Isomerism and Magnetic properties Questions in English

(A) To determine the magnetic nature,we look at the electronic configuration of the central metal ion:
$1$. In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. $H_2O$ is a weak field ligand,so the electrons remain unpaired in the $d$-orbitals,resulting in $2$ unpaired electrons. Thus,it is paramagnetic.
$2$. In $[Ni(CO)_4]$,$Ni$ is in the $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons. Thus,it is diamagnetic.
$3$. In $[Zn(NH_3)_4]^{2+}$,$Zn$ is in the $+2$ oxidation state $(3d^{10})$. All electrons are paired. Thus,it is diamagnetic.
$4$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in the $+3$ oxidation state $(3d^6)$. $NH_3$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons. Thus,it is diamagnetic.
Therefore,the correct option is $A$.

(A) In ${K_3}[FeF_6]$,the oxidation state of $Fe$ is $+3$.
$Fe^{3+}$ has the electronic configuration $[Ar]3d^5$.
Since $F^-$ is a weak field ligand,the $5$ electrons in the $3d$ orbitals remain unpaired.
Number of unpaired electrons $(n)$ $= 5$.
The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ B.M.$

(C) To determine paramagnetism,we check for unpaired electrons in the central metal ion:
$(i)$ $K_4[Fe(CN)_6]$: $Fe^{2+}$ is $d^6$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^6 e_g^0$. No unpaired electrons (diamagnetic).
(ii) $K_3[Cr(CN)_6]$: $Cr^{3+}$ is $d^3$. Configuration: $t_{2g}^3 e_g^0$. Three unpaired electrons (paramagnetic).
(iii) $K_3[Fe(CN)_6]$: $Fe^{3+}$ is $d^5$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^5 e_g^0$. One unpaired electron (paramagnetic).
(iv) $K_2[Ni(CN)_4]$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing in square planar geometry. Configuration: $d_{xy}^2 d_{yz}^2 d_{xz}^2 d_{z^2}^2 d_{x^2-y^2}^0$. No unpaired electrons (diamagnetic).
Thus,$(ii)$ and $(iii)$ are paramagnetic.

(C) The complex $[Co(en)_2Cl_2]Cl$ is an octahedral complex ($d^2sp^3$ hybridization).
It exhibits geometrical isomerism due to the presence of $cis$ and $trans$ forms.
The $cis$-isomer is optically active and shows optical isomerism.
Ionic isomerism occurs when the counter ion in the coordination compound is a potential ligand that can exchange places with a ligand in the coordination sphere. In $[Co(en)_2Cl_2]Cl$,the chloride ion is the counter ion,but there is no other isomer possible where the $Cl^-$ ion enters the coordination sphere and a ligand from the coordination sphere becomes the counter ion without changing the chemical formula or structure in a way that qualifies as ionic isomerism. Thus,statement $(c)$ is false.

(A) The 'spin only' magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
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 $d^4$ configuration with a strong ligand field,the electrons pair up in the $t_{2g}$ orbitals,resulting in $2$ unpaired electrons $(t_{2g}^4, e_g^0)$.
Thus,$n = 2$ and $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.828 \approx 2.84 \ BM$.

(D) $[Co(en)_2Cl_2]^+$ has the largest number of isomers.
It exhibits geometrical isomerism (cis and trans forms).
The cis-isomer is optically active and exists as two enantiomers ($d$ and $l$ forms).
Thus,it shows both geometrical and optical isomerism,resulting in a total of $3$ isomers.

(B) The complex $Co(NH_3)_4Br_2Cl$ exhibits both geometrical and ionization isomerism.
$1$. Geometrical Isomerism: The complex can exist in $cis$ and $trans$ forms due to the different spatial arrangements of the $Br$ ligands around the central $Co$ atom.
$2$. Ionization Isomerism: The $Cl^-$ ion can be inside the coordination sphere while a $Br^-$ ion is outside,or vice versa,leading to different ions in solution (e.g.,$[Co(NH_3)_4Br_2]Cl$ and $[Co(NH_3)_4BrCl]Br$).
Therefore,the correct option is $B$.

(D) Optical isomerism is exhibited by coordination compounds that lack a plane of symmetry or a center of inversion.
$1$. $cis-[Pt(NH_3)_2Cl_2]$ and $trans-[Pt(NH_3)_2Cl_2]$ are square planar complexes,which generally do not exhibit optical isomerism due to the presence of a molecular plane of symmetry.
$2$. $trans-[Co(en)_2Cl_2]^+$ has a plane of symmetry passing through the $Co$ atom and the two $Cl$ atoms,making it achiral.
$3$. $cis-[Co(en)_2Cl_2]^+$ lacks a plane of symmetry and a center of inversion,meaning it exists as a pair of non-superimposable mirror images (enantiomers). Therefore,it exhibits optical isomerism.

(B) Optical isomerism is shown by complexes that lack a plane of symmetry and a center of inversion (i.e.,they are chiral).
$1$. $[Co(en)_3]Cl_3$: This complex contains three bidentate ethylenediamine $(en)$ ligands. It exists as a pair of non-superimposable mirror images (enantiomers) and thus shows optical isomerism.
$2$. $cis-[Co(en)_2Cl_2]Cl$: In the $cis$ isomer,the two $Cl$ ligands are adjacent. This configuration lacks a plane of symmetry,making it chiral and optically active.
Therefore,both complexes in option $B$ exhibit optical isomerism.

(C) Optical isomerism is shown by coordination compounds that lack a plane of symmetry and a center of symmetry.
$A$. $[Cu(NH_3)_4]^{2+}$ is a square planar complex,which is achiral.
$B$. $[ZnCl_4]^{2-}$ is a tetrahedral complex with identical ligands,which is achiral.
$C$. $[Cr(C_2O_4)_3]^{3-}$ is an octahedral complex with three bidentate oxalate ligands. It exists as a pair of non-superimposable mirror images (enantiomers),thus showing optical isomerism.
$D$. $[Co(CN)_6]^{3-}$ is an octahedral complex with identical ligands,which is achiral.
Therefore,the correct option is $C$.

(C) In the nitroprusside ion $(Fe(CN)_5NO)^{2-}$,the oxidation state of iron is confirmed by measuring the magnetic moment in the solid state.
For $Fe^{II}$ $(3d^6)$,the magnetic moment corresponds to the presence of unpaired electrons in the low-spin complex,which distinguishes it from the $Fe^{III}$ configuration.

(A) The magnetic moment $(\mu)$ of a transition metal ion is calculated using the spin-only formula:
$\mu = \sqrt{n(n+2)} \text{ B.M.}$
where $n$ represents the number of unpaired electrons in the $d$-orbitals.

(A) The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$.
For $Zn^{2+}$,the configuration is $[Ar] 3d^{10}$.
Since all $10$ electrons in the $3d$ subshell are paired,there are no unpaired electrons.
Therefore,$Zn^{2+}$ is diamagnetic.

(C) The theoretical magnetic moment is calculated using the spin-only formula,$\mu_{so} = \sqrt{n(n+2)} \ BM$.
However,the experimental value often deviates from this because of the contribution from the orbital angular momentum of the electrons.
This phenomenon is known as spin-orbit coupling,which leads to a difference between the theoretical and experimental values.

(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.
$1$. In $[MnCl_4]^{2-}$,$Mn$ is in $+2$ oxidation state: $[Ar] 3d^5$. Since $Cl^-$ is a weak field ligand,$n = 5$. Thus,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, BM$.
$2$. In $[CoCl_4]^{2-}$,$Co$ is in $+2$ oxidation state: $[Ar] 3d^7$. Since $Cl^-$ is a weak field ligand,$n = 3$. Thus,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, BM$.
$3$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state: $[Ar] 3d^6$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in $n = 0$. Thus,$\mu = 0 \, BM$.
Therefore,the correct order is $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$.

(D) The formula for magnetic moment is $\mu = \sqrt{n(n + 2)} \, B.M.$,where $n$ is the number of unpaired electrons.
Given $\mu = \sqrt{15} \, B.M.$
So,$\sqrt{15} = \sqrt{n(n + 2)}$.
Squaring both sides,we get $15 = n(n + 2)$.
Solving for $n$,$n^2 + 2n - 15 = 0$.
$(n + 5)(n - 3) = 0$.
Since $n$ must be positive,$n = 3$.

(B) In $K_2MnO_4$,the oxidation state of $Mn$ is calculated as: $2(+1) + x + 4(-2) = 0$,which gives $x = +6$.
The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$.
The electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1$.
Since there is $1$ unpaired electron $(n=1)$,the magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$.

(A) To determine if a compound is colourless,we check for the presence of unpaired electrons ($d-d$ transition). If the number of unpaired electrons $(n)$ is $0$,the compound is colourless.

$TiF_6^{2-}$	$Cu_2Cl_2$
$Ti^{4+}: 3d^0$	$Cu^{+}: 3d^{10}$
$n = 0$	$n = 0$

For $CoF_6^{3-}$,$Co^{3+}$ is $3d^6$ ($n=4$,paramagnetic,coloured).
For $NiCl_4^{2-}$,$Ni^{2+}$ is $3d^8$ ($n=2$,paramagnetic,coloured).
Thus,$TiF_6^{2-}$ and $Cu_2Cl_2$ are colourless.

(D) The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
For $Ni^{2+}$,the configuration is $[Ar] 3d^8$.
In the $3d^8$ configuration,there are $2$ unpaired electrons.
The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \ B.M.$

(B) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 2.83 \, BM$.
$2.83 = \sqrt{n(n+2)}$
Squaring both sides: $8 = n(n+2)$
$n^2 + 2n - 8 = 0$
$(n+4)(n-2) = 0$
Since $n$ must be positive,$n = 2$.
Thus,there are $2$ unpaired electrons present in the ion.

(A) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu = \sqrt{15} \ BM$,we have $n(n+2) = 15$,which gives $n = 3$.
Let us analyze the electronic configurations:
$1$. $Co^{2+}$ $(Z=27)$: $[Ar] 3d^7$,unpaired electrons $n=3$.
$2$. $Cr^{3+}$ $(Z=24)$: $[Ar] 3d^3$,unpaired electrons $n=3$.
Both $Co^{2+}$ and $Cr^{3+}$ have $3$ unpaired electrons,so both have a magnetic moment of $\sqrt{3(3+2)} = \sqrt{15} \ BM$.

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Mn^{2+}$ $(Z=25)$,the electronic configuration is $[Ar] \ 3d^5$. Here,$n = 5$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Thus,$Mn^{2+}$ has a magnetic moment of approximately $5.93 \ BM$.

(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
All the given complexes are octahedral complexes with $H_2O$ as a weak field ligand,so the number of unpaired electrons remains the same as in the free metal ions.
For $[Cr(H_2O)_6]^{2+}$,$Cr^{2+}$ is $d^4$,$n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \text{ BM}$.
For $[Mn(H_2O)_6]^{2+}$,$Mn^{2+}$ is $d^5$,$n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \text{ BM}$.
For $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $d^6$,$n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \text{ BM}$.
For $[Co(H_2O)_6]^{2+}$,$Co^{2+}$ is $d^7$,$n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \text{ BM}$.
Comparing the values,$[Co(H_2O)_6]^{2+}$ has the lowest magnetic moment.

(A) To determine the magnetic property,we look at the electronic configuration of the metal ion:
$1$. In $ZnCl_2$,$Zn$ is in the $+2$ oxidation state. The configuration is $[Ar] 3d^{10}$. Since all electrons are paired,it is diamagnetic.
$2$. In $CrCl_3$,$Cr$ is in the $+3$ oxidation state. The configuration is $3d^3$,which has unpaired electrons,making it paramagnetic.
$3$. In $CuSO_4$,$Cu$ is in the $+2$ oxidation state. The configuration is $3d^9$,which has one unpaired electron,making it paramagnetic.
$4$. In $[NiCl_4]^{2-}$,$Ni$ is in the $+2$ oxidation state. The configuration is $3d^8$,which has two unpaired electrons,making it paramagnetic.
Therefore,the correct answer is $ZnCl_2$.

(C) In $VOSO_4$,the oxidation state of Vanadium is $V^{+4}$.
The electronic configuration of $V^{+4}$ is $[Ar] 3d^1 4s^0$.
Since it has one unpaired electron,it is paramagnetic and colored.

(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.
Given $\mu = 3.87 \, B.M$.
Substituting the value: $3.87 = \sqrt{n(n+2)}$.
Squaring both sides: $14.9769 \approx n(n+2)$.
$n^2 + 2n - 15 \approx 0$.
Solving the quadratic equation: $(n+5)(n-3) = 0$.
Since $n$ must be positive,$n = 3$.
Therefore,there are $3$ unpaired electrons.

(B) In the complex $[Ag(CN)_2]^{-}$,the oxidation state of $Ag$ is $+1$.
The electronic configuration of $Ag^{+}$ is $[Kr] 4d^{10}$.
Since all electrons are paired in the $d$-subshell,there are no unpaired electrons $(n = 0)$.
The magnetic moment $\mu = \sqrt{n(n+2)} \ BM$ becomes $\sqrt{0(0+2)} = 0 \ BM$.
Therefore,$[Ag(CN)_2]^{-}$ has a zero magnetic moment.

(A) In $KMnO_4$,the oxidation state of $Mn$ is $+7$. The electronic configuration of $Mn^{7+}$ is $[Ar] 3d^0$. Although it has no unpaired electrons,it is intensely purple due to charge transfer spectra (ligand to metal charge transfer).
In $K_2MnO_4$,$Mn$ is in $+6$ state $(3d^1)$,which has one unpaired electron.
In $MnSO_4$,$Mn$ is in $+2$ state $(3d^5)$,which has five unpaired electrons.
In $MnO$,$Mn$ is in $+2$ state $(3d^5)$,which has five unpaired electrons.
Thus,$KMnO_4$ is colored despite having no unpaired electrons.

(B) The magnetic moment is determined by the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.
$1$. $Cr^{3+}$ $([Ar]3d^3)$: $n = 3$
$2$. $Mn^{2+}$ $([Ar]3d^5)$: $n = 5$
$3$. $Cr^{2+}$ $([Ar]3d^4)$: $n = 4$
$4$. $Fe^{2+}$ $([Ar]3d^6)$: $n = 4$
$5$. $V^{2+}$ $([Ar]3d^3)$: $n = 3$
$6$. $Sc^{3+}$ $([Ar]3d^0)$: $n = 0$
$7$. $Ti^{2+}$ $([Ar]3d^2)$: $n = 2$
Since $Cr^{2+}$ and $Fe^{2+}$ both have $n = 4$ unpaired electrons,they have the same magnetic moment.

(A) The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
For $Ni^{2+}$,the configuration is $[Ar] 3d^8$.
In the $3d$ subshell,there are $2$ unpaired electrons $(n = 2)$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M$.
Substituting $n = 2$: $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \ B.M$.

(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu = 2.82 \ BM$,$n$ must be $2$.
In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons,resulting in $n=0$.
In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,so electrons do not pair up,resulting in $n=2$.
In $Ni(PPh_3)_4$,$Ni$ is in $0$ oxidation state,resulting in $n=0$.
In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $n=0$.
Thus,$[NiCl_4]^{2-}$ has $2$ unpaired electrons and a magnetic moment of $2.82 \ BM$.

(B) Ordinary light consists of light waves vibrating in all possible planes perpendicular to the direction of propagation.
When this light is passed through a $Nicol$ prism,it allows only those light waves to pass through which vibrate in a single plane.
This resulting light is known as plane-polarized light.
Therefore,a $Nicol$ prism is used to convert ordinary light into plane-polarized light.

(C) The complex $[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O$ is known to be green in color.
It contains one ionizable chloride ion $(Cl^-)$ outside the coordination sphere.
When it reacts with $AgNO_3$,the ionizable $Cl^-$ reacts to form $AgCl$ precipitate:
$[Cr(H_2O)_4Cl_2]Cl + AgNO_3 \rightarrow [Cr(H_2O)_4Cl_2]NO_3 + AgCl(s)$.
Since only one $Cl^-$ ion is available for precipitation,it produces one mole of $AgCl$.

(A) The complex $[Pt(NH_3)_2Cl_2]$ is a square planar complex of the type $[MA_2B_2]$.
It exhibits geometric isomerism due to the different spatial arrangements of the ligands around the central metal atom.
The two geometric isomers are:
$1$. $cis$-isomer: In this isomer,the two identical ligands ($NH_3$ or $Cl$) are adjacent to each other.
$2$. $trans$-isomer: In this isomer,the two identical ligands ($NH_3$ or $Cl$) are opposite to each other.
Thus,there are $2$ geometric isomers.

(B) Optical isomerism is shown by coordination compounds that lack a plane of symmetry and a center of symmetry.
$1$. $trans-[Cr(en)_2(CN)_2]Cl$ has a plane of symmetry,so it is optically inactive.
$2$. $[Co(en)_3]Br_3$ is a tris-chelated complex of the type $[M(AA)_3]$. It does not possess a plane of symmetry or a center of symmetry,making it optically active.
$3$. $[Co(NH_3)_5(NO_2)]I_2$ is a simple octahedral complex with high symmetry,making it optically inactive.
$4$. $[Pt(NH_3)_2Cl_2]$ is a square planar complex,which is inherently achiral (optically inactive) due to the presence of a molecular plane of symmetry.

(C) In the complex $[Cr(CN)_6]^{3-}$, the oxidation state of $Cr$ is $+3$.
$Cr$ $(Z=24)$ has the electronic configuration $[Ar] 3d^5 4s^1$.
$Cr^{3+}$ has the configuration $[Ar] 3d^3$.
Since $CN^-$ is a strong field ligand, the $3$ electrons remain in the $t_{2g}$ orbitals.
The number of unpaired electrons $(n)$ is $3$.
The magnetic moment $(\mu)$ is calculated as $\mu = \sqrt{n(n+2)} \ BM$.
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.

(D) The given complex is a square planar complex of the type $[MA_2B_2]$.
In a square planar complex of the type $[MA_2B_2]$,there are two possible geometric isomers:
$1$. $cis$-isomer: In this isomer,the two identical ligands ($NH_3$ or $NO_2$) are adjacent to each other.
$2$. $trans$-isomer: In this isomer,the two identical ligands ($NH_3$ or $NO_2$) are opposite to each other.
Therefore,the total number of geometric isomers is $2$.

(B) The complex ions $[Co(NH_3)_5(NO_2)]^{2+}$ and $[Co(NH_3)_5(ONO)]^{2+}$ exhibit linkage isomerism.
Linkage isomerism occurs in coordination compounds containing ambidentate ligands.
An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different donor atoms.
In this case,the nitrite ion $(NO_2^-)$ can coordinate through the nitrogen atom $(-NO_2)$ or through the oxygen atom $(-ONO)$.

(A) In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$.
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Thus,there are $2$ unpaired electrons in the $3d$ orbitals.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$
Substituting $n = 2$,we get $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ B.M.$

(D) $1$. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state $(3d^6)$. Since $F^-$ is a weak field ligand,electrons do not pair up,resulting in $4$ unpaired electrons.
$2$. In $[FeF_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. Since $F^-$ is a weak field ligand,there are $5$ unpaired electrons.
$3$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. Since $CN^-$ is a strong field ligand,electrons pair up,resulting in $1$ unpaired electron.
$4$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. Since $CN^-$ is a strong field ligand,all electrons pair up,resulting in $0$ unpaired electrons.
$5$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. Since $CN^-$ is a strong field ligand,electrons pair up,resulting in $0$ unpaired electrons.
$6$. In $[Fe(H_2O)_6]^{2+}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. Since $H_2O$ is a weak field ligand,there are $4$ unpaired electrons.
Comparing the pairs:
- Pair $A$: $4$ and $5$ (Not equal)
- Pair $B$: $1$ and $0$ (Not equal)
- Pair $C$: $1$ and $0$ (Not equal)
- Pair $D$: $4$ and $4$ (Equal)
Therefore,the correct option is $D$.

(C) The complex $[Co(NH_3)_4Cl_2]^+$ exhibits geometrical isomerism.
It exists in two forms: $cis$ and $trans$.
The $cis$-isomer has the two $Cl^-$ ligands adjacent to each other,while the $trans$-isomer has the two $Cl^-$ ligands opposite to each other.
These two isomers have different physical properties,including color,due to their different spatial arrangements.

(A) In $Cr(CO)_6$,the oxidation state of $Cr$ is $0$ because $CO$ is a neutral ligand.
The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
$CO$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
All electrons become paired in the presence of $CO$,resulting in $n = 0$ unpaired electrons.
The magnetic moment $\mu = \sqrt{n(n+2)} \ BM = \sqrt{0(0+2)} = 0 \ BM$.

(D) $1$. $[Co(Py)_3(NH_3)_3]^{3+}$ is of the type $[M(A)_3(B)_3]$,which shows facial $(fac)$ and meridional $(mer)$ geometrical isomers. It does not show optical isomerism. Total isomers = $2$.
$2$. $[Ni(en)(NH_3)_4]^{2+}$ is of the type $[M(AA)(B)_4]$,which shows only geometrical isomerism (cis-trans). Total isomers = $2$.
$3$. $[Fe(C_2O_4)(en)_2]^{2-}$ is of the type $[M(AA)_3]$,which shows optical isomerism (d and l forms). Total isomers = $2$.
$4$. $[Cr(NO_2)_2(NH_3)_4]^{+}$ is of the type $[M(A)_4(B)_2]$,which shows geometrical isomerism (cis and trans). The cis-isomer is optically active (exists as d and l forms),while the trans-isomer is optically inactive. Total isomers = $3$ (cis-d,cis-l,and trans).
Therefore,$[Cr(NO_2)_2(NH_3)_4]^{+}$ exhibits the highest number of isomers.

(B) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:
$1$. $[Cr(H_2O)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. Number of unpaired electrons $(n)$ = $3$.
$2$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. With $H_2O$ as a weak field ligand,$n = 4$.
$3$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ is $3d^9$. Number of unpaired electrons $(n)$ = $1$.
$4$. $[Zn(H_2O)_6]^{2+}$: $Zn^{2+}$ is $3d^{10}$. Number of unpaired electrons $(n)$ = $0$.
Since the magnetic moment $\mu = \sqrt{n(n+2)} \ BM$,the complex with the highest number of unpaired electrons will have the highest magnetic moment.
Thus,$[Fe(H_2O)_6]^{2+}$ with $n = 4$ has the highest magnetic moment.

(A) Ionization isomerism occurs when the counter ion in a coordination compound is a potential ligand and can displace a ligand which then becomes the counter ion.
In the pair $[Co(NH_3)_5Br]SO_4$ and $[Co(NH_3)_5SO_4]Br$,the $SO_4^{2-}$ ion and $Br^-$ ion exchange positions between the coordination sphere and the ionization sphere.
Therefore,they are ionization isomers.

(B) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:
$1$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$. With a weak field ligand $(H_2O)$,$n = 4$.
$2$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. With a weak field ligand $(H_2O)$,$n = 4$.
$3$. $[CoCl_4]^{2-}$: $Co^{2+}$ is $3d^7$. With a weak field ligand $(Cl^-)$,$n = 3$.
$4$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ is $3d^5$. With a weak field ligand $(H_2O)$,$n = 5$.
Since $[Cr(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{2+}$ both have $n = 4$,they will have the same magnetic moment.

(A) In the complex $[CoF_6]^{3-}$,the oxidation state of $Co$ is $+3$.
$Co$ has an atomic number of $27$,so the electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$F^-$ is a weak field ligand,so it does not cause pairing of electrons.
In the $3d$ orbital,the $6$ electrons are arranged as $t_{2g}^4 e_g^2$.
This results in $4$ unpaired electrons.

(B) An optically active isomer must lack a plane of symmetry and a center of inversion.
$1$. $trans-[CoCl_2(en)_2]^+$ has a plane of symmetry,so it is optically inactive.
$2$. $[Cr(ox)_3]^{3-}$ is a tris-chelated complex with $D_3$ symmetry,which is chiral and optically active.
$3$. $cis-[CoCl_2(en)_2]^+$ is also chiral and optically active.
$4$. $[Co(en)_3]^{3+}$ is also chiral and optically active.
However,in standard multiple-choice questions of this type,$[Cr(ox)_3]^{3-}$ is a classic example of an optically active complex due to the absence of any symmetry elements.