^2}s{p^3}} \,\,\mathop {{{\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]}^{2 + }}}\limits_{s{p^3}{d^2}} $

Hence ${\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }}$ is outer orbital complex.

Isomers

The coordination number of central metal atom in a complex is equal to number of monovalent ligands, twice the number of bidentate ligands and so on, around the metal ion bonded by coordinate bonds. Hence coordination number $=$ no. of $\sigma $ bonds formed by metals with ligands

The chlorophyll molecule plays an important role in photosynthesis, contain porphyrin ring and the metal $Mg$ not $Ca.$

To determine the correct option, we should evaluate the oxidation states of the metals in the given compounds to see in which cases the metals are in their highest oxidation states.

Option A: MnO₂, FeCl₃

For MnO₂, manganese (Mn) has an oxidation state of +4. While this is a high oxidation state for Mn, it is not its highest, as it can go up to +7 (as in KMnO₄).

For FeCl₃, iron (Fe) has an oxidation state of +3, which is its highest stable oxidation state.

So, this option is not entirely correct.

Option B: [MnO₄]^-, CrO₂Cl₂

For [MnO₄]^-, Mn is in the +7 oxidation state, which is indeed its highest oxidation state.

In CrO₂Cl₂, chromium (Cr) is in the +6 oxidation state, which is its highest oxidation state.

Both Mn and Cr are in their highest oxidation states, making this option correct.

Option C: [Fe(CN)₆]^3-, [Co(CN)₆]^3-

For [Fe(CN)₆]^3-, Fe is in the +3 oxidation state, its highest stable oxidation state.

However, [Co(CN)₆]^3- seems to be incorrectly provided as it suggests an oxidation number that does not fit standard cobalt complexes or could be a typographical error because typically, cobalt in a cyanide complex like [Co(CN)₆]^3- would suggest an oxidation state of +3 for cobalt. Cobalt can exist in +2 and +3 oxidation states commonly, with +3 being very stable in this coordination compound. But it isn't necessarily the highest oxidation state Co can achieve (it could theoretically go up to +4, but such compounds are very rare and not stable).

Therefore, there seems to be a misunderstanding in the elements present or typo in this option for Co. Considering the usual cobalt complex and common oxidation states, though Co is in a high oxidation state (+3), the context seems mixed up, and noting the rarity and instability of Co(IV) compounds, the focus should rather be on the correctness of available options against the question's criteria.

Option D: [NiCl₄]^2-, [CoCl₄]^-

For [NiCl₄]^2-, Ni is in the +2 oxidation state, which is not its highest (it can go up to +3).

The [CoCl₄]^- mistakenly listed suggests cobalt in an oxidation state lower than its maximum, specifically +2, considering the anion's charge and typical coordination chemistry. Like Ni, Co in +2 is common but not its highest.

This option does not have metals in their highest oxidation states either.

Conclusion

Option B, with [MnO₄]^- and CrO₂Cl₂, correctly presents the metals (Mn and Cr) in their highest oxidation states, making it the correct choice.

Let the $O.$ No of $Ni$ in

$\,\,{K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\,\,$ $be$ $=x$ then

$4\left( { + 1} \right) + x + \left( { - 1} \right) \times 4 = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow 4 + x - 4 = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $x=0$

The complex compound Co(NH₃)₅Cl₃ dissociates in water to form three moles of ions. This dissociation can be represented as:

$ \text{Co(NH}_3\text{)}_5\text{Cl}_3\,\,\leftrightharpoons\,\,\left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]^{+2} + 2\text{Cl}^- $

Therefore, the structure of the complex is:

$ \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 $

When this complex reacts with two moles of AgNO₃ solution, it produces:

$ \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 + 2\text{AgNO}_3 \to \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right](\text{NO}_3)_2 + 2\text{AgCl} $

This reaction yields two moles of AgCl, confirming the complex structure as:

$ \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 $