Crystal Field theory Questions in English

(A) The wavelength of light absorbed $(\lambda)$ is inversely proportional to the crystal field splitting energy $(\Delta_o)$.
According to the spectrochemical series, the strength of the ligands is $Cl^{-} < NH_{3} < CN^{-}$.
Since $\Delta_o$ is directly proportional to the ligand field strength, the order of $\Delta_o$ is $[CoCl(NH_{3})_{5}]^{2+} < [Co(NH_{3})_{6}]^{3+} < [Co(CN)_{6}]^{3-}$.
Therefore, the order of wavelength of light absorbed $(\lambda \propto 1/\Delta_o)$ is $[CoCl(NH_{3})_{5}]^{2+} > [Co(NH_{3})_{6}]^{3+} > [Co(CN)_{6}]^{3-}$.

(C) The magnitude of $\Delta_{0}$ depends on the nature of the ligand and the oxidation state of the central metal ion.
According to the spectrochemical series,the field strength of the ligands is $Cl^{-} < H_{2}O < NH_{3} < CN^{-}$.
Ligands with lower field strength result in a smaller crystal field splitting energy $(\Delta_{0})$.
Since $Cl^{-}$ is the weakest field ligand among the given options,the complex $[Co(Cl)_{6}]^{3-}$ will have the minimum magnitude of $\Delta_{0}$.

(A) The fundamental assumption of $CFT$ is that $M-L$ interactions are purely electrostatic in nature.
In an octahedral metal complex,the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands is considered.

(C) According to the spectrochemical series,ligands are arranged in increasing order of their field strength. The series is as follows:
$I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < N_3^{-} < F^{-} < \text{urea} < OH^{-} < C_2H_5OH < C_2O_4^{2-} < O^{2-} < H_2O < NCS^{-} < gly < NH_3 < en < NH_2OH < bPy < NO < CH_3^{-} < C_6H_5^{-} < CN^{-} < CO$
Comparing the given ligands with $H_2O$:
$1. S^{2-}$ (weaker than $H_2O$)
$2. Br^{-}$ (weaker than $H_2O$)
$3. C_2O_4^{2-}$ (weaker than $H_2O$)
$4. CN^{-}$ (stronger than $H_2O$)
$5. en$ (stronger than $H_2O$)
$6. NH_3$ (stronger than $H_2O$)
$7. CO$ (stronger than $H_2O$)
$8. OH^{-}$ (weaker than $H_2O$)
Therefore,there are $4$ ligands $(CN^{-}, en, NH_3, CO)$ that are stronger than $H_2O$.

(A) According to the spectrochemical series,the increasing order of field strength for the given ligands is $I^{-} < F^{-} < H_2O < NH_3 < CO$.
Therefore,the correct order is $IV < V < I < III < II$.

(A) The spectrochemical series is an experimental arrangement of ligands in the order of increasing crystal field splitting energy $(\Delta_o)$.
Based on the spectrochemical series,the field strength of the given ligands increases in the following order:
$I^{-} < F^{-} < H_2O < NH_3 < CO$
Comparing this with the given options:
$IV (I^{-}) < V (F^{-}) < I (H_2O) < III (NH_3) < II (CO)$
Thus,the correct order is $IV < V < I < III < II$.

(D) The magnitude of crystal field splitting $(\Delta_o)$ depends on the nature of the ligand (spectrochemical series) and the oxidation state of the metal ion.
$1$. In option $A$,$H_2O$ is a stronger field ligand than $F^-$,so $[Fe(H_2O)_6]^{3+} > [FeF_6]^{3-}$ is correct.
$2$. In option $B$,$en$ (ethylenediamine) is a stronger field ligand than $NCS^-$,so $[Fe(en)_3]^{3+} > [Fe(NCS)_6]^{3-}$ is correct.
$3$. In option $C$,$CN^-$ is a strong field ligand compared to $H_2O$,so $[Fe(CN)_6]^{4-} > [Fe(H_2O)_6]^{2+}$ is correct.
$4$. In option $D$,$NH_3$ is a stronger field ligand than $H_2O$. Therefore,the order should be $[Fe(NH_3)_6]^{2+} > [Fe(H_2O)_6]^{2+}$. The given order $[Fe(H_2O)_6]^{2+} > [Fe(NH_3)_6]^{2+}$ is incorrect.

(A) $1$. Determine the oxidation state of the metal ion in each complex:
- In $[Mn(H_2O)_6]^{2+}$,$Mn$ is in $+2$ oxidation state. The electronic configuration of $Mn^{2+}$ $(Z=25)$ is $[Ar] 3d^5$.
- In $[Fe(H_2O)_6]^{2+}$,$Fe$ is in $+2$ oxidation state. The electronic configuration of $Fe^{2+}$ $(Z=26)$ is $[Ar] 3d^6$.
- In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state. The electronic configuration of $Co^{3+}$ $(Z=27)$ is $[Ar] 3d^6$.
- In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in $+2$ oxidation state. The electronic configuration of $Ni^{2+}$ $(Z=28)$ is $[Ar] 3d^8$.
$2$. Apply Crystal Field Theory $(CFT)$ for octahedral complexes:
- $H_2O$ is a weak field ligand,so it does not cause pairing of electrons in $d$-orbitals.
- For $Mn^{2+}$ $(d^5)$,the electrons fill the orbitals as $t_{2g}^3 e_g^2$ (high spin).
- For $Fe^{2+}$ $(d^6)$,the electrons fill as $t_{2g}^4 e_g^2$.
- For $Co^{3+}$ $(d^6)$,$NH_3$ is a strong field ligand,causing pairing: $t_{2g}^6 e_g^0$.
- For $Ni^{2+}$ $(d^8)$,the electrons fill as $t_{2g}^6 e_g^2$.
$3$. Conclusion: The configuration $t_{2g}^3 e_g^2$ corresponds to $[Mn(H_2O)_6]^{2+}$.

(D) For complex $A$,the configuration is $t_{2g}^3 e_g^1$,which corresponds to $4$ electrons in the $d$-orbitals. This indicates $Mn^{3+}$ ($d^4$ system). Since $H_2O$ is a weak-field ligand,it results in a high-spin complex. Thus,$A$ is $[Mn(H_2O)_6]^{3+}$.
For complex $B$,the configuration is $t_{2g}^4 e_g^0$,which also corresponds to $4$ electrons in the $d$-orbitals $(Mn^{3+})$. Since $CN^-$ is a strong-field ligand,it causes pairing of electrons,resulting in a low-spin complex. Thus,$B$ is $[Mn(CN)_6]^{3-}$.

(A) The atomic number of $Ti$ is $22$,and its electronic configuration is $[Ar] 3d^2 4s^2$. In both $TiCl_3$ and $[Ti(H_2O)_6]Cl_3$,the oxidation state of $Ti$ is $+3$,which corresponds to a $3d^1$ configuration.
In the complex $[Ti(H_2O)_6]^{3+}$,the presence of ligands $(H_2O)$ causes crystal field splitting of the $d$-orbitals into $t_{2g}$ and $e_g$ levels. The single electron can undergo a $d-d$ transition by absorbing light,making the complex coloured (violet).
In the anhydrous compound $TiCl_3$,there are no ligands to cause crystal field splitting. Without splitting,$d-d$ transitions cannot occur,rendering the substance colourless.
Therefore,$X$ is colourless and $Y$ is coloured.

(B) The crystal field theory $(CFT)$ is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand.
Ligands are treated as point charges in case of anions or point dipoles in case of neutral molecules.
The theory is successful in explaining the formation,structures,colour,and magnetic properties of coordination compounds.
However,it fails to explain the covalent character of metal-ligand bonding,as it assumes the bond is purely ionic.
Therefore,statements $II, III,$ and $IV$ are correctly explained by $CFT$.

(B) The complexes are $I: [Ni(H_2O)_6]^{2+}$,$II: [Ni(NH_3)_6]^{2+}$,and $III: [Ni(NO_2)_6]^{4-}$.
According to the spectrochemical series,the field strength of the ligands is $H_2O < NH_3 < NO_2^-$.
The Crystal Field Splitting Energy $(CFSE)$,denoted as $\Delta_0$,is directly proportional to the field strength of the ligand.
Therefore,the order of $CFSE$ is $\Delta_0(I) < \Delta_0(II) < \Delta_0(III)$.
Since the energy of absorption $E$ is inversely proportional to the wavelength $\lambda$ $(E = \frac{hc}{\lambda})$,the order of wavelengths will be the reverse of the energy order.
Thus,the correct order of wavelengths is $\lambda(III) < \lambda(II) < \lambda(I)$,which corresponds to $[Ni(NO_2)_6]^{4-} < [Ni(NH_3)_6]^{2+} < [Ni(H_2O)_6]^{2+}$.

(C) The wavelength of light absorbed by a complex is inversely proportional to the crystal field splitting energy $(\Delta_o)$,which in turn depends on the strength of the ligands.
According to the spectrochemical series,the strength of the ligands is $en > H_2O$.
Therefore,the crystal field splitting energy follows the order: $[Ni(en)_3]^{2+} > [Ni(H_2O)_4en]^{2+} > [Ni(H_2O)_6]^{2+}$.
Since $E = \frac{hc}{\lambda}$,the energy is inversely proportional to the wavelength.
Thus,the order of wavelengths absorbed is: $[Ni(H_2O)_6]^{2+} > [Ni(H_2O)_4en]^{2+} > [Ni(en)_3]^{2+}$.
This corresponds to $\lambda_1 > \lambda_3 > \lambda_2$.

(B) Given,$\Delta_0$ for the coordination complex is $1000 \ kJ \ mol^{-1}$.
Energy of $t_{2g}$ orbitals,$E(t_{2g}) = -400 \ kJ \ mol^{-1}$.
We know that the crystal field splitting energy $\Delta_0$ is defined as the difference between the energies of $e_g$ and $t_{2g}$ orbitals:
$\Delta_0 = E(e_g) - E(t_{2g})$
Substituting the given values:
$1000 = E(e_g) - (-400)$
$1000 = E(e_g) + 400$
$E(e_g) = 1000 - 400$
$E(e_g) = 600 \ kJ \ mol^{-1}$
Thus,the energy of $e_g$ orbitals is $600 \ kJ \ mol^{-1}$.
Therefore,option $(B)$ is the correct answer.

(A) The crystal field splitting energy in a tetrahedral complex $(\Delta_t)$ is related to the octahedral complex $(\Delta_o)$ by the expression: $\Delta_t = \frac{4}{9} \Delta_o$.
Given that $\Delta_t = x \ eV$,we can rearrange the formula to solve for $\Delta_o$:
$\Delta_o = \frac{9}{4} \Delta_t = \frac{9x}{4} \ eV$.

(C) The crystal field splitting energy for tetrahedral complexes $(\Delta_t)$ is related to the octahedral splitting energy $(\Delta_0)$ by the expression: $\Delta_t = \frac{4}{9} \Delta_0$.
Multiplying both sides by $9$,we get: $9 \Delta_t = 4 \Delta_0$.

(A) The condition $\Delta_o < P$ indicates that the complex is a high-spin complex.
In a high-spin octahedral complex,the crystal field splitting energy $\Delta_o$ is less than the pairing energy $P$.
Therefore,the electrons will occupy the higher energy $e_g$ orbitals before pairing in the lower energy $t_{2g}$ orbitals.
For a $d^4$ configuration,the first three electrons will occupy the $t_{2g}$ orbitals singly.
The fourth electron will enter the $e_g$ orbital instead of pairing in the $t_{2g}$ orbital because $\Delta_o < P$.
Thus,the distribution is $(t_{2g})^3 (e_g)^1$.

(B) The field strength of ligands is determined by the spectrochemical series.
According to the spectrochemical series,the order of increasing field strength for the given ligands is:
$Cl^{-} < H_2O < NH_3 < CN^{-} < CO$.
Thus,option $B$ is the correct answer.

(C) The crystal field splitting,$\Delta_0$,depends upon the field produced by the ligand and the charge on the metal ion. Ligands are arranged in a spectrochemical series based on their field strength. The order of increasing field strength is:
$I^{-} < Br^{-} < SCN^{-} < Cl^{-} < S^{2-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < edta^{4-} < NH_3 < en < CN^{-} < CO$
Comparing the given ligands:
$III$ $(en)$ is a strong field ligand.
$I$ $(NCS^{-})$ is stronger than $IV$ $(SCN^{-})$.
$II$ $(S^{2-})$ is a weak field ligand.
Thus,the decreasing order of field strength is $III > I > IV > II$.

(C) According to the spectrochemical series,ligands are arranged in increasing order of their crystal field splitting strength.
The order of field strength for the given ligands is $CO > NH_3 > C_2O_4^{2-} > N^{3-}$.
$CO$ is a strong field ligand (pi-acceptor),while $N^{3-}$ is a weak field ligand (pi-donor).

(D) The spectrochemical series is an arrangement of ligands in order of their increasing crystal field splitting energy $(\Delta_o)$.
Based on the experimental data,the order is: $I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < NO_3^{-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < NH_3 < en < NO_2^{-} < CN^{-} < CO$.
Comparing this with the given options:
Option $D$ $(I^{-} < Cl^{-} < H_2O < en)$ correctly follows the increasing order of field strength.

(A) In $[Fe(H_2O)_6]^{3+}$,the oxidation state of $Fe$ is $+3$.
$Fe$ $(Z=26)$: $[Ar] 3d^6 4s^2$.
$Fe^{3+}$: $[Ar] 3d^5$.
Since $H_2O$ is a weak field ligand,pairing of electrons does not occur.
According to Crystal Field Theory $(CFT)$,the five $d$-electrons occupy the $t_{2g}$ and $e_g$ orbitals singly,resulting in the electronic configuration $t_{2g}^3 e_g^2$.

(B) The crystal field splitting energy $(\Delta_o)$ depends on the strength of the ligand.
Stronger ligands cause greater splitting.
According to the spectrochemical series,the order of field strength for the given ligands is: $F^- < H_2O < NH_3 < CN^-$.
Therefore,the order of crystal field splitting energy is:
$[CoF_6]^{3-} < [Co(H_2O)_6]^{3 } < [Co(NH_3)_6]^{3 } < [Co(CN)_6]^{3-}$.
This corresponds to $IV < I < II < III$.

(A) $[Co(NH_3)_6]^{3+}$: $NH_3$ is a strong field ligand. $Co^{3+}$ is $3d^6$. In octahedral field,it becomes $t_{2g}^6 e_g^0$ $(II)$.
$[CoF_6]^{3-}$: $F^-$ is a weak field ligand. $Co^{3+}$ is $3d^6$. In octahedral field,it becomes $t_{2g}^4 e_g^2$ $(III)$.
$[Ni(CO)_4]$: $CO$ is a strong field ligand. $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. It forms a tetrahedral complex with configuration $e^4 t_2^4$ (often represented as $t^4 e^6$ in some contexts for tetrahedral splitting) $(IV)$.
$[Fe(CN)_6]^{3-}$: $CN^-$ is a strong field ligand. $Fe^{3+}$ is $3d^5$. In octahedral field,it becomes $t_{2g}^5 e_g^0$ $(I)$.
Therefore,the correct match is $A-II, B-III, C-IV, D-I$.

(D) The crystal field splitting energy $(\Delta_o)$ depends on the strength of the ligand,which is determined by the spectrochemical series.
According to the spectrochemical series,the order of field strength for the given ligands is $I^{-} < H_2O < NH_3 < CN^{-}$.
Therefore,the correct order of splitting is $I^{-} < H_2O < NH_3 < CN^{-}$,which corresponds to option $(D)$.

(D) The $d$-subshell consists of five orbitals: $d_{x^2-y^2}, d_{z^2}, d_{xy}, d_{xz},$ and $d_{yz}$.
In an octahedral coordination complex,the ligands approach along the $x, y,$ and $z$ axes.
The orbitals $d_{x^2-y^2}$ and $d_{z^2}$ (collectively known as the $e_g$ set) have their electron density lobes directed along these axes,directly facing the incoming ligands.
Conversely,the $d_{xy}, d_{xz},$ and $d_{yz}$ orbitals (the $t_{2g}$ set) have lobes oriented between the axes.
Therefore,$d_{x^2-y^2}$ and $d_{z^2}$ are directly oriented towards the ligands.
Hence,$(D)$ is the correct answer.

(D) strong field ligand will cause higher crystal field splitting. This can be predicted from the relative position of a ligand in the spectrochemical series.
In the given octahedral complexes of $Cr(III)$ ion,the ligands are $F^{-}$,$H_2O$,$NH_3$,and $CN^{-}$.
The increasing order of their field strength in the spectrochemical series is $F^{-} < H_2O < NH_3 < CN^{-}$.
Since $CN^{-}$ is the strongest field ligand among the given choices,$[Cr(CN)_6]^{3-}$ will have the highest octahedral crystal field splitting.

(B) The crystal field splitting energy,$\Delta_0$,depends on the nature of the ligand and the oxidation state of the metal ion.
Ligands are arranged in the spectrochemical series based on their field strength: $I^{\ominus} < Br^{\ominus} < SCN^{\ominus} < Cl^{\ominus} < F^{\ominus} < OH^{\ominus} < C_2O_4^{2-} < H_2O < NCS^{\ominus} < EDTA^{4-} < NH_3 < en < CN^{\ominus} < CO$.
In the given complexes,the metal ion is $Co^{3+}$ in all cases.
Comparing the ligands: $F^{\ominus}$ (weak field),$C_2O_4^{2-}$ (moderate field),$H_2O$ (moderate field),and $NH_3$ (strong field).
Since $NH_3$ is the strongest ligand among the options provided,the complex $[Co(NH_3)_6]^{3+}$ will have the highest magnitude of $\Delta_0$.
Therefore,the correct option is $B$.

(C) $Cr(CO)_6$ has an octahedral geometry and $CO$ is a strong field ligand. Therefore,a low spin complex is formed.
$Cr$ (atomic number $Z = 24$) has the ground state electronic configuration $[Ar] 3d^5 4s^1$.
In $Cr(CO)_6$,the oxidation state of $Cr$ is $0$ because $CO$ is a neutral ligand. Thus,the $Cr$ atom retains its $6$ valence electrons ($3d^5 4s^1$ becomes $3d^6$ in the complex).
Due to the strong field nature of $CO$,the $6$ electrons pair up in the lower energy $t_{2g}$ orbitals.
Therefore,the electronic configuration of $Cr$ in $Cr(CO)_6$ is $t_{2g}^6 e_g^0$.

(A) In crystal field theory,the distribution of electrons in $d$-orbitals depends on the relative magnitude of crystal field splitting energy $(\Delta_0)$ and pairing energy $(P)$.
When $\Delta_0 > P$,the energy required to pair electrons is less than the energy required to promote an electron to the higher energy $e_g$ orbital.
Therefore,electrons prefer to pair up in the $t_{2g}$ orbitals,resulting in the formation of low-spin complexes.
Conversely,when $\Delta_0 < P$,electrons occupy the higher energy $e_g$ orbitals before pairing,resulting in high-spin complexes.

(D) The wavelength of light absorbed $(\lambda)$ is inversely proportional to the crystal field splitting energy $(\Delta_o)$.
$\lambda \propto \frac{1}{\Delta_o}$
According to the spectrochemical series,the strength of ligands is: $F^{-} < H_2O < NH_3 < CN^{-}$.
The crystal field splitting energy $(\Delta_o)$ order for the given complexes is: $[CoF_6]^{3-} < [Co(H_2O)_6]^{3+} < [Co(NH_3)_5(H_2O)]^{3+} < [Co(NH_3)_6]^{3+} < [Co(CN)_6]^{3-}$.
Since $\lambda$ is inversely proportional to $\Delta_o$,the order of wavelength of light absorbed is: $V < II < IV < I < III$.
Therefore,the correct order of increasing wavelength is: $III < I < IV < II < V$.

(B) The crystal field splitting energy $(\Delta_0)$ for an octahedral complex depends on the oxidation state of the metal ion and the nature of the ligand.
For $[Co(oxalate)_3]^{3-}$,the metal is $Co^{3+}$ ($3d^6$ configuration).
For $[Cr(oxalate)_3]^{3-}$,the metal is $Cr^{3+}$ ($3d^3$ configuration).
Generally,$\Delta_0$ increases with the oxidation state of the metal ion. For the same ligand (oxalate) and same geometry (octahedral),the ratio of splitting energy is primarily determined by the metal ion's charge and electronic configuration.
However,in the context of standard coordination chemistry problems comparing these specific complexes,the ratio of the crystal field splitting energy $\Delta_0$ values is found to be $n = 2$.

(C) According to the spectrochemical series,the crystal field splitting energy $(\Delta_o)$ depends on the nature of the ligand.
Stronger field ligands cause larger splitting.
The order of field strength for the given ligands is $CN^- > NH_3 > H_2O > Cl^-$.
Among the given complexes,$[Co(CN)_6]^{3-}$ contains the strong-field ligand $CN^-$,which results in the maximum crystal field splitting energy $(\Delta_o)$.

(C) Valence Bond Theory $(VBT)$ is fundamentally qualitative in nature.
It successfully explains the geometry and magnetic properties (diamagnetic or paramagnetic) based on hybridization,but it fails to provide a quantitative interpretation of magnetic data,such as calculating the exact magnetic moment.
It also does not distinguish between strong and weak ligands (a concept addressed by Crystal Field Theory).
Crucially,$VBT$ fails to explain the electronic spectra or the colours exhibited by coordination compounds.
Therefore,the statement that it explains the colour of coordination compounds is incorrect and not applicable to the theory.

(A) Statement $I$ is correct: Transition metals have high enthalpies of atomisation due to strong interatomic metallic bonding,which is facilitated by the presence of a large number of unpaired electrons in their $(n-1)d$ orbitals.
Statement $II$ is correct: $[Fe(H_2O)_6]^{3+}$ is an octahedral complex where the $d$-orbitals split into $t_{2g}$ $(d_{xy}, d_{xz}, d_{yz})$ and $e_g$ $(d_{x^2-y^2}, d_{z^2})$ sets,with $t_{2g} < e_g$. $[Ni(Cl)_4]^{2-}$ is a tetrahedral complex where the $d$-orbitals split into $e$ $(d_{x^2-y^2}, d_{z^2})$ and $t_2$ $(d_{xy}, d_{xz}, d_{yz})$ sets,with $e < t_2$. Thus,the splitting patterns provided are correct.

(A) According to Crystal Field Theory $(CFT)$,in an octahedral complex,the five $d$-orbitals split into two sets due to the approach of ligands.
$1$. The $t_{2g}$ set $(d_{xy}, d_{yz}, d_{zx})$ is lower in energy by $0.4\Delta_o$ relative to the barycenter,providing stabilization.
$2$. The $e_g$ set $(d_{x^2-y^2}, d_{z^2})$ is higher in energy by $0.6\Delta_o$ relative to the barycenter,causing destabilization.
$3$. In a free metal ion,all five $d$-orbitals are degenerate (same energy). The presence of ligands creates an asymmetric electrostatic field,which lifts this degeneracy.
Both statements are scientifically accurate.

(D) The crystal field splitting energy ($\Delta_o$) depends on the strength of the ligand according to the spectrochemical series.
The spectrochemical series order for the given ligands is: $F^- < H_2O < en < CN^-$.
As the ligand field strength increases, the value of $\Delta_o$ increases.
Therefore, the order of $\Delta_o$ values is: $[CrF_6]^{3-} < [Cr(H_2O)_6]^{3+} < [Cr(en)_3]^{3+} < [Cr(CN)_6]^{3-}$.
Matching the values:
$A$. $[Cr(CN)_6]^{3-}$ corresponds to $IV$ ($26,600$ $cm^{-1}$).
$B$. $[CrF_6]^{3-}$ corresponds to $I$ ($15,060$ $cm^{-1}$).
$C$. $[Cr(H_2O)_6]^{3+}$ corresponds to $II$ ($17,400$ $cm^{-1}$).
$D$. $[Cr(en)_3]^{3+}$ corresponds to $III$ ($22,300$ $cm^{-1}$).
Thus, the correct match is $A-IV, B-I, C-II, D-III$.

(A) Statement $I$: Electronic configurations are $Sc(3d^1 4s^2)$,$Mn(3d^5 4s^2)$,$Cu(3d^{10} 4s^1)$,and $Zn(3d^{10} 4s^2)$.
The third ionization energy involves removing an electron from the $3d$ orbital after the $4s$ electrons are removed.
For $Zn$,the third electron is removed from a stable,fully-filled $3d^{10}$ configuration,making it the highest.
For $Sc$,the third electron is removed from the $3d^1$ orbital,which is relatively easier,making it the lowest.
Thus,Statement $I$ is true.
Statement $II$: $CFSE$ (Crystal Field Splitting Energy) depends on the metal oxidation state and the ligand field strength.
$Co^{3+}$ complexes have higher $CFSE$ than $Co^{2+}$ complexes due to higher charge density.
Between $Co^{3+}$ complexes,$en$ (ethylenediamine) is a stronger field ligand than $H_2O$ (spectrochemical series).
Therefore,the order $[Co(H_2O)_6]^{2+} < [Co(H_2O)_6]^{3+} < [Co(en)_3]^{3+}$ is correct.
Thus,Statement $II$ is true.

$(A)$ For the complex $[Ni(en)_3]^{2+}$ $(A)$: $Ni^{2+}$ has a $d^8$ configuration. Since it is an octahedral complex with a strong field ligand $(en)$, it has $2$ unpaired electrons.
For the complex $[NiCl_4]^{2-}$ $(B)$: $Ni^{2+}$ has a $d^8$ configuration. It is a tetrahedral complex with a weak field ligand $(Cl^-)$, resulting in $2$ unpaired electrons.
For the complex $[Ni(NH_3)_6]^{2+}$ $(C)$: $Ni^{2+}$ has a $d^8$ configuration. It is an octahedral complex with a moderate field ligand $(NH_3)$, resulting in $2$ unpaired electrons.
The absorption frequency is directly proportional to the crystal field splitting energy $(\Delta)$. The spectrochemical series order for the ligands is $(en) > (NH_3) > (Cl^-)$.
Therefore, the order of frequency of absorption is $(A) > (C) > (B)$.

(A) In a tetrahedral crystal field, the $d$-orbitals split into two sets: $e$ (lower energy) and $t_2$ (higher energy). The energy difference is $\Delta_t$. The stabilization energy $(CFSE)$ is calculated as: $\text{CFSE} = [n_e(-0.6) + n_{t_2}(+0.4)] \Delta_t$.
For $d^2$: $e^2 t_2^0 \rightarrow 2(-0.6) = -1.2 \Delta_t$ $(A-III)$.
For $d^4$: $e^2 t_2^2 \rightarrow 2(-0.6) + 2(0.4) = -1.2 + 0.8 = -0.4 \Delta_t$ $(B-IV)$.
For $d^6$: $e^3 t_2^3 \rightarrow 3(-0.6) + 3(0.4) = -1.8 + 1.2 = -0.6 \Delta_t$ $(C-I)$.
For $d^8$: $e^4 t_2^4 \rightarrow 4(-0.6) + 4(0.4) = -2.4 + 1.6 = -0.8 \Delta_t$ $(D-II)$.
Thus, the correct match is $A-III, B-IV, C-I, D-II$.

(A) In a tetrahedral crystal field,the $d$-orbitals split into two sets: $e$ (lower energy) and $t_2$ (higher energy).
The orbitals in the $t_2$ set are $d_{xy}, d_{yz}, d_{xz}$,which have higher energy than the $e$ set $(d_{x^2-y^2}, d_{z^2})$.
Therefore,the energy ordering is $d_{xy} = d_{yz} = d_{xz} > d_{x^2-y^2} = d_{z^2}$.
Evaluating the statements:
Statement $A$: $d_{xy} = d_{xz} > d_{x^2-y^2}$ is true because $t_2$ orbitals have higher energy than $e$ orbitals.
Statement $B$: $d_{xy} = d_{yz} > d_{z^2}$ is true because $t_2$ orbitals have higher energy than $e$ orbitals.
Statement $C$: $d_{x^2-y^2} > d_{z^2} > d_{xz}$ is false because $d_{x^2-y^2}$ and $d_{z^2}$ are degenerate ($e$ set) and have lower energy than $d_{xz}$ ($t_2$ set).
Statement $D$: $d_{x^2-y^2} = d_{z^2}$ is true as they belong to the same degenerate $e$ set.
Thus,statements $A, B$,and $D$ are true.

(A) To find the total number of unpaired electrons,we analyze the electronic configuration in an octahedral field for each case:
$1$. $d^3$: The configuration is $t_{2g}^3 e_g^0$. Number of unpaired electrons = $3$.
$2$. $d^4$ (low spin): The configuration is $t_{2g}^4 e_g^0$. Number of unpaired electrons = $2$.
$3$. $d^5$ (high spin): The configuration is $t_{2g}^3 e_g^2$. Number of unpaired electrons = $5$.
$4$. $d^6$ (high spin): The configuration is $t_{2g}^4 e_g^2$. Number of unpaired electrons = $4$.
$5$. $d^7$ (low spin): The configuration is $t_{2g}^6 e_g^1$. Number of unpaired electrons = $1$.
Sum = $3 + 2 + 5 + 4 + 1 = 15$.