Isomerism and Magnetic properties Questions in English

(B) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$1$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $d^5$. $CN^-$ is a strong field ligand,so $n = 1$. $\mu = \sqrt{1(3)} = 1.73 \ B.M.$
$2$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $d^4$. $CN^-$ is a strong field ligand,so $n = 2$. $\mu = \sqrt{2(4)} = 2.83 \ B.M.$
$3$. $[CoF_6]^{3-}$: $Co^{3+}$ is $d^6$. $F^-$ is a weak field ligand,so $n = 4$. $\mu = \sqrt{4(6)} = 4.90 \ B.M.$
$4$. $[MnBr_4]^{2-}$: $Mn^{2+}$ is $d^5$. $Br^-$ is a weak field ligand,so $n = 5$. $\mu = \sqrt{5(7)} = 5.92 \ B.M.$
The correct order is $[Fe(CN)_6]^{3-} < [Mn(CN)_6]^{3-} < [CoF_6]^{3-} < [MnBr_4]^{2-}$.

(A) For option $(A)$: $[Cr(CN)_6]^{3-}$,$Cr^{3+}$ is $3d^3$. Since $CN^-$ is a strong field ligand,the electrons remain in $t_{2g}$ orbitals. Number of unpaired electrons $= 3$.
For option $(B)$: $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $3d^6$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^4 e_g^2$. Number of unpaired electrons $= 4$.
For option $(C)$: $[Co(NH_3)_6]^{3+}$,$Co^{3+}$ is $3d^6$. Since $NH_3$ is a strong field ligand,all electrons are paired. Number of unpaired electrons $= 0$.
For option $(D)$: $[Ni(NH_3)_6]^{2+}$,$Ni^{2+}$ is $3d^8$. The configuration is $t_{2g}^6 e_g^2$. Number of unpaired electrons $= 2$.
Thus,the correct matching is $A-II, B-IV, C-I, D-III$.

(A) The magnetic moment $\mu = \sqrt{n(n+2)} \ BM$.
Given $\mu = 6.06 \ BM$,which corresponds to $n = 5$ unpaired electrons.
For $Mn$ $(Z=25)$,the electronic configuration is $[Ar] 3d^5 4s^2$.
To have $5$ unpaired electrons in the complex,$Mn$ must be in the $+2$ oxidation state $(Mn^{2+})$,which has the configuration $[Ar] 3d^5$.
Let the oxidation state of $Mn$ be $y$. The charge on the complex is given by $y + 6 \times (-1) = -x$.
Since $y = +2$,we have $2 - 6 = -x$,which gives $-4 = -x$,so $x = 4$.

(B) . $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $d^5$. $CN^-$ is a strong field ligand,so pairing occurs. $n = 1$.
$B$. $[FeF_6]^{3-}$: $Fe^{3+}$ is $d^5$. $F^-$ is a weak field ligand,so no pairing. $n = 5$.
$C$. $[CoF_6]^{3-}$: $Co^{3+}$ is $d^6$. $F^-$ is a weak field ligand,so no pairing. $n = 4$.
$D$. $[Cr(oxalate)_3]^{3-}$: $Cr^{3+}$ is $d^3$. $n = 3$.
$E$. $[Ni(CO)_4]$: $Ni^0$ is $d^{10}$. $CO$ is a strong field ligand,all electrons paired. $n = 0$.
The number of unpaired electrons are: $A=1, B=5, C=4, D=3, E=0$.
Ordering them: $E(0) < A(1) < D(3) < C(4) < B(5)$.
Thus,the correct order is $E < A < D < C < B$.

(C) The magnetic moment $\mu$ is related to the number of unpaired electrons $n$ by the formula: $\mu = \sqrt{n(n+2)} \ BM$.
Given $\mu = 4.90 \ BM$,we have $4.90 = \sqrt{n(n+2)}$.
Squaring both sides,we get $24.01 = n(n+2)$,which is approximately $n^2 + 2n - 24 = 0$.
Solving this quadratic equation,$(n+6)(n-4) = 0$.
Since $n$ must be positive,$n = 4$.
Thus,the metal ion has $4$ unpaired electrons.

(A) Complexes of the type $[MA_3B_3]$ exhibit facial (fac) and meridional (mer) isomerism.
In the given options,$[Co(NH_3)_3(NO_2)_3]$ is of the type $[MA_3B_3]$,where $M = Co$,$A = NH_3$,and $B = NO_2$.
Therefore,it can exist as both facial and meridional isomers.

(D) For $[Cr(CN)_6]^{3-}$,the oxidation state of $Cr$ is $+3$. The electronic configuration is $3d^3$. Since $CN^-$ is a strong field ligand,the configuration is $t_{2g}^3 e_g^0$. The number of unpaired electrons $n = 3$.
$\mu_1 = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \ BM$.
For $[Cr(H_2O)_6]^{3+}$,the oxidation state of $Cr$ is $+3$. The electronic configuration is $3d^3$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^3 e_g^0$. The number of unpaired electrons $n = 3$.
$\mu_2 = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \ BM$.
The ratio $\frac{\mu_1}{\mu_2} = \frac{\sqrt{15}}{\sqrt{15}} = 1$.

(D) For $[Fe(H_2O)_6]^{3+}$:
$Fe^{3+}$ has $d^5$ configuration. $H_2O$ is a weak field ligand,so no pairing occurs.
Number of unpaired electrons $(n)$ = $5$.
$\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
For $[Fe(CN)_6]^{3-}$:
$Fe^{3+}$ has $d^5$ configuration. $CN^-$ is a strong field ligand,so pairing occurs.
Number of unpaired electrons $(n)$ = $1$.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ BM$.
Thus,the magnetic moments are $5.92 \ BM$ and $1.732 \ BM$ respectively.

(B) The complex $[Cr(ox)_2ClBr]^{3-}$ is of the type $[M(AA)_2a_2]$ where $AA$ is a bidentate ligand (oxalate) and $a$ represents monodentate ligands ($Cl^-$ and $Br^-$).
$1$. Geometrical Isomerism: There are two geometrical isomers:
- $trans$-isomer: The two monodentate ligands ($Cl$ and $Br$) are at $180^{\circ}$ to each other. This isomer has a plane of symmetry and is optically inactive.
- $cis$-isomer: The two monodentate ligands ($Cl$ and $Br$) are at $90^{\circ}$ to each other. This isomer lacks a plane of symmetry and is chiral.
$2$. Optical Isomerism: The $cis$-isomer exists as a pair of enantiomers ($d$ and $l$ forms).
Total stereoisomers = $1$ $(trans)$ + $2$ ($cis$-$d$ and $cis$-$l$) = $3$.

(B) The complex with the maximum number of unpaired electrons will exhibit the maximum attraction to an applied magnetic field (paramagnetism).
$1$. $[Zn(H_2O)_6]^{2+}$: $Zn^{2+}$ is a $d^{10}$ system,which has $0$ unpaired electrons.
$2$. $[Co(H_2O)_6]^{2+}$: $Co^{2+}$ is a $d^7$ system. In an octahedral field with a weak field ligand $(H_2O)$,the configuration is $t_{2g}^5 e_g^2$,resulting in $3$ unpaired electrons.
$3$. $[Co(en)_3]^{3+}$: $Co^{3+}$ is a $d^6$ system. Since $en$ is a strong field ligand,the configuration is $t_{2g}^6 e_g^0$,resulting in $0$ unpaired electrons.
$4$. $[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ is a $d^8$ system,which has $2$ unpaired electrons.
Since $[Co(H_2O)_6]^{2+}$ has the highest number of unpaired electrons $(3)$,it will exhibit the maximum attraction.

(C) $[FeF_6]^{3-}$: $Fe^{3+}$ is $[Ar] 3d^5$. $F^-$ is a weak field ligand,so the number of unpaired electrons is $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$
$[V(H_2O)_6]^{2+}$: $V^{2+}$ is $3d^3$. The number of unpaired electrons is $3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \ B.M.$
$[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,so the number of unpaired electrons is $4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ B.M.$
Comparing the values: $\sqrt{15} < \sqrt{24} < \sqrt{35}$.
Therefore,the correct order is $Q < R < P$.

(D) The complex is $[Pt(NH_3)_2 Cl(NH_2 CH_3)] Cl$.
In this complex,the oxidation state of $Pt$ is $+2$.
$Pt$ $(Z=78)$ has an electronic configuration of $[Xe] 4f^{14} 5d^9 6s^1$. Thus,$Pt^{2+}$ has a $5d^8$ configuration.
Since the complex is square planar,it undergoes $dsp^2$ hybridization.
In a $5d^8$ square planar complex,the electrons pair up in the $5d$ orbitals,leaving no unpaired electrons.
Therefore,the number of unpaired electrons $(n)$ is $0$.
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$
Substituting $n=0$,we get $\mu = \sqrt{0(0+2)} = 0 \ B.M.$

(C) $1. cis-[Cr(ox)_2 Cl_2]^{3-}$: This complex lacks both a plane of symmetry $(POS)$ and a center of symmetry $(COS)$,so it shows optical isomerism.
$2. [Co(en)_3]^{3+}$: This complex is chiral and lacks $POS$ and $COS$,so it shows optical isomerism.
$3. cis-[Pt(en)_2 Cl_2]^{2+}$: This complex lacks $POS$ and $COS$,so it shows optical isomerism.
$4. cis-[Co(en)_2 Cl_2]^{+}$: This complex lacks $POS$ and $COS$,so it shows optical isomerism.
$5. trans-[Pt(en)_2 Cl_2]^{2+}$: This complex possesses both $POS$ and $COS$,so it is optically inactive.
$6. trans-[Cr(ox)_2 Cl_2]^{3-}$: This complex possesses both $POS$ and $COS$,so it is optically inactive.
Therefore,the total number of complexes showing optical isomerism is $4$.

(B) $NH_3$ acts as a weak field ligand with $Ni^{2+}$.
The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
For $Ni^{2+}$,the configuration is $3d^8$.
The $3d$ orbitals are filled as follows: two orbitals are doubly occupied,and two orbitals are singly occupied (total $2$ unpaired electrons).
The spin-only magnetic moment is calculated as:
$\mu = \sqrt{n(n+2)} \ BM$
Where $n = 2$ (number of unpaired electrons).
$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ BM$.
To express this as $x \times 10^{-1} \ BM$,we have $2.82 = 28.2 \times 10^{-1}$.
Rounding to the nearest integer,$x = 28$.

(D) In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in $+2$ oxidation state with $d^8$ configuration. $H_2O$ is a weak field ligand,so it forms an octahedral complex with two unpaired electrons,which undergoes $d-d$ transition,making the solution green.
In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state with $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. The complex is square planar and diamagnetic. However,it is yellow in colour due to charge transfer or specific electronic transitions,but it is often cited in textbooks as being colourless or having very pale colour compared to other complexes; however,strictly speaking,it is yellow. Given the standard context of such questions,Statement $I$ is correct and Statement $II$ is incorrect.

(A) To determine the number of unpaired $d$ electrons,we analyze the electronic configuration of the central metal ions in the given octahedral complexes:
$1$. $[V(H_2O)_6]^{3+}$: $V^{3+}$ is $3d^2$. Number of unpaired electrons $(n)$ = $2$ (Even).
$2$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$. Since $H_2O$ is a weak field ligand,electrons fill according to Hund's rule: $t_{2g}^3 e_g^1$. Number of unpaired electrons $(n)$ = $4$ (Even).
$3$. $[Fe(H_2O)_6]^{3+}$: $Fe^{3+}$ is $3d^5$. With a weak field ligand,$t_{2g}^3 e_g^2$. Number of unpaired electrons $(n)$ = $5$ (Odd).
$4$. $[Ni(H_2O)_6]^{3+}$: $Ni^{3+}$ is $3d^7$. With a weak field ligand,$t_{2g}^5 e_g^2$. Number of unpaired electrons $(n)$ = $3$ (Odd).
$5$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ is $3d^9$. With a weak field ligand,$t_{2g}^6 e_g^3$. Number of unpaired electrons $(n)$ = $1$ (Odd).
The complexes with an even number of unpaired electrons are $[V(H_2O)_6]^{3+}$ and $[Cr(H_2O)_6]^{2+}$.
Therefore,the total number of such complexes is $2$.

(A) To determine the paramagnetic behaviour,we calculate the number of unpaired electrons $(n)$ for each complex. Since $H_2O$ is a weak field ligand,the electrons remain unpaired in the $d$-orbitals.

Complex	Electronic Configuration and Unpaired Electrons $(n)$
$[Co(H_2O)_6]^{2+}$	$Co^{2+} (d^7): t_{2g}^5 e_g^2, n=3$
$[Fe(H_2O)_6]^{2+}$	$Fe^{2+} (d^6): t_{2g}^4 e_g^2, n=4$
$[Mn(H_2O)_6]^{2+}$	$Mn^{2+} (d^5): t_{2g}^3 e_g^2, n=5$
$[Cr(H_2O)_6]^{2+}$	$Cr^{2+} (d^4): t_{2g}^3 e_g^1, n=4$

Paramagnetic behaviour is directly proportional to the number of unpaired electrons $(n)$.
Since $[Co(H_2O)_6]^{2+}$ has the minimum number of unpaired electrons $(n=3)$,it exhibits the least paramagnetic behaviour.

(B) The complex $MABXL$ involves $sp^3$ hybridization,which indicates a tetrahedral geometry.
Tetrahedral complexes with four different unidentate ligands do not exhibit geometrical isomerism because all positions in a tetrahedron are equivalent relative to each other.
Therefore,the number of geometrical isomers is $0$.

(B) The oxidizing ability of these oxoanions depends on the oxidation state and the stability of the lower oxidation states of the central metal atom. The order of oxidizing power is $VO_2^{+} < Cr_2O_7^{2-} < MnO_4^{-}$.
Thus,$VO_2^{+}$ has the least oxidizing ability.
In $VO_2^{+}$,the oxidation state of $V$ is $+5$.
The electronic configuration of $V$ $(Z=23)$ is $[Ar] 3d^3 4s^2$.
For $V^{+5}$,the configuration is $[Ar] 3d^0 4s^0$.
Since there are no unpaired electrons $(n=0)$,the spin-only magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \ BM$.

(A) The metallic radii of the given elements generally decrease across the period due to increasing effective nuclear charge,but $Zn$ has a larger radius due to metallic bonding characteristics. Among the transition metals $Sc$ to $Mn$,the atomic radius decreases. However,the question asks for the least metallic radii among the listed elements. $Cr$ has a smaller radius than $Sc, Ti, V$ and $Mn$.
In $CrO_4^{2-}$,the oxidation state of $Cr$ is $x + 4(-2) = -2$,so $x = +6$.
The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
The electronic configuration of $Cr^{6+}$ is $[Ar] 3d^0$.
Since there are no unpaired electrons $(n=0)$,the spin-only magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \ BM$.

(A) 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.
$(A) [Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. Number of unpaired electrons $n=3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M. \ (II)$.
$(B) [NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. In a tetrahedral field,$n=2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ B.M. \ (IV)$.
$(C) [CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so $n=4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ B.M. \ (I)$.
$(D) [Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing,so $n=0$. $\mu = 0 \ B.M. \ (III)$.
Thus,the correct match is $A-II, B-IV, C-I, D-III$.

(D) For $[Co(NH_3)_6]^{3+}$: $Co$ is in $+3$ oxidation state. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$. Since $NH_3$ is a strong field ligand,it causes pairing of electrons in $d$-orbitals,resulting in $t_{2g}^6 e_g^0$. Thus,the number of unpaired electrons is $0$.
For $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. Since $Cl^-$ is a weak field ligand,the electrons remain unpaired in the $d$-orbitals. The configuration is $e^4 t_2^4$,which results in $2$ unpaired electrons.
Total number of unpaired electrons = $0 + 2 = 2$.

(C) The complex ion $[Co(en)_2 Cl_2]^{+}$ has an octahedral geometry.
It exhibits geometrical isomerism,specifically $cis$ and $trans$ isomers.
Therefore,the total number of geometrical isomers is $2$,not $3$.
Thus,Assertion $A$ is incorrect,while Reason $R$ is correct.

(D) The correct matches are:
$A$. $[Co(NH_3)_5(NO_2)]Cl_2$ exhibits $II$. Linkage isomerism because the $NO_2^-$ ligand is an ambidentate ligand that can coordinate through $N$ or $O$.
$B$. $[Co(NH_3)_5(SO_4)]Br$ exhibits $III$. Ionization isomerism as the $SO_4^{2-}$ and $Br^-$ ions can exchange positions between the coordination sphere and the counter ion.
$C$. $[Co(NH_3)_6][Cr(CN)_6]$ exhibits $IV$. Coordination isomerism because the ligands are distributed between two metal centers.
$D$. $[Co(H_2O)_6]Cl_3$ exhibits $I$. Solvate (hydrate) isomerism because water molecules can act as ligands or be present as solvent molecules of crystallization.
Therefore,the correct sequence is $A-II, B-III, C-IV, D-I$.

(D) In $[Co(NH_3)_6]^{3+}$,$Co^{3+}$ ion has a $3d^6$ configuration.
In the presence of the strong field ligand $NH_3$,pairing of electrons occurs,resulting in a diamagnetic complex ion with $d^2sp^3$ hybridization.
In the case of $[CoF_6]^{3-}$,$Co^{3+}$ is in a $3d^6$ configuration. In the presence of the weak field ligand $F^-$,pairing does not occur,resulting in an $sp^3d^2$ hybridized paramagnetic complex with four unpaired electrons.
Therefore,both Statement $I$ and Statement $II$ are true.

(C) In $CuCl$,$Cu$ is in the $+1$ oxidation state ($d^{10}$ configuration),which is colourless.
In $K_3[Cu(CN)_4]$,$Cu$ is in the $+1$ oxidation state ($d^{10}$ configuration),which is colourless.
In $[Cu(CH_3CN)_4]BF_4$,$Cu$ is in the $+1$ oxidation state ($d^{10}$ configuration),which is colourless.
In $CuF_2$,$Cu$ is in the $+2$ oxidation state ($d^9$ configuration). Due to the presence of an unpaired electron,$d-d$ transitions occur,making it blue-coloured.

(A) The complex $[M(NH_3)_4Cl_2]$ exhibits two geometrical isomers: $cis$ and $trans$.
In the $trans$-isomer,the two $Cl$ atoms are at $180^{\circ}$ to each other,and the molecule possesses a plane of symmetry.
In the $cis$-isomer,the two $Cl$ atoms are at $90^{\circ}$ to each other,and the molecule also possesses a plane of symmetry.
Since both isomers possess a plane of symmetry,they are achiral and therefore optically inactive.
Thus,both statements are true and $STATEMENT-2$ is the correct explanation for $STATEMENT-1$.

(B) Geometrical isomerism in coordination compounds occurs when ligands can be arranged in different spatial orientations around the central metal atom.
$A$. $[Pt(en)Cl_2]$ is a square planar complex of the type $[M(AA)X_2]$. It does not show geometrical isomerism because the chelating ligand $(en)$ occupies adjacent positions,and the two $Cl$ ligands are also adjacent.
$B$. $[Pt(en)_2]Cl_2$ is a square planar complex of the type $[M(AA)_2]$. It does not show geometrical isomerism as all possible arrangements are identical.
$C$. $[Pt(en)_2Cl_2]Cl_2$ is an octahedral complex of the type $[M(AA)_2X_2]$. It exhibits geometrical isomerism,existing as $cis$ and $trans$ isomers.
$D$. $[Pt(NH_3)_2Cl_2]$ is a square planar complex of the type $[MA_2X_2]$. It exhibits geometrical isomerism,existing as $cis$ and $trans$ isomers.
Thus,compounds $C$ and $D$ exhibit geometrical isomerism.

(A) In $Cr(CO)_6$, the central metal atom $Cr$ is in the $0$ oxidation state.
The ground state electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
$CO$ is a strong field ligand, which causes the pairing of all $6$ electrons ($5$ from $3d$ and $1$ from $4s$) into the $3d$ orbitals.
As a result, the number of unpaired electrons $(n)$ is $0$.
The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 0$, we get $\mu = \sqrt{0(0+2)} = 0 \ BM$.

(C) Let us analyze the magnetic nature of each compound:
$1$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. All electrons are paired,so it is diamagnetic.
$2$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,so it is paramagnetic (has $2$ unpaired electrons).
$3$. $[Co(NH_3)_4Cl_2]Cl$: $Co^{3+}$ is $3d^6$. In an octahedral complex with $NH_3$ and $Cl^-$,$Co^{3+}$ is paramagnetic (has $4$ unpaired electrons).
$4$. $Na_3[CoF_6]$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so it is paramagnetic (has $4$ unpaired electrons).
$5$. $Na_2O_2$: Contains peroxide ion $O_2^{2-}$,which has all electrons paired,so it is diamagnetic.
$6$. $CsO_2$: Contains superoxide ion $O_2^-$,which has an odd number of electrons,so it is paramagnetic.
The paramagnetic compounds are $[NiCl_4]^{2-}$,$[Co(NH_3)_4Cl_2]Cl$,$Na_3[CoF_6]$,and $CsO_2$. The total number is $4$.

(C) The complex is $[Co(AB)_2Cl_2]^{-}$,where $AB$ is an unsymmetrical bidentate ligand $(H_2NCH_2CH_2O^{-})$.
Here,$A$ represents the $N$ donor atom and $B$ represents the $O$ donor atom.
For an octahedral complex of the type $[M(AB)_2X_2]$,the possible geometric isomers are:
$1$. Trans-$Cl$,Trans-$A$,Trans-$B$
$2$. Trans-$Cl$,Cis-$A$,Cis-$B$
$3$. Cis-$Cl$,Trans-$A$,Cis-$B$
$4$. Cis-$Cl$,Cis-$A$,Trans-$B$
$5$. Cis-$Cl$,Cis-$A$,Cis-$B$
Thus,there are $5$ possible geometric isomers.

(B) The complex $[Pt(NH_3)_2 Br_2]$ exhibits geometrical isomerism (cis-trans isomerism).
$(A)$ $[Pt(en)(SCN)_2]$: This is a square planar complex with a bidentate ligand $(en)$. It cannot show cis-trans isomerism because the two donor atoms of the $en$ ligand are fixed in adjacent positions.
$(B)$ $[Zn(NH_3)_2 Cl_2]$: This is a tetrahedral complex. Tetrahedral complexes do not exhibit geometrical isomerism because all positions are adjacent to each other.
$(C)$ $[Pt(NH_3)_2 Cl_4]$: This is an octahedral complex of the type $[MA_2 B_4]$. It can exhibit cis-trans isomerism.
$(D)$ $[Cr(en)_2(H_2 O)(SO_4)]^{+}$: This is an octahedral complex of the type $[M(AA)_2 BC]$. It can exhibit cis-trans isomerism.
Thus,both $(C)$ and $(D)$ exhibit geometrical isomerism.

(B) Ionization isomers are coordination compounds that produce different ions in solution.
In the complex $[Cr(H_2O)_4Cl(NO_2)]Cl$,the $Cl^-$ ion is present in the ionization sphere.
To form an ionization isomer,the $Cl^-$ ion in the ionization sphere is exchanged with the $NO_2^-$ ligand inside the coordination sphere.
Thus,the isomer is $[Cr(H_2O)_4Cl_2](NO_2)$.

(B) The spin-only magnetic moment $\mu$ is given by $\sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons. For $\mu = 2.82 \ B.M.$,$n=2$.
$(A)$ $Ni(CO)_4$: $Ni$ is in $0$ oxidation state. $Ni(0) = 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of electrons. All electrons are paired $(n=0)$.
$(B)$ $\left[NiCl_4\right]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+} = 3d^8$. $Cl^-$ is a weak field ligand,so no pairing occurs. The $3d$ orbitals have $2$ unpaired electrons $(n=2)$. Thus,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ B.M.$
$(C)$ $Ni\left(PPh_3\right)_4$: $Ni$ is in $0$ oxidation state. $Ni(0) = 3d^8 4s^2$. $PPh_3$ is a strong field ligand,causing pairing. All electrons are paired $(n=0)$.
$(D)$ $\left[Ni(CN)_4\right]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+} = 3d^8$. $CN^-$ is a strong field ligand,causing pairing. All electrons are paired $(n=0)$.

(C) $K \Rightarrow K_3[Fe(CN)_6], Fe^{3+} = 3d^5$ (has $5$ unpaired electrons). It is paramagnetic.
$L \Rightarrow [Co(NH_3)_6]Cl_3, Co^{3+} = 3d^6$. $NH_3$ is a strong field ligand,causing pairing. It is diamagnetic.
$M \Rightarrow Na_3[Co(ox)_3], Co^{3+} = 3d^6$. Oxalate $(ox^{2-})$ acts as a strong field ligand with $Co^{3+}$,causing pairing. It is diamagnetic.
$N \Rightarrow [Ni(H_2O)_6]Cl_2, Ni^{2+} = 3d^8$ (has $2$ unpaired electrons). It is paramagnetic.
$O \Rightarrow K_2[Pt(CN)_4], Pt^{2+} = 5d^8$. $CN^{-}$ is a strong field ligand and $5d$ orbitals have large splitting. It is diamagnetic.
$P \Rightarrow [Zn(H_2O)_6](NO_3)_2, Zn^{2+} = 3d^{10}$. All electrons are paired. It is diamagnetic.
Therefore,the diamagnetic complexes are $L, M, O, P$.

(A) $1$. $H$ atom: $1s^1$ (Paramagnetic)
$2$. $NO_2$ monomer: Odd electron (Paramagnetic)
$3$. $O_2^{-}$ (superoxide): $13$ electrons (Paramagnetic)
$4$. $S_2$ (vapour): Similar to $O_2$,has two unpaired electrons in $\pi^*$ orbitals (Paramagnetic)
$5$. $Mn_3O_4$: Mixed oxide of $MnO \cdot Mn_2O_3$,contains $Mn^{2+}$ and $Mn^{3+}$ (Paramagnetic)
$6$. $(NH_4)_2[FeCl_4]$: $Fe^{2+}$ is $d^6$ (Paramagnetic)
$7$. $(NH_4)_2[NiCl_4]$: $Ni^{2+}$ is $d^8$ (Paramagnetic)
$8$. $K_2MnO_4$: $Mn^{6+}$ is $d^1$ (Paramagnetic)
$9$. $K_2CrO_4$: $Cr^{6+}$ is $d^0$ (Diamagnetic)
Only $K_2CrO_4$ is diamagnetic. The total number of diamagnetic species is $1$.

(A) In the $cis-[Mn(en)_2Cl_2]$ complex,the central metal ion $Mn$ is octahedrally coordinated.
There are two $Cl$ ligands and two $en$ (ethylenediamine) ligands.
Each $en$ ligand provides two $N$ donor atoms.
Let the $Cl$ atoms be $Cl_{(a)}$ and $Cl_{(b)}$,and the $N$ atoms be $N_{(1)}, N_{(2)}, N_{(3)}, N_{(4)}$.
By analyzing the octahedral geometry,the cis $N-Mn-Cl$ bond angles are:
$Cl_{(a)}-Mn-N_{(1)}$,$Cl_{(a)}-Mn-N_{(2)}$,$Cl_{(a)}-Mn-N_{(4)}$,$Cl_{(b)}-Mn-N_{(1)}$,$Cl_{(b)}-Mn-N_{(3)}$,and $Cl_{(b)}-Mn-N_{(4)}$.
Counting these,we find a total of $6$ such cis bond angles.

(B) Paramagnetic compounds are attracted towards a magnetic field,causing the pan to be deflected downwards (figure $III$).
Diamagnetic compounds are repelled by a magnetic field,causing the pan to be deflected upwards (figure $II$).
$(A)$ $H_2O_{(l)}$ is diamagnetic. Therefore,the deflection is upwards. (Correct)
$(B)$ $K_4[Fe(CN)_6]_{(s)}$ contains $Fe^{2+}$ with a strong field ligand $(CN^-)$,resulting in a $d^6$ configuration $[t_{2g}^6, e_g^0]$. It is diamagnetic. Therefore,the deflection is upwards. (Correct)
$(C)$ $O_{2(g)}$ is paramagnetic due to two unpaired electrons in its $\pi^*$ antibonding orbitals. Therefore,the deflection is downwards. (Correct)
$(D)$ $C_6H_{6(l)}$ (benzene) is diamagnetic as it has no unpaired electrons. Therefore,the deflection should be upwards,not downwards. (Incorrect)
Thus,statements $(A), (B),$ and $(C)$ are correct.

(A) For $[Cr(NH_3)_6]^{3+}$:
$Cr$ $(Z=24)$ has configuration $[Ar] 3d^5 4s^1$. Thus,$Cr^{3+}$ is $[Ar] 3d^3$.
Number of unpaired electrons $(n)$ = $3$.
$\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ B.M.}$
For $[CuF_6]^{3-}$:
$Cu$ $(Z=29)$ has configuration $[Ar] 3d^{10} 4s^1$. Thus,$Cu^{3+}$ is $[Ar] 3d^8$.
In an octahedral field,$3d^8$ configuration has $2$ unpaired electrons in the $e_g$ orbitals.
Number of unpaired electrons $(n)$ = $2$.
$\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \text{ B.M.}$
Therefore,the values are $3.87 \text{ B.M.}$ and $2.84 \text{ B.M.}$ respectively.

(B) The complex $[Pt(NH_3)_4Cl_2]Br_2$ exhibits ionization isomerism.
Possible ionization isomers are:
$1$. $[Pt(NH_3)_4Cl_2]Br_2$ (Geometrical isomers: $cis$ and $trans$,so $2$ isomers)
$2$. $[Pt(NH_3)_4ClBr]ClBr$ (Geometrical isomers: $cis$ and $trans$,so $2$ isomers)
$3$. $[Pt(NH_3)_4Br_2]Cl_2$ (Geometrical isomers: $cis$ and $trans$,so $2$ isomers)
Each ionization isomer has $2$ geometrical isomers.
Total isomers = $2 + 2 + 2 = 6$.

(B) $P = [FeF_6]^{3-}$,oxidation number of $Fe = +3$,configuration: $3d^5$.
Since $F^-$ is a weak field ligand,no pairing occurs,resulting in $5$ unpaired electrons.
$Q = [V(H_2O)_6]^{2+}$,oxidation number of $V = +2$,configuration: $3d^3$.
It has $3$ unpaired electrons.
$R = [Fe(H_2O)_6]^{2+}$,oxidation number of $Fe = +2$,configuration: $3d^6$.
Since $H_2O$ is a weak field ligand,no pairing occurs,resulting in $4$ unpaired electrons.
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
Number of unpaired electrons: $Q(3) < R(4) < P(5)$.
Therefore,the order of magnetic moments is $Q < R < P$.
Thus,the correct option is $B$.

(A) $[Cr(NH_3)_5Cl]Cl_2$ and $[Cr(NH_3)_4Cl_2]Cl$ exhibit ionisation isomerism.
$(B)$ $[Co(NH_3)_4Cl_2]^+$ (octahedral,$Ma_4b_2$ type) and $[Pt(NH_3)_2(H_2O)Cl]^+$ (square planar,$Ma_2bc$ type) both exhibit geometrical isomerism.
$(C)$ $[CoBr_2Cl_2]^{2-}$ (tetrahedral,$Ma_2b_2$ type) does not show geometrical isomerism,whereas $[PtBr_2Cl_2]^{2-}$ (square planar,$Ma_2b_2$ type) does.
$(D)$ $[Pt(NH_3)_3(NO_3)]Cl$ and $[Pt(NH_3)_3Cl]Br$ both exhibit ionisation isomerism.
Therefore,pairs $(B)$ and $(D)$ exhibit the same kind of isomerism.

(A) The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Case-$1$: In the presence of $SCN^{-}$ (a weak field ligand),no pairing occurs in the $d$-orbitals. The number of unpaired electrons $(n)$ is $5$.
$\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ BM$.
Case-$2$: In the presence of $CN^{-}$ (a strong field ligand),pairing occurs in the $d$-orbitals. The number of unpaired electrons $(n)$ is $1$.
$\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
Difference in spin-only magnetic moments $= 5.91 - 1.73 = 4.18 \ BM$.
Rounding to the nearest integer,the difference is $4$.

(D) $1$. $[Co(en)_2Cl_2]^{+}$: This is of the type $[M(AA)_2a_2]$,which shows cis-trans isomerism.
$2$. $[CrCl_2(C_2O_4)_2]^{3-}$: This is of the type $[M(AA)_2a_2]$,which shows cis-trans isomerism.
$3$. $[Fe(H_2O)_4(OH)_2]^{+}$: This is of the type $[Ma_4b_2]$,which shows cis-trans isomerism.
$4$. $[Fe(NH_3)_2(CN)_4]^{-}$: This is of the type $[Ma_4b_2]$,which shows cis-trans isomerism.
$5$. $[Co(en)_2(NH_3)Cl]^{2+}$: This is of the type $[M(AA)_2ab]$,which shows cis-trans isomerism.
$6$. $[Co(NH_3)_4(H_2O)Cl]^{2+}$: This is of the type $[Ma_4bc]$. It shows geometrical isomerism,but specifically,it does not have a 'cis-trans' pair in the same sense as the others (it has facial/meridional or specific geometric arrangements,but is often excluded from the simple cis-trans definition).
Thus,there are $5$ complex ions that show cis-trans isomerism.

(C) Set-$I$: $[Co(NH_3)_3(NO_2)_3]$ exhibits facial and meridional geometrical isomers. $[Co(en)_2Cl_2]$ exhibits cis and trans geometrical isomers.
Set-$II$: $[Co(NH_3)_5Cl]SO_4$ and $[Co(NH_3)_5(SO_4)]Cl$ are ionization isomers because they exchange the counter ion $(SO_4^{2-})$ with the ligand $(Cl^-)$ inside the coordination sphere.

(A) $Mn^{3+} (3d^4)$: In both strong and weak fields,it has unpaired electrons. Hence,$[Mn(NH_3)_6]^{3+}$ and $[MnCl_6]^{3-}$ are paramagnetic.
$Fe^{3+} (3d^5)$: In both strong and weak fields,it has unpaired electrons. Hence,$[FeF_6]^{3-}$ and $[Fe(NH_3)_6]^{3+}$ are paramagnetic.
$Co^{3+} (3d^6)$:
$1$. $[CoF_6]^{3-}$: $F^-$ is a weak field ligand,so it forms a high-spin complex with $t_{2g}^4 e_g^2$ configuration,which is paramagnetic.
$2$. $[Co(en)_3]^{3+}$: $en$ is a strong field ligand,so it forms a low-spin complex with $t_{2g}^6 e_g^0$ configuration,which is diamagnetic.
Thus,only $[Co(en)_3]^{3+}$ is diamagnetic. The total number of diamagnetic species is $1$.

(C) homoleptic complex contains only one type of ligand.
$(A) \ [Fe(CN)_5NO]^{2-}$: Heteroleptic (contains $CN^-$ and $NO$ ligands).
$(B) \ [CoF_6]^{3-}$: Homoleptic,$Co^{3+}$ $(3d^6)$,$F^-$ is a weak field ligand,so it is high spin.
$(C) \ [Fe(CN)_6]^{4-}$: Homoleptic,$Fe^{2+}$ $(3d^6)$,$CN^-$ is a strong field ligand,so it is low spin $(t_{2g}^6 e_g^0)$.
$(D) \ [Co(NH_3)_6]^{3+}$: Homoleptic,$Co^{3+}$ $(3d^6)$,$NH_3$ is a strong field ligand,so it is low spin $(t_{2g}^6 e_g^0)$.
$(E) \ [Cr(H_2O)_6]^{2+}$: Homoleptic,$Cr^{2+}$ $(3d^4)$,$H_2O$ is a weak field ligand,so it is high spin $(t_{2g}^3 e_g^1)$.
Thus,$(C)$ and $(D)$ are homoleptic and low spin complexes.

(A) $Ma_3b_3$ type complexes show Facial - Meridional isomerism.
$(i)$ $[Co(NH_3)_3Cl_3] \Rightarrow Ma_3b_3$
$(ii)$ $[Co(NH_3)_4Cl_2]^{+} \Rightarrow Ma_4b_2$
$(iii)$ $[Co(en)_3]^{3+} \Rightarrow M(AA)_3$
$(iv)$ $[Co(en)_2Cl_2]^{+} \Rightarrow M(AA)_2b_2$
Here,$a = NH_3$,$b = Cl^{-}$,and $AA = en$ (ethylenediamine).
Thus,only $[Co(NH_3)_3Cl_3]$ corresponds to the $Ma_3b_3$ type.

(B) $Ni^{2+}$ gives a violet-coloured bead in a non-luminous flame under hot conditions.
$Ni^{2+}$ has a $d^8$ configuration,which remains unaffected by the nature of the ligand in an octahedral complex.
$Ni^{2+}: t_{2g}^{6} e_{g}^{2}$

(B) In $K_3[Fe(OH)_6]$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $OH^-$ is a weak field ligand,the electrons remain unpaired. Thus,$n = 5$. The magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ B.M.$
In $K_4[Fe(OH)_6]$,the oxidation state of $Fe$ is $+2$. The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$. Since $OH^-$ is a weak field ligand,the electrons remain unpaired. Thus,$n = 4$. The magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ B.M.$
Therefore,the values are $5.92 \ B.M.$ and $4.90 \ B.M.$ respectively.