Crystal Field theory Questions in English

(B) The spectrochemical series is an arrangement of ligands in order of their increasing field strength.
According to the spectrochemical series,the field strength order is $I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < N_{3} < F^{-} < OH^{-} < C_{2}O_{4}^{2-} < H_{2}O < NCS^{-} < EDTA^{4-} < NH_{3} < en < NO_{2}^{-} < CN^{-} < CO$.
Comparing the given ligands: $CN^{-} > en > NCS^{-} > Cl^{-}$.
Thus,the correct order is $CN^{-} > en > NCS^{-} > Cl^{-}$.

(A) The colour observed is the complementary colour of the light absorbed by the complex.
$ \Delta_0 $ is calculated from the energy of the absorbed light,not the emitted (observed) light.
Since $(A)$ is violet,it absorbs yellow light,and since $(B)$ is yellow,it absorbs violet light.
Therefore,the statement that $ \Delta_0 $ values are calculated from the energies of violet and yellow light respectively is incorrect.

(D) The crystal field splitting energy $(\Delta)$ depends on the nature of the ligand and the oxidation state of the central metal ion.
Strong field ligands cause larger splitting compared to weak field ligands.
Among the given ligands,$CN^-$ is a strong field ligand (spectrochemical series).
In $K_3[Co(CN)_6]$,the cobalt ion is in the $+3$ oxidation state and is coordinated to six strong $CN^-$ ligands,resulting in the highest crystal field splitting energy.

(A) The crystal field splitting energy $(\Delta_o)$ is inversely proportional to the wavelength of light absorbed $(\Delta_o = \frac{hc}{\lambda})$.
The order of wavelengths for the given colors is $\lambda_{blue} < \lambda_{green} < \lambda_{red}$.
Thus,the energy order is $E_{blue} > E_{green} > E_{red}$.
Since ligand strength is directly proportional to $\Delta_o$,the order of ligand strength is $L_2 > L_1 > L_3$ or $L_3 < L_1 < L_2$.

(B) $K_3[Co(CN)_6]$ is an octahedral coordination complex.
In an octahedral geometry,the ligands approach the central metal ion along the $x, y,$ and $z$ axes.
The $d_{x^2 - y^2}$ and $d_{z^2}$ orbitals (collectively known as $e_g$ orbitals) have their electron density lobes directed along the axes.
Therefore,these orbitals directly face the incoming ligands,leading to greater electrostatic repulsion and higher energy compared to the $t_{2g}$ orbitals $(d_{xy}, d_{xz}, d_{yz})$,which have lobes directed between the axes.

(A) In an octahedral complex like $[Cr(H_2O)_6]^{3+}$,the five $d$-orbitals split into two sets due to the crystal field: the $t_{2g}$ set and the $e_g$ set.
The $t_{2g}$ set consists of three degenerate orbitals: $d_{xy}$,$d_{yz}$,and $d_{xz}$.
The $e_g$ set consists of two degenerate orbitals: $d_{x^2-y^2}$ and $d_{z^2}$.
Among the given options,$d_{xz}$ and $d_{yz}$ belong to the same degenerate set $(t_{2g})$.

(C) $(I)$ Valence Bond Theory $(VBT)$ does not explain the color exhibited by transition metal complexes because the splitting of $d-$orbitals is explained by Crystal Field Theory $(CFT)$.
$(II)$ $VBT$ cannot predict the magnetic properties of transition metal complexes quantitatively; it only provides qualitative information.
$(III)$ $VBT$ does not distinguish between strong field and weak field ligands.
Therefore,statements $I$ and $III$ are correct.

(C) The energy of absorbed light $(E)$ is inversely proportional to the wavelength $(\lambda)$ of the absorbed light,i.e.,$E = \frac{hc}{\lambda}$.
The energy of absorption corresponds to the Crystal Field Splitting Energy ($CFSE$ or $\Delta_o$),which depends on the strength of the ligands.
The spectrochemical series order for the ligands is $Cl^- < H_2O < NH_3$.
Comparing the complexes:
$(I) [CoCl(NH_3)_5]^{2 }$ has $Cl^-$ as a ligand.
$(II) [Co(NH_3)_5H_2O]^{3 }$ has $H_2O$ as a ligand.
$(III) [Co(NH_3)_6]^{3 }$ has $NH_3$ as a ligand.
Since $NH_3$ is a stronger ligand than $H_2O$,and $H_2O$ is stronger than $Cl^-$,the order of $CFSE$ is $(III) > (II) > (I)$.
Since $E \propto \Delta_o$,the order of energy absorbed is $(III) > (II) > (I)$.
Since $\lambda \propto \frac{1}{E}$,the order of wavelength absorbed is $(I) > (II) > (III)$.

(A) $1$. For $[Fe(H_2O)_6]Cl_2$: The central metal ion is $Fe^{2+}$,which has a $d^6$ configuration. $H_2O$ is a weak field ligand,so the configuration is $t_{2g}^4 e_g^2$. The $CFSE$ is calculated as: $CFSE = [-0.4 \times 4 + 0.6 \times 2] \, \Delta_o = [-1.6 + 1.2] \, \Delta_o = -0.4 \, \Delta_o$.
$2$. For $K_2[NiCl_4]$: The central metal ion is $Ni^{2+}$,which has a $d^8$ configuration. It forms a tetrahedral complex $[NiCl_4]^{2-}$. The configuration is $e^4 t_2^4$. The $CFSE$ for a tetrahedral complex is calculated as: $CFSE = [-0.6 \times n_e + 0.4 \times n_{t2}] \, \Delta_t = [-0.6 \times 4 + 0.4 \times 4] \, \Delta_t = [-2.4 + 1.6] \, \Delta_t = -0.8 \, \Delta_t$.
$3$. Therefore,the values are $-0.4 \, \Delta_o$ and $-0.8 \, \Delta_t$.

(D) The Crystal Field Stabilization Energy $(CFSE)$ for an octahedral complex is given by $CFSE = (-0.4n_{t2g} + 0.6n_{eg})\Delta_o$.
For $[Fe(phen)_3]^{2+}$,the central metal is $Fe^{2+}$ ($d^6$ configuration). Since $phen$ is a strong field ligand,it forms a low-spin complex with $t_{2g}^6 e_g^0$ configuration.
$CFSE = (-0.4 \times 6 + 0.6 \times 0)\Delta_o = -2.4\Delta_o$.
Upon oxidation to $Fe^{3+}$ ($d^5$ configuration),the complex becomes $[Fe(phen)_3]^{3+}$ with $t_{2g}^5 e_g^0$ configuration.
$CFSE = (-0.4 \times 5 + 0.6 \times 0)\Delta_o = -2.0\Delta_o$.
However,the question implies a significant loss or change in stability. Let's check $[Co(phen)_3]^{2+}$ $(d^7)$: $t_{2g}^6 e_g^1$ $(CFSE = -1.8\Delta_o)$. Oxidation to $Co^{3+}$ $(d^6)$: $t_{2g}^6 e_g^0$ $(CFSE = -2.4\Delta_o)$.
Actually,for $Fe^{2+}$ $(d^6)$ to $Fe^{3+}$ $(d^5)$,the loss of an electron from the $t_{2g}$ orbital significantly reduces the stabilization energy. Thus,$[Fe(phen)_3]^{2+}$ is the correct answer.

(B) In an octahedral complex,the ligands are present along the $x, y,$ and $z$ axes.
When the two ligands along the $z-$ axis are removed,the complex becomes square planar.
The ligands now exert repulsion only along the $x$ and $y$ axes.
Consequently,the orbitals with components along the $x$ and $y$ axes ($d_{x^2-y^2}$ and $d_{xy}$) will experience higher repulsion and thus have higher energy.
The $d_{z^2}$ orbital,having no ligand interaction along the $z-$ axis,will decrease significantly in energy.
The $d_{xz}$ and $d_{yz}$ orbitals will also decrease in energy but remain higher than $d_{z^2}$ due to their interaction with the $x$ and $y$ ligands.
The correct splitting pattern for a square planar complex is $d_{x^2-y^2} > d_{xy} > d_{z^2} > d_{xz}, d_{yz}$,which corresponds to the pattern shown in image $821-b1769$.

(C) In $[Co(NH_3)_6]Cl_3$,the oxidation state of $Co$ is $+3$.
The electronic configuration of $Co^{3+}$ is $3d^6$.
Since $NH_3$ is a strong field ligand,it causes pairing of electrons in the $d$-orbitals.
Thus,all $6$ electrons occupy the $t_{2g}$ orbitals.
The crystal field stabilisation energy $(CFSE)$ is calculated as:
$CFSE = (n_{t2g} \times -0.4 \, \Delta_0) + (n_{eg} \times +0.6 \, \Delta_0) + mP$
$CFSE = (6 \times -0.4 \, \Delta_0) + (0 \times +0.6 \, \Delta_0) + 3P$
$CFSE = -2.4 \, \Delta_0 + 3P$
Considering only the crystal field splitting energy term,the value is $-2.4 \, \Delta_0$.

(A) In $[Co(CN)_6]^{3-}$,the central metal ion is $Co^{3+}$. The electronic configuration of $Co^{3+}$ is $3d^6$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in a low-spin complex.
All $6$ electrons occupy the $t_{2g}$ orbitals,and $0$ electrons occupy the $e_g$ orbitals.
The formula for $CFSE$ is: $CFSE = (n_{t_{2g}} \times -0.4\Delta_0) + (n_{e_g} \times 0.6\Delta_0) + mP$,where $m$ is the number of extra electron pairs formed.
For $d^6$ in a strong field,$CFSE = (6 \times -0.4\Delta_0) + (0 \times 0.6\Delta_0) = -2.4\Delta_0$.
Including the pairing energy $2p$ (for the two extra pairs formed compared to the free ion),the total energy is $-2.4\Delta_0 + 2p$.

(C) $H_2O$ is generally considered a weak field ligand in the spectrochemical series.
However,its behavior can be influenced by the oxidation state of the central metal ion.
In $[Mn(H_2O)_6]^{+2}$,$Mn$ is in the $+2$ oxidation state,and $H_2O$ acts as a weak field ligand.
In $[Co(H_2O)_6]^{+3}$,$Co$ is in the $+3$ oxidation state,where $H_2O$ acts as a strong field ligand due to the high charge density of the metal ion.
In $[Pt(H_2O)_4]^{+2}$,$Pt$ is a $5d$ series element,and complexes of $5d$ metals are almost always low-spin (strong field behavior).
Therefore,$H_2O$ acts as a weak field ligand specifically in $[Mn(H_2O)_6]^{+2}$.

(C) For a tetrahedral complex,the $d$-orbitals split into two sets: the lower energy $e$ set and the higher energy $t_2$ set.
The energy of the $e$ set is $-0.6 \, \Delta_t$ and the energy of the $t_2$ set is $+0.4 \, \Delta_t$.
For a high-spin $d^4$ configuration,the electrons fill the orbitals according to Hund's rule: $e^2 t_2^2$.
$CFSE = (n_e \times -0.6 \, \Delta_t) + (n_{t_2} \times +0.4 \, \Delta_t)$
$CFSE = (2 \times -0.6 \, \Delta_t) + (2 \times +0.4 \, \Delta_t)$
$CFSE = -1.2 \, \Delta_t + 0.8 \, \Delta_t = -0.4 \, \Delta_t$

(A) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
For a $d^3$ configuration,the three electrons occupy the three $t_{2g}$ orbitals singly according to Hund's rule.
Since the $e_g$ orbitals are higher in energy,these three electrons remain unpaired regardless of whether the ligand is a weak field or a strong field ligand.
For $d^4, d^5,$ and $d^6$ configurations,the number of unpaired electrons changes depending on the crystal field splitting energy (pairing vs. excitation to $e_g$ orbitals).

(D) The crystal field splitting energy $\Delta_0$ depends on the nature of the ligand,the oxidation state of the metal,and the principal quantum number of the metal ion.
For the pair $[Rh(H_2O)_6]^{3+}$ and $[Co(H_2O)_6]^{3+}$,both have the same ligand $(H_2O)$ and the same oxidation state $(+3)$.
However,$Rh$ belongs to the $4d$ series while $Co$ belongs to the $3d$ series.
As we move down a group,the size of the $d$-orbitals increases,leading to greater overlap with ligand orbitals and thus a larger $\Delta_0$ value.
Therefore,$\Delta_0$ for $[Rh(H_2O)_6]^{3+} > \Delta_0$ for $[Co(H_2O)_6]^{3+}$.

(D) In a square planar complex,the ligands approach along the $x$ and $y$ axes.
This leads to a significant increase in the energy of the $d_{x^2-y^2}$ orbital.
The $d_{xy}$ orbital also experiences repulsion,but less than $d_{x^2-y^2}$.
The $d_{z^2}$ orbital is lower in energy than $d_{xy}$ because the ligands are absent along the $z$-axis.
The $d_{yz}$ and $d_{zx}$ orbitals are the lowest in energy and are degenerate.
Thus,the correct order of energy is $d_{x^2-y^2} > d_{xy} > d_{z^2} > d_{yz} = d_{zx}$.

(C) In $[Co(NH_3)_6]Cl_3$,the oxidation state of $Co$ is $+3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Since $NH_3$ is a strong field ligand,it causes pairing of electrons in the $d$-orbitals.
Thus,the $6$ electrons occupy the $t_{2g}$ orbitals,and the $e_g$ orbitals remain empty.
The $CFSE$ is calculated as: $CFSE = (n_{t_{2g}} \times -0.4 \Delta_0) + (n_{e_g} \times 0.6 \Delta_0)$.
$CFSE = (6 \times -0.4 \Delta_0) + (0 \times 0.6 \Delta_0) = -2.4 \Delta_0$.

(A) The magnitude of crystal field splitting energy $(\Delta_o)$ in an octahedral complex depends on several factors:
$1.$ The nature of the ligand: Strong field ligands cause larger splitting than weak field ligands.
$2.$ The charge on the metal ion: Higher oxidation states of the metal ion lead to greater splitting.
$3.$ The transition series: The splitting energy increases as we move down a group (from $3d$ to $4d$ to $5d$ series) because the $d$-orbitals become more extended and interact more strongly with ligands.
Therefore,all three factors $(I, II, III)$ influence the magnitude of the crystal field stabilisation energy.

(C) According to Crystal Field Theory,the degeneracy of $d-$orbitals is lifted when a metal ion is surrounded by ligands.
This splitting occurs regardless of whether the ligand is a strong field ligand,a weak field ligand,a mixed field ligand,or a chelating ligand.
Therefore,in all the given cases $(I, II, III, IV)$,the degeneracy of $d-$orbitals is lost.
The correct option is $C$.

(C) In an octahedral complex,the $d$-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).
For a $d^4$ configuration,if the crystal field splitting energy $\Delta_0$ is less than the pairing energy $P$ (i.e.,$\Delta_0 < P$),the complex is a high-spin complex.
In this case,the fourth electron will occupy the higher energy $e_g$ orbital rather than pairing up in the $t_{2g}$ orbital because the energy required to pair is greater than the energy required to promote the electron to the $e_g$ level.
Thus,the first three electrons occupy the $t_{2g}$ orbitals singly,and the fourth electron occupies one of the $e_g$ orbitals.
The electronic configuration is $t^3_{2g}e^1_g$.

(B) The energy of absorption is given by $E = \frac{hc}{\lambda}$, where $\lambda$ is the wavelength.
Since $NH_3$ is a stronger ligand than $H_2O$ in the spectrochemical series, the crystal field splitting energy $(\Delta_o)$ for $[M(NH_3)_6]^{2+}$ is greater than that for $[M(H_2O)_6]^{2+}$.
Because $E \propto \frac{1}{\lambda}$, a higher splitting energy corresponds to a smaller wavelength of absorption.
Therefore, the absorption wavelength for $[M(NH_3)_6]^{2+}$ must be less than $600 \ nm$.
Among the given options, $580 \ nm$ is the only value smaller than $600 \ nm$ and reasonably close to it, as the difference in splitting power between $NH_3$ and $H_2O$ is not extremely large.

(D) The crystal field splitting energy $\Delta_o$ is directly proportional to the strength of the ligand.
Stronger ligands cause larger splitting,leading to the absorption of light with shorter wavelengths.
The order of ligand field strength is $Br^{-} < H_2O < en$.
Consequently,the order of $\Delta_o$ is $[MBr_6]^{4-} < [M(H_2O)_6]^{2+} < [M(en)_3]^{2+}$.
As the energy of absorbed light increases,the wavelength of the absorbed light decreases,shifting the observed (complementary) color from green to red to blue.
Therefore,$[M(H_2O)_6]^{2+}$ is blue,$[M(en)_3]^{2+}$ is red,and $[MBr_6]^{4-}$ is green.

(D) The relationship between octahedral splitting energy $(\Delta_0)$ and tetrahedral splitting energy $(\Delta_t)$ is given by the formula: $\Delta_t = \frac{4}{9} \Delta_0$.
Given that for the octahedral complex $[CoCl_6]^{4-}$,$\Delta_0 = 18000 \ cm^{-1}$.
Substituting this value into the formula: $\Delta_t = \frac{4}{9} \times 18000 \ cm^{-1}$.
$\Delta_t = 4 \times 2000 \ cm^{-1} = 8000 \ cm^{-1}$.

(B) The $CFSE$ (Crystal Field Splitting Energy) depends on the nature of the ligand and the metal ion.
For the same metal ion,the order of field strength is $en > NH_3 > H_2O$,so option $A$ is correct.
For $Ni^{2+}$ complexes,$[Ni(DMG)_2]$ is a square planar complex with strong field ligands,whereas $[Ni(en)_2]^{2+}$ is tetrahedral/octahedral depending on conditions,but generally,the $CFSE$ of $[Ni(DMG)_2]$ is higher than $[Ni(en)_2]^{2+}$.
Thus,the statement $[Ni(DMG)_2] < [Ni(en)_2]^{2+}$ is incorrect.

(D) The $CFSE$ (Crystal Field Stabilization Energy) for an octahedral complex is calculated as $CFSE = (-0.4n_{t_{2g}} + 0.6n_{e_g}) \Delta_o$.
For $CFSE$ to be zero,the configuration must be $d^0$,$d^5$ (high spin),or $d^{10}$.
Let us analyze the oxidation states and $d$-electron configurations:
$A$: $[Cr(H_2O)_6]^{3+}$ $\rightarrow Cr^{3+} (d^3)$ $\rightarrow CFSE = -1.2 \Delta_o$.
$B$: $[CoF_6]^{3-}$ $\rightarrow Co^{3+} (d^6)$ $\rightarrow \text{High spin } t_{2g}^4 e_g^2$ $\rightarrow CFSE = -0.4 \Delta_o$.
$C$: $[MnF_6]^{2-}$ $\rightarrow Mn^{4+} (d^3)$ $\rightarrow CFSE = -1.2 \Delta_o$.
$D$: $[FeF_6]^{3-}$ $\rightarrow Fe^{3+} (d^5)$ $\rightarrow \text{High spin } t_{2g}^3 e_g^2$ $\rightarrow CFSE = (-0.4 \times 3 + 0.6 \times 2) \Delta_o = (-1.2 + 1.2) \Delta_o = 0$.
Thus,$[FeF_6]^{3-}$ has a $CFSE$ of zero.

(C) The magnitude of crystal field splitting energy $(\Delta_o)$ depends on several factors:
$1$. Nature of the ligand: Strong field ligands (like $CN^-$) cause larger splitting than weak field ligands (like $H_2O$). The spectrochemical series order is $H_2O < NO_2^- < CN^-$.
$2$. Oxidation state of the metal: Higher oxidation states lead to larger $\Delta_o$ values.
$3$. Principal quantum number $(n)$: For the same group,$\Delta_o$ increases down the group $(3d < 4d < 5d)$.
Evaluating the options:
- Option $A$ is incorrect because the order should be $3d < 4d < 5d$,i.e.,$[Co(NH_3)_6]^{3+} < [Rh(NH_3)_6]^{3+} < [Ir(NH_3)_6]^{3+}$.
- Option $B$ is incorrect because octahedral complexes generally have larger $\Delta_o$ than tetrahedral complexes.
- Option $C$ is correct based on the spectrochemical series: $H_2O$ is a weaker ligand than $NO_2^-$,which is weaker than $CN^-$.
- Option $D$ is correct because higher oxidation state of the metal ion $(M^{3+} > M^{2+})$ results in greater $\Delta_o$.

(B) The crystal field splitting energy $(\Delta_o)$ depends on the nature of the ligand,the oxidation state of the metal ion,and the principal quantum number of the metal ($d$-orbitals).
$1$. Ligand strength: $CN^-$ is a strong field ligand,while $H_2O$ and $NH_3$ are weaker.
$2$. Metal identity: For complexes with the same geometry and similar ligands,$\Delta_o$ increases down a group in the periodic table. $Ru$ is in the $4d$ series,while $Fe$ is in the $3d$ series.
$3$. Oxidation state: Higher oxidation states generally lead to larger $\Delta_o$.
Comparing the options,$[Ru(CN)_6]^{3-}$ involves a $4d$ metal $(Ru^{3+})$ and a very strong field ligand $(CN^-)$,which results in a significantly larger $\Delta_o$ compared to the $3d$ complexes $[Fe(H_2O)_6]^{3+}$,$[Fe(H_2O)_6]^{2+}$,and $[Fe(NH_3)_6]^{3+}$.

(B) The spectrochemical series is an arrangement of ligands in order of their increasing crystal field splitting power $(\Delta_o)$.
According to the spectrochemical series, the order for the given ligands is: $I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < S_2O_3^{2-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < EDTA^{4-} < NH_3 < en < CN^{-} < CO$.
Comparing the given ligands $Cl^{-}$, $OH^{-}$, and $CN^{-}$, the order of increasing splitting power is $Cl^{-} < OH^{-} < CN^{-}$.
Therefore, the correct option is $B$.

(C) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
When the crystal field splitting energy $\Delta_0$ is less than the pairing energy $P$ (weak field ligand),the electrons prefer to occupy higher energy orbitals rather than pairing up.
For a $d^4$ configuration,the first three electrons occupy the $t_{2g}$ orbitals singly.
The fourth electron enters the higher energy $e_g$ orbital because $\Delta_0 < P$.
Therefore,the electronic configuration is $t_{2g}^3 e_g^1$.

(A) The crystal field splitting strength depends on the ability of the ligand to donate electron density to the metal center.
Carbon is the least electronegative among $C$,$N$,and $O$,allowing it to donate electron density more effectively,often involving $\pi$-backbonding.
Nitrogen is less electronegative than oxygen,making it a better donor than oxygen.
Therefore,the order of field strength is $C^{-} \text{ donor} > N^{-} \text{ donor} > O^{-} \text{ donor}$.

(A) In a tetrahedral complex,the $d$-orbitals split into $e$ and $t_2$ sets,where the $e$ set is lower in energy and the $t_2$ set is higher in energy.
For a $d^4$ configuration,the electrons fill the orbitals according to the energy levels.
Since tetrahedral complexes are generally high-spin (because $\Delta_t < P$),the electrons fill the orbitals as $e^2 t_2^2$.
The $CFSE$ formula for tetrahedral complexes is $CFSE = [(-0.6 \times n_e) + (0.4 \times n_{t_2})] \Delta_t$.
Substituting $n_e = 2$ and $n_{t_2} = 2$:
$CFSE = [(-0.6 \times 2) + (0.4 \times 2)] \Delta_t = [-1.2 + 0.8] \Delta_t = -0.4 \Delta_t$.
$-0.4 \Delta_t$ is equivalent to $-\frac{2}{5} \Delta_t$.

(A) The colour of coordination compounds arises due to $d-d$ transitions.
The energy gap $(\Delta_o)$ between $t_{2g}$ and $e_g$ orbitals depends on the strength of the ligand.
$CN^{-}$ is a strong field ligand,which causes a larger splitting $(\Delta_o)$,while $H_2O$ is a weak field ligand,causing smaller splitting.
Since the energy gap is different,they absorb different wavelengths of light from the visible spectrum,resulting in different observed colours.

(D) The crystal field splitting energy $(\Delta)$ depends on the oxidation state of the central metal ion and the nature of the ligand.
$1$. Higher oxidation state of the central metal ion leads to greater splitting energy.
$2$. Strong field ligands (like $NH_3$) cause greater splitting compared to weak field ligands (like $F^-$,$Cl^-$,$H_2O$).
$3$. In $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$ and $NH_3$ is a strong field ligand.
$4$. In the other complexes,the oxidation state of $Co$ is $+2$ and the ligands are weaker.
Therefore,$[Co(NH_3)_6]^{3+}$ has the highest crystal field stabilisation energy.

(A) According to the spectrochemical series,the order of field strength for the given ligands is $I^{-} < Cl^{-} < F^{-} < H_2O$.
Crystal field splitting energy $(\Delta)$ is directly proportional to the field strength of the ligand.
Since $I^{-}$ is the weakest ligand among the given options,it will cause the smallest crystal field splitting.

(B) The magnitude of crystal field splitting energy $(\Delta_o)$ depends on the nature of the ligand and the oxidation state of the metal ion.
$1$. Strong field ligands like $CN^{-}$ cause larger splitting compared to weak field ligands like $H_2O$ or $NH_3$.
$2$. For the same ligand,$\Delta_o$ increases as we move down a group in the periodic table because the $4d$ and $5d$ orbitals are more extended than $3d$ orbitals.
$3$. Comparing $[Fe(CN)_6]^{3-}$ ($3d$ metal) and $[Ru(CN)_6]^{3-}$ ($4d$ metal),the complex with the $4d$ metal $(Ru)$ will have a significantly higher $\Delta_o$ value.
Therefore,$[Ru(CN)_6]^{3-}$ has the highest crystal field splitting energy.

(A) According to the spectrochemical series,the strength of ligands is determined by their ability to cause splitting in the $d$-orbitals of the central metal ion.
The order of field strength for the given ligands is $NH_3 < en < NO_2^- < CN^-$.
Among the given options,$CN^-$ is the strongest field ligand as it is a strong $\pi$-acceptor ligand.

(D) According to the spectrochemical series,ligands are arranged in increasing order of their crystal field splitting energy $(CFSE)$.
$CO$,$CN^-$,and $NO_2^-$ are all strong field ligands that cause large crystal field splitting.
Among the given options,$CO$ is a very strong $\pi$-acid ligand,but $CN^-$ and $NO_2^-$ also exhibit significant splitting.
In the context of standard chemistry problems,all these ligands are considered strong field ligands that produce high crystal field splitting compared to halides or water.
Therefore,the correct answer is $D$.

(B) In crystal field theory,the relationship between tetrahedral splitting energy $(\Delta_t)$ and octahedral splitting energy $(\Delta_o)$ is given by the expression: $\Delta_t = \frac{4}{9} \Delta_o$.
Therefore,the tetrahedral crystal field splitting is $4/9$ of the octahedral crystal field splitting.

(A) Halide ligands ($F^-$,$Cl^-$,$Br^-$,$I^-$) are weak field ligands according to the spectrochemical series.
Weak field ligands cause a small crystal field splitting energy $(\Delta_o < P)$,where $P$ is the pairing energy.
Because the splitting energy is small,electrons prefer to occupy higher energy orbitals rather than pairing up in lower energy orbitals.
Therefore,complexes containing halide ligands are generally high spin complexes.

(B) $CN^{-}$ is a strong field ligand according to the spectrochemical series.
Strong field ligands cause a large splitting of $d$-orbitals ($\Delta_o > P$,where $P$ is the pairing energy).
Due to this large splitting,electrons prefer to pair up in the lower energy $t_{2g}$ orbitals rather than occupying the higher energy $e_g$ orbitals.
Therefore,complexes containing $CN^{-}$ ligands are generally low spin complexes.

(C) In the complex $[Ti(H_2O)_6]^{3+}$,the central metal ion is $Ti^{3+}$.
The electronic configuration of $Ti^{3+}$ is $3d^1$.
When white light falls on the complex,the single electron in the $t_{2g}$ orbital absorbs a photon of light and gets excited to the higher energy $e_g$ orbital.
This process is known as $d \to d$ transition.
The energy absorbed corresponds to the yellow-green region,and the complex appears purple due to the transmission of the complementary color.

(C) The crystal field splitting energy $\Delta_0$ depends on the oxidation state of the metal ion,the nature of the ligand,and the principal quantum number of the metal ($3d$ vs $4d$ vs $5d$).
$(a)$ $[Cr(H_2O)_6]^{2+}$ vs $[Cr(H_2O)_6]^{3+}$: Higher oxidation state $(+3 > +2)$ leads to larger $\Delta_0$. Thus,$II > I$.
$(b)$ $[Fe(H_2O)_6]^{3+}$ vs $[Fe(CN)_6]^{3-}$: $CN^-$ is a stronger field ligand than $H_2O$,leading to larger $\Delta_0$. Thus,$II > I$.
$(c)$ $[Fe(CN)_6]^{3-}$ vs $[Ru(CN)_6]^{3-}$: $Ru$ is a $4d$ metal while $Fe$ is a $3d$ metal. $\Delta_0$ increases down the group. Thus,$II > I$.
$(d)$ $[NiF_6]^{4-}$ vs $[NiF_6]^{2-}$: Higher oxidation state $(+4 > +2)$ leads to larger $\Delta_0$. Thus,$II > I$.
All pairs satisfy the condition $\Delta_0(II) > \Delta_0(I)$.

(C) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
When $\Delta_{0} < P$ (where $\Delta_{0}$ is the crystal field splitting energy and $P$ is the pairing energy),the complex is a high-spin complex.
For a $d^4$ configuration,the first three electrons occupy the lower energy $t_{2g}$ orbitals singly according to Hund's rule.
The fourth electron will occupy the higher energy $e_{g}$ orbital because the energy required to pair it in the $t_{2g}$ orbital is greater than the energy required to promote it to the $e_{g}$ orbital.
Therefore,the electronic arrangement is $t_{2g}^3 e_g^1$.

(C) In potassium ferrocyanide,$K_4[Fe(CN)_6]$,$Fe$ is in the $+2$ oxidation state ($Fe^{2+}$: $3d^6$). Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in $t_{2g}^6 e_g^0$ configuration with $n=0$ unpaired electrons,making it diamagnetic.
In potassium ferricyanide,$K_3[Fe(CN)_6]$,$Fe$ is in the $+3$ oxidation state ($Fe^{3+}$: $3d^5$). $CN^-$ causes pairing,resulting in $t_{2g}^5 e_g^0$ configuration with $n=1$ unpaired electron,making it paramagnetic.
The Assertion is correct.
The Reason states that crystal field splitting in ferrocyanide is greater than in ferricyanide. This is incorrect because crystal field splitting energy $(\Delta_o)$ increases with the oxidation state of the metal ion. Since $Fe^{3+}$ has a higher oxidation state than $Fe^{2+}$,the crystal field splitting in the ferricyanide ion is greater than in the ferrocyanide ion. Thus,the Reason is incorrect.

(B) In the complex $K_{4}[Fe(CN)_{6}]$,the oxidation state of $Fe$ is $+2$.
The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^{6} 4s^{2}$.
For $Fe^{2+}$,the configuration is $3d^{6}$.
$CN^{-}$ is a strong field ligand $(SFL)$,which causes pairing of electrons in the $d$-orbitals.
According to crystal field theory for an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_{g}$ sets.
Due to the strong field,all $6$ electrons pair up in the lower energy $t_{2g}$ orbitals,resulting in the configuration $t_{2g}^{6} e_{g}^{0}$.

(D) The relationship between the crystal field splitting energy of tetrahedral complexes $(\Delta_{t})$ and octahedral complexes $(\Delta_{o})$ is given by the formula: $\Delta_{t} = \frac{4}{9} \times \Delta_{o}$.
Given that $\Delta_{o}$ for $[CoCl_{6}]^{4-}$ is $18000 \; cm^{-1}$.
Therefore,$\Delta_{t}$ for $[CoCl_{4}]^{2-}$ is calculated as: $\Delta_{t} = \frac{4}{9} \times 18000 \; cm^{-1} = 8000 \; cm^{-1}$.

(B) The complex $[Pd(F)(Cl)(Br)(I)]^{2-}$ is a square planar complex of the type $[M(abcd)]$. It has $3$ geometrical isomers. Thus,$n = 3$.
The formula for the second complex becomes $[Fe(CN)_6]^{3- - 6} = [Fe(CN)_6]^{3-}$.
In $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $3d^5$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals. The configuration becomes $t_{2g}^5 e_g^0$.
Number of unpaired electrons $(n_{u}) = 1$.
Spin-only magnetic moment $= \sqrt{n_{u}(n_{u}+2)} \ BM = \sqrt{1(1+2)} \ BM = \sqrt{3} \ BM \approx 1.73 \ BM$.
$CFSE = [(-0.4 \times n_{t2g}) + (0.6 \times n_{eg})] \ \Delta_{0} = [(-0.4 \times 5) + (0.6 \times 0)] \ \Delta_{0} = -2.0 \ \Delta_{0}$.

(N/A) In both $[Fe(H_2O)_6]^{3+}$ and $[Fe(CN)_6]^{3-}$,$Fe$ exists in the $+3$ oxidation state,i.e.,in $d^5$ configuration.
Since $CN^{-}$ is a strong field ligand,it causes the pairing of unpaired electrons. Therefore,there is only one unpaired electron left in the $d$-orbital.
Therefore,
$\mu = \sqrt{n(n+2)}$
$= \sqrt{1(1+2)}$
$= \sqrt{3}$
$= 1.732 \, BM$
On the other hand,$H_2O$ is a weak field ligand. Therefore,it cannot cause the pairing of electrons. This means that the number of unpaired electrons is $5$.
Therefore,
$\mu = \sqrt{n(n+2)}$
$= \sqrt{5(5+2)}$
$= \sqrt{35}$
$\simeq 5.92 \, BM$
Thus,it is evident that $[Fe(H_2O)_6]^{3+}$ is strongly paramagnetic (with $5$ unpaired electrons),while $[Fe(CN)_6]^{3-}$ is weakly paramagnetic (with $1$ unpaired electron).