Isomerism and Magnetic properties Questions in English

(B) 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.
$2$. $trans$-isomer: In this form,the two $Cl^-$ ligands are opposite to each other.
Since there are no chiral centers,optical isomerism is not observed.
Therefore,there are a total of $2$ isomeric forms.

(D) The complex is $[Cr(NH_3)_2(OH)_2Cl_2]^-$.
$1.$ It does not show ionization isomerism as there are no counter ions.
$2.$ It does not show hydrate isomerism as there is no water molecule outside the coordination sphere.
$3.$ It does not show linkage isomerism as there are no ambidentate ligands.
$4.$ It shows geometrical isomerism due to the presence of different ligands $(NH_3, OH^-, Cl^-)$ in an octahedral geometry.
$5.$ It shows optical isomerism because the complex can exist in chiral forms (e.g.,the cis-isomer can be optically active depending on the arrangement).
Therefore,the complex exhibits geometrical and optical isomerism ($4$ and $5$).

(D) The two complexes are identical because they represent the same chemical structure viewed from different orientations or simply written in a different order. In coordination chemistry,if the connectivity of ligands to the central metal atom remains the same and the spatial arrangement is equivalent,the complexes are considered identical.

(B) The given compounds are $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5(SO_4)]Cl$.
These complexes differ in the counter ions ($Br^-$ and $Cl^-$) present outside the coordination sphere.
When these complexes dissolve in water,they produce different ions in the solution.
This type of isomerism,where the counter ion is exchanged with a ligand from the coordination sphere or simply differs in the counter ion,is known as ionization isomerism.

(B) $1.$ $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5ONO]Cl_2$ exhibit linkage isomerism because the ligand $NO_2^-$ can coordinate through either $N$ or $O$ (as $ONO^-$). This is correct.
$2.$ $[Cu(NH_3)_4] [PtCl_4]$ and $[Pt(NH_3)_4] [CuCl_4]$ exhibit coordination isomerism due to the interchange of ligands between cationic and anionic coordination spheres. This is correct.
$3.$ $[Pt(NH_3)_4Cl_2]Br_2$ and $[Pt(NH_3)_4Br_2]Cl_2$ exhibit ionization isomerism because they produce different ions in solution ($Br^-$ vs $Cl^-$). This is correct.
Therefore,all three pairs are correctly matched.

(B) The given complexes are $[Co(NO_2)(NH_3)_5]Cl_2$ and $[Co(ONO)(NH_3)_5]Cl_2$.
In these complexes,the ligand $NO_2^-$ is an ambidentate ligand,which can bind to the central metal atom through either the nitrogen atom $(-NO_2)$ or the oxygen atom $(-ONO)$.
This type of isomerism,where an ambidentate ligand coordinates through different atoms,is known as linkage isomerism.
Therefore,the correct option is $B$.

(B) Coordination isomerism occurs in compounds containing both cationic and anionic coordination entities,where the ligands can be exchanged between the metal centers.
In the complex $[Cr(NH_3)_6] [Co(CN)_6]$,both the cation and the anion are coordination entities.
This compound can exhibit coordination isomerism by exchanging ligands to form $[Co(NH_3)_6] [Cr(CN)_6]$.

(D) The complex $[Co(NH_3)_3(NO_2)_3]$ is of the type $[MA_3B_3]$.
For octahedral complexes of the type $[MA_3B_3]$,there are two possible geometrical isomers:
$1$. 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 geometrical isomers is $2$.

(C) Geometric isomerism is commonly observed in coordination compounds with coordination numbers $4$ (square planar geometry) and $6$ (octahedral geometry).
Among the given options,both $4$ and $6$ allow for geometric isomerism.
However,in the context of standard chemistry curriculum questions of this type,$4$ and $6$ are the primary coordination numbers exhibiting this phenomenon.
Given the typical structure of such multiple-choice questions,if a single answer is required,$4$ and $6$ are both valid,but $4$ is frequently cited for square planar complexes like $[M(A)_2(B)_2]$.
Since both $4$ and $6$ are correct,and $4$ is a fundamental case,we select $4$ as a representative answer.

(C) Optical activity is exhibited by coordination complexes that lack a plane of symmetry or center of inversion (i.e.,they are chiral).
$1$. $Trans-[Co(NH_3)_4Cl_2]^+$ has a plane of symmetry and is achiral.
$2$. $[Cr(H_2O)_6]^{3+}$ is a symmetric octahedral complex and is achiral.
$3$. $Cis-[Co(NH_3)_2(en)_2]^{3+}$ lacks a plane of symmetry and is chiral,thus it exhibits optical activity.
$4$. $Trans-[Co(NH_3)_2(en)_2]^{3+}$ has a plane of symmetry and is achiral.
Therefore,the correct option is $C$.

(B) Optical isomerism is exhibited by complexes that lack a plane of symmetry or center of inversion.
For the complex $[Co(en)_2Cl_2]Cl$,the cis-isomer exists in two enantiomeric forms (non-superimposable mirror images) because it lacks a plane of symmetry.
The trans-isomer,however,possesses a plane of symmetry and is optically inactive.
Since the cis-form is chiral,the complex $[Co(en)_2Cl_2]^+$ exhibits optical isomerism.

(D) $1$. $[Co(NH_3)_4Cl_2]^+$ is an $MA_4B_2$ type complex,which shows $2$ geometrical isomers (cis and trans). The cis-isomer is optically active,giving $2$ enantiomers,so total isomers = $3$.
$2$. $[Ni(en)(NH_3)_4]^{2+}$ is an $M(AA)B_4$ type complex,which shows no geometrical or optical isomerism.
$3$. $[Ni(C_2O_4)(en)_2]$ is an $M(AA)_3$ type complex,which shows optical isomerism ($2$ enantiomers).
$4$. $[Cr(SCN)_2(NH_3)_4]^+$ is an $MA_4B_2$ type complex (with linkage isomerism due to $SCN^-$),which shows $2$ geometrical isomers. The cis-isomer is optically active ($2$ enantiomers). Additionally,$SCN^-$ can show linkage isomerism ($SCN^-$ and $NCS^-$). Thus,it exhibits the maximum number of isomers.

(D) The chemical formula for pentaamminenitrocobalt $(III)$ ion is $[Co(NH_3)_5(NO_2)]^{2+}$.
In this complex,the ligand $NO_2^-$ is an ambidentate ligand,meaning it can coordinate to the central metal atom through either the nitrogen atom (nitro,$-NO_2$) or the oxygen atom (nitrito,$-ONO$).
Due to the presence of an ambidentate ligand,the complex exhibits linkage isomerism.

(C) Geometrical isomerism is shown by coordination complexes that have different spatial arrangements of ligands around the central metal atom.
For octahedral complexes of the type $[MA_4B_2]$,geometrical isomerism (cis and trans) is possible.
For octahedral complexes of the type $[MA_5B]$,geometrical isomerism is $NOT$ possible because all positions are equivalent relative to the unique ligand $B$.
In $[Co(NH_3)_5 NO_2]Cl_2$,the complex ion is $[Co(NH_3)_5 NO_2]^{2+}$,which is of the type $[MA_5B]$.
Therefore,it does not exhibit geometrical isomerism.

(A) Optical isomerism is exhibited by complexes that lack a plane of symmetry or center of inversion (chiral complexes).
$1$. $[Co(NH_3)_3 Cl_3]$ exists as facial $(fac)$ and meridional $(mer)$ isomers,both of which have planes of symmetry,so it is optically inactive.
$2$. $[Co(en)_3]Cl_3$ is a tris-chelated complex and is chiral.
$3$. $[Co(en)_2 Cl_2]Cl$ has a cis-isomer that is chiral (optically active).
$4$. $[Co(en)(NH_3)_2Cl_2]Cl$ has geometric isomers,and the cis-form is chiral.
Therefore,$[Co(NH_3)_3 Cl_3]$ does not exhibit optical isomerism.

(C) For a square planar complex of the type $[Mabcd]$,where all four ligands are different,the number of geometric isomers is $3$.
These isomers are formed by fixing one ligand (e.g.,$a$) and arranging the other three ligands $(b, c, d)$ in different positions relative to it.
The three possible arrangements are:
$1$. $a$ 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,the total number of geometric isomers is $3$.

(A) The complex $[Co(en)(NH_3)_2Cl_2]Cl$ has the formula $[M(AA)(a)_2(b)_2]$.
This complex exists as geometric isomers (cis and trans).
For the cis-isomer,it can further exist as optical isomers (d and l forms).
Thus,the cis-isomer has two optical isomers,and the trans-isomer is achiral.
However,considering the specific coordination sphere $[Co(en)(NH_3)_2Cl_2]^+$,it exhibits geometric isomerism (cis and trans) and the cis-form is optically active,leading to a total of $4$ isomers (cis-d,cis-l,trans-d,trans-l is not correct,rather it is cis-d,cis-l,trans-isomer,and another geometric arrangement).
Actually,for $[M(AA)(a)_2(b)_2]$,the cis-form is optically active ($2$ isomers) and the trans-form is optically inactive ($1$ isomer),but with different ligands,the count varies.
$[Co(en)(NH_3)_2Cl_2]^+$ specifically allows for $4$ isomers: $cis-trans$ arrangements and optical activity.

(D) For a complex to exhibit both geometrical and optical isomerism,it must have geometrical isomers where at least one isomer is chiral (optically active).
$1$. $[Pt(NH_3)_2Cl_2]$ is a square planar complex,which shows geometrical isomerism but not optical isomerism.
$2$. $[Pt(NH_3)_2Cl_4]$ is an octahedral complex,which shows geometrical isomerism but not optical isomerism.
$3$. $[Pt(en)_3]^{4+}$ is an octahedral complex,which shows optical isomerism but not geometrical isomerism.
$4$. $[Pt(en)_2Cl_2]$ is an octahedral complex. It exists as $cis$ and $trans$ isomers. The $cis$-isomer is optically active because it lacks a plane of symmetry,while the $trans$-isomer is optically inactive. Thus,it exhibits both geometrical and optical isomerism.

(B) To determine if a complex is colorless,we check the electronic configuration of the central metal ion. $A$ complex is colorless if the metal ion has a $d^{10}$ configuration (no unpaired electrons and no $d-d$ transitions) or a $d^0$ configuration.
$1$. In $Na_2[CuCl_4]$,$Cu$ is in the $+2$ oxidation state $(Cu^{2+})$,which is $3d^9$. It has one unpaired electron and is colored.
$2$. In $Na_2[CdCl_4]$,$Cd$ is in the $+2$ oxidation state $(Cd^{2+})$. The electronic configuration of $Cd$ is $[Kr] 4d^{10} 5s^2$. Thus,$Cd^{2+}$ is $[Kr] 4d^{10}$. Since the $d$-subshell is completely filled,no $d-d$ transition is possible,making it colorless.
$3$. In $K_4[Fe(CN)_6]$,$Fe$ is in the $+2$ oxidation state $(Fe^{2+})$,which is $3d^6$. It is colored.
$4$. In $K_3[Fe(CN)_6]$,$Fe$ is in the $+3$ oxidation state $(Fe^{3+})$,which is $3d^5$. It is colored.
Therefore,the correct option is $B$.

(C) . $K[PtCl_3(\pi-C_2H_4)]$ contains a square planar platinum center,which is correct.
$B$. $[Co(NH_3)_3Cl_3]$ exists as $fac$ and $mer$ geometrical isomers,neither of which is optically active,so this is correct.
$C$. $[Fe(EDTA)]^-$ is an octahedral complex where $EDTA$ is a hexadentate ligand,so the coordination number is $6$. However,$[Fe(EDTA)]^-$ is not optically active because it has a plane of symmetry. Thus,this statement is incorrect.
$D$. $[Co(NH_3)_5NO_2]Cl$ gives $1$ mole of $AgCl$ precipitate,while $[Pt(NH_3)_4]Cl_2$ gives $2$ moles of $AgCl$ precipitate with $AgNO_3$. They can be differentiated,so this is correct.

(C) The given complex $[Cu(NH_3)_4][PtCl_4]$ exhibits coordination isomerism.
The possible coordination isomers are formed by the exchange of ligands between the cationic and anionic coordination spheres:
$1. [Cu(NH_3)_4][PtCl_4]$
$2. [Cu(NH_3)_3Cl][Pt(NH_3)Cl_3]$
$3. [Cu(NH_3)_2Cl_2][Pt(NH_3)_2Cl_2]$
$4. [Cu(NH_3)Cl_3][Pt(NH_3)_3Cl]$
$5. [CuCl_4][Pt(NH_3)_4]$
Thus,there are a total of $5$ possible isomers.

(B) In $Na_2[CdCl_4]$,the oxidation state of $Cd$ is $+2$.
The electronic configuration of $Cd^{+2}$ is $[Kr] 4d^{10}$.
Since all $d$-orbitals are completely filled,there are no unpaired electrons.
Therefore,$Na_2[CdCl_4]$ is colourless.

(A) To determine the number of stereoisomers:
$(i)$ $[Cr(NO_3)_3(NH_3)_3]$ is of the type $[MA_3B_3]$,which exhibits facial $(fac)$ and meridional $(mer)$ isomerism. It has $2$ stereoisomers.
$(ii)$ $K_3[Co(C_2O_4)_3]$ is of the type $[M(AA)_3]$. It exhibits optical isomerism ($d$ and $l$ forms). It has $2$ stereoisomers.
$(iii)$ $K_3[CoCl_2(C_2O_4)_2]$ is of the type $[M(AA)_2B_2]$. It exhibits cis-trans isomerism. The cis-form is optically active ($d$ and $l$),while the trans-form is optically inactive. Total stereoisomers = $3$ $(cis-d, cis-l, trans)$.
$(iv)$ $[CoBrCl(en)_2]$ is of the type $[M(AA)_2BC]$. It exhibits cis-trans isomerism. The cis-form is optically active ($d$ and $l$),while the trans-form is optically inactive. Total stereoisomers = $3$ $(cis-d, cis-l, trans)$.
Thus,$(iii)$ and $(iv)$ have $3$ stereoisomeric forms.

(D) 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 $\mu = \sqrt{15} \ BM$.
Equating the two expressions: $\sqrt{15} = \sqrt{n(n+2)}$.
Squaring both sides: $15 = n(n+2)$.
$n^2 + 2n - 15 = 0$.
Solving the quadratic equation: $(n+5)(n-3) = 0$.
Since $n$ must be a positive integer,$n = 3$.
Therefore,the number of unpaired electrons is $3$.

(B) To have the same magnetic moment,the species must have the same number of unpaired electrons $(n)$.
$1$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$. It has $4$ unpaired electrons.
$2$. $[CoCl_4]^{2-}$: $Co^{2+}$ is $3d^7$. In a tetrahedral field,it has $3$ unpaired electrons.
$3$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. It has $4$ unpaired electrons.
$4$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ is $3d^5$. It has $5$ unpaired electrons.
Comparing the number of unpaired electrons,$[Cr(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{2+}$ both have $4$ unpaired electrons. Therefore,they have the same magnetic moment.

(C) The complex $[Co(C_2O_4)_2(NH_3)_2]^-$ is of the type $[M(AA)_2a_2]$,where $AA$ is a bidentate ligand (oxalate) and $a$ is a monodentate ligand (ammonia).
This complex exhibits geometrical isomerism,having two forms: $cis-$ and $trans-$.
The $cis-$ isomer is optically active and exists as a pair of enantiomers ($d$ and $l$ forms).
The $trans-$ isomer is optically inactive (achiral) due to the presence of a plane of symmetry.
Therefore,the total number of isomers is $3$ ($cis-d$,$cis-l$,and $trans-$).

(C) $Cu^{2+}$ ($3d^9$ configuration) forms complexes with the same magnetic moment and geometry irrespective of the nature of the ligand.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. The $3d$ subshell contains one unpaired electron.
Since the $3d$ subshell is almost completely filled,the incoming ligands cannot cause any pairing or redistribution of electrons in the $3d$ orbitals. Therefore,the number of unpaired electrons remains $1$ (magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$) regardless of whether the ligand is strong or weak.
Furthermore,the geometry of $Cu^{2+}$ complexes is typically determined by the number of ligands (e.g.,square planar for $4$ ligands due to Jahn-Teller distortion),which remains consistent for a given coordination number.

(A) The given compounds contain the ambidentate ligand $-NO_2^-$,which can coordinate to the central metal atom through either the nitrogen atom (nitro,$-NO_2$) or the oxygen atom (nitrito,$-ONO$).
Because the point of attachment of the ligand to the metal center differs between the two complexes,they exhibit linkage isomerism.

(D) The magnetic moment (spin only) is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. Zero magnetic moment implies $n = 0$.
$1$. $[Ni(NH_3)_6]Cl_2$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand,resulting in $sp^3d^2$ hybridization with $2$ unpaired electrons.
$2$. $Na_3[FeF_6]$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,resulting in $sp^3d^2$ hybridization with $5$ unpaired electrons.
$3$. $[Cr(H_2O)_6]SO_4$: $Cr^{2+}$ is $3d^4$. $H_2O$ is a weak field ligand,resulting in $sp^3d^2$ hybridization with $4$ unpaired electrons.
$4$. $K_4[Fe(CN)_6]$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $d^2sp^3$ hybridization with $0$ unpaired electrons.
Thus,$K_4[Fe(CN)_6]$ has a zero magnetic moment.

(B) For coordination complexes with a coordination number of $4$,the geometry can be either tetrahedral or square planar.
Square planar complexes typically exhibit low-spin configurations because the crystal field splitting energy is large enough to cause pairing.
Tetrahedral complexes,due to smaller crystal field splitting,almost always exhibit high-spin configurations.
Therefore,measuring the magnetic moment to determine the number of unpaired electrons helps distinguish between these two geometries.

(A) For an octahedral complex of the type $[M(AB)(CD)ef]$,where $AB$ and $CD$ are unsymmetrical bidentate ligands and $e, f$ are monodentate ligands,the total number of geometrical isomers is $20$.
Each geometrical isomer is chiral because the complex lacks a plane of symmetry and a center of inversion.
Since each geometrical isomer exists as a pair of enantiomers (a $d$ and $l$ form),the total number of enantiomeric pairs is equal to the total number of geometrical isomers.
Therefore,there are $20$ pairs of enantiomers possible.

(D) The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 4.5 \ BM$,we have $\sqrt{n(n+2)} \approx 4.5$,which implies $n(n+2) \approx 20.25$. Solving for $n$,we get $n \approx 4$.
For $Co$ in $+3$ oxidation state $(3d^6)$,if the ligand $L$ is weak,the electrons will be arranged as shown in the diagram,resulting in $4$ unpaired electrons.
Thus,$Co$ must be in $+3$ oxidation state.

(D) These two compounds are coordination isomers.
In the first complex,$[Co(NH_3)_6][Cr(NO_2)_6]$,the cation is $[Co(NH_3)_6]^{3+}$ and the anion is $[Cr(NO_2)_6]^{3-}$.
In the second complex,$[Cr(NH_3)_6][Co(NO_2)_6]$,the cation is $[Cr(NH_3)_6]^{3+}$ and the anion is $[Co(NO_2)_6]^{3-}$.
Upon electrolysis of their aqueous solutions,the metal ions present in the cationic complex will migrate to the cathode and be deposited.
For $[Co(NH_3)_6][Cr(NO_2)_6]$,$Co$ metal is deposited at the cathode.
For $[Cr(NH_3)_6][Co(NO_2)_6]$,$Cr$ metal is deposited at the cathode.
Thus,they can be distinguished by the electrolysis of their aqueous solutions.

(D) To determine if a complex is paramagnetic,we check for the presence of unpaired electrons in the central metal ion.
$1$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons (diamagnetic).
$2$. In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing,resulting in $0$ unpaired electrons (diamagnetic).
$3$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing,resulting in $0$ unpaired electrons (diamagnetic).
$4$. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak field ligand,so no pairing occurs. The configuration is $t_{2g}^4 e_g^2$,which has $4$ unpaired electrons,making it paramagnetic.

(D) Optical isomerism is exhibited by coordination compounds that lack a plane of symmetry and a center of symmetry.
$A$. $[Co(NH_3)_5(NO_2)]I_2$ has a plane of symmetry and is achiral.
$B$. $[Pt(NH_3)_2Cl_2]$ is a square planar complex,which is inherently achiral.
$C$. The $trans$ isomer of $[Cr(en)_2(CN)_2]^+$ has a plane of symmetry and is achiral.
$D$. $[Co(en)_3]Br_3$ contains three bidentate ethylenediamine $(en)$ ligands. This complex does not have a plane of symmetry or a center of symmetry,making it chiral and optically active. It exists as a pair of non-superimposable mirror images (enantiomers).

(A) The complex $[Co(NO_2)_3(NH_3)_3]$ exists in two geometric isomeric forms: $fac$ (facial) and $mer$ (meridional).
Both of these isomers possess a plane of symmetry,which makes them achiral.
Since optical isomerism requires the molecule to be chiral (lacking a plane of symmetry),this complex does not exhibit optical isomerism.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(B) In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state ($3d^5$ configuration). Due to the strong field ligand $CN^-$,the electrons pair up,leaving $1$ unpaired electron,making it paramagnetic.
In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state ($3d^6$ configuration). Due to the strong field ligand $CN^-$,all $6$ electrons pair up in the $t_{2g}$ orbitals,leaving $0$ unpaired electrons,making it diamagnetic.
Both statements are true,but the magnetic property depends on the number of unpaired electrons resulting from the crystal field splitting and the nature of the ligand,not just the oxidation state. Thus,the Reason is not the correct explanation of the Assertion.

(A) The complex $[CoCl_2(en)_2]$ contains two identical ligands $(Cl^-)$ and two bidentate ligands $(en)$.
It exists in two forms: $cis$ and $trans$.
In the $trans$ form,the two $Cl$ atoms are at $180^{\circ}$ to each other.
In the $cis$ form,the two $Cl$ atoms are at $90^{\circ}$ to each other.
Since the arrangement of ligands around the central metal atom differs,this is an example of geometrical isomerism.

(A) For a complex $MA_{2}B_{2}$ with $sp^{3}$ hybridisation,the geometry is tetrahedral. Tetrahedral complexes of the type $MA_{2}B_{2}$ possess a plane of symmetry,making them optically inactive.
Optical isomers $= 0$.
For a complex $MA_{2}B_{2}$ with $dsp^{2}$ hybridisation,the geometry is square planar. Square planar complexes are planar and possess a plane of symmetry (the molecular plane itself),making them optically inactive.
Optical isomers $= 0$.
Thus,the number of optical isomers for both cases is $0$ and $0$.

(D) Geometrical isomerism in coordination complexes depends on the arrangement of ligands around the central metal atom.
$(a) [Pt(NH_3)_3 Cl]^+$ is a square planar complex of the type $[MA_3B]$,which does not show geometrical isomerism.
$(b) [Pt(NH_3) Cl_5]^-$ is an octahedral complex of the type $[MA_5B]$,which does not show geometrical isomerism.
$(c) [Pt(NH_3)_2 Cl(NO_2)]$ is a square planar complex of the type $[MA_2BC]$. It can exist in cis and trans forms.
$(d) [Pt(NH_3)_4 ClBr]^{2+}$ is an octahedral complex of the type $[MA_4BC]$. It can exist in cis and trans forms.
Therefore,complexes $(c)$ and $(d)$ exhibit geometrical isomerism.

(A) $1$. $Ni(CO)_{4}$: $Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand,causing pairing of electrons. All electrons are paired,so $\mu_{m} = 0 \ B.M.$
$2$. $[Ni(H_{2}O)_{6}]Cl_{2}$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $H_{2}O$ is a weak field ligand. It has $2$ unpaired electrons. $\mu_{m} = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ B.M.$
$3$. $Na_{2}[Ni(CN)_{4}]$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $CN^{-}$ is a strong field ligand,causing pairing of electrons. All electrons are paired,so $\mu_{m} = 0 \ B.M.$
$4$. $PdCl_{2}(PPh_{3})_{2}$: $Pd^{2+}$ is a $4d^{8}$ ion. Due to the large size of $4d$ orbitals,the crystal field splitting energy is high,leading to pairing of electrons. Thus,$\mu_{m} = 0 \ B.M.$
Comparing the values: $A (0) \approx C (0) \approx D (0) < B (2.83)$.
Therefore,the correct order is $A \approx C \approx D < B$.

(C) Complexes of the type $[Ma_{3}b_{3}]$ exhibit facial $(fac)$ and meridional $(mer)$ isomerism.
In the $fac$-isomer,the three identical ligands occupy one face of the octahedron.
In the $mer$-isomer,the three identical ligands occupy a meridian plane of the octahedron.
The complex $[Co(NH_{3})_{3}(NO_{2})_{3}]$ is of the type $[Ma_{3}b_{3}]$,where $a = NH_{3}$ and $b = NO_{2}^{-}$.
Therefore,it can show $fac-$ and $mer-$isomerism.

(B) $(I) [Cr(H_2O)_6]^{2+}$: $Cr^{2+} \Rightarrow [Ar] 3d^4$. $H_2O$ is a weak field ligand. Unpaired $e^- = 4$. Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM \approx 4.89 \ BM$.
$(II) [Fe(CN)_6]^{4-}$: $Fe^{2+} \Rightarrow [Ar] 3d^6$. $CN^-$ is a strong field ligand. Unpaired $e^- = 0$. Magnetic moment $\mu = 0 \ BM$.
$(III) [Fe(C_2O_4)_3]^{3-}$: $Fe^{3+} \Rightarrow [Ar] 3d^5$. Given $\Delta_0 > P$,it is a low spin complex. Unpaired $e^- = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM \approx 1.73 \ BM$.
$(IV) [CoCl_4]^{2-}$: $Co^{2+} \Rightarrow [Ar] 3d^7$. Tetrahedral complex,$Cl^-$ is a weak field ligand. Unpaired $e^- = 3$. Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM \approx 3.87 \ BM$.
Comparing the values: $4.89 (I) > 3.87 (IV) > 1.73 (III) > 0 (II)$.
Therefore,the correct order is $(I) > (IV) > (III) > (II)$.

(D) The complex $[Co(NH_{3})_{4}Cl_{2}]^{+}$ exhibits geometrical isomerism,existing as $cis$ and $trans$ isomers.
In the $trans$ isomer,the two $Cl$ ligands are at opposite positions ($180^{\circ}$ angle).
In the $cis$ isomer,the two $Cl$ ligands are adjacent to each other,resulting in a $Cl-Co-Cl$ bond angle of $90^{\circ}$.

(A) The spin-only magnetic moment of $3.83 \ BM$ corresponds to $n = 3$ unpaired electrons,indicating $Cr$ is in the $+3$ oxidation state. Thus,the formula is $Cr(H_{2}O)_{6}Cl_{3}$.
Possible isomers are:
$1$. $[Cr(H_{2}O)_{6}]Cl_{3}$: Reacts with $AgNO_{3}$ but shows no geometrical isomerism.
$2$. $[Cr(H_{2}O)_{5}Cl]Cl_{2} \cdot H_{2}O$: Reacts with $AgNO_{3}$ but shows no geometrical isomerism.
$3$. $[Cr(H_{2}O)_{4}Cl_{2}]Cl \cdot 2H_{2}O$: Reacts with $AgNO_{3}$ and shows geometrical isomerism (cis and trans forms).
$4$. $[Cr(H_{2}O)_{3}Cl_{3}] \cdot 3H_{2}O$: Does not react with $AgNO_{3}$ (as no $Cl^-$ is outside the coordination sphere) and shows geometrical isomerism.
The complex $[Cr(H_{2}O)_{4}Cl_{2}]Cl \cdot 2H_{2}O$ satisfies both conditions. Its $IUPAC$ name is Tetraaquadichloridochromium $(III)$ chloride dihydrate.

(D) The atomic number of the element is $25$, which corresponds to Manganese $(Mn)$.
Its electronic configuration is $[Ar] 3d^{5} 4s^{2}$.
A divalent ion $(Mn^{2+})$ is formed by losing two electrons from the $4s$ orbital, resulting in a $3d^{5}$ configuration.
This configuration has $n = 5$ unpaired electrons.
The magnetic moment $(\mu)$ is calculated using the spin-only formula: $\mu = \sqrt{n(n+2)} \, BM$.
Substituting $n = 5$: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, BM$.

(N/A) In a tetrahedral complex,all four positions are equivalent with respect to each other.
Because the bond angles between any two ligands are identical $(109.5^{\circ})$,the relative positions of the unidentate ligands remain the same regardless of how they are arranged.
Therefore,no two ligands can be considered 'cis' or 'trans' to each other,making geometrical isomerism impossible in tetrahedral complexes.

(N/A) The complex $[Fe(NH_3)_2(CN)_4]^-$ is an octahedral complex of the type $[MA_2B_4]$.
In the $cis$-isomer,the two $NH_3$ ligands are adjacent to each other (at $90^{\circ}$ angle).
In the $trans$-isomer,the two $NH_3$ ligands are opposite to each other (at $180^{\circ}$ angle).
The structures are as follows:
$cis$-isomer: The two $NH_3$ groups are at $90^{\circ}$ to each other.
$trans$-isomer: The two $NH_3$ groups are at $180^{\circ}$ to each other.

(A) coordination entity is chiral if it lacks a plane of symmetry and a center of inversion.
$(a)$ The $cis-[CrCl_{2}(ox)_{2}]^{3-}$ isomer lacks both a plane of symmetry and a center of inversion,making it chiral (optically active).
$(b)$ The $trans-[CrCl_{2}(ox)_{2}]^{3-}$ isomer possesses a plane of symmetry (and a center of inversion),making it achiral (optically inactive).
Therefore,the $cis$ isomer is the chiral one.