Isomerism and Magnetic properties Questions in English

(B) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.
$I. [Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing. All electrons are paired,so $n = 0$.
$II. [Mn(CN)_6]^{4-}$: $Mn$ is in $+2$ oxidation state $(3d^5)$. $CN^-$ is a strong field ligand,causing pairing. $n = 1$.
$III. [Cr(NH_3)_6]^{3+}$: $Cr$ is in $+3$ oxidation state $(3d^3)$. $n = 3$.
$IV. [CoF_6]^{3-}$: $Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak field ligand,no pairing. $n = 4$.
Comparing the number of unpaired electrons: $I (0) < II (1) < III (3) < IV (4)$.
Thus,the order of magnetic moments is $I < II < III < IV$.

(B) $I. K_4[Fe(CN)_6]$: $Fe^{2+}$ is $d^6$. $CN^-$ is a strong field ligand,so electrons pair up. $t_{2g}^6 e_g^0$. Diamagnetic.
$II. K_3[Cr(CN)_6]$: $Cr^{3+}$ is $d^3$. $t_{2g}^3$. Paramagnetic due to $3$ unpaired electrons.
$III. K_3[Co(CN)_6]$: $Co^{3+}$ is $d^6$. $CN^-$ is a strong field ligand,so electrons pair up. $t_{2g}^6 e_g^0$. Diamagnetic.
$IV. K_2[Ni(CN)_4]$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,leading to $dsp^2$ hybridization. All electrons are paired. Diamagnetic.
Thus,$I, III$ and $IV$ are diamagnetic.

(C) The spin-only magnetic moment is calculated using the formula $\mu_{eff} = \sqrt{n(n+2)} \ B.M.$
In the complex $Hg[Co(SCN)_4]$,$Hg$ is in the $+2$ oxidation state,so the complex ion is $[Co(SCN)_4]^{2-}$.
Let the oxidation state of $Co$ be $x$. Then $x + 4(-1) = -2$,which gives $x = +2$.
The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.
In a tetrahedral field,$3d^7$ configuration results in $3$ unpaired electrons $(n=3)$.
Substituting $n=3$ into the formula: $\mu_{eff} = \sqrt{3(3+2)} = \sqrt{15} \ B.M.$
Thus,option $C$ is correct.

(D) The magnetic behavior of a coordination compound depends on the number of unpaired electrons present in the central metal ion.
$1$. In $[Cr(CN)_6]^{3-}$,$Cr^{3+}$ is $3d^3$. Number of unpaired electrons = $3$.
$2$. In $[Mn(CN)_6]^{3-}$,$Mn^{3+}$ is $3d^4$. In the presence of a strong field ligand $CN^-$,electrons pair up. Number of unpaired electrons = $2$.
$3$. In $[Fe(CN)_6]^{3-}$,$Fe^{3+}$ is $3d^5$. In the presence of a strong field ligand $CN^-$,electrons pair up. Number of unpaired electrons = $1$.
$4$. In $[Co(CN)_6]^{3-}$,$Co^{3+}$ is $3d^6$. In the presence of a strong field ligand $CN^-$,all electrons pair up. Number of unpaired electrons = $0$.
Since $[Co(CN)_6]^{3-}$ has $0$ unpaired electrons,it is diamagnetic and has the lowest magnetic behavior.

(C) To determine if a complex is paramagnetic and low spin,we analyze the electronic configuration of the central metal ion and the nature of the ligand:
$(I) \, K_3[Fe(CN)_6]$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^5 e_g^0$. It has one unpaired electron (paramagnetic) and is a low spin complex.
$(II) \, [Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand. Configuration: $3d^{10} 4s^0$. It has no unpaired electrons (diamagnetic).
$(III) \, [Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. Configuration: $t_{2g}^3 e_g^0$. It has three unpaired electrons (paramagnetic). Since it is a $d^3$ system,the concept of high/low spin does not apply.
$(IV) \, [Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^5 e_g^0$. It has one unpaired electron (paramagnetic) and is a low spin complex.
Thus,both $(I)$ and $(IV)$ are paramagnetic and low spin complexes.

(C) To determine the magnetic property,we calculate the number of unpaired electrons in each complex:
$1$. In $[Co(H_2O)_6]^{2+}$,$Co$ is in $+2$ oxidation state $(d^7)$. $H_2O$ is a weak field ligand,so it has $3$ unpaired electrons.
$2$. In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in $+2$ oxidation state $(d^8)$. $H_2O$ is a weak field ligand,so it has $2$ unpaired electrons.
$3$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state $(d^6)$. $NH_3$ is a strong field ligand,causing pairing of electrons. The configuration becomes $t_{2g}^6 e_g^0$,resulting in $0$ unpaired electrons. Thus,it is diamagnetic.
$4$. In $[Ni(NH_3)_6]^{2+}$,$Ni$ is in $+2$ oxidation state $(d^8)$. $NH_3$ is a strong field ligand,but for $d^8$ configuration,it still has $2$ unpaired electrons.

(A) The reaction is: $[Ni(H_2O)_6]^{2+} + 6NH_3 \rightleftarrows [Ni(NH_3)_6]^{2+} + 6H_2O$.
In both complexes,$Ni^{2+}$ has a $d^8$ electronic configuration.
Both $[Ni(H_2O)_6]^{2+}$ and $[Ni(NH_3)_6]^{2+}$ are octahedral complexes with $sp^3d^2$ hybridization.
Both contain $2$ unpaired electrons.
Thus,the effective magnetic moment remains $\mu_{eff} = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, BM$.

(A) The complex is $[PtCl_2(H_2O)(NH_3)]$.
In this complex,the oxidation state of $Pt$ is $+2$.
The valence electronic configuration of $Pt^{2+}$ is $5d^8$.
Since $Pt^{2+}$ is a $5d$ metal ion,it forms square planar complexes with strong field ligands,which are diamagnetic.
Square planar complexes of the type $[M(a)(b)(c)(d)]$ or $[M(a)_2(b)(c)]$ exhibit geometrical isomerism.
Therefore,the complex exhibits geometrical isomerism.

(C) The complex $[Cr(NCS)(NH_3)_5][ZnCl_4]$ consists of a cationic part $[Cr(NCS)(NH_3)_5]^{2+}$ and an anionic part $[ZnCl_4]^{2-}$.
In the cationic part,$Cr$ is in the $+3$ oxidation state ($d^3$ configuration),which has three unpaired electrons in the $t_{2g}$ orbitals,making it paramagnetic and responsible for the green colour.
Since both the cationic and anionic parts are complex ions,this compound exhibits coordination isomerism.

(B) The magnetic moment is given as $0 \, B.M.$,which implies that there are no unpaired electrons in the titanium ion.
For an octahedral complex,the coordination number is $6$.
In the complex $[Ti(H_2O)_6]Cl_4$,the oxidation state of $Ti$ is calculated as $x + 6(0) = +4$,so $Ti$ is in the $+4$ oxidation state.
The electronic configuration of $Ti$ $(Z=22)$ is $[Ar] 3d^2 4s^2$.
For $Ti^{4+}$,the configuration is $[Ar] 3d^0$,which has $0$ unpaired electrons.
Therefore,the magnetic moment is $0 \, B.M.$

(D) Molar conductivity depends on the number of ions produced in an aqueous solution.
For $[CoBr(NH_3)_5]SO_4$,the dissociation is $[CoBr(NH_3)_5]SO_4 \rightarrow [CoBr(NH_3)_5]^{2+} + SO_4^{2-}$,which gives $2$ ions.
For $[Co(SO_4)(NH_3)_5]Br$,the dissociation is $[Co(SO_4)(NH_3)_5]Br \rightarrow [Co(SO_4)(NH_3)_5]^+ + Br^-$,which gives $2$ ions.
Wait,let us re-evaluate the options.
Option $D$ involves ionization isomers where the ions produced are different ($SO_4^{2-}$ vs $Br^-$).
Although both produce $2$ ions,the charge density and mobility of $SO_4^{2-}$ and $Br^-$ differ,leading to different molar conductivities.
Therefore,the pair $[CoBr(NH_3)_5]SO_4$ and $[Co(SO_4)(NH_3)_5]Br$ are ionization isomers with different molar conductivities.

(D) In the given compounds,the number of water molecules present inside the coordination sphere and outside the coordination sphere (as lattice water) is different.
Specifically,the exchange of $H_2O$ molecules and $Cl^-$ ions between the coordination sphere and the ionization sphere leads to hydrate isomerism.
This is a specific type of solvate isomerism where water acts as the solvent molecule.

(B) $(i)$ $[Co(NH_3)_5(NO_2)]Cl_2$ and $[Co(NH_3)_5(ONO)]Cl_2$ exhibit linkage isomerism because the $NO_2^-$ ligand can bind through either $N$ or $O$ atoms.
$(ii)$ $[Cu(NH_3)_4][PtCl_4]$ and $[Pt(NH_3)_4][CuCl_4]$ exhibit coordination isomerism because the ligands are exchanged between the cationic and anionic coordination spheres.
$(iii)$ $[PtCl_2(NH_3)_4]Br_2$ and $[PtBr_2(NH_3)_4]Cl_2$ exhibit ionization isomerism because the counter ions ($Br^-$ and $Cl^-$) are exchanged with ligands inside the coordination sphere.
All three pairs are correctly matched.

(D) The chemical formulas for the given compounds are $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5(SO_4)]Cl$.
These compounds have different counter ions ($Br^-$ and $Cl^-$) but the same coordination sphere.
However,for ionization isomerism,the counter ion must be capable of exchanging positions with a ligand inside the coordination sphere.
In these specific compounds,the sulphate ion $(SO_4^{2-})$ is acting as a ligand,and the halide ions are counter ions.
Since these two compounds contain different counter ions ($Br^-$ vs $Cl^-$),they are distinct chemical substances but do not exhibit isomerism with each other because they have different molecular formulas (different anions).
Therefore,they do not represent any type of isomerism.

(D) Paramagnetic complexes are attracted by a magnetic field.
$1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It is diamagnetic.
$2$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,no pairing occurs. It has $2$ unpaired electrons,so it is paramagnetic.
$3$. $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing. It is diamagnetic.
$4$. $[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $H_2O$ is a weak field ligand,no pairing occurs. It has $2$ unpaired electrons,so it is paramagnetic.
Thus,complexes $(II)$ and $(IV)$ are paramagnetic and will be attracted by a magnetic field.

(C) In $[Fe(H_2O)_6]^{2+}$, $Fe$ is in $+2$ oxidation state $(3d^6)$. $H_2O$ is a weak field ligand, so no pairing occurs. Hybridization is $sp^3d^2$ (outer orbital complex), it is paramagnetic $(\mu = 4.9 \ B.M.)$, and pale green in colour.
In $[Fe(CN)_6]^{4-}$, $Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand, causing pairing of electrons. Hybridization is $d^2sp^3$ (inner orbital complex), it is diamagnetic $(\mu = 0 \ B.M.)$, and yellow in colour.
Both have octahedral geometry and the same number of $d-$ electrons $(6)$.
Therefore, they differ in magnetic moment and colour.

(A) The spin only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.
Given $\mu = \sqrt{24} \, BM$,we have $n(n+2) = 24$,which implies $n = 4$.
Let us analyze the species:
$A$: In $[CoF_3(H_2O)_3]$,$Co$ is in $+3$ oxidation state $(d^6)$. With weak field ligands,$n = 4$.
$B$: In $[CoCl_4]^{2-}$,$Co$ is in $+2$ oxidation state $(d^7)$. With weak field ligands,$n = 3$.
$C$: In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(d^8)$. With weak field ligands,$n = 2$.
$D$: In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in $+2$ oxidation state $(d^8)$. With weak field ligands,$n = 2$.
Thus,$[CoF_3(H_2O)_3]$ has $4$ unpaired electrons.

(B) Paramagnetism is determined by the number of unpaired electrons $(n)$. The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$.
$1$. In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ is $3d^3$,so $n = 3$.
$2$. In $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $3d^6$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^4 e_g^2$,resulting in $n = 4$.
$3$. In $[Cu(H_2O)_6]^{2+}$,$Cu^{2+}$ is $3d^9$,so $n = 1$.
$4$. In $[Zn(H_2O)_6]^{2+}$,$Zn^{2+}$ is $3d^{10}$,so $n = 0$.
Since $[Fe(H_2O)_6]^{2+}$ has the highest number of unpaired electrons $(n = 4)$,it exhibits the highest paramagnetism.

(C) Let us calculate the number of unpaired electrons for each complex:
$1$. $[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,so electrons remain unpaired. $n = 5$. (Correct)
$2$. $[Cr(en)_3]^{2+}$: $Cr^{2+}$ is $3d^4$. $en$ is a strong field ligand,so electrons pair up. $t_{2g}^4 e_g^0$ gives $n = 2$. (Correct)
$3$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,so all electrons pair up in $t_{2g}$ orbitals. $n = 0$. (Incorrect,given as $4$)
$4$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ is $3d^5$. $H_2O$ is a weak field ligand,so $n = 5$. (Correct)

(D) $1$. In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. Since $CO$ is a strong field ligand,all electrons pair up,resulting in $0$ unpaired electrons.
$2$. In $[Co(NH_3)_4(NO_2)_2]^+$,$Co$ is in $+3$ oxidation state $(3d^6)$. $NH_3$ and $NO_2^-$ are strong field ligands,causing pairing of electrons,resulting in $0$ unpaired electrons.
$3$. In $[Ag(CN)_2]^-$,$Ag^+$ has $d^{10}$ configuration,resulting in $0$ unpaired electrons.
$4$. In $[CuBr_4]^{2-}$,$Cu$ is in $+2$ oxidation state $(3d^9)$. There is $1$ unpaired electron in the $d$-orbital.
Comparing all,$[CuBr_4]^{2-}$ has the maximum number of unpaired electrons $(1)$.

(B) The complex $[CrCl_2(OH)_2(NH_3)_2]^-$ has the general formula $[MA_2B_2C_2]$.
This type of complex exhibits geometrical isomerism due to the different possible arrangements of the ligands around the central metal ion $(Cr^{3+})$.
Specifically,it can exist in various cis and trans configurations.

(B) For a coordination complex of the type $Ma_2b_2c_2$ with coordination number $6$,the geometrical isomers are determined by the relative positions of the ligands.
There are $5$ possible geometrical isomers for this composition:
$1$. All pairs of identical ligands are trans to each other (all-trans).
$2$. Two pairs are trans and one pair is cis.
$3$. One pair is trans and two pairs are cis.
$4$. All pairs are cis to each other (two distinct arrangements).
Thus,the total number of geometrical isomers is $5$.

(B) For the square planar complex $[Pt(abcd)]^+$,the number of geometrical isomers is $3$. These are the arrangements where pairs of ligands are trans to each other: $(a,b), (a,c), (a,d)$.
For the octahedral complex $[Pt(abcdef)]$,the number of geometrical isomers is $15$. Since all ligands are different,each geometrical isomer exists as a pair of enantiomers (optical isomers). Therefore,the total number of stereoisomers is $15 \times 2 = 30$.

(D) The coordination compounds with the general formula $[MA_3B_3]$ exhibit $fac$ (facial) and $mer$ (meridional) isomerism.
In the $fac$ isomer,the three identical ligands occupy the corners of one triangular face of the octahedron.
In the $mer$ isomer,the three identical ligands occupy the positions around the meridian of the octahedron.

(B) To determine the number of stereoisomeric forms:
$I. [Cr(NO_3)_3(NH_3)_3]$ is of the type $MA_3B_3$. It exhibits facial $(fac)$ and meridional $(mer)$ isomers. Thus,it has $2$ stereoisomers.
$II. K_3[Fe(C_2O_4)_3]$ is of the type $M(AA)_3$. It exists as a pair of enantiomers ($d$ and $l$ forms). Thus,it has $2$ stereoisomers.
$III. [CoCl_2(en)_2]^+$ is of the type $M(AA)_2B_2$. It exhibits $cis$ and $trans$ isomers. The $cis$ form is optically active ($d$ and $l$ forms),while the $trans$ form is optically inactive. Thus,it has $3$ stereoisomers.
$IV. [CoBrCl(ox)_2]^{3-}$ is of the type $M(AA)_2BC$. It exhibits $cis$ and $trans$ isomers. The $cis$ form is optically active ($d$ and $l$ forms),while the $trans$ form is optically inactive. Thus,it has $3$ stereoisomers.
Therefore,only $I$ and $II$ have exactly $2$ stereoisomeric forms.

(C) For a metal complex to exhibit optical activity,it must be chiral,meaning it should not possess a plane of symmetry or a center of inversion.
$1$. Complexes of the type $[MA_4X_2]$ (both cis and trans) possess a plane of symmetry and are therefore optically inactive.
$2$. In the case of $[M(AA)_2X_2]$ complexes:
- The $trans$ isomer has a plane of symmetry and is optically inactive.
- The $cis$ isomer lacks a plane of symmetry and exists as a pair of enantiomers (non-superimposable mirror images),thus exhibiting optical activity.
Therefore,the correct option is $cis-[M(AA)_2X_2]$.

(C) coordination compound is optically active if it lacks a plane of symmetry or a center of symmetry (i.e.,it is chiral).
$1$. $[Cr(NH_3)_6]^{3+}$ is an octahedral complex with six identical ligands,possessing multiple planes of symmetry. It is optically inactive.
$2$. $[Co(CN)_6]^{3-}$ is an octahedral complex with six identical ligands,possessing multiple planes of symmetry. It is optically inactive.
$3$. $[Ru(NH_3)_6]^{3+}$ is an octahedral complex with six identical ligands,possessing multiple planes of symmetry. It is optically inactive.
$4$. $[Co(gly)_3]$ contains three bidentate glycinate $(gly^-)$ ligands. This complex exists as facial $(fac)$ and meridional $(mer)$ isomers. Both isomers lack a plane of symmetry and are chiral,making the species optically active.

(D) $Cis-trans$ isomerism in coordination compounds is a type of geometrical isomerism.
For square planar complexes of the type $[MA_2B_2]$,$cis-trans$ isomerism is possible.
In the complex $[PtCl_2(NH_3)_2]$,the central metal $Pt$ is bonded to two $Cl^-$ ligands and two $NH_3$ ligands.
When the two identical ligands are adjacent,it is the $cis$-isomer,and when they are opposite to each other,it is the $trans$-isomer.
The other given complexes $[PtCl(NH_3)_3]^+$,$[Pt(NH_3)_4]^{2+}$,and $[PtCl_4]^{2-}$ are of the type $[MA_3B]$ or $[MA_4]$,which do not exhibit geometrical isomerism.

(A) $I$. $\text{cis}-[Co(NH_3)_2(en)_2]^{3+}$ lacks a plane of symmetry and is optically active. It exists as $d$ and $l$ enantiomers.
$II$. $\text{trans}-[IrCl_2(C_2O_4)_2]^{3-}$ possesses a plane of symmetry and is optically inactive.
$III$. $[Rh(en)_3]^{3+}$ lacks a plane of symmetry and is optically active. It exists as $d$ and $l$ enantiomers.
$IV$. $\text{cis}-[Ir(H_2O)_3Cl_3]$ possesses a plane of symmetry and is optically inactive.

(D) $1$. $[PtCl_3(C_2H_4)]^-$ is a square planar complex of type $[M(A)_3(B)]$,which shows no stereoisomerism.
$2$. $[CuBr_2Cl_2]^{2-}$ is a tetrahedral complex,which does not show stereoisomerism.
$3$. $[Co(C_2O_4)_3]^{3-}$ is an octahedral complex of type $[M(AA)_3]$,which shows $2$ optical isomers (enantiomers).
$4$. $[Cr(NH_3)_2(en)_2]^{3+}$ is an octahedral complex of type $[M(A)_2(AA)_2]$. It exists as $cis$ and $trans$ geometric isomers. The $cis$ isomer is optically active and exists as $2$ enantiomers,while the $trans$ isomer is optically inactive. Thus,it has a total of $3$ stereoisomers ($cis-d$,$cis-l$,and $trans$).
$5$. Therefore,$[Cr(NH_3)_2(en)_2]^{3+}$ has the maximum number of stereoisomers.

(B) To find the magnetic moment $(\mu)$,we use the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$I. [FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,so $n=5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.93 \ BM$ $(B)$.
$II. [Ti(H_2O)_6]^{3+}$: $Ti^{3+}$ is $3d^1$. $n=1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$ $(A)$.
$III. [Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. $n=3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.88 \ BM$ $(E)$.
$IV. [Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $n=2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$ $(D)$.
$V. [Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing,so $n=0$. $\mu = 0.00 \ BM$ $(C)$.
Thus,the correct matching is $I-B, II-A, III-E, IV-D, V-C$.

(B) The formula for 'spin only' magnetic moment is $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\sqrt{n(n+2)} = 2.84$,squaring both sides gives $n(n+2) \approx 8$,which implies $n = 2$.
For a $d^4$ configuration in a strong field ligand,the electrons pair up in the $t_{2g}$ orbitals,leaving $2$ unpaired electrons $(t_{2g}^4 e_g^0)$.
Thus,$n = 2$ corresponds to the $d^4$ strong field configuration.

(D) $Cl^-$ is a weak field ligand,while $CN^-$ is a strong field ligand.
$1$. In $[MnCl_4]^{2-}$,$Mn$ is in $+2$ oxidation state ($d^5$ configuration). Since $Cl^-$ is a weak ligand,it has $5$ unpaired electrons.
$2$. In $[CoCl_4]^{2-}$,$Co$ is in $+2$ oxidation state ($d^7$ configuration). Since $Cl^-$ is a weak ligand,it has $3$ unpaired electrons.
$3$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state ($d^6$ configuration). Since $CN^-$ is a strong ligand,it forms a low spin complex with $0$ unpaired electrons.
The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Thus,the order of magnetic moments is $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$.

(D) $[Cr(en)_3]^{3+}$: Hybridization is $d^2sp^3$. It shows optical isomerism but not geometrical isomerism.
$[IrF_3(H_2O)_2(NH_3)]$: Hybridization is $d^2sp^3$. It shows geometrical isomerism but not optical isomerism.
$[NiCl_2(en)_2]$: Hybridization is $sp^3d^2$ (outer orbital complex). It shows both geometrical and optical isomerism.
$[Co(CN)_2(ox)_2]^{3-}$: Hybridization is $d^2sp^3$ (inner orbital complex). It shows both geometrical and optical isomerism.

(D) $FALSE$: Peroxide ion $(O_2^{2-})$ is diamagnetic,while dioxygen $(O_2)$ is paramagnetic.
$(b)$ $FALSE$: Conc. $H_2SO_4$ acts as a dehydrating agent and removes water of crystallization from $[CrCl(H_2O)_5]Cl_2 \cdot H_2O$,but it cannot remove water molecules coordinated as ligands in $[Cr(H_2O)_6]Cl_3$.
$(c)$ $TRUE$: In $NO$ ($15$ electrons),the bond order is $2.5$ and it is paramagnetic. In $NO^+$ ($14$ electrons),the bond order is $3.0$ (bond length decreases) and it is diamagnetic.
$(d)$ $TRUE$: Ethers do not form intermolecular hydrogen bonds,whereas alcohols do. Therefore,ethers are more volatile than isomeric alcohols.

(D) For $[Co(en)_3] [Cr(C_2O_4)_3]$,the possible coordination isomers are:
$1$. $[Co(en)_3] [Cr(C_2O_4)_3]$
$2$. $[Co(C_2O_4)(en)_2] [Cr(C_2O_4)_2(en)]$
$3$. $[Cr(C_2O_4)(en)_2] [Co(C_2O_4)_2(en)]$
$4$. $[Cr(en)_3] [Co(C_2O_4)_3]$
Total coordination isomers = $4$.
For $[Cu(NH_3)_4] [CuCl_4]$,the possible coordination isomers are:
$1$. $[Cu(NH_3)_4] [CuCl_4]$
$2$. $[CuCl(NH_3)_3] [CuCl_3(NH_3)]$
Total coordination isomers = $2$.
For $[Ni(en)_3] [Co(NO_2)_6]$,the possible coordination isomers are:
$1$. $[Ni(en)_3] [Co(NO_2)_6]$
$2$. $[Ni(NO_2)_2(en)_2] [Co(NO_2)_4(en)]$
$3$. $[Co(NO_2)_2(en)_2] [Ni(NO_2)_4(en)]$
$4$. $[Co(en)_3] [Ni(NO_2)_6]$
Total coordination isomers = $4$.

(A) In option $A$,the pair $[Cr(NO_2)(NH_3)_5][ZnCl_4]$ and $[Cr(NO_3)(NH_3)_5][ZnCl_4]$ are not isomers because the ligands $NO_2^-$ and $NO_3^-$ are different chemical species.
Linkage isomerism occurs when an ambidentate ligand (like $NO_2^-$ or $SCN^-$) coordinates through different donor atoms.
$NO_3^-$ is not an ambidentate ligand.
Therefore,$A$ is the incorrect match.

(C) The complex $[Cr(NH_3)_5(NO_2)][Zn(SCN)_4]$ consists of a cationic part $[Cr(NH_3)_5(NO_2)]^+$ and an anionic part $[Zn(SCN)_4]^{2-}$.
$1$. It shows coordination isomerism because the ligands can be exchanged between the cation and anion.
$2$. It shows linkage isomerism due to the presence of the ambidentate ligand $NO_2^-$.
$3$. The complex does not show optical activity because the coordination sphere of the chromium atom $[Cr(NH_3)_5(NO_2)]$ is not chiral (it has a plane of symmetry).
$4$. The $IUPAC$ name is correct as stated.
Therefore,the incorrect statement is that it shows optical activity.

(B) The number of stereoisomers for each complex is calculated as follows:
$1$. For $[M(AA)_2b_2]^{n \pm}$: This complex has $3$ stereoisomers (cis-isomer which is optically active,and trans-isomer which is optically inactive).
$2$. For $[Ma_3b_3]^{n \pm}$: This complex has $2$ stereoisomers (facial and meridional isomers).
$3$. For $[Ma_3bcd]^{n \pm}$: This complex has $5$ stereoisomers.
$4$. For $[Ma_2bcde]^{n \pm}$: This complex has $5$ stereoisomers.
Among the given options,only $[Ma_3b_3]^{n \pm}$ has an even number of stereoisomers $(2)$.

(A) $(i) \ Pt(SCN)_2 \cdot 3PEt_3$ is a square planar complex $(Pt^{II})$,which generally does not exhibit optical isomerism.
$(ii) \ CoBr \cdot SO_4 \cdot 5NH_3$ can show ionisation isomerism (exchange of $Br^-$ and $SO_4^{2-}$).
$(iii) \ FeCl_3 \cdot 6H_2O$ can show hydrate isomerism (exchange of $H_2O$ and $Cl^-$).
Linkage isomerism is possible in $(i)$ due to the ambidentate ligand $SCN^-$.
Therefore,optical isomerism is the type that is not universally possible across these specific complexes.

(D) The complex is $[PdCl_2(H_2O)_2(NH_3)_2]^{2+}$.
Here,$Pd$ is in the $+4$ oxidation state $(Pd^{4+})$,which has a $4d^6$ configuration.
Since $Pd$ is a $4d$ series metal,it forms low-spin complexes with all ligands,leading to pairing of electrons.
Thus,the complex is diamagnetic $(m.m. = 0 \; B.M.)$.
It exhibits geometrical isomerism due to the $Ma_2b_2c_2$ type structure.
However,$fac$ and $mer$ isomerism is characteristic of octahedral complexes of the type $Ma_3b_3$,not $Ma_2b_2c_2$.
Therefore,the unmatched characteristic is the $Fac.$ and $Mer.$ form.

(A) The complex $[Co(en)_2Cl_2]^+$ exhibits the largest number of isomers among the given options.
It shows geometrical isomerism ($cis$ and $trans$ forms).
The $cis$ form is optically active and exists as a pair of enantiomers ($d$ and $l$ forms),while the $trans$ form is optically inactive.
Thus,it has a total of $3$ isomers ($cis-d$,$cis-l$,and $trans$).
$en$ (ethylenediamine) is a bidentate ligand,which allows for these stereochemical variations.

(A) The complex $[Cr(ox)_3]^{3-}$ is an octahedral complex with three bidentate oxalate ligands. It lacks any plane or center of symmetry,making it chiral. It exists as two non-superimposable mirror images,known as the $d$ and $l$ forms.
The other complexes,cis-$[PtCl_2(en)]$,cis-$[RhCl_2(NH_3)_4]^+$ and mer-$[Co(NO_2)_3(dien)]$,possess planes of symmetry and are therefore achiral.

(D) The complex $[Cr(NH_3)_2(OH)_2Cl_2]^-$ is an octahedral complex.
$1$. Ionization isomerism requires an exchange of ligands between the coordination sphere and the counter ion,which is not possible here as there is no counter ion.
$2$. Hydrate isomerism requires water molecules to be present as ligands or in the crystal lattice,which is not applicable here.
$3$. Coordination isomerism requires a complex salt with both cationic and anionic parts,which is not applicable here.
$4$. Geometrical isomerism: The complex $[Ma_2b_2c_2]$ can exhibit geometrical isomerism (cis and trans forms).
$5$. Optical isomerism: The complex can exist in various geometrical isomers,some of which are chiral and exhibit optical activity.
Therefore,the complex exhibits geometrical and optical isomerism.

(C) Optical activity in coordination complexes is observed when the complex lacks a plane of symmetry or a center of inversion (i.e.,it is chiral).
$1$. $trans-[Co(NH_3)_4Cl_2]^+$ has a plane of symmetry and is achiral.
$2$. $[Cr(H_2O)_6]^{3+}$ is an octahedral complex with identical ligands,possessing multiple planes of symmetry,and is achiral.
$3$. $cis-[Co(NH_3)_2(en)_2]^{3+}$ lacks a plane of symmetry and is chiral,thus it shows optical activity.
$4$. $trans-[Co(NH_3)_2(en)_2]^{3+}$ has a plane of symmetry and is achiral.
Therefore,the correct option is $C$.

(C) $1$. In $[PtCl_2(NH_3)_2]^{2+}$,both $cis$ and $trans$ isomers are square planar and possess a plane of symmetry,making them optically inactive.
$2$. $Mabcd$ square planar complexes show geometrical isomerism ($3$ isomers) but do not show optical isomerism due to the presence of a molecular plane of symmetry.
$3$. $[Fe(C_2O_4)_3]^{3-}$ contains three bidentate ligands. It does not show geometrical isomerism because all positions are equivalent for bidentate ligands,but it exists as two non-superimposable mirror images (enantiomers),thus showing optical isomerism.
$4$. $Mabcd$ tetrahedral complexes are chiral and can exhibit optical isomerism.

(D) Analysis of optical activity:
$1$. $\text{trans}-[Co(NH_3)_4Cl_2]^+$ has a plane of symmetry,so it is optically inactive.
$2$. $\text{cis}-[Co(NH_3)_2(en)_2]^{3+}$ lacks a plane of symmetry and is chiral,so it is optically active.
$3$. $\text{trans}-[Co(NH_3)_2(en)_2]^{3+}$ has a plane of symmetry,so it is optically inactive.
Analysis of colour:
$4$. $[NiCl_4]^{2-}$ ($Ni^{2+}$: $d^8$ configuration) is coloured due to $d-d$ transitions.
$5$. $[TiF_6]^{2-}$ ($Ti^{4+}$: $d^0$ configuration) is colourless due to the absence of $d-d$ transitions.
$6$. $[CoF_6]^{3-}$ ($Co^{3+}$: $d^6$ configuration) is coloured due to $d-d$ transitions.
Comparing with options,option $D$ correctly states that $2$ is optically active,while $1$ and $3$ are optically inactive.

(A) Geometrical isomerism is shown by coordination complexes that can exist in different spatial arrangements of ligands around the central metal ion.
$1$. The complex $[Co(en)_2 NH_3Cl]^{2+}$ is of the type $[M(AA)_2 a b]$,where $en$ is a bidentate ligand. It can exist in $cis$ and $trans$ forms,where the $NH_3$ and $Cl^-$ ligands are either adjacent $(cis)$ or opposite $(trans)$ to each other.
$2$. The complex $[Pt(bn)_2]^{2+}$ (where $bn$ is $2,3-diaminobutane$,i.e.,$NH_2-CH(CH_3)-CH(CH_3)-NH_2$) is of the type $[M(AA)_2]$. Due to the presence of chiral centers in the ligand,it can show geometrical isomerism ($cis$ and $trans$ forms) as shown in the provided structures.
Since both complexes exhibit geometrical isomerism,the correct option is $(A)$.

(C) Cis-trans isomerism in octahedral complexes occurs when ligands are arranged differently relative to each other.
$1$. $[CoCl(NH_3)_4(H_2O)]^{2+}$ (type $MA_4BC$) shows cis-trans isomerism.
$2$. $[CoCl_3(NH_3)_3]$ (type $MA_3B_3$) shows facial (fac) and meridional (mer) isomerism,which is a form of geometric isomerism similar to cis-trans.
$3$. $[CoCl_2(NH_3)_4]^+$ (type $MA_4B_2$) shows cis-trans isomerism.
Since all the given complexes exhibit geometric isomerism,the correct option is $C$.

(C) To determine the number of stereoisomeric forms,we analyze each complex:
$(i) \ [Cr(NO_3)_3(NH_3)_3]$ is of the type $[MA_3B_3]$. It exhibits facial $(fac)$ and meridional $(mer)$ geometric isomerism. It does not show optical isomerism. Thus,it has $2$ stereoisomers.
$(ii) \ K_3[Co(C_2O_4)_3]$ is of the type $[M(AA)_3]$. It shows optical isomerism ($d$ and $l$ forms). It has $2$ stereoisomers.
$(iii) \ K_3[CoCl_2(C_2O_4)_2]$ is of the type $[M(AA)_2B_2]$. The cis-isomer is optically active ($d$ and $l$ forms),while the trans-isomer is optically inactive (meso-form). Thus,it has $3$ stereoisomers $(d, l, \text{and } trans)$.
$(iv) \ [CoBrCl(en)_2]$ is of the type $[M(AA)_2BC]$. The cis-isomer is optically active ($d$ and $l$ forms),while the trans-isomer is optically inactive (meso-form). Thus,it has $3$ stereoisomers $(d, l, \text{and } trans)$.
Therefore,$(iii)$ and $(iv)$ have three stereoisomeric forms.