Isomerism and Magnetic properties Questions in English

(D) The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
For $Fe^{3+}$,the configuration is $[Ar] 3d^5$.
$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
In $Fe^{3+}$ $(3d^5)$,the five electrons are paired such that four electrons form two pairs and one electron remains unpaired $(n=1)$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(C) Geometrical isomerism is shown by complexes that have different spatial arrangements of ligands around the central metal ion.
For octahedral complexes of the type $[MA_6]$,$[MA_5B]$,$[MA_4B_2]$,$[MA_3B_3]$,and $[MA_2B_4]$,geometrical isomerism is possible.
Option $A$: $[Co(NH_3)_4Cl_2]^+$ is of the type $[MA_4B_2]$,which shows $cis$ and $trans$ isomers.
Option $B$: $[Fe(NH_3)_2(CN)_4]^-$ is of the type $[MA_2B_4]$,which shows $cis$ and $trans$ isomers.
Option $C$: $[Cr(OX)_3]^{3-}$ contains a bidentate ligand $(OX^{2-})$. Complexes of the type $[M(AA)_3]$ do not show geometrical isomerism because all positions are equivalent.
Option $D$: $[Co(NH_3)_3(NO_2)_3]$ is of the type $[MA_3B_3]$,which shows facial $(fac)$ and meridional $(mer)$ isomers.
Therefore,the correct answer is $C$.

(C) Optical isomerism is shown by complexes that lack a plane of symmetry and a center of inversion.
$Cis-[CrCl_2(ox)_2]^{3-}$ contains two bidentate oxalate $(ox^{2-})$ ligands and two chloride ligands in a cis configuration.
This complex lacks a plane of symmetry and is chiral,meaning it exists as a pair of non-superimposable mirror images (enantiomers).
In contrast,$Trans-[PtCl_2(en)_2]^{2+}$ has a plane of symmetry,and the other options are either achiral or do not exhibit optical isomerism due to their specific geometry.

(A) The given complexes $[Co(NH_3)_5Cl]SO_4$ and $[Co(NH_3)_5SO_4]Cl$ differ in the counter ion present outside the coordination sphere.
In the first complex,the $SO_4^{2-}$ ion is the counter ion,while in the second complex,the $Cl^-$ ion is the counter ion.
When these complexes are dissolved in water,they give different ions in the solution.
This type of isomerism,where the counter ion is exchanged with a ligand from the coordination sphere,is known as ionisation isomerism.
Therefore,the correct option is $A$.

(A) The complex $[Cr(H_2O)_2(C_2O_4)_2]^{-}$ is an octahedral complex of the type $[M(AA)_2(a)_2]$.
Here,$AA$ represents the bidentate oxalate ligand $(C_2O_4^{2-})$ and $a$ represents the monodentate aqua ligand $(H_2O)$.
This type of complex exhibits geometric isomerism:
$1$. $cis$-isomer: The two $H_2O$ ligands are adjacent to each other. The $cis$-isomer is optically active and exists as a pair of enantiomers ($d$ and $l$ forms).
$2$. $trans$-isomer: The two $H_2O$ ligands are opposite to each other. The $trans$-isomer is optically inactive due to the presence of a plane of symmetry.
Therefore,the total number of isomers is $3$ ($cis$-$d$,$cis$-$l$,and $trans$).

(D) The complex $[PtCl_2(en)_2]$ contains two monodentate ligands $(Cl^-)$ and two bidentate ligands $(en)$.
It exhibits geometrical isomerism due to the different spatial arrangements of the $Cl^-$ ligands (cis and trans forms).
The cis-isomer of $[PtCl_2(en)_2]$ is optically active because it lacks a plane of symmetry,while the trans-isomer is optically inactive.
Therefore,the complex exhibits both geometrical and optical isomerism.

(C) The complexes $[Co(NH_3)_5SO_4]Br$ and $[Co(NH_3)_5Br]SO_4$ exhibit ionisation isomerism.
Ionisation isomerism occurs when the counter ion in a coordination compound is a potential ligand and can displace a ligand which can then become the counter ion.
In this case,the $SO_4^{2-}$ ion and the $Br^-$ ion exchange positions between the coordination sphere and the counter ion position.

(C) Optical activity in coordination compounds requires the absence of a plane of symmetry or a center of inversion.
$Cis-[CrCl_2(ox)_2]^{3-}$ has no plane of symmetry and is optically active.
$Cis-[CoBr_2(en)_2]^{+}$ has no plane of symmetry and is optically active.
$Cis-[Fe(NH_3)_2(CN)_4]^{-}$ possesses a plane of symmetry (the plane containing the $Fe$,$2NH_3$,and $2CN$ ligands),making it achiral and optically inactive.
$Cis-[PtCl_2(en)_2]^{2+}$ is optically active due to the lack of a plane of symmetry.
Therefore,the correct option is $C$.

(D) Linkage isomerism occurs in coordination compounds containing ambidentate ligands,which can coordinate to the central metal atom through two different donor atoms.
In the pair $[Co(NO_2)(NH_3)_5]Cl_2$ and $[Co(ONO)(NH_3)_5]Cl_2$,the ligand $NO_2^-$ is an ambidentate ligand.
In the first complex,the nitrogen atom $(N)$ is the donor atom $(Co-NO_2)$,
while in the second complex,the oxygen atom $(O)$ is the donor atom $(Co-ONO)$.
Therefore,this pair represents linkage isomerism.

(C) $1$. Coordination number: $NH_3$ is a monodentate ligand $(4 \times 1 = 4)$ and $CO_3^{2-}$ is a bidentate ligand $(1 \times 2 = 2)$. Total coordination number = $4 + 2 = 6$.
$2$. Oxidation number: Let the oxidation state of $Co$ be $x$. The complex is $[Co(NH_3)_4CO_3]ClO_4$. The charge on $ClO_4$ is $-1$,so the complex ion $[Co(NH_3)_4CO_3]^+$ has a charge of $+1$. Thus,$x + 4(0) + 1(-2) = +1$,which gives $x = +3$.
$3$. Number of electrons in $d$-orbital: $Co$ $(Z=27)$ has the configuration $[Ar] 3d^7 4s^2$. $Co^{3+}$ has the configuration $[Ar] 3d^6$. Thus,there are $6$ electrons in the $d$-orbital.
$4$. Number of unpaired electrons: In the presence of strong field ligands like $NH_3$ and $CO_3^{2-}$,the $6$ electrons in the $3d$ orbitals of $Co^{3+}$ pair up in the $t_{2g}$ set. Therefore,the number of unpaired electrons is $0$.

(A) Ionisation isomerism occurs when the counter ion in a coordination compound is a potential ligand and can displace a ligand that is currently bonded to the central metal atom.
In the complex $[Pt(NH_3)_4 Cl_2] Br_2$,the $Br^-$ ions are outside the coordination sphere and can exchange places with the $Cl^-$ ligands inside the coordination sphere to form $[Pt(NH_3)_4 Cl Br] Cl Br$ or $[Pt(NH_3)_4 Br_2] Cl_2$.
Therefore,the correct option is $A$.

(B) $1$. In the complex ion $[Ni(Cl)_4]^{2-}$, the oxidation state of $Ni$ is $+2$.
$2$. The electronic configuration of $Ni^{2+}$ $(Z=28)$ is $[Ar] 3d^8$.
$3$. Since $Cl^-$ is a weak field ligand, it does not cause pairing of electrons in the $3d$ orbitals.
$4$. The $3d^8$ configuration has two unpaired electrons $(n=2)$.
$5$. The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
$6$. Substituting $n=2$, we get $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.

(B) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Co^{2+}$ $(3d^7)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$2$. For $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$3$. For $Ti^{2+}$ $(3d^2)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$4$. For $Fe^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
Comparing the values,$Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,hence it shows the highest magnetic moment.

(D) The element with atomic number $26$ is Iron $(Fe)$.
Its electronic configuration is $[Ar] 3d^6 4s^2$.
The divalent ion $Fe^{2+}$ has the configuration $[Ar] 3d^6$.
In the $3d$ subshell, there are $4$ unpaired electrons.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.
Substituting $n = 4$, we get $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.

(A) The theoretical magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1$. For $Ti^{3+}$ $(3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$.
$2$. For $Co^{3+}$ $(3d^6)$: In an octahedral field,$n = 4$ (high spin) or $n = 0$ (low spin). Assuming high spin,$n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \text{ BM}$.
$3$. For $Cr^{3+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \text{ BM}$.
$4$. For $V^{3+}$ $(3d^2)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \text{ BM}$.
Comparing the values,$Ti^{3+}$ has the least number of unpaired electrons $(n=1)$,hence it has the least magnetic moment.

(B) The atomic number of the element is $28$, which corresponds to Nickel $(Ni)$.
The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
A divalent ion $(Ni^{2+})$ is formed by the loss of two electrons from the $4s$ orbital, resulting in the configuration $[Ar] 3d^8$.
In the $3d^8$ configuration, there are $2$ unpaired electrons $(n=2)$.
The magnetic moment $(\mu)$ is calculated using the spin-only formula: $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=2$: $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \ BM$.

(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
For $\mu = 4.90 \ B.M.$,we solve $\sqrt{n(n+2)} = 4.90$,which gives $n = 4$.
In $FeSO_4$,the iron ion is $Fe^{2+}$.
The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
In the $3d^6$ configuration,there are $4$ unpaired electrons.
Therefore,$FeSO_4$ has a magnetic moment of $4.90 \ B.M.$

(B) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each metal ion:
$1$. In $Ni(NO_3)_2$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. It has $n = 2$ unpaired electrons.
$2$. In $MnSO_4$,$Mn$ is in the $+2$ oxidation state $(3d^5)$. It has $n = 5$ unpaired electrons.
$3$. In $CrCl_3$,$Cr$ is in the $+3$ oxidation state $(3d^3)$. It has $n = 3$ unpaired electrons.
$4$. In $FeSO_4$,$Fe$ is in the $+2$ oxidation state $(3d^6)$. It has $n = 4$ unpaired electrons.
The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$. Since $Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,it will have the highest magnetic moment.

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Ti^{3+}$ $(3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$2$. For $Co^{3+}$ $(3d^6)$: In the presence of weak field ligands,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
$3$. For $Cr^{3+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$4$. For $Fe^{3+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Comparing the values,$Ti^{3+}$ has the lowest number of unpaired electrons $(n=1)$,therefore it has the lowest magnetic moment.

(A) The electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$.
For $Co^{2+}$,the configuration is $[Ar] 3d^7$.
In the $3d$ subshell,there are $3$ unpaired electrons $(n=3)$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=3$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.

(D) The complex compounds $[Co(NH_3)_5 SO_4] Br$ and $[Co(NH_3)_5 Br] SO_4$ are ionisation isomers of each other.
- Ionisation isomerism occurs when the counter ion in a coordination compound is a potential ligand and can exchange places with a ligand present in the coordination sphere.
- In $[Co(NH_3)_5 SO_4] Br$,the $Br^-$ ion is the counter ion,while in $[Co(NH_3)_5 Br] SO_4$,the $SO_4^{2-}$ ion is the counter ion.

(D) The complex $[Co(en)_{2} Cl_{2}]^{+}$ exhibits geometrical isomerism and optical isomerism.
$1$. Geometrical isomerism: It exists in two forms,the $cis$-isomer and the $trans$-isomer.
$2$. Optical isomerism: The $cis$-isomer is optically active and exists as a pair of enantiomers ($d$ and $l$ forms),while the $trans$-isomer is optically inactive (achiral) due to the presence of a plane of symmetry.
Therefore,the total number of stereoisomers is $2$ (from $cis$-enantiomers) $+ 1$ (from $trans$-isomer) $= 3$.

(A) $[Fe(NO_{2})_{3} Cl_{3}]$ and $[Fe(ONO)_{3} Cl_{3}]$ exhibit linkage isomerism.
This occurs because the nitrite ion $(NO_{2}^-)$ is an ambidentate ligand.
It can coordinate to the central metal atom through either the nitrogen atom $(-NO_{2})$ or the oxygen atom $(-ONO)$.

(B) The complex $[CoCl_{2}(en)(NH_{3})_{2}]^{+}$ has the general formula $[M(AA)a_{2}b_{2}]$,where $M = Co$,$AA = en$,$a = Cl$,and $b = NH_{3}$.
For this type of complex,there are $3$ geometrical isomers possible:
$1$. Trans-isomer: Both $Cl$ atoms are trans to each other,and both $NH_{3}$ molecules are trans to each other.
$2$. Cis-isomer $(I)$: $Cl$ atoms are cis to each other,and $NH_{3}$ molecules are cis to each other.
$3$. Cis-isomer $(II)$: $Cl$ atoms are cis to each other,and $NH_{3}$ molecules are trans to each other.
Thus,a total of $3$ geometrical isomers are possible.

(C) Tetrahedral complexes do not show geometrical isomerism because,due to their symmetrical structure,the relative positions of the ligands are the same with respect to each other.
Square planar complexes of the type $M A_{3} B$ do not show geometrical isomerism because all possible spatial arrangements are equivalent.
Square planar complexes of the type $M A B C D$ show three geometrical isomers. These can be obtained by fixing the position of one ligand and placing any of the remaining three ligands at the trans-position one by one.

(D) The 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 $\sqrt{n(n+2)} = \sqrt{24}$.
Squaring both sides,$n(n+2) = 24$,which simplifies to $n^2 + 2n - 24 = 0$.
Solving the quadratic equation: $(n+6)(n-4) = 0$.
Since the number of electrons cannot be negative,$n = 4$.
The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
For $Fe^{2+}$,the configuration is $[Ar] 3d^6$,which has $4$ unpaired electrons.
Thus,$x = 2$.

(C) The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^{6}$.
It has $4$ unpaired electrons as shown in the $d$-orbital diagram:
$d^{6} = [\uparrow\downarrow] [\uparrow] [\uparrow] [\uparrow] [\uparrow]$
The spin-only magnetic moment is calculated using the formula:
$\mu = \sqrt{n(n+2)} \ BM$
Where $n$ is the number of unpaired electrons.
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ BM$
Rounding to the nearest integer,the value is $5 \ BM$.

(C) The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
$Ni^{2+}$ ion has the configuration $[Ar] 3d^8$.
In an aqueous solution,$Ni^{2+}$ forms the octahedral complex $[Ni(H_2O)_6]^{2+}$.
In the $3d$ subshell,the $8$ electrons are arranged as follows: three orbitals are doubly occupied and two orbitals are singly occupied.
Therefore,the number of unpaired electrons $(n)$ $= 2$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n + 2)} \ BM$.
Substituting $n = 2$,we get $\mu = \sqrt{2(2 + 2)} = \sqrt{8} \ BM$.

(A) The magnetic moment $(\mu)$ is related to the number of unpaired electrons $(n)$ by the formula: $\mu = \sqrt{n(n+2)} \ BM$.
Greater the number of unpaired electrons,higher is the magnetic moment.
Let us determine the number of unpaired electrons for each ion:
$Cr^{3+}$ $(3d^3)$: $n = 3$
$Fe^{2+}$ $(3d^6)$: $n = 4$
$Co^{2+}$ $(3d^7)$: $n = 3$
$Ni^{2+}$ $(3d^8)$: $n = 2$
Since $Fe^{2+}$ has the maximum number of unpaired electrons $(n=4)$,it exhibits the highest magnetic moment.

(A) For a square planar complex of the type $M(abcd)$,there are $3$ possible geometric isomers.
In the case of $MAXBL$,we can fix one ligand (e.g.,$M$) and arrange the others.
There are $2$ isomers where two specific ligands are in the $cis$ position (adjacent) and $1$ isomer where they are in the $trans$ position (opposite).
Thus,the complex shows $2$ $cis$ and $1$ $trans$ isomers.

(A) The magnetic moment $(\mu)$ is calculated using the spin-only formula: $\mu = \sqrt{n(n+2)} \ \text{BM}$,where $n$ is the number of unpaired electrons.
Given $\mu = \sqrt{15} \ \text{BM}$.
$\sqrt{n(n+2)} = \sqrt{15}$
$n(n+2) = 15$
$n^2 + 2n - 15 = 0$
$(n+5)(n-3) = 0$
Since $n$ cannot be negative,$n = 3$.

(C) $(i)$ $[Fe(CN)_6]^{3-}$: $Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ is a strong-field ligand,causing pairing. Unpaired electrons = $1$.
$(ii)$ $[MnCl_6]^{3-}$: $Mn$ is in $+3$ oxidation state $(3d^4)$. $Cl^-$ is a weak-field ligand. Unpaired electrons = $4$.
$(iii)$ $[FeF_6]^{3-}$: $Fe$ is in $+3$ oxidation state $(3d^5)$. $F^-$ is a weak-field ligand. Unpaired electrons = $5$.
$(iv)$ $[Co(NH_3)_6]^{3+}$: $Co$ is in $+3$ oxidation state $(3d^6)$. $NH_3$ is a strong-field ligand,causing pairing. Unpaired electrons = $0$.
Increasing order of unpaired electrons: $(iv) < (i) < (ii) < (iii)$.

(C) Geometrical isomerism in octahedral complexes occurs when ligands are arranged in different spatial orientations (cis and trans forms).
$I$) $[Co(en)(NH_3)_2 Cl_2] Cl$: This complex has the form $[M(AA)(a)_2(b)_2]$,which exhibits geometrical isomerism.
$II$) $[Co(NH_3)_4 Cl_2] Cl$: This complex has the form $[M(a)_4(b)_2]$,which exhibits cis and trans geometrical isomers.
$III$) $[Co(en)_3] Cl_3$: This complex has the form $[M(AA)_3]$. It does not show geometrical isomerism because all positions are equivalent due to the symmetry of the bidentate ligands.
$IV$) $[Co(en)_2 Cl_2] Br$: This complex has the form $[M(AA)_2(b)_2]$,which exhibits cis and trans geometrical isomers.
Therefore,complexes $I, II,$ and $IV$ exhibit geometrical isomerism.

(A) Ionization isomerism occurs when the counter ion in a coordination complex is a potential ligand and can displace a ligand from the coordination sphere.
$I) [Cr(NH_3)_4Cl_2]Cl$ can form $[Cr(NH_3)_4Cl_3]$ (No,$Cl$ is already inside).
$II) [Ti(H_2O)_5Cl](NO_3)_2$ can form $[Ti(H_2O)_5(NO_3)]Cl(NO_3)$ or $[Ti(H_2O)_5(NO_3)_2]Cl$. This exhibits ionization isomerism.
$III) [Pt(en)(NH_3)Cl]NO_3$ can form $[Pt(en)(NH_3)(NO_3)]Cl$. This exhibits ionization isomerism.
$IV) [Co(NH_3)_4(NO_3)_2]NO_3$ can form $[Co(NH_3)_4(NO_3)_3]$. This exhibits ionization isomerism.
However,looking at the options provided,the most appropriate pair exhibiting this behavior clearly is $II$ and $III$.

(C) To be optically active,a molecule must not possess any of the following symmetry elements:
$1-$ Centre of symmetry
$2-$ Plane of symmetry $(POS)$
Analysis of the given complexes:
$A) Cis-[Cr Cl_2(C_2 O_4)_2]^{3-}$: This complex lacks a plane of symmetry and is optically active.
$B) [Pt Cl_2(en)_2]^{2+}$: The $cis$-isomer of this complex lacks a plane of symmetry and is optically active.
$C) [Co(NH_3)_3(NO_2)_3]$: This complex exists in $fac$ and $mer$ forms. Both forms possess a plane of symmetry,making them optically inactive.
$D) [Co(en)_3]^{3+}$: This complex is a tris-chelated species and is optically active (chiral).
Therefore,the complex that does not show optical isomerism is $[Co(NH_3)_3(NO_2)_3]$.

(A) The complex $[Co(NH_3)_5(NO_2)](NO_3)_2$ does not exhibit optical isomerism because it possesses a plane of symmetry.
Linkage isomerism is exhibited due to the ambidentate ligand $NO_2^-$,which can coordinate through $N$ or $O$ (forming $[Co(NH_3)_5(NO_2)]^{2+}$ and $[Co(NH_3)_5(ONO)]^{2+}$).
Ionization isomerism is exhibited because the $NO_2^-$ ligand inside the coordination sphere can exchange with the $NO_3^-$ ion outside the sphere (forming $[Co(NH_3)_5(NO_2)](NO_3)_2$ and $[Co(NH_3)_5(NO_3)](NO_2)(NO_3)$).
Coordination isomerism requires both cationic and anionic entities to be coordination complexes,which is not the case here.
Therefore,the complex exhibits linkage and ionization isomerism.

(A) The complex ion $\left[Cr(H_2O)_4Cl_2\right]^{+}$ exhibits geometrical isomerism.
It is an octahedral complex of the type $\left[Ma_4b_2\right]$,where $M = Cr$,$a = H_2O$,and $b = Cl$.
In this type of complex,the two $b$ ligands can be adjacent to each other (cis-isomer) or opposite to each other (trans-isomer).
Other options do not show geometrical isomerism:
$\left[Pt(NH_3)_3Cl\right]^{+}$ is of the type $\left[Ma_3b\right]$ (square planar),which does not show geometrical isomerism.
$\left[Co(NH_3)_6\right]^{3+}$ is of the type $\left[Ma_6\right]$,which is highly symmetrical and does not show isomerism.
$\left[Co(CN)_5(NC)\right]^{3-}$ is of the type $\left[Ma_5b\right]$,which does not show geometrical isomerism.

(C) Optical isomerism is exhibited by complexes that lack a plane of symmetry and a center of symmetry (chiral complexes).
For octahedral complexes of the type $[M(AA)_3]$,where $AA$ is a bidentate ligand like ethylenediamine $(en)$,the complex exists as a pair of enantiomers (non-superimposable mirror images).
In the given options,$[Co(en)_3]Cl_3$ contains three bidentate ligands,which creates a chiral environment around the central metal ion,thus exhibiting optical isomerism.
The other complexes listed are either of the type $[MA_5B]$ or $[MA_4BC]$,which generally exhibit geometrical isomerism but not optical isomerism.

(A) Geometrical isomerism is observed in coordination complexes where ligands can be arranged in different spatial positions relative to each other.
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 the complex $[Co(NH_3)_5 Cl] Cl_2$,the coordination sphere is $[Co(NH_3)_5 Cl]^{2+}$,which is of the type $[MA_5B]$.
Therefore,it does not exhibit geometrical isomerism.
In contrast,$[Co(NH_3)_4 Cl_2]^+$ (type $[MA_4B_2]$),$[Co(NH_3)_3(NO_2)_3]$ (type $[MA_3B_3]$),and $[Co(en)_2 Cl_2]^+$ (type $[M(AA)_2B_2]$) all exhibit geometrical isomerism (cis-trans or fac-mer).

(A) The given compounds are:
$1. [Co(NH_3)_5SO_4]Br \rightleftharpoons [Co(NH_3)_5SO_4]^+ + Br^-$
$2. [Co(NH_3)_5Br]SO_4 \rightleftharpoons [Co(NH_3)_5Br]^{2+} + SO_4^{2-}$
Since both compounds have the same molecular formula but produce different ions in an aqueous solution,they exhibit ionisation isomerism.

(D) The spin-only magnetic moment $(\mu_{eff})$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu = 4.90 \ BM$,$n$ must be $4$ (since $\sqrt{4(4+2)} = \sqrt{24} \approx 4.90$).
$A$: $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $d^6$. $NH_3$ is a strong field ligand,causing pairing. $n = 0$.
$B$: $[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $d^3$. $n = 3$.
$C$: $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $d^4$. $CN^{-}$ is a strong field ligand,causing pairing. $n = 2$.
$D$: $[MnCl_6]^{3-}$: $Mn^{3+}$ is $d^4$. $Cl^{-}$ is a weak field ligand,no pairing occurs. The configuration is $t_{2g}^3 e_g^1$,resulting in $n = 4$ unpaired electrons.
Therefore,$[MnCl_6]^{3-}$ has a magnetic moment of $4.90 \ BM$.

(A) In $[Mn(CN)_6]^{3-}$,the oxidation state of $Mn$ is $+3$. The electronic configuration of $Mn^{3+}$ is $3d^4$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons in $d$-orbitals. Thus,the configuration becomes $t_{2g}^4 e_g^0$,resulting in $n = 2$ unpaired electrons.
$\mu_{\text{spin only}} = \sqrt{n(n+2)} \ BM = \sqrt{2(2+2)} \ BM = \sqrt{8} \ BM \approx 2.84 \ BM$.
In $[Co(C_2O_4)_3]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $3d^6$. Since $C_2O_4^{2-}$ is a chelating ligand,in the octahedral field,$Co^{3+}$ $(d^6)$ forms a low-spin complex with $t_{2g}^6 e_g^0$ configuration,resulting in $n = 0$ unpaired electrons.
$\mu_{\text{spin only}} = \sqrt{0(0+2)} \ BM = 0 \ BM$.

(C) The paramagnetism of a complex depends on the number of unpaired electrons $(n)$.
$1.$ In $[Co(NH_3)_6]^{2+}$,$Co$ is in $+2$ oxidation state. $Co^{2+} = [Ar] 3d^7$. $NH_3$ is a strong field ligand,so $n = 1$.
$2.$ In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state. $Co^{3+} = [Ar] 3d^6$. $NH_3$ is a strong field ligand,so $n = 0$.
$3.$ In $[Co(H_2O)_6]^{2+}$,$Co$ is in $+2$ oxidation state. $Co^{2+} = [Ar] 3d^7$. $H_2O$ is a weak field ligand,so $n = 3$.
$4.$ In $[Co(H_2O)_6]^{3+}$,$Co$ is in $+3$ oxidation state. $Co^{3+} = [Ar] 3d^6$. $H_2O$ acts as a strong field ligand for $Co^{3+}$,so $n = 0$.
Since $[Co(H_2O)_6]^{2+}$ has the maximum number of unpaired electrons $(n=3)$,it is the most paramagnetic.

(A) $I. [Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state $(d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons. Number of unpaired electrons = $0$.
$II. [Fe(H_2O)_6]^{2+}$: $Fe$ is in $+2$ oxidation state $(d^6)$. $H_2O$ is a weak field ligand,no pairing occurs. Configuration is $t_{2g}^4 e_g^2$. Number of unpaired electrons = $4$.
$III. [Co(H_2O)_6]^{2+}$: $Co$ is in $+2$ oxidation state $(d^7)$. $H_2O$ is a weak field ligand. Configuration is $t_{2g}^5 e_g^2$. Number of unpaired electrons = $3$.
Thus,the correct order is $II (4) > III (3) > I (0)$.

(D) To determine which complex is paramagnetic,we analyze the electronic configuration of the central metal ion in each complex:
$1$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing. All electrons are paired $(t_{2g}^6 e_g^0)$,so it is diamagnetic.
$2$. $[Co(CN)_6]^{3-}$: $Co^{3+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing. All electrons are paired $(t_{2g}^6 e_g^0)$,so it is diamagnetic.
$3$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing. All electrons are paired $(t_{2g}^6 e_g^0)$,so it is diamagnetic.
$4$. $[Cr(CN)_6]^{3-}$: $Cr^{3+}$ is $3d^3$. The configuration is $t_{2g}^3 e_g^0$. There are $3$ unpaired electrons in the $3d$ orbitals. Thus,it is paramagnetic.
Therefore,the correct option is $D$.

(B) Paramagnetic character depends on the number of unpaired electrons in the $d$-subshell.
In these complex ions,$H_2O$ is a weak field ligand,so the electrons remain unpaired according to Hund's rule:
$(a)$ In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ has a $3d^3$ configuration. Number of unpaired electrons $= 3$.
$(b)$ In $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ has a $3d^6$ configuration. In an octahedral field,this corresponds to $t_{2g}^4 e_g^2$,resulting in $4$ unpaired electrons.
$(c)$ In $[Cu(H_2O)_6]^{2+}$,$Cu^{2+}$ has a $3d^9$ configuration. Number of unpaired electrons $= 1$.
$(d)$ In $[Zn(H_2O)_6]^{2+}$,$Zn^{2+}$ has a $3d^{10}$ configuration. Number of unpaired electrons $= 0$.
Since paramagnetic character is directly proportional to the number of unpaired electrons,$[Fe(H_2O)_6]^{2+}$ is the most paramagnetic.

(A) The complexes are octahedral and involve weak field ligands,resulting in high spin (outer orbital) complexes.
$1$. $[MnCl_6]^{3-}$: $Mn^{3+}$ is $d^4$. With weak field ligands,all $4$ electrons remain unpaired $(n = 4)$.
$2$. $[FeF_6]^{3-}$: $Fe^{3+}$ is $d^5$. With weak field ligands,all $5$ electrons remain unpaired $(n = 5)$.
$3$. $[CoF_6]^{3-}$: $Co^{3+}$ is $d^6$. With weak field ligands,the configuration is $t_{2g}^4 e_g^2$,resulting in $4$ unpaired electrons $(n = 4)$.
$4$. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $d^8$. The configuration is $t_{2g}^6 e_g^2$,resulting in $2$ unpaired electrons $(n = 2)$.
Thus,complexes $1$ and $3$ have the same number of unpaired electrons $(n = 4)$.

(D) To determine the spin only magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:
$1$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^5, e_g^0$. Unpaired electrons $(n)$ = $1$.
$2$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand. Configuration: $t_{2g}^6, e_g^0$. Unpaired electrons $(n)$ = $0$.
$3$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,forcing pairing to form $dsp^2$ hybridisation. Unpaired electrons $(n)$ = $0$.
$4$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,no pairing occurs. Configuration: $3d^8$ with $sp^3$ hybridisation. Unpaired electrons $(n)$ = $2$.
Since magnetic moment $\mu = \sqrt{n(n+2)} \ BM$,the complex with the highest number of unpaired electrons $(n=2)$ has the highest magnetic moment.
Therefore,$[NiCl_4]^{2-}$ has the highest value.

(D) To find the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex. The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$.
$(I)$ $\left[Cr(CN)_6\right]^{3-}$: $Cr^{3+}$ is $3d^3$. With $CN^-$ as a strong field ligand,all $3$ electrons remain unpaired in the $t_{2g}$ orbitals. $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM$.
$(II)$ $\left[Mn(CN)_6\right]^{3-}$: $Mn^{3+}$ is $3d^4$. With $CN^-$ as a strong field ligand,electrons pair up,leaving $n = 2$ unpaired electrons. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM$.
$(III)$ $\left[Fe(CN)_6\right]^{3-}$: $Fe^{3+}$ is $3d^5$. With $CN^-$ as a strong field ligand,electrons pair up,leaving $n = 1$ unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM$.
$(IV)$ $\left[Co(CN)_6\right]^{3-}$: $Co^{3+}$ is $3d^6$. With $CN^-$ as a strong field ligand,all electrons pair up in the $t_{2g}$ orbitals $(t_{2g}^6 e_g^0)$. $n = 0$,$\mu = 0 \ BM$.
Thus,$\left[Co(CN)_6\right]^{3-}$ has the lowest magnetic moment.