Isomerism and Magnetic properties Questions in English

(D) In the complex $[CoF_6]^{3-}$,the oxidation state of $Co$ is $x + 6(-1) = -3$,which gives $x = +3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Since $F^-$ is a weak field ligand,it does not cause pairing of electrons.
Thus,the $3d$ orbitals have $4$ unpaired electrons $(n = 4)$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 4$,we get $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.

(C) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Cr^{3+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$. (Correct)
$2$. For $Cu^{2+}$ $(3d^9)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$. (Correct)
$3$. For $Co^{3+}$ (hydrated,i.e.,$[Co(H_2O)_6]^{3+}$): $H_2O$ is a weak field ligand. The configuration is $t_{2g}^4 e_g^2$,so $n = 4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$. The given value $0$ is incorrect.
$4$. For $Fe^{2+}$ $(3d^6)$: In aqueous solution,$[Fe(H_2O)_6]^{2+}$ has $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$. (Correct)
Thus,$Co^{3+}$ is not correctly matched.

(C) To determine the number of unpaired electrons,we analyze the oxidation state of the metal and the nature of the ligand:
$I.$ $[MnCl_6]^{3-}$: $Mn^{3+}$ $(3d^4)$,$Cl^-$ is a weak field ligand. Unpaired electrons = $4$.
$II.$ $[FeF_6]^{3-}$: $Fe^{3+}$ $(3d^5)$,$F^-$ is a weak field ligand. Unpaired electrons = $5$.
$III.$ $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ $(3d^4)$,$CN^-$ is a strong field ligand. Unpaired electrons = $2$.
$IV.$ $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ $(3d^5)$,$CN^-$ is a strong field ligand. Unpaired electrons = $1$.
Comparing the number of unpaired electrons: $IV (1) < III (2) < I (4) < II (5)$.
Thus,the increasing order is $IV < III < I < II$.

(D) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:
$I.$ $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong field ligand,causing pairing. $n = 2$. Magnetic moment $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$II.$ $[MnCl_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $Cl^-$ is a weak field ligand,no pairing. $n = 4$. Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
$III.$ $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. $n = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$IV.$ $[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,no pairing. $n = 5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Comparing the values: $1.73 (III) < 2.83 (I) < 4.90 (II) < 5.92 (IV)$.
Thus,the increasing order is $III < I < II < IV$.

(B) The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
For $Fe^{2+}$ ion,the configuration is $[Ar] 3d^6$.
In the $3d^6$ configuration,there are $4$ unpaired electrons $(n=4)$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=4$,we get $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.

(C) The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n + 2)} \text{ B.M. }$,where $n$ is the number of unpaired electrons.
The atomic number of $Fe$ is $26$,and its electronic configuration is $[Ar] 3d^6 4s^2$.
For $Fe^{2+}$,the configuration is $[Ar] 3d^6$.
In the $3d$ subshell,there are $5$ orbitals. According to Hund's rule,the $6$ electrons occupy these orbitals as follows: one orbital has $2$ electrons (paired),and the remaining $4$ orbitals have $1$ electron each (unpaired).
Thus,the number of unpaired electrons $n = 4$.
Substituting $n = 4$ into the formula: $\mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.9 \text{ B.M. }$

(C) The 'spin only' magnetic moment $(\mu_{so})$ is calculated using the number of unpaired electrons $(n)$ in an atom or ion. The formula is given by:
$\mu_{so} = \sqrt{n(n+2)} \ BM$
where $BM$ stands for Bohr Magneton.
Hence,the correct option is $(C)$.

(C) The magnetic moment (spin only) is given by $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1. [CoCl_4]^{2-}: Co^{2+} (d^7)$ is a tetrahedral complex. The configuration is $e^4 t_2^3$,so $n = 3$.
$2. [Fe(H_2O)_6]^{2+}: Fe^{2+} (d^6)$ is an octahedral complex. The configuration is $t_{2g}^4 e_g^2$,so $n = 4$.
$3. [Mn(H_2O)_6]^{2+}: Mn^{2+} (d^5)$ is an octahedral complex. The configuration is $t_{2g}^3 e_g^2$,so $n = 5$.
$4. [Cr(H_2O)_6]^{2+}: Cr^{2+} (d^4)$ is an octahedral complex. The configuration is $t_{2g}^3 e_g^1$,so $n = 4$.
Comparing the number of unpaired electrons,both $[Fe(H_2O)_6]^{2+}$ and $[Cr(H_2O)_6]^{2+}$ have $n = 4$,thus they have the same magnetic moment.

(C) $1$. $[NiCl_4]^{2-}$: This is a tetrahedral complex ($sp^3$ hybridization). Tetrahedral complexes do not exhibit geometrical isomerism. Number of isomers = $0$.
$2$. $[CoCl_2(NH_3)_4]^{+}$: This is an octahedral complex of the type $[MA_4B_2]$. It shows $2$ geometrical isomers ($cis$ and $trans$).
$3$. $[Co(NH_3)_3(NO_2)_3]$: This is an octahedral complex of the type $[MA_3B_3]$. It shows $2$ geometrical isomers ($fac$ and $mer$).
$4$. $[Co(NH_3)_5Cl]^{2+}$: This is an octahedral complex of the type $[MA_5B]$. It does not show geometrical isomerism. Number of isomers = $0$.
Total number of geometrical isomers = $0 + 2 + 2 + 0 = 4$.

(A) Geometrical isomerism in coordination complexes occurs when ligands can be arranged in different spatial positions relative to each other.
For the complex $\left[ Co(Cl)_2(en)_2 \right]^{+}$,the two $Cl^-$ ligands can be either adjacent to each other (cis-isomer) or opposite to each other (trans-isomer).
Therefore,$\left[ Co(Cl)_2(en)_2 \right]^{+}$ exhibits geometrical isomerism.

(A) $Pd(II)$ complexes are square planar.
$I$. $[Pd(NH_3)_2 ClBr]$ is of the type $[M(a)_2bc]$,which shows $2$ geometrical isomers (cis and trans).
$II$. $[Pd(NH_3)_2 Cl_2]$ is of the type $[M(a)_2b_2]$,which shows $2$ geometrical isomers (cis and trans).
$III$. $[Pd(en) Cl_2]$ is of the type $[M(AA)b_2]$,which shows $0$ geometrical isomers.
$IV$. $[Pd(en) ClBr]$ is of the type $[M(AA)bc]$,which shows $0$ geometrical isomers.
$V$. $[Pd(en)_2 Cl_2]$ is of the type $[M(AA)_2b_2]$,which shows $0$ geometrical isomers.
Thus,the number of geometrical isomers of $(I)$ is the same as that of $(II)$.

(C) The complexes are evaluated for optical activity:
$1$. $[Co(en)_3]^{3+}$: Optically active (contains $en$ ligand,no plane of symmetry).
$2$. $[Co(NH_3)_4Cl_2]^{+}$: Optically inactive ($cis$-form has a plane of symmetry; $trans$-form has a plane of symmetry).
$3$. $[CoCl_2(en)_2]^{+}$: Optically active ($cis$-form is chiral; $trans$-form is achiral).
$4$. $[Co(NH_3)_3(NO_2)_3]$: Optically inactive ($fac$ and $mer$ isomers are achiral).
Thus,there are $2$ optically active complexes.

(C) Geometrical isomerism is observed in complexes where the relative positions of ligands around the central metal atom can change.
$1$. Octahedral complexes of the type $[MX_2L_4]$ exhibit geometrical isomerism (cis and trans forms).
$2$. Square planar complexes of the type $[MX_2L_2]$ exhibit geometrical isomerism (cis and trans forms).
$3$. Tetrahedral complexes of the type $[MABXL]$ do not exhibit geometrical isomerism because all four positions in a tetrahedron are adjacent to each other,making the relative positions of ligands identical.
$4$. Octahedral complexes of the type $[MX_2(L^{-}L)_2]$ exhibit geometrical isomerism due to the arrangement of $X$ ligands.
Therefore,the correct answer is the tetrahedral complex.

(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 is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$I$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,so electrons pair up. $n = 0$,$\mu = 0 \ B.M.$
$II$. $[MnCl_4]^{2-}$: $Mn^{2+}$ is $3d^5$. $Cl^-$ is a weak field ligand,so electrons do not pair. $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \ B.M.$
$III$. $[Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand,so electrons pair up. $n = 1$,$\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73 \ B.M.$
$IV$. $[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. $n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \ B.M.$
Comparing the values: $0 (I) < 1.73 (III) < 3.87 (IV) < 5.92 (II)$.
Thus,the increasing order is $I < III < IV < II$.

(B) To determine the magnetic nature,we analyze the electronic configuration of the central metal ion in each complex:
$1$. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak field ligand,so electrons remain unpaired. It is paramagnetic.
$2$. In $[Co(ox)_3]^{3-}$,$Co$ is in $+3$ oxidation state $(3d^6)$. $ox^{2-}$ (oxalate) is a strong field ligand,causing pairing of electrons in $t_{2g}$ orbitals. The configuration becomes $t_{2g}^6 e_g^0$,meaning all electrons are paired. Thus,it is diamagnetic.
$3$. In $[Mn(CN)_6]^{3-}$,$Mn$ is in $+3$ oxidation state $(3d^4)$. It has unpaired electrons. It is paramagnetic.
$4$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. It has one unpaired electron. It is paramagnetic.
Therefore,the correct option is $B$.

(D) To determine the number of unpaired electrons,we analyze the electronic configuration of the central metal ion in each complex:
$1$. $[Mn(CN)_6]^{3-}$ $(A)$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^4 e_g^0$. Unpaired electrons = $2$.
$2$. $[Fe(CN)_6]^{3-}$ $(B)$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand. Configuration: $t_{2g}^5 e_g^0$. Unpaired electrons = $1$.
$3$. $[CoF_6]^{3-}$ $(C)$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand. Configuration: $t_{2g}^4 e_g^2$. Unpaired electrons = $4$.
$4$. $[Co(C_2O_4)_3]^{3-}$ $(D)$: $Co^{3+}$ is $3d^6$. $C_2O_4^{2-}$ is a strong field ligand. Configuration: $t_{2g}^6 e_g^0$. Unpaired electrons = $0$.
Comparing the number of unpaired electrons: $D (0) < B (1) < A (2) < C (4)$.
The correct increasing order is $D < B < A < C$.

(C) In $[Cr(H_2O)_6]Br_2$,$Cr$ is in $+2$ oxidation state $(3d^4)$,which has $4$ unpaired electrons,so it is paramagnetic.
In $Na_4[Fe(CN)_6]$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons,so it is diamagnetic.
In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of all electrons,resulting in $0$ unpaired electrons,so it is diamagnetic.
In $Na_2[NiCl_4]$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,resulting in $2$ unpaired electrons,so it is paramagnetic.
Therefore,the incorrect match is $[Ni(CO)_4]$ - Paramagnetic.

(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. The lower the number of unpaired electrons,the lower the magnetic moment.

Ion	Electronic Configuration	Number of Unpaired Electrons $(n)$
$[Ti(H_2O)_6]^{3+}$	$3d^1$	$1$
$[Mn(H_2O)_6]^{2+}$	$3d^5$	$5$
$[Ni(H_2O)_6]^{2+}$	$3d^8$	$2$
$[Co(H_2O)_6]^{2+}$	$3d^7$	$3$

Since $[Ti(H_2O)_6]^{3+}$ has the lowest number of unpaired electrons $(n=1)$,it has the lowest spin-only magnetic moment.

(B) . Tetrahedral $NiCl_2Br_2^{2-}$: Tetrahedral complexes do not exhibit geometric isomerism because all positions are equivalent.
$B$. Square planar $Pt(NH_3)_2Cl_2$: This is of the type $Ma_2b_2$,which exhibits cis-trans geometric isomerism.
$C$. Octahedral $Co(NH_3)_3Cl_3$: This is of the type $Ma_3b_3$,which exhibits facial (fac) and meridional (mer) geometric isomerism.
$D$. Square planar $Pd(NH_3)_3Br^{+}$: This is of the type $Ma_3b$,which does not exhibit geometric isomerism.
$E$. Octahedral $Co(en)_3^{3+}$: This is of the type $M(AA)_3$,which exhibits optical isomerism (enantiomers).
Therefore,$B, C,$ and $E$ exhibit isomerism.

(A) The colour observed is the complementary colour of the light absorbed by the compound.
According to the colour wheel, the wavelength range $490-500 \,nm$ corresponds to the blue-green region of the visible spectrum.
The complementary colour of blue-green is red.
Therefore, the compound will appear red.

(C) The spin-only magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu = 2.86 \ BM$,we have $\sqrt{n(n+2)} \approx 2.86$,which implies $n(n+2) \approx 8$,so $n = 2$.
In $[NiCl_4]^{2-}$,$Ni$ is in the $+2$ oxidation state ($3d^8$ configuration).
For $Ni^{2+}$ $(3d^8)$,the electrons are arranged as $t_{2g}^6 e_g^2$ in a tetrahedral field,resulting in $n = 2$ unpaired electrons.
Thus,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$,which matches the given value.

(B) $1$. For $[Zn(OH_2)_6]^{2+}$: $Zn^{2+}$ is $3d^{10}$,which has no unpaired electrons,so it is diamagnetic.
$2$. For $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing of electrons. It forms $d^2sp^3$ hybridization with no unpaired electrons,making it diamagnetic.
$3$. For $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so no pairing occurs. It has $4$ unpaired electrons,making it paramagnetic.
$4$. For $[V(OH_2)_6]^{2+}$: $V^{2+}$ is $3d^3$,which has $3$ unpaired electrons,making it paramagnetic.
Therefore,the correct match is $[Co(NH_3)_6]^{3+}$; diamagnetic.

(A) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.
$1$. $Ni(CN)_4^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing. $n = 0$,$\mu = 0 \ BM$.
$2$. $Cu(H_2O)_6^{2+}$: $Cu^{2+}$ is $d^9$. $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$3$. $NiCl_4^{2-}$: $Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand,no pairing. $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$4$. $Fe(H_2O)_6^{2+}$: $Fe^{2+}$ is $d^6$. $H_2O$ is a weak field ligand. $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
The increasing order of magnetic moments is $Ni(CN)_4^{2-} < Cu(H_2O)_6^{2+} < NiCl_4^{2-} < Fe(H_2O)_6^{2+}$.

(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$Cr^{2+} (Z=24): [Ar] 3d^4 4s^0$; $n = 4$ unpaired electrons.
$Fe^{2+} (Z=26): [Ar] 3d^6 4s^0$; $n = 4$ unpaired electrons.
Since $Cr^{2+}$ and $Fe^{2+}$ have the same number of unpaired electrons $(n=4)$,they have the same calculated value of magnetic moment.

(A) $[CoF_6]^{3-}: Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak-field ligand,so electrons remain unpaired $(n=4)$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ B.M.$
$[Co(C_2O_4)_3]^{3-}: Co$ is in $+3$ oxidation state $(3d^6)$. $C_2O_4^{2-}$ is a strong-field ligand,so electrons pair up $(n=0)$. $\mu = 0 \ B.M.$
$[FeF_6]^{3-}: Fe$ is in $+3$ oxidation state $(3d^5)$. $F^-$ is a weak-field ligand,so electrons remain unpaired $(n=5)$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$
$[Mn(CN)_6]^{3-}: Mn$ is in $+3$ oxidation state $(3d^4)$. $CN^-$ is a strong-field ligand,causing pairing $(n=2)$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M.$
Thus,the correct match is $A-II, B-I, C-IV, D-III$.

(C) Optical isomerism is exhibited by coordination compounds that lack a plane of symmetry and a center of symmetry.
$trans-[CrCl_2(ox)_2]^{3-}$ has a plane of symmetry,making it superimposable on its mirror image,and thus it is optically inactive.
$cis-[CrCl_2(ox)_2]^{3-}$,$[Co(en)_3]^{3+}$,and $[Co(ox)(en)_2]^{+}$ lack a plane of symmetry and exhibit optical isomerism.

(B, D) To determine if a complex is diamagnetic,we check for the presence of unpaired electrons in the metal ion's $d$-orbitals.
$1$. $[CoF_6]^{3-}$: $Co^{3+}$ is a $d^6$ ion. $F^-$ is a weak field ligand,resulting in a high-spin complex with $4$ unpaired electrons. It is paramagnetic.
$2$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is a $d^6$ ion. $NH_3$ is a strong field ligand,causing pairing of electrons. The configuration is $t_{2g}^6 e_g^0$,meaning $0$ unpaired electrons. It is diamagnetic.
$3$. $[Fe(OH_2)_6]^{2+}$: $Fe^{2+}$ is a $d^6$ ion. $H_2O$ is a weak field ligand,resulting in a high-spin complex with $4$ unpaired electrons. It is paramagnetic.
$4$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is a $d^6$ ion. $CN^-$ is a strong field ligand,causing pairing of electrons. The configuration is $t_{2g}^6 e_g^0$,meaning $0$ unpaired electrons. It is diamagnetic.
Thus,both $[Co(NH_3)_6]^{3+}$ and $[Fe(CN)_6]^{4-}$ are diamagnetic.

(A) In both complexes,the central metal ion is $Fe^{3+}$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
For $[Fe(CN)_6]^{3-}$: $CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals. This results in one unpaired electron $(t_{2g}^5 e_g^0)$,making it paramagnetic.
For $[FeF_6]^{3-}$: $F^-$ is a weak field ligand,which does not cause pairing of electrons. This results in five unpaired electrons $(t_{2g}^3 e_g^2)$,making it paramagnetic.
Therefore,both complexes are paramagnetic.

(D) Let us analyze each complex:
$1$. In $[VF_6]^{3-}$,$V$ is in $+3$ oxidation state. $V^{3+}$ is $3d^2$. It has $2$ unpaired electrons,so it is paramagnetic. This statement is correct.
$2$. In $[CuCl_4]^{2-}$,$Cu$ is in $+2$ oxidation state. $Cu^{2+}$ is $3d^9$. It has $1$ unpaired electron,so it is paramagnetic. This statement is correct.
$3$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state. $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing of electrons. All $6$ electrons are paired,so it is diamagnetic. This statement is correct.
$4$. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state. $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so no pairing occurs. The $3d$ orbitals have $4$ unpaired electrons. Thus,it is paramagnetic with $4$ unpaired electrons. The statement claiming $2$ unpaired electrons is incorrect.

(A) In $K_3[Fe(CN)_6]$,the oxidation state of $Fe$ is $+3$.
Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $CN^-$ is a strong field ligand,pairing occurs,resulting in $t_{2g}^5 e_g^0$ configuration with $1$ unpaired electron.
In $K_4[Fe(CN)_6]$,the oxidation state of $Fe$ is $+2$.
Electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
Due to the strong field ligand $CN^-$,pairing occurs,resulting in $t_{2g}^6 e_g^0$ configuration with $0$ unpaired electrons.

(D) The magnetic moment depends on the number of unpaired electrons $(n)$. The formula is $\mu = \sqrt{n(n+2)} \ BM$.
For $[Cr(H_{2}O)_{6}]^{3+}$: $Cr$ is in $+3$ oxidation state. Electronic configuration of $Cr^{3+}$ is $[Ar] 3d^{3}$. Number of unpaired electrons $(n)$ = $3$.
Now,let us check the number of unpaired electrons for each option:
$A$) $[Cu(H_{2}O)_{6}]^{2+}$: $Cu^{2+}$ is $[Ar] 3d^{9}$,$n = 1$.
$B$) $[Mn(H_{2}O)_{6}]^{3+}$: $Mn^{3+}$ is $[Ar] 3d^{4}$,$n = 4$.
$C$) $[Fe(H_{2}O)_{6}]^{3+}$: $Fe^{3+}$ is $[Ar] 3d^{5}$,$n = 5$.
$D$) $[Mn(H_{2}O)_{6}]^{4+}$: $Mn^{4+}$ is $[Ar] 3d^{3}$,$n = 3$.
Since $[Mn(H_{2}O)_{6}]^{4+}$ has $3$ unpaired electrons,it has the same magnetic moment as $[Cr(H_{2}O)_{6}]^{3+}$.

(B, C, D) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons. For $n=5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$.
$(A)$ In $K_{4}[Mn(CN)_{6}]$,$Mn$ is in $+2$ oxidation state $(3d^{5})$. $CN^{-}$ is a strong field ligand,causing pairing of electrons. Thus,$n=1$.
$(B)$ In $[Fe(H_{2}O)_{6}]Cl_{3}$,$Fe$ is in $+3$ oxidation state $(3d^{5})$. $H_{2}O$ is a weak field ligand,so no pairing occurs. Thus,$n=5$.
$(C)$ In $K_{3}[FeF_{6}]$,$Fe$ is in $+3$ oxidation state $(3d^{5})$. $F^{-}$ is a weak field ligand,so no pairing occurs. Thus,$n=5$.
$(D)$ In $K_{4}[MnF_{6}]$,$Mn$ is in $+2$ oxidation state $(3d^{5})$. $F^{-}$ is a weak field ligand,so no pairing occurs. Thus,$n=5$.
Therefore,the compounds with five unpaired electrons are $(B)$,$(C)$,and $(D)$.

(A) Species having unpaired electrons are paramagnetic.
$Ni$ has atomic number $28$,electronic configuration $[Ar] 3d^{8} 4s^{2}$.
$1$. $[Ni(H_{2}O)_{6}]^{2+}$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $H_{2}O$ is a weak field ligand,so no pairing occurs. It has $2$ unpaired electrons (Paramagnetic).
$2$. $[Ni(PPh_{3})_{2}Cl_{2}]$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. It is a tetrahedral complex. Due to the presence of $Cl^{-}$ ligands,it remains high spin with $2$ unpaired electrons (Paramagnetic).
$3$. $[Ni(CO)_{4}]$: $Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand,causing pairing of all electrons. It has $0$ unpaired electrons (Diamagnetic).
$4$. $[Ni(CN)_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $CN^{-}$ is a strong field ligand,causing pairing of electrons. It has $0$ unpaired electrons (Diamagnetic).
Thus,$[Ni(H_{2}O)_{6}]^{2+}$ and $[Ni(PPh_{3})_{2}Cl_{2}]$ are paramagnetic.

(A) For both complexes,the central metal ion is $Fe^{3+}$.
Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
In $[FeCl_4]^-$,$Cl^-$ is a weak field ligand,so no pairing of electrons occurs. The number of unpaired electrons $(n)$ is $5$.
The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \ BM$.
In $[Fe(CN)_6]^{3-}$,$CN^-$ is a strong field ligand,causing pairing of electrons. The configuration becomes $t_{2g}^5 e_g^0$,leaving $n = 1$ unpaired electron.
The spin-only magnetic moment is $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ BM$.
Thus,the values are $5.9 \ BM$ and $1.732 \ BM$.

(B) $1$. In $[Ni(CO)_{4}]$,$Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand,causing pairing of electrons. Hybridisation is $sp^{3}$,unpaired electrons $= 0$.
$2$. In $[NiCl_{4}]^{2-}$,$Ni^{2+}$ is $3d^{8}$. $Cl^{-}$ is a weak field ligand,no pairing occurs. Hybridisation is $sp^{3}$,unpaired electrons $= 2$.
$3$. In $[PtCl_{2}(NH_{3})_{2}]$,$Pt^{2+}$ is $5d^{8}$. $Pt$ is a $5d$ series metal,so pairing occurs even with weak ligands. Hybridisation is $dsp^{2}$,unpaired electrons $= 0$.
$4$. In $[Ni(CN)_{4}]^{2-}$,$Ni^{2+}$ is $3d^{8}$. $CN^{-}$ is a strong field ligand,causing pairing. Hybridisation is $dsp^{2}$,unpaired electrons $= 0$.
$5$. In $[Pt(CN)_{4}]^{2-}$,$Pt^{2+}$ is $5d^{8}$. $CN^{-}$ is a strong field ligand. Hybridisation is $dsp^{2}$,unpaired electrons $= 0$.
Total unpaired electrons $= 0 + 2 + 0 + 0 + 0 = 2$.

(A) The spin-only magnetic moment of $3.87 \ BM$ corresponds to $3$ unpaired electrons,indicating $Cr^{3+}$ in an octahedral field. The $1:2$ electrolyte behavior implies the formula is $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$.
When passed through a cation exchanger,the complex cation $[Cr(H_2O)_5Cl]^{2+}$ is retained,and the $2Cl^-$ ions are released into the solution.
The molar mass of $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O = 52 + (5 \times 18) + (3 \times 35.5) + 18 = 266.5 \ g \ mol^{-1}$.
Moles of complex = $\frac{2.75 \ g}{266.5 \ g \ mol^{-1}} \approx 0.0103 \ mol$.
Since each mole of complex releases $2$ moles of $Cl^-$,moles of $AgCl$ formed = $2 \times 0.0103 = 0.0206 \ mol$.
Mass of $AgCl = 0.0206 \ mol \times 143.5 \ g \ mol^{-1} = 2.956 \ g \approx 3 \ g$.

(A) To determine the spin-only magnetic moment $(\mu = \sqrt{n(n+2)} \ BM)$, we calculate the number of unpaired electrons $(n)$ for each complex:
$1$. $[MnBr_{4}]^{2-}$: $Mn^{2+}$ is $3d^{5}$. Since $Br^{-}$ is a weak field ligand, it forms a high-spin tetrahedral complex with $n = 5$.
$2$. $[Cu(H_{2}O)_{6}]^{2+}$: $Cu^{2+}$ is $3d^{9}$. It has $n = 1$ unpaired electron.
$3$. $[Ni(CN)_{4}]^{2-}$: $Ni^{2+}$ is $3d^{8}$. Since $CN^{-}$ is a strong field ligand, it forms a square planar complex with $n = 0$.
$4$. $[Ni(H_{2}O)_{6}]^{2+}$: $Ni^{2+}$ is $3d^{8}$. It forms an octahedral complex with $n = 2$ unpaired electrons.
Comparing the number of unpaired electrons: $C (n=0) < B (n=1) < D (n=2) < A (n=5)$.
Thus, the increasing order of magnetic moments is $C < B < D < A$.

(C) $X$: For square planar complex $[M(abcd)]$,the number of geometrical isomers is $3$ (three pairs of trans ligands).
$Y$: For octahedral complex $[M(aa)_2b_2]$,the $cis$-isomer is optically active (chiral) and the $trans$-isomer is optically inactive (achiral). Thus,$Y = 1$.
$Z$: For octahedral complex $[Ma_3b_3]$,the geometrical isomers are $fac$ (facial) and $mer$ (meridional). Thus,$Z = 2$.
$X+Y+Z = 3+1+2 = 6$.

(A) Statement-$I$:
$1$. $[CoF_6]^{3-}$: $Co^{3+}$ $(d^6)$,$F^-$ is a weak field ligand,configuration is $t_{2g}^4 e_g^2$,$4$ unpaired electrons (Paramagnetic).
$2$. $[TiF_6]^{3-}$: $Ti^{3+}$ $(d^1)$,$1$ unpaired electron (Paramagnetic).
$3$. $V_2O_5$: $V^{5+}$ $(d^0)$,$0$ unpaired electrons (Diamagnetic).
$4$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ $(d^5)$,$CN^-$ is a strong field ligand,configuration is $t_{2g}^5 e_g^0$,$1$ unpaired electron (Paramagnetic).
Total paramagnetic species = $3$. Statement-$I$ is true.
Statement-$II$:
$1$. $K_4[Fe(CN)_6]$: $Fe^{2+}$ $(d^6)$,strong field,$0$ unpaired electrons.
$2$. $K_3[Fe(CN)_6]$: $Fe^{3+}$ $(d^5)$,strong field,$1$ unpaired electron.
$3$. $[Fe(H_2O)_6]SO_4 \cdot H_2O$: $Fe^{2+}$ $(d^6)$,weak field,$4$ unpaired electrons.
$4$. $[Fe(H_2O)_6]Cl_3$: $Fe^{3+}$ $(d^5)$,weak field,$5$ unpaired electrons.
Order: $0 < 1 < 4 < 5$. Statement-$II$ is true.

(A) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu > 3.0 \ BM$,we need $\sqrt{n(n+2)} > 3.0$,which implies $n(n+2) > 9$. This holds for $n \geq 3$.
Electronic configurations:
$V^{3+} (3d^2): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$
$Ti^{2+} (3d^2): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$
$Ni^{2+} (3d^8): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$
$Fe^{2+} (3d^6): n = 4, \mu = \sqrt{4(6)} = 4.90 \ BM$
$Co^{2+} (3d^7): n = 3, \mu = \sqrt{3(5)} = 3.87 \ BM$
Thus,$Fe^{2+}$ and $Co^{2+}$ have $\mu > 3.0 \ BM$.
In high spin octahedral complexes:
$Fe^{2+} (3d^6)$ has $4$ unpaired electrons.
$Co^{2+} (3d^7)$ has $3$ unpaired electrons.
Sum of unpaired electrons = $4 + 3 = 7$.

(D) Facial $(fac)$ and meridional $(mer)$ isomerism is a specific type of geometric isomerism observed in octahedral complexes with the general formula $MA_3B_3$.
In this type of isomerism,three identical ligands occupy either one face of the octahedron ($fac$-isomer) or a plane passing through the metal atom ($mer$-isomer).
For the given complex $[Co(NH_3)_x(NO_2)_y]$,to exhibit $fac$ and $mer$ isomerism,the coordination number must be $6$ and the stoichiometry must match the $MA_3B_3$ pattern.
Therefore,$x$ must be $3$ and $y$ must be $3$,resulting in the complex $[Co(NH_3)_3(NO_2)_3]$.

(D) The spin-only magnetic moment $(mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$A$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ has a $d^4$ configuration. Number of unpaired electrons $(n)$ = $4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$ $(III)$.
$B$. $[Co(H_2O)_6]^{2+}$: $Co^{2+}$ has a $d^7$ configuration. Number of unpaired electrons $(n)$ = $3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$ $(I)$.
$C$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ has a $d^9$ configuration. Number of unpaired electrons $(n)$ = $1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$ $(IV)$.
$D$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ has a $d^5$ configuration. Number of unpaired electrons $(n)$ = $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$ $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) Coordination isomerism occurs in coordination compounds where both the cation and the anion are complex ions,and the ligands can be exchanged between the metal centers.
For coordination isomerism to occur,both the metal ions must be different or the ligands must be different,allowing for the formation of isomers like $[M_1(L_1)_n][M_2(L_2)_m] \rightleftharpoons [M_1(L_2)_m][M_2(L_1)_n]$.
In option $A$,the metal ions are the same $(Ag^+)$,so exchanging ligands does not result in a new isomer.
In options $B, C, D,$ and $E$,the complexes consist of two different metal centers (or different combinations of ligands),which allows for the exchange of ligands between the coordination spheres.
Specifically,complexes $B, C, D,$ and $E$ all satisfy the condition of having complex cations and complex anions with different metal centers or ligand sets,thus exhibiting coordination isomerism.
Therefore,the correct set is $B, C, D,$ and $E$. However,based on the provided options,the most appropriate choice is $D$ ($C, D$ and $E$ Only) as it includes the valid complexes.

(C) To determine the number of paramagnetic complexes,we analyze the electronic configuration of the central metal ion in each complex:
$1$. $[MnBr_4]^{2-}$: $Mn^{2+}$ $(d^5)$,weak field ligand,$5$ unpaired electrons. (Paramagnetic)
$2$. $[NiCl_4]^{2-}$: $Ni^{2+}$ $(d^8)$,weak field ligand,$2$ unpaired electrons. (Paramagnetic)
$3$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ $(d^8)$,strong field ligand,$dsp^2$ hybridization,$0$ unpaired electrons. (Diamagnetic)
$4$. $[Ni(CO)_4]$: $Ni^0$ $(d^{10})$,strong field ligand,$sp^3$ hybridization,$0$ unpaired electrons. (Diamagnetic)
$5$. $[CoF_6]^{3-}$: $Co^{3+}$ $(d^6)$,weak field ligand,$4$ unpaired electrons. (Paramagnetic)
$6$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ $(d^6)$,strong field ligand,$t_{2g}^6 e_g^0$,$0$ unpaired electrons. (Diamagnetic)
$7$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ $(d^4)$,strong field ligand,$t_{2g}^3 e_g^1$,$2$ unpaired electrons. (Paramagnetic)
$8$. $[Ti(CN)_6]^{3-}$: $Ti^{3+}$ $(d^1)$,$1$ unpaired electron. (Paramagnetic)
$9$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ $(d^9)$,$1$ unpaired electron. (Paramagnetic)
$10$. $[Co(C_2O_4)_3]^{3-}$: $Co^{3+}$ $(d^6)$,strong field ligand,$t_{2g}^6 e_g^0$,$0$ unpaired electrons. (Diamagnetic)
Total paramagnetic complexes = $6$.

(C) . $[Pt(NH_3)_2Cl_2]$ shows geometrical isomerism (cis-trans).
$B$. $[Co(en)_3]^{3+}$ shows optical isomerism due to the lack of a plane of symmetry.
$C$. $[Co(NH_3)_5NO_2]Cl_2$ exhibits linkage isomerism because of the ambidentate $NO_2^-$ ligand.
$D$. $[Cr(H_2O)_6]Cl_3$ exhibits solvate (hydrate) isomerism.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(C) The formula for the spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ $BM$,where $n$ is the number of unpaired electrons.
For the $Ti^{2+}$ ion with electronic configuration $3d^2$,the number of unpaired electrons $n = 2$.
Substituting the value of $n$ into the formula:
$\mu = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83$ $BM$.
Rounding to two decimal places,we get $2.84$ $BM$.
Thus,the correct option is $C$.