Isomerism and Magnetic properties Questions in English

(C) Statement-$I$: $[Fe(ox)_3]^{3-}$ is an octahedral complex with three bidentate oxalate ligands. It lacks a plane of symmetry and a center of inversion, making it optically active and chiral. Thus, Statement-$I$ is correct.
Statement-$II$: $trans-[Cr(H_2O)_2(ox)_2]^-$ has a trans-configuration where the two water molecules are at $180^{\circ}$ to each other. This configuration possesses a plane of symmetry and a center of inversion, making it achiral (optically inactive). Thus, Statement-$II$ is incorrect.

(A) To find the spin-only magnetic moment, we calculate the number of unpaired electrons $(n)$ in each complex.
$1$. $[Mn(CN)_6]^{3-}$: $Mn$ is in $+3$ oxidation state $(3d^4)$. $CN^-$ is a strong field ligand, causing pairing. Configuration: $t_{2g}^4 e_g^0$. Unpaired electrons $n = 2$. Magnetic moment $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$2$. $[Fe(CN)_6]^{3-}$: $Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ is a strong field ligand. Configuration: $t_{2g}^5 e_g^0$. Unpaired electrons $n = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$3$. $[Co(NH_3)_6]^{3+}$: $Co$ is in $+3$ oxidation state $(3d^6)$. $NH_3$ is a strong field ligand. Configuration: $t_{2g}^6 e_g^0$. Unpaired electrons $n = 0$. Magnetic moment $\mu = 0 \ BM$.
$4$. $[Ti(H_2O)_6]^{3+}$: $Ti$ is in $+3$ oxidation state $(3d^1)$. $H_2O$ is a weak field ligand. Configuration: $t_{2g}^1 e_g^0$. Unpaired electrons $n = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
Comparing the values, $[Mn(CN)_6]^{3-}$ has the highest number of unpaired electrons $(n=2)$, hence the highest magnetic moment.

(A) Facial $(fac)$ and meridional $(mer)$ isomerism is a type of geometric isomerism observed in octahedral complexes with the general formula $[MA_3B_3]$.
In the $fac$ isomer, the three identical ligands occupy the corners of one triangular face of the octahedron.
In the $mer$ isomer, the three identical ligands occupy the meridian of the octahedron.
Among the given options, $[Cr(py)_3(Cl)_3]$ and $[Co(NH_3)_3(H_2O)_3]^{3+}$ both follow the $[MA_3B_3]$ pattern. However, $[Cr(py)_3(Cl)_3]$ is the classic example provided in textbooks for $fac-mer$ isomerism. $[Co(NH_3)_3(H_2O)_3]^{3+}$ also exhibits this isomerism. Given the standard options, $[Cr(py)_3(Cl)_3]$ is the most frequently cited example for this specific coordination geometry.

(C) $1$. The atomic number of $Cu$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
$2$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
$3$. In $3d^9$, there is $1$ unpaired electron $(n = 1)$.
$4$. The formula for spin-only magnetic moment is $\mu = \sqrt{n(n+2)} \ BM$.
$5$. Substituting $n = 1$: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(A) Step $1$: Transition metal ions exhibit colour due to $d-d$ electronic transitions.
Step $2$: For a $d-d$ transition to occur, the $d$-orbital must be partially filled, meaning there must be at least one unpaired electron.
Step $3$: If a transition metal ion is colourless, it implies that no $d-d$ transition is possible.
Step $4$: This occurs when the $d$-orbital is either completely empty $(d^0)$ or completely filled $(d^{10})$, both of which correspond to $0$ unpaired electrons.

(B) $1$. The color observed by the human eye is the complementary color of the light absorbed by the substance.
$2$. According to the color wheel, the complementary color of $red$ is $green$ or $blue-green$.
$3$. Among the given options, $blue$ is the closest complementary color to $red$ in the context of color absorption theory.

(B) Step $1$: Ionization isomers are compounds that produce different ions in solution. This occurs when a counter ion in the coordination compound exchanges its position with a ligand present in the coordination sphere.
Step $2$: The given complex is $[Cr(H_2O)_4 Cl(NO_2)] Cl$. Here, $Cl^-$ is the counter ion.
Step $3$: To form an ionization isomer, the $Cl^-$ ion outside the coordination sphere must exchange with a ligand inside the sphere (e.g., $NO_2^-$).
Step $4$: Swapping $Cl^-$ with $NO_2^-$ gives the complex $[Cr(H_2O)_4 Cl_2](NO_2)$.
Step $5$: Thus, option $B$ is the correct ionization isomer.

(A) Step $1$: Identify the general formula $MA_4BC$. Here, $M$ is the central metal atom, $A$ is a monodentate ligand present in $4$ units, and $B$ and $C$ are two different monodentate ligands.
Step $2$: Analyze the options. Option $A$ is $[Pt(NH_3)_4 ClBr]^{2+}$. Here, $M = Pt$, $A = NH_3$ ($4$ units), $B = Cl$, and $C = Br$. This matches the $MA_4BC$ type.
Step $3$: Diastereoisomers (geometric isomers) for $MA_4BC$ type complexes exist in two forms: $cis$ (where $B$ and $C$ are adjacent) and $trans$ (where $B$ and $C$ are opposite).
Step $4$: Therefore, $[Pt(NH_3)_4 ClBr]^{2+}$ can exhibit geometric isomerism.