Isomerism and Magnetic properties Questions in English

(A) In ${K_3}[Fe(CN)_6]$,the oxidation state of $Fe$ is $+3$.
Electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
The $3d^5$ configuration becomes $t_{2g}^5 e_g^0$,which means there is $1$ unpaired electron.
The magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
For $n=1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ B.M.$
Thus,there is $1$ unpaired electron present per molecule.

(B) The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
For $Fe^{2+}$,the configuration is $[Ar] 3d^6$.
In the $3d$ subshell,there are $4$ unpaired electrons.
Since the ion contains unpaired electrons,it exhibits paramagnetism.

(C) The magnetic moment of an atom or ion is expressed in units of $Bohr \, Magneton$ $(BM)$.

(B) Compounds that contain one or more unpaired electrons exhibit paramagnetic properties due to the magnetic moment associated with the spin of these electrons.

(B) substance shows an increase in weight in a magnetic field if it is paramagnetic (contains unpaired electrons).
$A$) $TiO_2$: $Ti^{+4}$ is $3d^0$ (diamagnetic).
$B$) $Fe_2(SO_4)_3$: $Fe^{+3}$ is $3d^5$ ($5$ unpaired electrons,paramagnetic).
$C$) $KMnO_4$: $Mn^{+7}$ is $3d^0$ (diamagnetic).
$D$) $ScCl_3$: $Sc^{+3}$ is $3d^0$ (diamagnetic).
Since $Fe_2(SO_4)_3$ is paramagnetic,it will be attracted to the magnetic field,resulting in an apparent increase in weight.

(D) To determine the colourless species,we check for the presence of unpaired electrons in the metal ions:
$1$. In $TiF_6^{2-}$,$Ti$ is in the $+4$ oxidation state. The electronic configuration of $Ti^{4+}$ is $[Ar] 3d^0$. Since there are no $d$-electrons,it is colourless.
$2$. In $CoF_6^{3-}$,$Co$ is in the $+3$ oxidation state. The configuration of $Co^{3+}$ is $[Ar] 3d^6$. It has unpaired electrons,so it is coloured.
$3$. In $Cu_2Cl_2$,$Cu$ is in the $+1$ oxidation state. The configuration of $Cu^+$ is $[Ar] 3d^{10}$. Since the $d$-subshell is completely filled,there are no unpaired electrons,making it colourless.
$4$. In $NiCl_4^{2-}$,$Ni$ is in the $+2$ oxidation state. The configuration of $Ni^{2+}$ is $[Ar] 3d^8$. It has unpaired electrons,so it is coloured.
Therefore,$TiF_6^{2-}$ and $Cu_2Cl_2$ are the colourless species.

(B) The colour of transition metal complexes is generally due to $d-d$ transitions,which require the presence of unpaired $d$-electrons.
In $Na_2CdCl_4$,the cadmium ion is $Cd^{2+}$.
The electronic configuration of $Cd$ $(Z=48)$ is $[Kr] 4d^{10} 5s^2$.
Thus,$Cd^{2+}$ has the configuration $[Kr] 4d^{10}$.
Since the $d$-subshell is completely filled,there are no unpaired electrons available for $d-d$ transitions,making the compound colourless.
Therefore,the correct option is $(B)$.

(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$1$. For $[MnCl_4]^{2-}$: $Mn^{2+}$ is $3d^5$. Since $Cl^-$ is a weak field ligand,electrons do not pair up. $n = 5$.
$2$. For $[CoCl_4]^{2-}$: $Co^{2+}$ is $3d^7$. Since $Cl^-$ is a weak field ligand,electrons do not pair up. $n = 3$.
$3$. For $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. Since $CN^-$ is a strong field ligand,electrons pair up. $n = 0$.
Comparing the number of unpaired electrons: $5 > 3 > 0$.
Therefore,the order of magnetic moments is $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$.

(A) In $[FeF_6]^{3-}$,the oxidation state of $Fe$ is $+3$.
The electronic configuration of $Fe^{3+}$ is $[Ar] \ 3d^5$.
Since $F^-$ is a weak field ligand,it does not cause pairing of electrons.
Thus,there are $5$ unpaired electrons $(n = 5)$.
The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.

(C) In the complex $Hg[Co(SCN)_4]$,the oxidation state of $Co$ is $+2$.
The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.
In a tetrahedral field,$Co^{2+}$ has $3$ unpaired electrons $(n = 3)$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n + 2)} \ B.M.$
$\mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \ B.M.$

(D) To determine the magnetic character,we look at the electronic configuration of the central metal ion in the presence of the ligand:
$A$. $[NiCl_4]^{2-}$: $Ni^{2+}$ $(3d^8)$. $Cl^-$ is a weak field ligand,so it does not cause pairing. It has $2$ unpaired electrons,making it paramagnetic.
$B$. $[CoF_6]^{3-}$: $Co^{3+}$ $(3d^6)$. $F^-$ is a weak field ligand. It has $4$ unpaired electrons,making it paramagnetic.
$C$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ $(3d^6)$. $H_2O$ is a weak field ligand. It has $4$ unpaired electrons,making it paramagnetic.
$D$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ $(3d^8)$. $CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals. The configuration becomes $3d^{10}$ (all paired). Thus,it is diamagnetic.

(A) The electronic configuration of neutral scandium is $_{21}Sc = [Ar] \; 3d^1 4s^2$.
For the $Sc^{3+}$ ion,three electrons are removed,resulting in the configuration $Sc^{3+} = [Ar] \; 3d^0 4s^0$.
Since there are no unpaired electrons in the $d$-subshell,the ion does not undergo $d-d$ transitions and is therefore colourless.
Additionally,the absence of unpaired electrons makes it diamagnetic.

(A) $1$. $[Cr(NH_3)_6]^{3+}$: $Cr$ is in $+3$ oxidation state. Electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$. It has $3$ unpaired electrons,so it is paramagnetic.
$2$. $[Fe(CO)_5]$: $Fe$ is in $0$ oxidation state. $CO$ is a strong field ligand,causing pairing of electrons. $Fe(0)$ is $3d^6 4s^2$. All electrons are paired,so it is diamagnetic.
$3$. $[Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state. $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing of electrons. All electrons are paired,so it is diamagnetic.
$4$. $[Cr(CO)_6]$: $Cr$ is in $0$ oxidation state. $Cr(0)$ is $3d^5 4s^1$. $CO$ is a strong field ligand,causing pairing of all electrons. It is diamagnetic.
Therefore,$[Cr(NH_3)_6]^{3+}$ is paramagnetic.

(C) The existence of $Fe^{2+}$ and $NO^+$ in the nitroprusside ion $[Fe(CN)_5NO]^{2-}$ can be established by measuring the magnetic moment of the solid compound.
If iron were $Fe^{III}$ $(3d^5)$,the magnetic properties would differ significantly from the $Fe^{II}$ $(3d^6)$ state present in the complex.
Measuring the magnetic moment allows for the determination of the oxidation state of the metal center.

(A) Geometric isomerism in octahedral complexes occurs when there are different spatial arrangements of ligands around the central metal atom.
For complexes of the type $[MA_5B]$,all positions are equivalent relative to the unique ligand $B$,meaning no geometric isomers can be formed.
Complexes of the type $[MA_4B_2]$ exhibit $cis$ and $trans$ isomers.
Complexes of the type $[MA_3B_3]$ exhibit $fac$ (facial) and $mer$ (meridional) isomers.
Complexes of the type $[MA_2B_4]$ exhibit $cis$ and $trans$ isomers.
Therefore,$[MA_5B]$ is the correct answer.

(A) Coordination isomerism occurs in compounds containing both cationic and anionic complex ions of different metal ions.
It arises from the interchange of ligands between the cationic and anionic entities of different metal ions.
In the given options,$[Cr(NH_3)_6] [Co(CN)_6]$ consists of both a cationic complex $[Cr(NH_3)_6]^{3+}$ and an anionic complex $[Co(CN)_6]^{3-}$.
This compound can exhibit coordination isomerism by interchanging the ligands to form $[Co(NH_3)_6] [Cr(CN)_6]$.

(C) The complexes $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5(ONO)]Cl_2$ contain an ambidentate ligand,$NO_2^-$.
An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different donor atoms.
In the first complex,the $NO_2$ group is bonded to the $Co$ atom through the nitrogen atom $(Co-NO_2)$.
In the second complex,the $ONO$ group is bonded to the $Co$ atom through the oxygen atom $(Co-ONO)$.
Since the mode of linkage of the ligand to the central metal atom is different,these complexes exhibit linkage isomerism.
Therefore,the correct option is $C$.

(C) The two given compounds have the same molecular formula but produce different ions in an aqueous solution.
$[Co(NH_3)_5Br]SO_4 \rightarrow [Co(NH_3)_5Br]^{2+} + SO_4^{2-}$
$[Co(NH_3)_5SO_4]Br \rightarrow [Co(NH_3)_5SO_4]^+ + Br^-$
Since the counter ions are different,this is an example of ionization isomerism.

(D) Ionization isomerism occurs when the counter ion in a coordination complex is itself a potential ligand and can displace a ligand that is currently part of the coordination sphere.
In the given pair:
$[Co(NH_3)_4Cl_2]NO_2$ dissociates to give $NO_2^-$ ions in the solution.
$[Co(NH_3)_4Cl.NO_2]Cl$ dissociates to give $Cl^-$ ions in the solution.
Since they produce different ions in the solution,they are ionization isomers.

(B) Ionisation isomerism occurs when the counter ion in a coordination complex is a potential ligand and can displace a ligand from the coordination sphere.
In the complex $[Co(NH_3)_5Br]SO_4$,the $SO_4^{2-}$ ion is outside the coordination sphere and the $Br^-$ ion is inside. These can exchange positions to form the isomer $[Co(NH_3)_5SO_4]Br$.
These isomers produce different ions in solution:
$[Co(NH_3)_5Br]SO_4 \rightleftharpoons [Co(NH_3)_5Br]^{2+} + SO_4^{2-}$
$[Co(NH_3)_5SO_4]Br \rightleftharpoons [Co(NH_3)_5SO_4]^{+} + Br^{-}$

(A) The atomic number of $Ti$ is $22$,and its electronic configuration is $[Ar] 3d^2 4s^2$.
In $[Ti(H_2O)_6]^{3+}$,the oxidation state of $Ti$ is $+3$.
Therefore,the electronic configuration of $Ti^{3+}$ is $[Ar] 3d^1$.
Since there is $1$ unpaired electron in the $d$-orbital,the complex is paramagnetic.

(A) Coordination isomerism occurs in compounds containing both complex cationic and complex anionic parts.
It is caused by the interchange of ligands between the $A$ complex cation and the complex anion.

(C) Geometrical isomerism is shown by complexes of the type $[MA_4B_2]$,$[M(AA)_2B_2]$,and $[MA_2B_2]$.
$1$. $[Cr(NH_3)_4Cl_2]Cl$ is of the type $[MA_4B_2]$,which shows geometrical isomerism (cis and trans forms).
$2$. $[Co(en)_2Cl_2]Cl$ is of the type $[M(AA)_2B_2]$,which shows geometrical isomerism.
$3$. $[Pt(NH_3)_2Cl_2]$ is of the type $[MA_2B_2]$,which shows geometrical isomerism (cis-platin and trans-platin).
$4$. $[Co(NH_3)_5NO_2]Cl_2$ is of the type $[MA_5B]$,which does not show geometrical isomerism because all positions are equivalent relative to the unique ligand $B$.

(D) Paramagnetic substances are those that are attracted by an external magnetic field. This property arises due to the presence of one or more unpaired electrons in the coordination entity. Therefore,paramagnetic coordination compounds contain unpaired electrons.

(D) Ionization isomerism occurs when the counter ion in a coordination complex is an exchangeable ligand with a ligand present inside the coordination sphere.
In option $D$,the sulfate ion $(SO_4^{2-})$ and the bromide ion $(Br^-)$ exchange positions between the coordination sphere and the counter ion position.
Thus,$[Co(NH_3)_5Br]SO_4$ and $[Co(NH_3)_5SO_4]Br$ are ionization isomers.

(B) To determine the paramagnetism,we calculate the number of unpaired electrons in each complex:
$1$. In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ is $3d^3$,which has $3$ unpaired electrons.
$2$. 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 $4$ unpaired electrons.
$3$. In $[Cu(H_2O)_6]^{2+}$,$Cu^{2+}$ is $3d^9$,which has $1$ unpaired electron.
$4$. In $[Zn(H_2O)_6]^{2+}$,$Zn^{2+}$ is $3d^{10}$,which has $0$ unpaired electrons.
Since $[Fe(H_2O)_6]^{2+}$ has the highest number of unpaired electrons $(4)$,it exhibits the highest paramagnetism.

(C) $1$. In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state. The configuration is $[Ar] 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$ orbitals,resulting in $sp^3$ hybridization with no unpaired electrons. Thus,it is diamagnetic.
$2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons in $3d$ orbitals,resulting in $dsp^2$ hybridization with no unpaired electrons. Thus,it is diamagnetic.
$3$. In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The configuration is $[Ar] 3d^8$. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons. This results in $sp^3$ hybridization with two unpaired electrons in $3d$ orbitals. Thus,it is paramagnetic.
Therefore,$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic,while $[NiCl_4]^{2-}$ is paramagnetic.

(A) $[Co(NH_3)_4Cl_2]^+$ is a complex of the type $Ma_4b_2$.
This type of complex exhibits geometrical isomerism,existing in two forms: $cis$ and $trans$.
In the $cis$ form,the two $Cl^-$ ligands are adjacent to each other,while in the $trans$ form,they are opposite to each other.

(B) . In $[Ag(NH_3)_2]Cl$,the central metal ion is $Ag^+$,which has a $d^{10}$ electronic configuration,meaning all electrons are paired,making it diamagnetic.
In $[Cu(NH_3)_4]Cl_2$,$Cu^{2+}$ has a $d^9$ configuration,which has one unpaired electron.
$NO$ and $NO_2$ are odd-electron molecules,which are paramagnetic.

(A) The coordination complex $[Pt(NH_3)_2Cl_2]$ is a square planar complex of the type $[MA_2B_2]$.
It exhibits $2$ geometrical isomers:
$1$. $\text{Cis-isomer}$: Where similar ligands ($NH_3$ or $Cl$) are adjacent to each other.
$2$. $\text{Trans-isomer}$: Where similar ligands ($NH_3$ or $Cl$) are opposite to each other.

(D) In the given compounds,the number of water molecules present in the coordination sphere and in the ionization sphere is different.
Specifically,in $[Cr(H_2O)_6]Cl_3$,all $6$ water molecules are coordinated to the metal ion.
In $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$,only $5$ water molecules are coordinated,while $1$ water molecule acts as a lattice water (solvent of crystallization).
This difference in the number of water molecules inside and outside the coordination sphere characterizes hydrate isomerism,which is a type of solvate isomerism.

(A) The given complex is $[Co(NO_2)_2(NH_3)_2]$. Assuming a square planar geometry (as it is a $4$-coordinate complex),it follows the form $[MA_2B_2]$.
For a square planar complex of the type $[MA_2B_2]$,there are $2$ possible geometrical isomers: $cis$ and $trans$.
In the $cis$ isomer,the identical ligands are adjacent to each other.
In the $trans$ isomer,the identical ligands are opposite to each other.
Therefore,the total number of geometrical isomers is $2$.

(B) Linkage isomerism is observed in coordination compounds containing ambidentate ligands.
In pentaammine nitro chromium$(III)$ chloride,the ligand $NO_2^-$ is an ambidentate ligand that can coordinate through either the nitrogen atom (nitro,$-NO_2$) or the oxygen atom (nitrito,$-ONO$).
Therefore,the complex exhibits linkage isomerism,forming isomers such as $[Cr(NH_3)_5(NO_2)]Cl_2$ and $[Cr(NH_3)_5(ONO)]Cl_2$.

(B) The complex $[Co(en)_2Cl_2]^+$ exhibits geometrical isomerism (cis and trans forms).
The cis-form is optically active and exists as two enantiomers ($d$-cis and $l$-cis).
The trans-form is optically inactive (achiral) and exists as a single meso form.
Therefore,the total number of stereoisomers is $2$ (cis-enantiomers) + $1$ (trans-isomer) = $3$.

(A) The magnetic moment is calculated using the spin-only formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
For the $[Cu(NH_3)_4]^{2+}$ ion,the central metal ion is $Cu^{2+}$,which has an electronic configuration of $[Ar] 3d^9$.
In the $3d^9$ configuration,there is $n=1$ unpaired electron.
Substituting $n=1$ into the formula: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$

(D) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$ and it has a $d^{10}$ configuration. Since $CO$ is a strong field ligand,all electrons are paired,making it diamagnetic.
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,making it diamagnetic.
In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons,making it diamagnetic.
In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(d^8)$. $Cl^-$ is a weak field ligand and cannot cause pairing of electrons. Therefore,it has $2$ unpaired electrons,making it paramagnetic.

(D) The given complex is a coordination isomer. The possible isomers are formed by the exchange of ligands between the two coordination spheres:
$1. [Cu(NH_3)_4][PtCl_4]$
$2. [Cu(NH_3)_3Cl][PtCl_3(NH_3)]$
$3. [Cu(NH_3)_2Cl_2][PtCl_2(NH_3)_2]$ (cis-isomer)
$4. [Cu(NH_3)_2Cl_2][PtCl_2(NH_3)_2]$ (trans-isomer)
$5. [Cu(NH_3)Cl_3][PtCl(NH_3)_3]$
$6. [Pt(NH_3)_4][CuCl_4]$
Thus,the total number of possible isomers is $6$.

(A) $(I) [Fe(CN)_6]^{3-}$
Here,the oxidation state of $Fe$ is $+3$.
The electronic configuration of $Fe$ (ground state) is $[Ar] 3d^6 4s^2$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5 4s^0$.
Number of unpaired electrons $= 5$.
$(II) [Fe(CN)_6]^{4-}$
Here,the oxidation state of $Fe$ is $+2$.
The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6 4s^0$.
Due to the strong field ligand $CN^-$,the electrons pair up,resulting in $0$ unpaired electrons.
$(III) [Cr(H_2O)_6]^{3+}$
Here,the oxidation state of $Cr$ is $+3$.
The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3 4s^0$.
Number of unpaired electrons $= 3$.
$(IV) [Cu(H_2O)_6]^{2+}$
Here,the oxidation state of $Cu$ is $+2$.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9 4s^0$.
Number of unpaired electrons $= 1$.
The paramagnetic character is directly proportional to the number of unpaired electrons.
Therefore,$[Fe(CN)_6]^{3-}$ has the maximum paramagnetic character.

(C) The complexes $[Co(NH_3)_6][Cr(C_2O_4)_3]$ and $[Cr(NH_3)_6][Co(C_2O_4)_3]$ exhibit coordination isomerism.
Coordination isomerism occurs in compounds where both the cation and the anion are complex ions,and the isomers differ in the distribution of ligands between the metal centers of the cation and the anion.

(A) The atomic number of $Mn$ is $25$,so its electronic configuration is $[Ar] 3d^5 4s^2$.
In $[Mn(H_2O)_6]^{2+}$,the oxidation state of $Mn$ is $+2$,so its configuration is $3d^5$.
$H_2O$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $3d^5$ configuration has $5$ unpaired electrons.
Since the ion contains unpaired electrons,it exhibits paramagnetism.

(C) The complex is $[PdClBr_2(SCN)]^{2-}$.
For a square planar complex of the type $[Mabcd]$,there are $3$ geometrical isomers.
However,here the ligand $SCN^-$ is an ambidentate ligand,which can coordinate through $S$ (thiocyanate) or $N$ (isothiocyanate).
For each geometrical isomer,linkage isomerism is possible.
Specifically,for the complex $[PdClBr_2(SCN)]^{2-}$,the geometrical isomers are $cis$ and $trans$ with respect to the $Br$ ligands.
Considering the linkage isomerism ($SCN$ vs $NCS$),the total number of isomers is $2 \times 2 = 4$.

(D) The magnetic moment $(\mu)$ is related to the number of unpaired electrons $(n)$ by the formula $\mu = \sqrt{n(n+2)} \text{ B.M.}$.
Given that the magnetic moment is $0 \text{ B.M.}$,we have $\sqrt{n(n+2)} = 0$.
This implies $n(n+2) = 0$,which gives $n = 0$ (since $n$ cannot be negative).
Therefore,the number of unpaired electrons is $0$.

(C) The given complexes $[Pt(NH_3)_4Cl_2]Br_2$ and $[Pt(NH_3)_4Br_2]Cl_2$ have the same molecular formula but produce different ions in an aqueous solution.
$[Pt(NH_3)_4Cl_2]Br_2$ dissociates to give $Br^-$ ions,whereas $[Pt(NH_3)_4Br_2]Cl_2$ dissociates to give $Cl^-$ ions.
This phenomenon is known as ionization isomerism.

(D) The complex $[Pt(en)_2Cl_2]$ has an octahedral geometry (assuming $Pt(IV)$).
It exhibits geometrical isomerism due to the existence of $cis$ and $trans$ forms.
The $cis$-isomer of $[Pt(en)_2Cl_2]$ is optically active because it lacks a plane of symmetry and a center of inversion,thus existing as a pair of enantiomers.
The $trans$-isomer is optically inactive due to the presence of a plane of symmetry.
Therefore,$[Pt(en)_2Cl_2]$ shows both geometrical and optical isomerism.

(A) In $Ni(CO)_4$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
Since $CO$ is a strong field ligand,it causes the pairing of electrons in the $3d$ and $4s$ orbitals,resulting in a $sp^3$ hybridized state with no unpaired electrons.
Therefore,$Ni(CO)_4$ is diamagnetic.
In contrast,$Ni(Cl)_4^{2-}$ and $Ni(Br)_4^{2-}$ involve weak field ligands ($Cl^-$ and $Br^-$),leading to $sp^3$ hybridization with two unpaired electrons,making them paramagnetic.

(C) Paramagnetism is directly proportional to the number of unpaired electrons $(n)$.
$1$. In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons,so $n = 0$.
$2$. In $[Ni(NH_3)_4]Cl_2$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $NH_3$ is a strong field ligand,causing pairing,so $n = 0$ (square planar).
$3$. In $[Ni(NH_3)_6]Cl_2$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $NH_3$ is a strong field ligand,but in octahedral geometry,$n = 2$.
$4$. In $[Cu(NH_3)_4]Cl_2$,$Cu$ is in $+2$ oxidation state $(3d^9)$. There is $1$ unpaired electron,so $n = 1$.
Comparing the values,$[Ni(NH_3)_6]Cl_2$ has the highest number of unpaired electrons $(n = 2)$,hence it shows the highest paramagnetism.

(B) Optical isomerism is exhibited by coordination compounds that lack a plane of symmetry and a center of symmetry (i.e.,they are chiral).
$A$. $trans-[Cr(en)_2(CN)_2]Cl$ has a plane of symmetry,so it is optically inactive.
$B$. $[Co(en)_3]Br_3$ (tris(ethylenediamine)cobalt$(III)$ bromide) contains three bidentate ligands $(en)$. It does not possess a plane of symmetry or a center of symmetry,making it chiral and optically active. It exists as $d$ and $l$ enantiomers.
$C$. $[Co(NH_3)_5(NO_2)]I_2$ is a simple octahedral complex with high symmetry,making it optically inactive.
$D$. $[Pt(NH_3)_2Cl_2]$ is a square planar complex,which is inherently achiral and optically inactive.
Therefore,the correct option is $B$.

(A) . $[Co(NH_3)_3Cl_3]$ does not exhibit optical isomerism because it follows the $MA_3B_3$ type geometry,which possesses a plane of symmetry in both its facial $(fac)$ and meridional $(mer)$ isomers.

(C) Change in the composition of the coordination sphere yields ionisation isomers.
For example,$[Cr(H_2O)_6]Cl_3$ and $[CrCl(H_2O)_5]Cl_2 \cdot H_2O$ are ionisation isomers.

(B) The complex $[Cr(NH_3)_5NO_2]Cl_2$ exhibits two types of isomerism:
$1$. Linkage isomerism: Due to the presence of the ambidentate ligand $NO_2^-$,which can coordinate through either the $N$ atom (nitro,$-NO_2$) or the $O$ atom (nitrito,$-ONO$).
$2$. Ionisation isomerism: Due to the exchange of the $Cl^-$ ion inside the coordination sphere with the $NO_2^-$ ligand,resulting in $[Cr(NH_3)_5Cl]Cl(NO_2)$.