Isomerism and Magnetic properties Questions in English

(A) The initial complex is $[Ni(H_2O)_6]^{2+}$. In this complex,$Ni^{2+}$ has a $d^8$ configuration. $H_2O$ is a weak field ligand,so the number of unpaired electrons is $2$. The magnetic moment is calculated as $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
When ammonia $(NH_3)$ is added,it acts as a stronger ligand than $H_2O$ and replaces it to form $[Ni(NH_3)_6]^{2+}$.
However,$NH_3$ is not strong enough to cause pairing of electrons in the $d^8$ configuration of $Ni^{2+}$. Thus,the number of unpaired electrons remains $2$.
Therefore,the magnetic moment remains $2.83 \ BM$.

(A) The complex $[Co(en)_2Cl_2]^+$ is of the type $[M(AA)_2X_2]$.
It exists in two geometric forms: $cis$ and $trans$.
The $cis$-isomer is optically active because it lacks a plane of symmetry,while the $trans$-isomer is optically inactive.
Therefore,$[Co(en)_2Cl_2]^+$ exhibits both geometric and optical isomerism.

(B) Geometrical isomerism in coordination compounds requires different spatial arrangements of ligands around the central metal atom.
For octahedral complexes of the type $[MA_5B]$,geometrical isomerism is not possible because all positions are equivalent relative to the single $B$ ligand.
In the complex $[Cr(NH_3)_5Cl]Cl_2$,the coordination sphere is $[Cr(NH_3)_5Cl]^{+2}$. This is an $MA_5B$ type complex,where $M = Cr$,$A = NH_3$,and $B = Cl$.
Since there is only one $Cl$ ligand,it cannot be placed in different positions (cis or trans) relative to other ligands,hence it does not show geometrical isomerism.
Other options like $[Pt(NH_3)_2(H_2O)_2]^{+2}$ ($MA_2B_2$ type),$[Co(en)_2Cl_2]Cl$ ($MA_2B_2$ type),and $[Co(NH_3)_4Cl_2]Cl$ ($MA_4B_2$ type) all exhibit geometrical isomerism.

(C) The complexes $[Co(NH_3)_5Br]SO_4$ and $[Co(NH_3)_5SO_4]Br$ are isomers because they produce different ions in aqueous solution.
In the first complex,the sulfate ion $(SO_4^{2-})$ is the counter ion,while in the second complex,the bromide ion $(Br^-)$ is the counter ion.
This type of isomerism,where the counter ion and the ligand exchange positions,is known as $Ionization$ isomerism.

(B) $H_2O$ is a weak field ligand. The magnetic moment is determined by the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.
For $[Cr(H_2O)_6]^{2+}$ $(3d^4)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.
For $[Mn(H_2O)_6]^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \ BM$.
For $[Fe(H_2O)_6]^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.
For $[Ni(H_2O)_6]^{2+}$ $(3d^8)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ BM$.
Since $[Ni(H_2O)_6]^{2+}$ has the lowest number of unpaired electrons $(n=2)$,it has the lowest magnetic moment.

(C) Optical isomerism is exhibited by coordination compounds that lack a plane of symmetry or a center of inversion (chiral molecules).
$1$. $[Co(en)_3]^{3+}$: This is a tris-chelated complex of the type $[M(AA)_3]$,which is chiral and shows optical isomerism.
$2$. $[Co(en)_2Cl_2]^{+}$: The cis-isomer of this complex lacks a plane of symmetry and is optically active.
$3$. $[Co(NH_3)_3Cl_3]$: This complex exists in facial $(fac)$ and meridional $(mer)$ forms. Both the $fac$ and $mer$ isomers of $[Co(NH_3)_3Cl_3]$ possess planes of symmetry,making them achiral and optically inactive.
$4$. $[Co(en)(NH_3)_2Cl_2]^{+}$: This complex can exist in chiral forms depending on the arrangement of ligands.
Therefore,$[Co(NH_3)_3Cl_3]$ does not exhibit optical isomerism.

(B) Optical isomerism is exhibited by coordination complexes that lack a plane of symmetry and a center of inversion.
For octahedral complexes of the type $[M(AA)_2a_2]^{n+}$,the cis-isomer is optically active because it lacks a plane of symmetry.
In the given options,$[Co(en)_2(NH_3)_2]^{3+}$ is of the type $[M(AA)_2a_2]^{n+}$,where $en$ is ethylenediamine (a bidentate ligand).
The cis-form of $[Co(en)_2(NH_3)_2]^{3+}$ exists as a pair of enantiomers (non-superimposable mirror images),thus exhibiting optical isomerism.

(A) The magnetic moment depends on the number of unpaired electrons $(n)$.
For $[Cr(H_2O)_6]^{2+}$,$Cr^{2+}$ is $3d^4$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^3 e_g^1$,so $n = 4$.
For $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $3d^6$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^4 e_g^2$,so $n = 4$.
Since both complexes have $n = 4$,they exhibit the same magnetic moment.

(A) The complex $[Pt(Py)(NH_3)BrCl]$ is a square planar complex of the type $[M(abcd)]$,where $M = Pt$,$a = Py$,$b = NH_3$,$c = Br$,and $d = Cl$.
For a square planar complex of the type $[M(abcd)]$,the number of geometric isomers is given by $3$.
These isomers correspond to the arrangements where any one ligand is fixed,and the other three are arranged in different positions relative to it.
Specifically,if we fix $Py$ at one position,the other three ligands $(NH_3, Br, Cl)$ can be placed in $3$ different ways relative to $Py$ (trans to $NH_3$,trans to $Br$,or trans to $Cl$).

(A) Coordination isomerism occurs in complexes where both the cation and the anion are complex ions.
In the complex $[Cr(NH_3)_6] [Co(CN)_6]$,the ligands can be exchanged between the two metal centers,which is the characteristic feature of coordination isomerism.

(A) In $Cr(CO)_6$,the oxidation state of $Cr$ is $0$.
$Cr$ has an electronic configuration of $[Ar] 3d^5 4s^1$.
Since $CO$ is a strong field ligand,it causes pairing of electrons.
All electrons in the $d$-orbitals are paired,resulting in $n = 0$ unpaired electrons.
The magnetic moment $\mu = \sqrt{n(n+2)} \ BM = \sqrt{0(0+2)} = 0 \ BM$.

(A) The magnetic moment $(\mu)$ is calculated using the spin-only formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
For a $d^7$ configuration in a high-spin octahedral field,the electrons are distributed as $t_{2g}^5 e_g^2$,resulting in $n = 3$ unpaired electrons.
Substituting $n = 3$ into the formula: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M.$

(C) For a square planar complex to exhibit $cis-trans$ isomerism,it must have at least two different types of ligands.
In $Ma_2b_2$ type complexes,the two identical ligands $a$ can be adjacent to each other $(cis)$ or opposite to each other $(trans)$.
$Ma_4$,$Ma_3b$,and $Mab_3$ do not show $cis-trans$ isomerism because the relative positions of the ligands remain equivalent due to symmetry.

(A) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:
$1$. In $[FeF_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. Since $F^-$ is a weak field ligand,electrons do not pair up. Thus,$n = 5$.
$2$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state $(3d^6)$. $NH_3$ is a strong field ligand,causing pairing. Thus,$n = 0$.
$3$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing. Thus,$n = 0$.
$4$. In $[Mn(CN)_6]^{4-}$,$Mn$ is in $+2$ oxidation state $(3d^5)$. $CN^-$ is a strong field ligand,causing pairing. Thus,$n = 1$.
Since magnetic moment $\mu = \sqrt{n(n+2)} \ BM$,the complex with the highest $n$ has the highest magnetic moment. Therefore,$[FeF_6]^{3-}$ has the highest magnetic moment.

(B) In $(1)$,the ligand $NO_2^-$ is an ambidentate ligand that can coordinate through $N$ or $O$,representing linkage isomerism. This is correct.
In $(2)$,the ligands are exchanged between the two metal centers,which is the definition of coordination isomerism. This is correct.
In $(3)$,the counter ions $Cl^-$ and $Br^-$ are exchanged between the coordination sphere and the ionization sphere,representing ionization isomerism. This is correct.
Therefore,all three pairs are correctly matched with their respective types of isomerism.

(B) The complex $[Co(NH_3)_3(NO_3)_3]$ is of the type $[MA_3B_3]$.
For octahedral complexes of the type $[MA_3B_3]$,there are two geometric isomers possible:
$1$. Facial $(fac)$ isomer: In this isomer,the three identical ligands occupy the corners of one triangular face of the octahedron.
$2$. Meridional $(mer)$ isomer: In this isomer,the three identical ligands occupy the meridian of the octahedron.
Therefore,the total number of geometric isomers is $2$.

(C) The complex $[Pt(NO_2)(py)(NH_3)(NH_2OH)]^{+}$ is a square planar complex of the type $[M(abcd)]^n+$.
In a square planar complex with four different ligands,the number of geometric isomers is given by $3$.
Let the ligands be $a, b, c, d$. We can fix one ligand (say $a$) and arrange the others in different positions relative to it.
$1$. $a$ trans to $b$ (with $c$ trans to $d$)
$2$. $a$ trans to $c$ (with $b$ trans to $d$)
$3$. $a$ trans to $d$ (with $b$ trans to $c$)
Thus,there are $3$ possible geometric isomers.

(D) Geometrical isomerism in octahedral complexes occurs when the ligands can be arranged in different spatial positions relative to each other (cis and trans).
For a complex of the type $[M A_5 B]$,all positions are equivalent relative to the ligand $B$.
Any ligand $A$ is always in a cis position to $B$,and there is no possibility of a trans arrangement for $B$ with respect to any $A$.
Therefore,$[M A_5 B]$ does not exhibit geometrical isomerism.
In contrast,$[M A_4 B_2]$ and $[M A_2 B_4]$ show cis and trans isomers,and $[M A_3 B_3]$ shows facial $(fac)$ and meridional $(mer)$ isomers.

(D) The complex $[Cu^{II}(NH_3)_4][Pt^{II}Cl_4]$ exhibits coordination isomerism by exchanging ligands between the cationic and anionic parts.
The possible coordination isomers are:
$1. [Cu(NH_3)_4][PtCl_4]$
$2. [Cu(NH_3)_3Cl][PtCl_3(NH_3)]$
$3. [Cu(NH_3)_2Cl_2][PtCl_2(NH_3)_2]$ (cis-isomer)
$4. [Cu(NH_3)_2Cl_2][PtCl_2(NH_3)_2]$ (trans-isomer)
$5. [Cu(NH_3)Cl_3][PtCl(NH_3)_3]$
$6. [CuCl_4][Pt(NH_3)_4]$
Thus,there are a total of $6$ possible isomers.

(A) The complex ion $[CO(NH_3)_4Cl_2]^{+}$ is of the type $[MA_4B_2]^n+$.
This type of octahedral complex exhibits geometrical isomerism.
The two possible geometrical isomers are:
$1$. $cis$-isomer: In this form,the two $Cl^-$ ligands are adjacent to each other (at $90^\circ$ angle).
$2$. $trans$-isomer: In this form,the two $Cl^-$ ligands are opposite to each other (at $180^\circ$ angle).
Since both isomers are achiral (they possess a plane of symmetry),they do not exhibit optical isomerism.
Therefore,the total number of possible isomeric forms is $2$.

(B) The complex formed by $Fe^{3+}$ and $CN^{-}$ is $[Fe(CN)_6]^{3-}$.
In this complex,the oxidation state of $Fe$ is $+3$,which has an electronic configuration of $[Ar] 3d^5$.
$CN^{-}$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
For $d^5$ configuration with strong field ligand,the electrons are arranged as $t_{2g}^5 e_g^0$,resulting in $1$ unpaired electron $(n = 1)$.
The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$
Substituting $n = 1$,we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$

(B) For a square planar complex of the type $[M(abcd)]$,where $a, b, c, d$ are four different ligands,the number of geometrical isomers is $3$.
In this case,the complex is $[Pt(NH_3)(Br)(Cl)(Py)]^{2+}$.
We can fix one ligand (e.g.,$NH_3$) at one position and then arrange the other three ligands $(Br, Cl, Py)$ in the remaining three positions.
$1$. $NH_3$ trans to $Br$ (with $Cl$ and $Py$ trans to each other).
$2$. $NH_3$ trans to $Cl$ (with $Br$ and $Py$ trans to each other).
$3$. $NH_3$ trans to $Py$ (with $Br$ and $Cl$ trans to each other).
Thus,there are $3$ possible geometrical isomers.

(D) Optical activity in coordination complexes is observed when the complex lacks a plane of symmetry and its mirror image is non-superimposable on the original structure.
Among the given options,the complex $[M(B)_2(en)_2]$ (where $en$ is ethylenediamine) exists in cis and trans forms.
The cis-isomer of $[M(B)_2(en)_2]$ lacks a plane of symmetry and is chiral,thus exhibiting optical activity.
The other complexes listed possess planes of symmetry in their most stable configurations,making them optically inactive.

(A) The complex $[Co(NH_3)_4Br_2]Cl$ exhibits two types of isomerism:
$1$. Geometrical Isomerism: The complex can exist in $cis$ and $trans$ forms due to the arrangement of $Br^-$ ligands around the central $Co^{3+}$ ion.
$2$. Ionization Isomerism: The $Cl^-$ ion can exchange positions with the $Br^-$ ligand inside the coordination sphere,resulting in the isomer $[Co(NH_3)_4BrCl]Br$.

(A) In the complex $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons.
The $3d$ orbitals have $1$ unpaired electron.
The magnetic moment $\mu$ is calculated as $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(B) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
For $\mu = 2.82 \ B.M.$,$n$ must be $2$.
In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $3d^8$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Thus,the $3d^8$ configuration has $2$ unpaired electrons $(n=2)$.
Therefore,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ B.M.$

(A) The spin-only magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu = 2.84 \ BM$,we have $\sqrt{n(n+2)} = 2.84$,which implies $n(n+2) \approx 8$,so $n = 2$.
In a $d^2$ configuration (weak ligand field),there are $2$ unpaired electrons.
Thus,$n = 2$ and $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.84 \ BM$.

(C) Geometrical isomerism in coordination compounds is typically exhibited by complexes with coordination numbers $4$ (square planar) and $6$ (octahedral).
For octahedral complexes of the type $[MA_3B_3]$,geometrical isomerism (facial and meridional) is possible.
However,the complex $[Co(NH_3)_3(NO_2)_3]$ is a common example used in textbooks to illustrate facial and meridional isomers.
Looking at the provided options,$[Co(NH_3)_3(NO_2)_3]$ is the correct choice as it is often contrasted with other types.
Actually,all listed complexes exhibit geometrical isomerism. If we must choose,$[Co(NH_3)_3(NO_2)_3]$ is the intended answer in many contexts where the question implies a specific structural limitation or typo in the original source.

(C) complex is colored if it contains a metal ion with an incomplete $d$-subshell,allowing for $d-d$ transitions.
In $[Cr(NH_3)_5Cl]^{2+}$,the oxidation state of $Cr$ is $+3$.
The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.
Since it has unpaired electrons in the $d$-orbitals,it undergoes $d-d$ transitions and exhibits color.
$Ti(NO_3)_4$ contains $Ti^{4+}$ $(3d^0)$,$[Cu(NCCH_3)_4]^{+}$ contains $Cu^{+}$ $(3d^{10})$,and $Li[AlH_4]$ contains $Al^{3+}$ $(3d^0)$,all of which are colorless due to the absence of $d-d$ transitions.

(D) For a complex to exhibit both geometric and optical isomerism,it must be able to form cis and trans isomers,and the cis isomer must be chiral (non-superimposable on its mirror image).
$1$. $[Pt(NH_3)_2Cl_2]$ is a square planar complex,which shows geometric isomerism but not optical isomerism.
$2$. $[Pt(en)_2Cl_2]^{2+}$ is an octahedral complex of the type $[M(AA)_2a_2]$.
$3$. The cis-isomer of $[Pt(en)_2Cl_2]^{2+}$ lacks a plane of symmetry and is optically active,while the trans-isomer is optically inactive.
$4$. Therefore,$[Pt(en)_2Cl_2]^{2+}$ exhibits both geometric and optical isomerism.

(C) Square planar complexes do not exhibit optical isomerism because they possess a plane of symmetry.
$trans-[Co(en)_2Cl_2]^+$ also does not exhibit optical isomerism as it has a plane of symmetry.
$cis-[Co(en)_2Cl_2]^+$ exists as a pair of enantiomers (non-superimposable mirror images) because it lacks a plane of symmetry and a center of inversion.

(B) The complex is of the type $[M(AA)a_2b_2]$,where $M = Co$,$AA = en$,$a = Cl$,and $b = NH_3$.
$1$. Geometric Isomers: The complex can exist in $cis$ and $trans$ forms regarding the $Cl$ ligands. For the $cis$ form,there are two arrangements based on the relative positions of $NH_3$ ligands,but specifically,for $[M(AA)a_2b_2]$,there are $2$ geometric isomers ($cis$ and $trans$).
$2$. Optical Isomers: The $trans$ isomer is achiral (has a plane of symmetry),while the $cis$ isomer is chiral (exists as a pair of enantiomers). Thus,there are $2$ optical isomers (the $d$ and $l$ forms of the $cis$ isomer).
$3$. Total Isomers: The $trans$ isomer $(1)$ + the $2$ enantiomers of the $cis$ isomer $(2)$ = $3$ total isomers.
Therefore,the number of geometric isomers is $2$,optical isomers is $2$,and total isomers is $3$.

(C) In the complex $Hg[Co(SCN)_4]$,the oxidation state of $Co$ is $+2$.
$Co^{2+}$ has a $3d^7$ electronic configuration.
In a tetrahedral field,$Co^{2+}$ $(3d^7)$ has $3$ unpaired electrons $(n = 3)$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n + 2)} \ BM$.
Substituting $n = 3$,we get $\mu = \sqrt{3(3 + 2)} = \sqrt{15} \ BM$.

(B) The chemical formula for nitropentaamminechromium $(III)$ chloride is $[Cr(NH_3)_5(NO_2)]Cl_2$.
In this complex,the ligand $NO_2^-$ is an ambidentate ligand,which can coordinate to the central metal atom through either the nitrogen atom (nitro) or the oxygen atom (nitrito).
This ability to coordinate through different donor atoms results in linkage isomerism.

(D) The complex $[CO(NO_2)(SCN)(en)_2]Br$ exhibits the following types of isomerism:
$1$. Geometrical isomerism: Due to the presence of two $(en)$ ligands and two different monodentate ligands $(NO_2^-, SCN^-)$,it can exist in $cis$ and $trans$ forms.
$2$. Linkage isomerism: The ligands $NO_2^-$ (nitro/nitrito-$N$) and $SCN^-$ (thiocyanato-$S$/isothiocyanato-$N$) are ambidentate ligands,which can coordinate through different donor atoms.
$3$. Ionization isomerism: The $Br^-$ ion is outside the coordination sphere and can exchange with the ligands inside the sphere (e.g.,$NO_2^-$ or $SCN^-$).
Therefore,the complex exhibits all of the above types of isomerism.

(B) Geometric isomerism is observed in square planar and octahedral complexes.
Tetrahedral complexes do not exhibit geometric isomerism because all positions in a tetrahedral geometry are adjacent to each other,making all ligand positions equivalent.

(D) Geometrical isomerism in coordination compounds is observed when ligands can occupy different positions around the central metal atom.
$(i) [Pt(en)Cl_2]$: This is a square planar complex of the type $[M(AA)X_2]$. It does not show geometrical isomerism because the chelating ligand $(en)$ occupies adjacent positions.
$(ii) [Pt(en)_2]Cl_2$: This is a square planar complex of the type $[M(AA)_2]$. It does not show geometrical isomerism.
$(iii) [Pt(en)_2Cl_2]Cl_2$: This is an octahedral complex of the type $[M(AA)_2X_2]$. It exhibits geometrical isomerism (cis and trans forms).
$(iv) [Pt(NH_3)_2Cl_2]$: This is a square planar complex of the type $[MA_2X_2]$. It exhibits geometrical isomerism (cis and trans forms).
Therefore,$(iii)$ and $(iv)$ exhibit geometrical isomerism.

(B) For an octahedral complex of the type $[Mabcdef]$,where all six ligands are different,the number of geometrical isomers is $15$.
Each of these $15$ geometrical isomers is chiral,meaning each exists as a pair of enantiomers (optical isomers).
Therefore,the total number of isomers is $15 \text{ (geometrical)} + 15 \text{ (enantiomeric pairs)} = 30$.

(D) The given complex is an octahedral complex of the type $[Mabcdef]$,where $M = Pt$ and the ligands are $py, NH_3, NO_2, Cl, Br, I$.
For an octahedral complex with six different ligands,the number of geometrical isomers is given by the formula $15$.
This is because there are $15$ possible ways to arrange six different ligands around a central metal atom in an octahedral geometry.

(B) The complex $[Co(en)_2Cl_2]^+$ exhibits geometrical isomerism and optical isomerism.
$1$. Geometrical Isomers: It has $2$ geometrical isomers,namely $cis$ and $trans$.
$2$. Optical Isomers: The $trans$ isomer is achiral (optically inactive) due to the presence of a plane of symmetry.
The $cis$ isomer is chiral and exists as a pair of enantiomers ($d$ and $l$ forms).
Therefore,the total number of isomers is $1$ (trans) + $2$ (cis enantiomers) = $3$ isomers.

(D) In all the given complexes,the central metal ion is $Co^{3+}$ ($3d^6$ configuration).
$F^-$ and $H_2O$ are weak field ligands.
Weak field ligands do not cause pairing of $d$-electrons in $Co^{3+}$,resulting in the presence of unpaired electrons.
Therefore,all these complexes are paramagnetic.

(A) To determine the magnetic property,we look at the electronic configuration of the central metal ion and the nature of the ligand:
$1$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons in the $3d$ orbitals,resulting in no unpaired electrons. Thus,it is diamagnetic.
$2$. In $[NiCl_4]^{2-}$,$Ni^{2+}$ $(3d^8)$ is bonded to $Cl^-$ (a weak field ligand),which does not cause pairing,resulting in two unpaired electrons. Thus,it is paramagnetic.
$3$. In $[CoCl_4]^{2-}$,$Co^{2+}$ $(3d^7)$ is bonded to $Cl^-$ (a weak field ligand),resulting in three unpaired electrons. Thus,it is paramagnetic.
$4$. In $[CoF_6]^{2-}$,$Co^{4+}$ $(3d^5)$ is bonded to $F^-$ (a weak field ligand),resulting in five unpaired electrons. Thus,it is paramagnetic.
Therefore,the correct option is $A$.

(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 2.83 \ BM$,we have:
$2.83 = \sqrt{n(n+2)}$
$(2.83)^2 = n(n+2)$
$8.00 \approx n^2 + 2n$
$n^2 + 2n - 8 = 0$
$(n+4)(n-2) = 0$
Since $n$ cannot be negative,$n = 2$.
Now,let us check the number of unpaired electrons for each ion:
$1. Ti^{3+} (Z=22): [Ar] 3d^1 \rightarrow n = 1$
$2. Ni^{2+} (Z=28): [Ar] 3d^8 \rightarrow n = 2$
$3. Cr^{3+} (Z=24): [Ar] 3d^3 \rightarrow n = 3$
$4. Mn^{2+} (Z=25): [Ar] 3d^5 \rightarrow n = 5$
Since $Ni^{2+}$ has $2$ unpaired electrons,it corresponds to a magnetic moment of $2.83 \ BM$.

(B) The complex is $[Co(en)_2Cl_2]Cl$.
$(1)$ Geometrical Isomerism: The complex exhibits two geometrical isomers: $cis$ and $trans$.
$(2)$ Optical Isomerism: The $trans$-isomer has a plane of symmetry and is optically inactive. The $cis$-isomer lacks a plane of symmetry and is optically active,existing as a pair of enantiomers ($d$ and $l$ forms).
Total stereoisomers = $1$ $(trans)$ + $2$ ($cis$ enantiomers) = $3$.

(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu = 1.73 \ BM$,we have $\sqrt{n(n+2)} = 1.73$,which implies $n = 1$.
$1$. $TiCl_4$: $Ti$ is in $+4$ oxidation state $(d^0)$,so $n = 0$.
$2$. $[CoCl_6]^{4-}$: $Co$ is in $+2$ oxidation state $(d^7)$. $Cl^-$ is a weak field ligand,so electrons remain unpaired,$n = 3$.
$3$. $[Cu(NH_3)_4]^{2+}$: $Cu$ is in $+2$ oxidation state $(d^9)$. In the square planar complex,there is $1$ unpaired electron in the $3d$ orbital,$n = 1$.
$4$. $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(d^8)$. $CN^-$ is a strong field ligand,causing pairing,$n = 0$.
Thus,$[Cu(NH_3)_4]^{2+}$ has $1$ unpaired electron and a magnetic moment of $1.73 \ BM$.

(B) The complex $[Co(NH_3)_4Cl_2]^0$ exhibits geometric isomerism.
In the $cis$-isomer,the two $Cl^-$ ligands are adjacent to each other,forming a $Cl-Co-Cl$ bond angle of $90^\circ$.
In the $trans$-isomer,the two $Cl^-$ ligands are opposite to each other,forming a $Cl-Co-Cl$ bond angle of $180^\circ$.
Since the given angle is $90^\circ$,the isomer is the $cis$-isomer.

(C) The magnetic property of a complex depends on the number of unpaired electrons $(n)$. $A$ complex is paramagnetic if $n > 0$.
$1. [Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing. All electrons are paired $(n=0)$,so it is diamagnetic.
$2. [Pt(en)Cl_2]$: $Pt^{2+}$ is a $5d^8$ system. $5d$ series metals always form low-spin complexes with strong field ligands like $en$. All electrons are paired $(n=0)$,so it is diamagnetic.
$3. [CoBr_4]^{2-}$: $Co^{2+}$ is $3d^7$. $Br^-$ is a weak field ligand,so no pairing occurs. The configuration is $e_g^4 t_{2g}^3$,resulting in $n=3$ unpaired electrons. Thus,it is paramagnetic.
$4. Mo(CO)_6$: $Mo$ is in $0$ oxidation state $(4d^6)$. $CO$ is a strong field ligand,causing pairing. All electrons are paired $(n=0)$,so it is diamagnetic.
Therefore,$[CoBr_4]^{2-}$ is the paramagnetic complex.

(B) complex is diamagnetic if all electrons are paired in the $d$-orbitals.
$(A)$ $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so electrons remain unpaired. It has $4$ unpaired electrons and is paramagnetic.
$(B)$ $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing. The configuration is $d_{sp^2}$ hybridization with all electrons paired. Thus,it is diamagnetic.
$(C)$ $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand. It has $2$ unpaired electrons and is paramagnetic.
$(D)$ $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,but $d^5$ configuration results in $1$ unpaired electron. Thus,it is paramagnetic.

(B) The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state,so its configuration is $Ni^{2+} = 3d^8$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
This results in $d^8$ configuration with all electrons paired,making the complex diamagnetic.
In contrast,$[NiCl_4]^{2-}$,$[CuCl_4]^{2-}$,and $[CoF_6]^{3-}$ contain unpaired electrons and are paramagnetic.