Isomerism and Magnetic properties Questions in English

(A) homoleptic complex is one in which the central metal atom or ion is bonded to only one type of donor atom.
$(A) \ [FeO_4]^{2-}$: Homoleptic. $Fe$ is in $+6$ oxidation state $(d^2)$. Even number of $d$ electrons.
$(B) \ [Fe(CN)_6]^{3-}$: Homoleptic. $Fe$ is in $+3$ oxidation state $(d^5)$. Odd number of $d$ electrons.
$(C) \ [Fe(CN)_5 NO]^{2-}$: Heteroleptic (contains $CN^-$ and $NO$ ligands).
$(D) \ [CoCl_4]^{2-}$: Homoleptic. $Co$ is in $+2$ oxidation state $(d^7)$. Odd number of $d$ electrons.
$(E) \ [Co(H_2O)_3 F_3]$: Heteroleptic (contains $H_2O$ and $F^-$ ligands).
Thus,the homoleptic complexes with an odd number of $d$ electrons are $(B)$ and $(D)$.

(A) The complexes are:
$1$. $K_3[Co(NO_2)_6]$: Contains $[Co(NO_2)_6]^{3-}$,where $Co^{3+}$ is $d^6$ low-spin. It is yellow in color. Magnetic moment $\mu = \sqrt{n(n+2)} = 0 \ B.M.$ (since $n=0$).
$2$. $K_4[Fe(CN)_6]$: Contains $[Fe(CN)_6]^{4-}$,where $Fe^{2+}$ is $d^6$ low-spin. It is pale yellow in color. Magnetic moment $\mu = 0 \ B.M.$ (since $n=0$).
$3$. $K_3[Fe(CN)_6]$: Contains $[Fe(CN)_6]^{3-}$,which is red.
$4$. $Cu_2[Fe(CN)_6]$: This is a brown/chocolate colored precipitate.
$5$. $Zn_2[Fe(CN)_6]$: This is a white precipitate.
The yellow complexes are $K_3[Co(NO_2)_6]$ and $K_4[Fe(CN)_6]$.
The sum of their spin-only magnetic moments is $0 + 0 = 0 \ B.M.$

(A) $1$. Identify the number of diamagnetic complexes $(n)$:
$K_2[NiCl_4]$: $Ni^{2+}$ is $d^8$, $sp^3$ hybridization, paramagnetic.
$[Zn(H_2O)_6]Cl_2$: $Zn^{2+}$ is $d^{10}$, $sp^3d^2$ hybridization, diamagnetic.
$K_3[Mn(CN)_6]$: $Mn^{3+}$ is $d^4$, $d^2sp^3$ hybridization, paramagnetic.
$[Cu(PPh_3)_3I]$: $Cu^+$ is $d^{10}$, $sp^3$ hybridization, diamagnetic.
Thus, $n = 2$.
$2$. Identify the metal with the least enthalpy of atomisation: Among $Ni$, $Zn$, $Mn$, and $Cu$, $Zn$ has the least enthalpy of atomisation due to the absence of unpaired $d$-electrons for metallic bonding.
$3$. Calculate the spin-only magnetic moment of $Zn^{2+}$:
$Zn^{2+}$ has the electronic configuration $[Ar] 3d^{10}$.
Since there are no unpaired electrons, the number of unpaired electrons $(x)$ is $0$.
$\mu = \sqrt{x(x+2)} = \sqrt{0(0+2)} = 0 \ BM$.

(D) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
Given $\mu = 4.9 \ B.M.$,we have $\sqrt{n(n+2)} = 4.9$,which implies $n = 4$.
Let us calculate the number of unpaired electrons for each ion:
$(A) {}_{24} Cr^{2+} \Rightarrow [Ar] 3d^4$ ($4$ unpaired $e^{-}$)
$(B) {}_{26} Fe^{2+} \Rightarrow [Ar] 3d^6$ ($4$ unpaired $e^{-}$)
$(C) {}_{26} Fe^{3+} \Rightarrow [Ar] 3d^5$ ($5$ unpaired $e^{-}$)
$(D) {}_{27} Co^{2+} \Rightarrow [Ar] 3d^7$ ($3$ unpaired $e^{-}$)
$(E) {}_{25} Mn^{3+} \Rightarrow [Ar] 3d^4$ ($4$ unpaired $e^{-}$)
Thus,ions $A, B,$ and $E$ have $4$ unpaired electrons and a magnetic moment of $4.9 \ B.M.$

(C) The reaction is: $FeCl_3 + 3KOH + 3H_2C_2O_4 \rightarrow K_3[Fe(C_2O_4)_3] + 3KCl + 6H_2O$.
The complex $(A)$ is $[Fe(C_2O_4)_3]^{3-}$.
This is an octahedral complex of the type $[M(AA)_3]$,where $AA$ is a bidentate ligand (oxalate ion).
Complexes of the type $[M(AA)_3]$ exhibit optical isomerism,existing as a pair of enantiomers (d-form and l-form).
Therefore,the total number of optical isomers is $2$.

(D) To determine if a complex is diamagnetic,we check the electronic configuration of the central metal ion and the effect of the ligand field strength on the $d$-orbital splitting.
$1$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^4 e_g^0$. It has two unpaired electrons,so it is paramagnetic.
$2$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand. Configuration: $t_{2g}^6 e_g^0$. All electrons are paired,so it is diamagnetic.
$3$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand. Configuration: $t_{2g}^6 e_g^0$. All electrons are paired,so it is diamagnetic.
$4$. $[Co(H_2O)_3F_3]$: $Co^{3+}$ is $3d^6$. $H_2O$ and $F^-$ are weak field ligands. Configuration: $t_{2g}^4 e_g^2$. It has unpaired electrons,so it is paramagnetic.
Thus,$B$ and $C$ are diamagnetic.

(C) $1$. For the complex $[CrCl_3(py)_3]$, which is of the type $[MA_3B_3]$, it exhibits geometrical isomerism with two isomers: facial $(fac)$ and meridional $(mer)$. Neither of these is optically active. Thus, the total number of stereoisomers is $2$.
$2$. For the complex $[CrCl_2(ox)_2]^{3-}$, which is of the type $[M(AA)_2B_2]$, it exhibits geometrical isomerism with two isomers: $cis$ and $trans$.
$3$. The $cis$ isomer is optically active and exists as a pair of enantiomers ($d$ and $l$ forms), while the $trans$ isomer is optically inactive (achiral). Thus, the total number of stereoisomers is $3$ ($cis-d$, $cis-l$, and $trans$).
$4$. Therefore, the number of stereoisomers for $[CrCl_3(py)_3]$ and $[CrCl_2(ox)_2]^{3-}$ are $2$ and $3$ respectively.

(C) $1$. The central atom $Br$ in $BrF_5$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair. Thus,$x = 1$.
$2$. Since $x = 1$,the complex $MCl_4 \cdot 3NH_3$ releases $1$ mole of $AgCl$ per mole of complex,indicating the formula is $[M(NH_3)_3Cl_3]Cl$.
$3$. The coordination sphere is $[M(NH_3)_3Cl_3]$,which is of the type $Ma_3b_3$.
$4$. Complexes of the type $Ma_3b_3$ exhibit $2$ geometrical isomers: facial (fac) and meridional (mer).

(C) The complex $A$ reacts with $AgNO_3$ to give a white precipitate,which indicates the presence of free $Cl^-$ ions outside the coordination sphere. Thus,$A$ is $[Co(NH_3)_5(SO_4)]Cl$.
The complex $B$ reacts with $BaCl_2$ to give a white precipitate,which indicates the presence of free $SO_4^{2-}$ ions outside the coordination sphere. Thus,$B$ is $[Co(NH_3)_5Cl]SO_4$.
Since the isomers differ in the ions they provide in solution,this is an example of Ionisation isomerism.

(D) Let us analyze the number of unpaired electrons $(n)$ for each complex:
$1.$ $[Co(NH_3)_6]^{3+}: Co^{3+} (3d^6)$,$\text{NH}_3$ is a strong field ligand $(SFL)$ $\rightarrow t_{2g}^6 e_g^0, n=0$ (Diamagnetic)
$2.$ $[Co(C_2O_4)_3]^{3-}: Co^{3+} (3d^6)$,$\text{C}_2\text{O}_4^{2-}$ is a $SFL$ with $Co^{3+} \rightarrow t_{2g}^6 e_g^0, n=0$ (Diamagnetic)
$3.$ $[MnCl_6]^{3-}: Mn^{3+} (3d^4)$,$\text{Cl}^{-}$ is a weak field ligand $(WFL)$ $\rightarrow t_{2g}^3 e_g^1, n=4$ (Paramagnetic)
$4.$ $[Mn(CN)_6]^{3-}: Mn^{3+} (3d^4)$,$\text{CN}^{-}$ is a $SFL$ $\rightarrow t_{2g}^4 e_g^0, n=2$ (Paramagnetic)
$5.$ $[CoF_6]^{3-}: Co^{3+} (3d^6)$,$\text{F}^{-}$ is a $WFL$ $\rightarrow t_{2g}^4 e_g^2, n=4$ (Paramagnetic)
$6.$ $[Fe(CN)_6]^{3-}: Fe^{3+} (3d^5)$,$\text{CN}^{-}$ is a $SFL$ $\rightarrow t_{2g}^5 e_g^0, n=1$ (Paramagnetic)
$7.$ $[FeF_6]^{3-}: Fe^{3+} (3d^5)$,$\text{F}^{-}$ is a $WFL$ $\rightarrow t_{2g}^3 e_g^2, n=5$ (Paramagnetic)
The paramagnetic species with the same number of unpaired electrons $(n=4)$ are $[MnCl_6]^{3-}$ and $[CoF_6]^{3-}$.
Therefore,the number of such species is $2$.

(D) Statement $I$ is true: $A$ homoleptic octahedral complex with only one type of monodentate ligand (e.g.,$[Ma_6]$) does not exhibit stereoisomerism.
Statement $II$ is false: $cis-$platin and $trans-$platin have the formula $[Pt(NH_3)_2Cl_2]$. These are heteroleptic complexes of $Pt$ (Platinum),not $Pd$ (Palladium).
Therefore,Statement $I$ is true but Statement $II$ is false.

(C) To determine paramagnetism,we check for the presence of unpaired electrons in the $Ni$ center:
$A$. $[NiCl_4]^{2-}$: $Ni^{2+}$ $(3d^8)$. $Cl^-$ is a weak field ligand,so no pairing occurs. It has $2$ unpaired electrons (Paramagnetic).
$B$. $Ni(CO)_4$: $Ni^0$ $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing. It is diamagnetic.
$C$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing. It is diamagnetic.
$D$. $[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ $(3d^8)$. $H_2O$ is a weak field ligand,so no pairing occurs. It has $2$ unpaired electrons (Paramagnetic).
$E$. $Ni(PPh_3)_4$: $Ni^0$ $(3d^8 4s^2)$. $PPh_3$ is a strong field ligand,causing pairing. It is diamagnetic.
Thus,only $A$ and $D$ are paramagnetic.

(C) Let us analyze each complex:
$A$. $[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,so no pairing occurs. Unpaired electrons $(n)$ = $5$. (Correctly matched)
$B$. $[Cr(en)_3]^{2+}$: $Cr^{2+}$ is $3d^4$. $en$ is a strong field ligand,causing pairing. The $3d^4$ configuration becomes $t_{2g}^4 e_g^0$,resulting in $n = 2$. (Correctly matched)
$C$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing all electrons to pair up in the $t_{2g}$ orbitals. Thus,$n = 0$. The given value $4$ is incorrect. (Incorrectly matched)
$D$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ is $3d^5$. $H_2O$ is a weak field ligand,so no pairing occurs. $n = 5$. (Correctly matched)
Therefore,the incorrectly matched option is $C$.

(D) The complex $trans-[CoCl_2(en)_2]^{+}$ has a plane of symmetry $(POS)$.
Due to the presence of a plane of symmetry,it is superimposable on its mirror image and is therefore optically inactive.
All other given complexes ($[Co(en)_3]^{3+}$,$[Cr(OX)_3]^{3-}$,and $cis-[CoCl_2(en)_2]^{+}$) lack a plane of symmetry and are optically active.

(B) The complex $[Cr(NO_2)_2(en)_2]^+$ contains the ambidentate ligand $NO_2^-$,which can coordinate through $N$ or $O$,thus exhibiting linkage isomerism.
Similarly,the complex $[Rh(SCN)(H_2O)(ox)_2]^{2-}$ contains the ambidentate ligand $SCN^-$,which can coordinate through $S$ or $N$,thus also exhibiting linkage isomerism.
Therefore,both complexes exhibit the same type of isomerism (linkage isomerism).

(B) In the given square planar or octahedral complex structure,the two identical ligands ($Cl$ atoms) are positioned opposite to each other at an angle of $180^{\circ}$.
When identical ligands are placed at opposite positions,the isomer is classified as a $Trans$ isomer.

(A) The given structures represent the geometrical isomers of the octahedral complex $[Co(NH_3)_3Cl_3]$.
In structure $(i)$,the three identical ligands ($NH_3$ or $Cl$) occupy the corners of one triangular face of the octahedron. This is known as the facial $(fac)$ isomer.
In structure $(ii)$,the three identical ligands lie in a plane passing through the metal atom,forming a meridian. This is known as the meridional $(mer)$ isomer.
Therefore,$(i)$ is the $fac$-isomer and $(ii)$ is the $mer$-isomer.

(B) The spin magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$(I) \ [FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,so electrons remain unpaired. $n = 5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ B.M$.
$(II) \ [V(H_2O)_6]^{2+}$: $V^{2+}$ is $3d^3$. $n = 3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M$.
$(III) \ [Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,so electrons are $t_{2g}^4 e_g^2$. $n = 4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ B.M$.
Comparing the values: $3.87 < 4.90 < 5.92$,which corresponds to $II < III < I$.

(D) Coordination isomerism occurs when there is an interchange of ligands between cationic and anionic entities of different metal ions in a complex salt.
For the complex $[Cr(NH_3)_6][CoCl_6]$,the total number of $NH_3$ ligands is $6$ and $Cl$ ligands is $6$.
In option $A$: $[Cr(NH_3)_5Cl][CoCl_5(NH_3)]$,total $NH_3 = 6$ and $Cl = 6$. This is a coordination isomer.
In option $B$: $[Cr(NH_3)_3Cl_3][CoCl_3(NH_3)_3]$,total $NH_3 = 6$ and $Cl = 6$. This is a coordination isomer.
In option $C$: $[Cr(NH_3)Cl_5][CoCl(NH_3)_5]$,total $NH_3 = 6$ and $Cl = 6$. This is a coordination isomer.
In option $D$: $[Cr(NH_3)_2Cl_4][CoCl_2(NH_3)_4]$,total $NH_3 = 6$ and $Cl = 6$. However,the coordination number of $Cr$ in the original complex is $6$. In option $D$,the complex $[Cr(NH_3)_2Cl_4]$ has a coordination number of $6$,but the ligand distribution does not follow the standard exchange pattern relative to the original stoichiometry of the metal centers in a way that maintains the specific isomerism definition compared to the others,or it is simply a valid isomer. Wait,checking the stoichiometry: all options maintain the same total ligand count. Actually,all options provided are technically coordination isomers. Given the standard nature of this question,there might be a typo in the question or options. Assuming the question asks for the one that is $NOT$,and checking the coordination numbers: all are $6$. If we assume the question is correct as provided,all are isomers. However,usually,such questions have one option that violates the stoichiometry. Since all options $A, B, C, D$ have $6$ $NH_3$ and $6$ $Cl$ total,they are all coordination isomers. If this is a multiple-choice question where one must be selected,there may be an error in the question source.

(D) To show all four types of isomerism (ionisation,linkage,geometrical,and optical),the complex must satisfy the following conditions:
$1$. Ionisation isomerism: It must have an exchangeable counter ion (e.g.,$Br^-$ or $NO_2^-$).
$2$. Linkage isomerism: It must contain an ambidentate ligand (e.g.,$NO_2^-$).
$3$. Geometrical isomerism: It must have a structure that allows cis-trans or facial-meridional arrangements (e.g.,$MA_2B_2$ or $MA_3B_3$ type).
$4$. Optical isomerism: It must be chiral (lacks a plane of symmetry).
Analyzing option $D$: $[Pt(NH_3)_3 Cl_3] NO_2$
- Ionisation: The $NO_2^-$ can exchange with $Cl^-$.
- Linkage: $NO_2^-$ is an ambidentate ligand.
- Geometrical: $MA_3B_3$ type complexes show facial $(fac)$ and meridional $(mer)$ isomerism.
- Optical: The $mer$ isomer is achiral,but the $fac$ isomer can exhibit optical activity in specific configurations,or the complex can be designed to show these properties depending on the coordination geometry.

(A) $(i)$ $[Cr(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_4(CN)_2][Cr(NH_3)_2(CN)_4]$ are coordination isomers because they involve the exchange of ligands between cationic and anionic coordination spheres. This statement is correct.
$(ii)$ $[Cr(py)_2(H_2O)_2Cl_2]Cl$ and $[Cr(py)_2(H_2O)Cl_3]H_2O$ are hydrate isomers,not linkage isomers. This statement is incorrect.
$(iii)$ $[Pt(NH_3)_4Br_2]Cl_2$ and $[Pt(NH_3)_4Cl_2]Br_2$ are ionization isomers,not linkage isomers. This statement is incorrect.
$(iv)$ Tetrahedral complexes of the type $[Ma_2b_2]$ do not exhibit geometrical isomerism because all positions in a tetrahedron are adjacent to each other. This statement is incorrect.
Therefore,only statement $(i)$ is correct.

(D) The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
Given $\mu = \sqrt{3} \ B.M.$,we have $\sqrt{n(n+2)} = \sqrt{3}$,which implies $n = 1$.
For $\Delta_0 > P$,the complex must be a low-spin complex.
Let's analyze the options:
$A$: $[Co(H_2O)_4(NH_3)_2]^{3+}$: $Co^{3+}$ is $d^6$. With strong field ligands,it forms a low-spin complex with $t_{2g}^6 e_g^0$,so $n = 0$.
$B$: $[RhF_6]^{3-}$: $Rh^{3+}$ is $4d^6$. $4d$ and $5d$ series elements always form low-spin complexes,but $F^-$ is a weak field ligand. However,$Rh^{3+}$ is $d^6$,so $t_{2g}^6 e_g^0$,$n = 0$.
$C$: $[Cr(CO)_6]$: $Cr^0$ is $d^6$. $CO$ is a strong field ligand,forming a low-spin complex with $t_{2g}^6 e_g^0$,$n = 0$.
$D$: $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $d^5$. $CN^-$ is a strong field ligand,forcing pairing. The configuration is $t_{2g}^5 e_g^0$,resulting in $n = 1$ unpaired electron. Thus,$\mu = \sqrt{1(1+2)} = \sqrt{3} \ B.M.$ and $\Delta_0 > P$.

(A) Ligands that can coordinate through two different donor atoms are called ambidentate ligands.
$SCN^{-}$ is an ambidentate ligand because it can coordinate through either the sulfur atom $(S)$ or the nitrogen atom $(N)$,thus exhibiting linkage isomerism.

(B) The general formula $MA_2BC$ represents a coordination complex with a central metal atom $M$,two identical ligands $A$,and two different ligands $B$ and $C$.
In the complex $[Pt(NH_3)(H_2O)Cl_2]$,the central metal is $Pt$,$A = Cl^-$,$B = NH_3$,and $C = H_2O$.
This complex exhibits geometric isomerism (cis and trans forms) and is a classic example of the $MA_2BC$ type.
Therefore,the correct option is $B$.

(A) Solvate isomerism (also known as hydrate isomerism when water is the solvent) occurs when the solvent molecule acts as a ligand in one isomer and as a lattice molecule (outside the coordination sphere) in another.
In the pair $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$,the first complex has $6$ water molecules as ligands,while the second has $5$ water molecules as ligands and $1$ water molecule as a lattice molecule.
Therefore,this pair exhibits solvate isomerism.

(D) Linkage isomerism is exhibited by ambidentate ligands.
An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different donor atoms.
Among the given options,$Nitro$ $(-NO_2^-)$ is an ambidentate ligand because it can coordinate through either the nitrogen atom $(-NO_2)$ or the oxygen atom $(-ONO)$.
Other options like $Aqua$ $(H_2O)$,$Iodo$ $(I^-)$,and $Ammine$ $(NH_3)$ are monodentate ligands that do not exhibit linkage isomerism.

(B) The complexes $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ differ in the number of water molecules directly coordinated to the central metal ion $Cr^{3+}$ versus those present as lattice water (solvent of crystallization).
This type of isomerism,where the solvent molecule (in this case,$H_2O$) acts as a ligand in one isomer and as a free solvent molecule in another,is known as solvate isomerism (also called hydrate isomerism).
Therefore,the correct answer is $B$.

(D) The given complexes are $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$.
In these complexes,the water molecule $(H_2O)$ acts as a ligand in the first complex,while in the second complex,one $H_2O$ molecule is present outside the coordination sphere as a solvent molecule.
This type of isomerism,where the solvent molecule (usually water) exchanges its position between the coordination sphere and the lattice,is known as solvate isomerism (also called hydrate isomerism).

(C) In coordination compounds where both the cation and the anion are complex ions,the interchange of ligands between the two metal centers results in coordination isomerism.
Since the ligands $NH_3$ and $CN^-$ are exchanged between the $Co$ and $Cr$ centers in the given pair,they exhibit coordination isomerism.

(C) Geometrical isomerism occurs when ligands can be arranged in different spatial configurations around the central metal ion.
For $[MA_6]$,all six ligands are identical,meaning any exchange of positions results in the same structure.
Therefore,$[MA_6]$ does not exhibit geometrical isomerism.

(A) The central metal ion is $Co^{3+}$,which has an electronic configuration of $[Ar] 3d^6$.
$NH_3$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
As a result,the $3d$ electrons are paired,leaving two $3d$ orbitals empty.
These two $3d$,one $4s$,and three $4p$ orbitals hybridize to form $d^2sp^3$ hybrid orbitals.
Since all electrons are paired,the complex is diamagnetic.
Because the electrons are paired in the lower energy $d$ orbitals,it is a low spin complex.

(C) The chemical formula for the manganate ion is $MnO_4^{2-}$.
In $MnO_4^{2-}$,the oxidation state of $Mn$ is $+6$. The electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1$.
Due to the presence of one unpaired electron in the $3d$ orbital,the manganate ion is paramagnetic.
The ionic charge of the manganate ion is $-2$.

(B) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
In the $+3$ oxidation state, $Cr^{3+}$ loses three electrons to form $[Ar] 3d^3$.
This configuration contains $n = 3$ unpaired electrons.
The effective magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=3$, we get $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.

(C) The atomic number of $Zn$ is $30$.
The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$.
The electronic configuration of $Zn^{2+}$ is $[Ar] 3d^{10}$.
Since all $10$ electrons in the $3d$ subshell are paired,the number of unpaired electrons $(n)$ is $0$.
The formula for spin-only magnetic moment is $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 0$,we get $\mu = \sqrt{0(0+2)} = 0 \ BM$.

(D) The atomic number of $Mn$ is $25$. The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$.
For $Mn^{2+}$,the configuration is $[Ar] 3d^5$.
This means there are $5$ unpaired electrons $(n = 5)$.
The spin only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 5$,we get $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.

(C) The atomic number of $Cu$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^{2+}$, the electronic configuration is $[Ar] 3d^9$.
This means there is $1$ unpaired electron in the $3d$ orbital $(n = 1)$.
The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 1$, we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1$. $Co^{2+}$ $(3d^7)$: $n=3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \text{ BM}$.
$2$. $Cu^{2+}$ $(3d^9)$: $n=1$,$\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73 \text{ BM}$.
$3$. $Fe^{2+}$ $(3d^6)$: $n=4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \text{ BM}$.
$4$. $Ni^{2+}$ $(3d^8)$: $n=2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \text{ BM}$.
Comparing the values,$Cu^{2+}$ has the lowest number of unpaired electrons $(n=1)$,hence it exhibits the lowest spin-only magnetic moment.

(D) The formula for spin-only magnetic moment is $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given that the number of unpaired electrons $n = 1$.
Substituting the value of $n$ into the formula: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(C) The spin-only magnetic moment $(\mu)$ of an atom or ion is calculated using the formula $\mu = \sqrt{n(n + 2)} \ BM$,where $n$ represents the number of unpaired electrons. The unit $BM$ stands for Bohr Magneton.

(C) The spin-only magnetic moment is calculated using the formula: $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For an element with $n = 1$ unpaired electron:
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1$. For $Mn^{6+}$ $([Ar]3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = 1.73 \text{ BM}$.
$2$. For $Ni^{2+}$ $([Ar]3d^8)$: $n = 2$,$\mu = \sqrt{2(2+2)} = 2.83 \text{ BM}$.
$3$. For $Fe^{3+}$ $([Ar]3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = 5.91 \text{ BM}$.
$4$. For $Ag^{+}$ $([Kr]4d^{10})$: $n = 0$,$\mu = 0 \text{ BM}$.
Thus,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$ and consequently shows the maximum magnetic moment.

(C) An optically active compound must be chiral,meaning it lacks a plane of symmetry or a center of symmetry.
In the complex $[Co(en)_3]Cl_3$,the ligand $en$ (ethylenediamine) is a bidentate ligand.
The complex $[Co(en)_3]^{3+}$ forms a tris-chelated octahedral structure that does not possess a plane of symmetry or a center of symmetry.
Therefore,it exists as a pair of non-superimposable mirror images (enantiomers),making it optically active.
The other complexes listed are either square planar or octahedral with planes of symmetry,rendering them optically inactive.

(B) The given complexes are $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$.
In these complexes,the water molecule acts as a ligand in the first case,while in the second case,one water molecule is present as a solvent of crystallization (lattice water) outside the coordination sphere.
This type of isomerism,where the solvent molecule (like $H_2O$) can be either inside or outside the coordination sphere,is known as solvate isomerism (also called hydrate isomerism).

(A) The complex $[Pt(NH_3)(Br)(Cl)(Py)]$ is of the type $[Mabcd]$,where $M = Pt$,$a = NH_3$,$b = Br$,$c = Cl$,and $d = Py$.
For a square planar complex of the type $[Mabcd]$,the number of geometrical isomers is $3$.
These isomers are formed by fixing one ligand (e.g.,$NH_3$) and arranging the other three ligands ($Br$,$Cl$,$Py$) in the $trans$ positions relative to the fixed ligand.

(B) The given isomers are $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$.
These compounds produce different ions in an aqueous solution.
$[Co(NH_3)_5(SO_4)]Br$ gives $Br^-$ ions,while $[Co(NH_3)_5Br]SO_4$ gives $SO_4^{2-}$ ions.
This type of isomerism,where the counter ion in the coordination sphere is exchanged with a ligand,is known as $Ionisation$ isomerism.

(B) The complex is $[Co(NH_3)_6][Cr(CN)_6]$.
This complex consists of both a cationic part $[Co(NH_3)_6]^{3+}$ and an anionic part $[Cr(CN)_6]^{3-}$.
Coordination isomerism occurs in complexes where both the cation and anion are complex ions,and the ligands can be exchanged between the metal centers.
Therefore,the isomer $[Cr(NH_3)_6][Co(CN)_6]$ is possible.
Thus,the correct answer is $B$.

(A) The meridional $(mer)$ isomer is a type of geometric isomerism found in octahedral complexes of the type $[MA_3B_3]$.
In this configuration,three identical ligands occupy the positions around the meridian of the octahedron.
Among the given options,$[Co(NH_3)_3Cl_3]$ is an octahedral complex of the type $[MA_3B_3]$.
It can exist in two geometric isomeric forms: facial $(fac)$ and meridional $(mer)$.
Therefore,the correct option is $A$.

(B) The facial $(fac)$ isomer is a type of geometric isomerism observed in octahedral complexes of the type $[MA_3B_3]$.
In the $fac$ isomer,the three identical ligands occupy the corners of one triangular face of the octahedron.
Among the given options,$[Co(NH_3)_3(NO_2)_3]$ is an octahedral complex of the type $[MA_3B_3]$,where $A = NH_3$ and $B = NO_2^-$.
Therefore,it can exhibit both $fac$ (facial) and $mer$ (meridional) isomerism.

(D) To determine the magnetic nature,we calculate the number of unpaired electrons in each complex:
$1$. In $[Co(NH_3)_6]^{3+}$,$Co^{3+}$ is $d^6$. $NH_3$ is a strong field ligand,causing pairing. All electrons are paired,so it is diamagnetic.
$2$. In $[Ni(CO)_4]$,$Ni$ is $d^8 s^2$. $CO$ is a strong field ligand,causing $4s$ electrons to pair into $3d$. All electrons are paired,so it is diamagnetic.
$3$. In $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing in $3d$ orbitals. All electrons are paired,so it is diamagnetic.
$4$. In $[NiCl_4]^{2-}$,$Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand,so no pairing occurs. There are $2$ unpaired electrons in the $3d$ orbitals,making it paramagnetic.

(D) Optical isomerism is shown by complexes that lack a plane of symmetry or center of inversion (i.e.,they are chiral).
$1$. $[Cr(C_2O_4)_3]^{3-}$ is an octahedral complex of the type $[M(AA)_3]$,which is chiral and shows optical isomerism.
$2$. $Cis-[Pt(Br)_2(en)_2]^{2+}$ is an octahedral complex of the type $Cis-[M(AA)_2X_2]$,which lacks a plane of symmetry and shows optical isomerism.
$3$. $[CrCl_2(NH_3)_2en]^+$ exists in various geometric isomers,and the specific isomer with the correct arrangement (lacking symmetry) shows optical isomerism.
$4$. $[Cr(NH_3)_4SO_4]^+$ is an octahedral complex of the type $[M(NH_3)_4(SO_4)]^+$. This complex possesses a plane of symmetry (the plane containing the $SO_4$ ligand and the two trans $NH_3$ ligands),making it achiral. Therefore,it does not show optical isomerism.