Hybridisation and Geometry Questions in English

(C) $1$. For $[Co(CO)_4]^-$,the oxidation state of $Co$ is $-1$. The electronic configuration is $3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons. The geometry is $sp^3$ (tetrahedral) and it is diamagnetic due to the pairing of all electrons.
$2$. For $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration is $3d^8 4s^2$. $CO$ causes pairing of electrons,resulting in $sp^3$ hybridization (tetrahedral) and it is diamagnetic.
$3$. Other options like $[NiCl_4]^{2-}$ are tetrahedral but paramagnetic,and $[Ni(CN)_4]^{2-}$ is square planar and diamagnetic.

(C) $[Ni(CN)_4]^{2-}$ is square-planar,diamagnetic ($0$ unpaired electrons) with $dsp^2$ hybridisation.
$[Ni(CO)_4]$ is tetrahedral,diamagnetic ($0$ unpaired electrons) with $sp^3$ hybridisation.
$[NiCl_4]^{2-}$ is tetrahedral,paramagnetic ($2$ unpaired electrons) with $sp^3$ hybridisation.
Hence,the option $(c)$ is the correct answer.

(C) The chemical formula of Wilkinson's catalyst is $[RhCl(PPh_3)_3].$
In this complex,the central metal ion is Rhodium $(Rh^+)$,which has a $d^8$ electronic configuration.
For a $d^8$ metal ion in a four-coordinate complex with strong field ligands like $PPh_3,$ the hybridization is $dsp^2.$
Consequently,the geometry of the complex is square planar.

(C) The atomic number of $Co$ is $27$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Since the complex is octahedral and diamagnetic,the ligands must be strong field ligands that cause pairing of electrons in the $3d$ orbitals.
This results in two empty $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals,which hybridize to form six $d^2sp^3$ hybrid orbitals.
Thus,the hybridisation involved is $d^2sp^3$.

(B) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.
In the complex $[NiL_4]^{2-}$,let the oxidation state of $Ni$ be $x$. Since $L$ is a uninegative ligand,its charge is $-1$.
$x + 4(-1) = -2 \implies x = +2$.
So,$Ni^{2+}$ has the configuration $[Ar] 3d^8$.
For a complex to be diamagnetic,all electrons must be paired.
In a $d^8$ system,if the complex is diamagnetic,the ligand must be a strong field ligand that causes pairing of electrons in the $3d$ orbitals,resulting in $dsp^2$ hybridisation.
With $dsp^2$ hybridisation,the $3d$ orbitals are arranged such that there are no unpaired electrons.
Therefore,the hybridisation is $dsp^2$ and the number of unpaired electrons is $0$.

(A) The complex $[PdCl_4]^{2-}$ involves $Pd^{2+}$ ion which has a $4d^8$ configuration.
Since $Pd$ is a $4d$ series element,the crystal field splitting energy is high,forcing the pairing of electrons.
This results in $dsp^2$ hybridization,leading to a square planar geometry.

(A) The complex $[PtCl_2(NH_3)_2]$ involves a $Pt^{2+}$ central metal ion with a $d^8$ electronic configuration.
It undergoes $dsp^2$ hybridization,which results in a square-planar geometry.
In contrast,$[NiCl_4]^{2-}$ has a tetrahedral geometry due to $sp^3$ hybridization,while $MnO_4^-$ and $CrO_4^{2-}$ exhibit tetrahedral geometry.

(A) The oxidation state of $Co$ in all given complexes is $+3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$1$. In $[Co(CN)_6]^{3-}$,$CN^-$ is a strong field ligand,causing pairing of electrons. It forms a low spin $d^2sp^3$ complex which is diamagnetic.
$2$. In $[Co(C_2O_4)_3]^{3-}$,$C_2O_4^{2-}$ is a strong field ligand,forming a low spin $d^2sp^3$ complex which is diamagnetic.
$3$. In $[Co(H_2O)_6]^{3+}$,$H_2O$ is a weak field ligand. It does not cause pairing of electrons. The configuration remains $t_{2g}^4 e_g^2$,resulting in $4$ unpaired electrons. Thus,it is a high spin paramagnetic complex.

(A) The complex $[Fe(CO)_5]$ has a trigonal bipyramidal geometry.
The hybridization involved in this complex is $dsp^3$.
In the $dsp^3$ hybridization for a trigonal bipyramidal geometry,the $d_{z^2}$ orbital of the central metal atom is involved.

(C) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry.
$2$. $[Rh(CO)_2(PPh_3)_2]^+$: $Rh^+$ has $d^8$ configuration. Being a $4d$ series metal,it forms square planar complexes with strong field ligands like $CO$ and $PPh_3$. Thus,it is square planar.
$3$. $[NiCl_4]^{2-}$ is tetrahedral $(sp^3)$.
$4$. $[Ni(CO)_4]$ is tetrahedral $(sp^3)$.
$5$. $[PtCl_4]^{2-}$ is square planar,but $[Ni(CN)_4]^{3-}$ is not a common stable species (usually $Ni(I)$ or $Ni(II)$ complexes are considered).
Therefore,the pair $[Ni(CN)_4]^{2-}$ and $[Rh(CO)_2(PPh_3)_2]^+$ consists of two square planar complexes.

(C) In the complex $K_3[Co(CN)_6]$,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.
Primary valency corresponds to the oxidation state,which is $3$,not $6$.
$CN^-$ is a strong field ligand,so it forms a low spin complex with $d^2sp^3$ hybridisation.
Since all electrons are paired,the magnetic moment is $0$.

(A) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In $[Co(NH_3)_6]^{+2}$,the oxidation state of $Co$ is $+2$. Thus,the configuration of $Co^{+2}$ is $[Ar] 3d^7$.
In the presence of $NH_3$ (a strong field ligand),the electrons in the $3d$ orbitals rearrange.
For $Co^{+2}$ $(3d^7)$,the distribution is $t_{2g}^5 e_g^2$.
This results in $3$ unpaired electrons in the $3d$ orbitals,making the complex paramagnetic.

(C) $1$. $[Ni(CN)_6]^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,forcing pairing. It forms $d^2sp^3$ hybridization (inner orbital complex) and is diamagnetic (magnetic moment = $0$).
$2$. $[Mn(NH_3)_6]^{2+}$: $Mn^{2+}$ is $d^5$. $NH_3$ is a strong field ligand,forcing pairing. It forms $d^2sp^3$ hybridization and has one unpaired electron (magnetic moment $\neq 0$).
$3$. $[Co(H_2O)_6]^{3+}$: $Co^{3+}$ is $d^6$. $H_2O$ is a weak field ligand. It forms $sp^3d^2$ hybridization (outer orbital complex) and has four unpaired electrons (magnetic moment $\neq 0$).
Therefore,$[Co(H_2O)_6]^{3+}$ satisfies both conditions.

(B) The dichromate ion $(Cr_2O_7^{2-})$ consists of two $CrO_4$ tetrahedra sharing one oxygen atom at the corner.
In this structure,there are two types of $Cr-O$ bonds:
$1$. The bridging $Cr-O-Cr$ bond,which involves two $Cr-O$ bonds that are not equivalent to the terminal ones.
$2$. The terminal $Cr-O$ bonds,which are $6$ in total.
Due to resonance,all $6$ terminal $Cr-O$ bonds are identical in length and strength.
Therefore,the number of identical $Cr-O$ bonds is $6$. Hence,option $B$ is the correct answer.

(B) The $MnO_4^-$ ion has a tetrahedral geometry.
In $MnO_4^-$,the central $Mn$ atom is in the $+7$ oxidation state $(3d^0 4s^0)$.
The bonding involves the overlap of $Mn$ orbitals with oxygen $2p$ orbitals.
Specifically,the $Mn$ atom uses its $3d$ orbitals $(3d_{xy}, 3d_{yz}, 3d_{zx})$ and $4s$ orbital to form four equivalent bonds with oxygen atoms,which corresponds to $d^3s$ hybridisation.

(C) $Pd(II)$ is a $4d$ transition metal, and due to the high crystal field splitting energy $(\Delta_o)$, it forms square planar complexes with strong field ligands, resulting in $0$ unpaired electrons.
$Ni(II)$ is a $3d$ transition metal with a $d^8$ configuration.
In the complex $[NiCl_2(PMe_3)_2]$, $Cl^-$ is a weak field ligand.
For $Ni(II)$ $(d^8)$ with weak field ligands, the complex typically adopts a tetrahedral geometry ($sp^3$ hybridization).
In a tetrahedral field, the $d$-orbitals split as $e < t_2$.
The $8$ electrons fill as $e^4 t_2^4$, resulting in $2$ unpaired electrons.

(D) In $[Co(NH_3)_5CO_3]ClO_4$,five $NH_3$ ligands (monodentate) and one $CO_3^{2-}$ ligand (bidentate) are attached to the central cobalt atom. However,the carbonate ion $CO_3^{2-}$ acts as a bidentate ligand,but in this specific complex,it is often considered to occupy two coordination sites. Wait,$NH_3$ is $5$ and $CO_3$ is $1$ (bidentate),so $5 + 2 = 7$ is incorrect; the coordination number is $6$ because $CO_3^{2-}$ is bidentate,but the complex is $[Co(NH_3)_5(CO_3)]^+$. Actually,$NH_3$ is $5$ and $CO_3$ is $1$ (bidentate),so $5 + 2 = 7$. Let us re-evaluate: $Co$ is bonded to $5$ $NH_3$ and $1$ $CO_3^{2-}$. $5 + 2 = 7$. Wait,the coordination number is $6$ if $CO_3$ is monodentate or if the formula implies $5+1=6$. Given the options,$6$ is the coordination number.
Let $x$ be the oxidation state of $Co$. $x + 5(0) + (-2) + (-1) = 0$,so $x = +3$.
The atomic number of $Co$ is $27$. Configuration: $[Ar] 3d^7 4s^2$. For $Co^{3+}$,it is $3d^6$.
Number of $d-$ electrons is $6$.
Since $NH_3$ is a strong field ligand,it causes pairing. For $d^6$ in an octahedral field with strong ligands,all electrons are paired. Thus,unpaired electrons = $0$.

(D) In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $+3$ ($3d^6$ configuration). $F^-$ is a weak field ligand,so it forms an outer orbital complex ($sp^3d^2$ hybridization) which is paramagnetic with $4$ unpaired electrons.
In $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$ ($3d^6$ configuration). $NH_3$ is a strong field ligand,so it causes pairing of electrons,forming an inner orbital complex ($d^2sp^3$ hybridization) which is diamagnetic.
Therefore,statement $D$ is correct: $[CoF_6]^{3-}$ is an outer orbital complex while $[Co(NH_3)_6]^{3+}$ is an inner orbital complex.

(C) To determine the number of unpaired electrons,we analyze the electronic configuration of the central metal ion in each complex:
$1$. In $[Co(CN)_6]^{4-}$,$Co$ is in $+2$ oxidation state $(d^7)$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^6 e_g^1$,resulting in $1$ unpaired electron.
$2$. In $[Cr(H_2O)_6]^{3+}$,$Cr$ is in $+3$ oxidation state $(d^3)$. Configuration: $t_{2g}^3 e_g^0$,resulting in $3$ unpaired electrons.
$3$. In $[FeCl_4]^{2-}$,$Fe$ is in $+2$ oxidation state $(d^6)$. $Cl^-$ is a weak field ligand. Configuration: $e^3 t_2^3$,resulting in $4$ unpaired electrons.
$4$. In $[Fe(H_2O)_6]^{3+}$,$Fe$ is in $+3$ oxidation state $(d^5)$. $H_2O$ is a weak field ligand. Configuration: $t_{2g}^3 e_g^2$,resulting in $5$ unpaired electrons.
Thus,the species with $4$ unpaired electrons is $[FeCl_4]^{2-}$.

(B) In the complex $K[PtCl_3(C_2H_4)]$,the central metal atom is $Pt^{2+}$.
It is a $4$-coordinate complex with four ligands attached to the central metal atom,which results in a square planar geometry.
For a square planar geometry in $d^8$ metal ions like $Pt^{2+}$,the hybridization involved is $dsp^2$.

(A) The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
In the presence of dimethyl glyoxime (dmg),which acts as a strong field ligand,the electrons in the $3d$ orbitals pair up.
This results in a $dsp^2$ hybridization,which corresponds to a square planar geometry.
As shown in the orbital diagram,all electrons are paired in the complex $[Ni(dmg)_2]$.
Since there are no unpaired electrons,the magnetic moment $\mu = 0$,making the complex diamagnetic.

(C) In the brown ring complex $[Fe(H_2O)_5NO]SO_4$,the iron is in the $+1$ oxidation state $(Fe^+)$.
The electronic configuration of $Fe^+$ is $[Ar] 3d^6 4s^1$.
In this complex,$NO$ acts as $NO^+$,and $H_2O$ is a weak field ligand. The $Fe^+$ ion undergoes $sp^3d^2$ hybridisation.
The $3d$ electrons are arranged as $t_{2g}^5 e_g^0$ (or similar distribution),resulting in $3$ unpaired electrons.
Magnetic moment $\mu = \sqrt{n(n+2)} \ BM = \sqrt{3(3+2)} = \sqrt{15} \ BM$.
Thus,the magnetic moment is $\sqrt{15} \ BM$ and the hybridisation is $sp^3d^2$.

(A) An outer orbital complex is one that uses the outer $d$-orbitals (e.g.,$4d$) for hybridization,typically resulting in $sp^3d^2$ hybridization.
$1$. $[Fe(CN)_6]^{3-}$ involves $d^2sp^3$ hybridization (inner orbital).
$2$. $[Co(NH_3)_6]^{3+}$ involves $d^2sp^3$ hybridization (inner orbital).
$3$. $[Fe(CN)_6]^{4-}$ involves $d^2sp^3$ hybridization (inner orbital).
$4$. $[Ni(NH_3)_6]^{2+}$ involves $sp^3d^2$ hybridization,which uses the $4d$ orbitals,making it an outer orbital complex.

(A) The magnetic moment is given as $5.9 \, BM$.
Using the spin-only formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons,we find $n = 5$.
In $[MnX_4]^{2-}$,the oxidation state of $Mn$ is $+2$.
The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.
Since there are $5$ unpaired electrons,the $d$-orbitals are $d^5$ (high spin).
For a $d^5$ configuration with $4$ ligands,the geometry is tetrahedral because the halide ligands are weak field ligands,leading to $sp^3$ hybridization.

(C) For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,leading to $sp^3$ hybridization. Thus,the geometry is tetrahedral.
For $[NiCl_2(PPh_3)_2]$: The oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. Both $Cl^-$ and $PPh_3$ are generally considered weak field ligands in this specific complex geometry,leading to $sp^3$ hybridization. Thus,the geometry is tetrahedral.
Therefore,both complexes are tetrahedral.

(A) For a coordination number of $5$,the central metal atom undergoes $dsp^3$ hybridization.
If the $d_{z^2}$ orbital is involved in the hybridization,the geometry is trigonal bipyramidal.
If the $d_{x^2-y^2}$ orbital is involved in the hybridization,the geometry is square pyramidal.
Therefore,the correct option is $A$.

(D) $1$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $d^2sp^3$ hybridization.
$2$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $d^2sp^3$ hybridization.
$3$. In $[Co(NO_2)_6]^{3-}$,$Co$ is in $+3$ oxidation state $(3d^6)$. $NO_2^-$ is a strong field ligand,causing pairing of electrons,resulting in $d^2sp^3$ hybridization.
$4$. Therefore,all three complexes exhibit $d^2sp^3$ hybridization.

(D) In $[Fe(CN)_6]^{4-}$,the oxidation state of $Fe$ is $+2$. The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
This results in $d^2sp^3$ hybridization,which corresponds to an octahedral geometry.
There are $6$ $Fe-C$ sigma bonds formed by the donation of electron pairs from $6$ $CN^-$ ligands.
The total number of electrons on $Fe^{2+}$ is $26 - 2 = 24$.
Thus,the statement regarding $6$ sigma bonds is correct.

(C) The complex $[Pt(NH_3)_2Cl_2]$ exhibits square planar geometry.
For a square planar complex with bond length $d = 2 \ \mathring{A}$,the distance between adjacent $Cl$ atoms (cis-position) is $d\sqrt{2} = 2 \times 1.414 = 2.828 \ \mathring{A} \approx 2.88 \ \mathring{A}$.
The distance between opposite $Cl$ atoms (trans-position) would be $2d = 2 \times 2 = 4 \ \mathring{A}$.
Since the given $Cl-Cl$ distance is $2.88 \ \mathring{A}$,which corresponds to the cis-configuration,the complex is cis-square planar.
Hence,the correct option is $C$.

(C) For a complex with the composition $[MA_3B]$,we analyze the possibility of geometrical isomerism in different geometries:
$1$. Tetrahedral complexes: In a tetrahedral geometry,all four positions are equivalent relative to each other. Therefore,any substitution of ligands does not lead to different geometrical arrangements. Thus,$[MA_3B]$ in a tetrahedral geometry does not show geometrical isomerism.
$2$. Square planar complexes: In a square planar geometry,for a complex of the type $[MA_3B]$,all positions are equivalent. Any arrangement of the three $A$ ligands and one $B$ ligand results in the same structure. Thus,$[MA_3B]$ in a square planar geometry also does not show geometrical isomerism.
Since both geometries do not exhibit geometrical isomerism for this specific composition,the correct answer is $(c)$.

(C) The magnetic moment of $[NiX_4]^{2-}$ is zero,which implies that all electrons in the $Ni^{2+}$ ion are paired.
$Ni^{2+}$ has a $3d^8$ electronic configuration.
For the electrons to be paired in a $4$-coordinate complex,the $Ni^{2+}$ ion must undergo $dsp^2$ hybridization,resulting in a square planar geometry.
This is characteristic of strong field ligands that force pairing of electrons.

(B) In the complex $[Ag(NH_3)_2]^{+}$,the central metal ion is $Ag^{+}$.
The electronic configuration of $Ag^{+}$ is $[Kr] 4d^{10}$.
Since the coordination number is $2$,the silver ion undergoes $sp$ hybridization to accommodate the two lone pairs donated by the two $NH_3$ ligands.
Thus,the hybridization is $sp$,resulting in a linear geometry.
Hence,the correct option is $B$.

(C) The complex $[Mn(CO)_4 NO]$ is diamagnetic because $NO$ acts as a $NO^+$ ligand (nitrosonium ion).
In this complex,the oxidation state of $Mn$ is $-1$.
The electronic configuration of $Mn^-$ is $3d^6 4s^2$,which becomes $3d^8$ after hybridization with $CO$ and $NO^+$ ligands.
Since all electrons are paired in the $d$-orbitals,the complex is diamagnetic.

(D) $(I)$ False: $Co(III)$ $(d^6)$ is stabilized by strong field ligands (low spin),and $Co(II)$ $(d^7)$ is also stabilized by strong field ligands.
$(II)$ True: $Pd^{2+}$ and $Pt^{2+}$ are $4d^8$ and $5d^8$ systems respectively. Due to high crystal field splitting energy,they always form square planar,diamagnetic complexes.
$(III)$ False: $[Ni(CO)_4]$ is tetrahedral $(sp^3)$ and diamagnetic. $[Ni(CN)_4]^{4-}$ is not a standard stable complex; $[Ni(CN)_4]^{2-}$ is square planar $(dsp^2)$ and diamagnetic.
$(IV)$ True: $Ni^{2+}$ $(d^8)$ in an octahedral field has configuration $t_{2g}^6 e_g^2$. It cannot form inner orbital complexes because it requires two empty $d$-orbitals,but only one is available in the $e_g$ set.

(B) In the permanganate ion $(MnO_4^-)$,the manganese atom is in the $+7$ oxidation state $(Mn^{VII})$,meaning it has a $3d^0$ configuration.
In a tetrahedral field,the $d-$ orbitals split into $e$ and $t_2$ sets.
The $t_2$ set consists of the $d_{xy}, d_{yz},$ and $d_{xz}$ orbitals.
These $t_2$ orbitals are involved in the $d^3s$ hybridization along with the $4s$ orbital to form the tetrahedral structure.
Therefore,the set of $d-$ orbitals used is $d_{xy}, d_{yz}, d_{xz}$.

(D) The reaction between $FeSO_4$ and $NO$ forms the brown ring complex: $[Fe(H_2O)_5(NO)]SO_4$.
In this complex,$Fe$ is in the $+1$ oxidation state ($d^7$ configuration).
$NO$ acts as a $NO^+$ ligand,making the metal center $Fe^+$,which has a $d^7$ configuration.
Due to the strong field nature of $NO^+$,the electrons pair up in the $t_{2g}$ orbitals.
The configuration is $t_{2g}^6 e_g^1$,resulting in $1$ unpaired electron in the $e_g$ orbital.
However,in the context of the brown ring test,the complex is often represented as $[Fe(H_2O)_5(NO)]^{2+}$ where $Fe$ is $+1$ and $NO$ is neutral,or $Fe$ is $+2$ and $NO$ is $NO^-$.
Based on the standard experimental magnetic moment of $3.89 \ BM$,the number of unpaired electrons is $3$.

(B) The magnetic moment $\mu = \sqrt{n(n+2)} \ B.M.$ where $n$ is the number of unpaired electrons.
For $\mu = \sqrt{3} \ B.M.$,$n = 1$.
$1$. $[Fe(NH_3)_6]^{3+}$: $Fe^{3+}$ is $3d^5$. $NH_3$ is a strong field ligand,causing pairing. $n=1$. Hybridization is $d^2sp^3$ (inner orbital).
$2$. $[CuCl_5]^{2-}$: $Cu^{2+}$ is $3d^9$. $Cl^-$ is a weak field ligand. $n=1$. Hybridization is $sp^3d$ (outer orbital).
$3$. $[Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand. $n=1$. Hybridization is $d^2sp^3$ (inner orbital).
$4$. $[Co(NH_3)_6]^{2+}$: $Co^{2+}$ is $3d^7$. $NH_3$ is a strong field ligand. $n=1$. Hybridization is $d^2sp^3$ (inner orbital).
Thus,$[CuCl_5]^{2-}$ uses outer $d-$orbitals $(sp^3d)$ and has a magnetic moment of $\sqrt{3} \ B.M.$

(B) In octahedral complexes,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
For $d^1, d^2, d^3$ configurations,electrons fill the $t_{2g}$ orbitals first,and there is no ambiguity regarding high or low spin.
For $d^4, d^5, d^6, d^7$ configurations,electrons can either pair up in $t_{2g}$ (low spin) or occupy $e_g$ (high spin) depending on the field strength of the ligand.
For $d^8, d^9, d^{10}$ configurations,the filling pattern is fixed by Hund's rule and the Pauli exclusion principle,resulting in only one possible spin state.
Therefore,$d^3$ and $d^8$ configurations do not exhibit high spin and low spin forms.

(D) The magnetic moment $\mu = \sqrt{n(n+2)} \ BM$. For $\mu = \sqrt{3} \ BM$,the number of unpaired electrons $n = 1$.
$1$. $[Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand,so $n=1$. Hybridization is $d^2sp^3$ (inner $d$-orbitals).
$2$. $[Fe(NH_3)_6]^{3+}$: $Fe^{3+}$ is $3d^5$. $NH_3$ is a strong field ligand,so $n=1$. Hybridization is $d^2sp^3$ (inner $d$-orbitals).
$3$. $[Co(CO)_4]$: $Co$ is $3d^7 4s^2$. $CO$ is a strong field ligand,$n=1$. Hybridization is $sp^3$.
$4$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ is $3d^9$. $H_2O$ is a weak field ligand,$n=1$. Hybridization is $sp^3d^2$ (outer $d$-orbitals).
Thus,$[Cu(H_2O)_6]^{2+}$ satisfies both conditions: $\mu = \sqrt{3} \ BM$ and uses outer $d$-orbitals $(sp^3d^2)$.

(C) $1$. In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. It is tetrahedral $(sp^3)$ and diamagnetic.
$2$. In $[Co(CO)_4]^{-}$,$Co$ is in $-1$ oxidation state $(3d^8 4s^2)$. It is isoelectronic with $Ni(CO)_4$,hence it is also tetrahedral $(sp^3)$ and diamagnetic.
$3$. $[Fe(CN)_6]^{3-}$ is paramagnetic due to one unpaired electron.
$4$. $[Ni(CN)_4]^{2-}$ is square planar $(dsp^2)$.
$5$. $[Co(H_2O)_6]^{3+}$ is $sp^3d^2$ (outer orbital complex) due to weak field ligand $H_2O$.

(A) $1. [Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand,so it forms an outer-orbital complex with $sp^3d^2$ hybridization.
$2. [PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d^8$ metal ion. $4d$ and $5d$ series metals always form square planar complexes with $dsp^2$ hybridization regardless of the ligand strength.
$3. [Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing,resulting in $sp^3$ hybridization.
$4. [Co(ox)_3]^{3-}$: $Co^{3+}$ is $3d^6$. $ox^{2-}$ is a strong field ligand,causing pairing,resulting in $d^2sp^3$ hybridization.
Therefore,the correct matching is $A-2, B-3, C-1, D-4$.

(B) The hybridization of the central atom in the given species is as follows:
$1$. $Ni(CN)_5^{3-}$: The central atom $Ni$ has a coordination number of $5$. It exhibits $sp^3d_{z^2}$ hybridization (trigonal bipyramidal geometry).
$2$. $CuCl_5^{3-}$: The central atom $Cu$ has a coordination number of $5$. It exhibits $d_{x^2-y^2}sp^3$ hybridization (square pyramidal geometry).
$3$. $AuCl_4^-$: The central atom $Au$ has a coordination number of $4$. It exhibits $dsp^2$ hybridization (square planar geometry).
$4$. $ClO_4^-$: The central atom $Cl$ has a coordination number of $4$. It exhibits $sp^3$ hybridization (tetrahedral geometry).
Thus,the correct matching is $A-3, B-4, C-2, D-1$.

(C) $1$. $[Cr(CO)_6]$: $Cr$ is in $0$ oxidation state $(3d^5 4s^1)$. $CO$ is a strong field ligand,causing pairing. It is diamagnetic and octahedral $(d^2sp^3)$.
$2$. $[Fe(CO)_5]$: $Fe$ is in $0$ oxidation state $(3d^6 4s^2)$. All electrons pair up. It is diamagnetic and trigonal bipyramidal $(dsp^3)$.
$3$. $[Co(CO)_4]^-$: $Co$ is in $-1$ oxidation state $(3d^8 4s^2)$. With $CO$ ligands,it becomes $3d^{10}$,which is diamagnetic and tetrahedral $(sp^3)$.
$4$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. It is diamagnetic and tetrahedral $(sp^3)$.
Therefore,option $C$ is correctly matched.

(C) $1$. Determine the oxidation state of $Cr$: Let the oxidation state of $Cr$ be $x$. The ligands are $Cl^-$ $(-1)$,$NO_2^-$ $(-1)$,and $NH_3$ $(0)$. The overall charge is $-1$.
$x + 2(-1) + 2(-1) + 2(0) = -1$
$x - 4 = -1$
$x = +3$
$2$. Electronic configuration of $Cr^{3+}$: The atomic number of $Cr$ is $24$. The ground state configuration is $[Ar] 3d^5 4s^1$. Thus,$Cr^{3+}$ is $[Ar] 3d^3$.
$3$. Coordination number: The complex has $6$ ligands,so the coordination number is $6$,implying an octahedral geometry.
$4$. Hybridization: Since $NO_2^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals. The $3d^3$ configuration has $3$ unpaired electrons in the $t_{2g}$ orbitals. For an octahedral complex with $d^3$ configuration,the inner $d$-orbitals are available for hybridization. Therefore,the hybridization is $d^2sp^3$.

(B) The reaction involves the substitution of $H_2O$ ligands by $NH_3$ ligands.
In the complex $[Cu(NH_3)_4]^{2+}$,the $Cu^{2+}$ ion ($d^9$ configuration) undergoes $dsp^2$ hybridization.
This results in a square planar geometry,not a tetrahedral structure.
The complex is paramagnetic due to the presence of one unpaired electron in the $3d$ orbital.
Therefore,the statement in option $B$ is incorrect.

(D) $[V(NH_3)_6]^{3+}$ involves $V^{3+}$ $(3d^2)$ which is $d^2sp^3$ hybridized.
$[CrCl_3(NMe_3)_3]$ involves $Cr^{3+}$ $(3d^3)$ which is $d^2sp^3$ hybridized.
$[Cu(CN)(NO_2)(NH_3)(py)]$ involves $Cu^{2+}$ $(3d^9)$ which is $dsp^2$ hybridized.
$K_3[Co(ox)_3]$ involves $Co^{3+}$ $(3d^6)$ with strong field ligand (oxalate),forming an inner orbital complex with $d^2sp^3$ hybridization,not $sp^3d^2$. Thus,option $D$ is incorrect.

(A) When an aqueous solution of copper $(II)$ sulphate $(CuSO_4 \cdot 5H_2O)$ is treated with excess ammonia,the deep blue complex $[Cu(NH_3)_4]SO_4 \cdot H_2O$ is formed.
This complex is known as tetraamminecopper $(II)$ sulphate monohydrate.
The geometry of this complex is square planar due to $dsp^2$ hybridization of the $Cu^{2+}$ ion.

(D) $1$. For $[NiCl_4]^{2-}$,$Ni^{2+}$ has $d^8$ configuration. $Cl^-$ is a weak field ligand,so it forms a tetrahedral complex due to $sp^3$ hybridization.
$2$. For $SF_4$,the central atom $S$ has $5$ electron pairs ($4$ bond pairs and $1$ lone pair),resulting in a see-saw shape.
$3$. For $XeF_4$,$Xe$ has $8$ valence electrons. It forms $4$ bond pairs with $F$ and has $2$ lone pairs. According to $VSEPR$ theory,$4$ bond pairs and $2$ lone pairs result in $sp^3d^2$ hybridization and a square planar geometry.
$4$. For $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons,leading to $dsp^2$ hybridization and a square planar geometry.
Both $C$ and $D$ have square planar shapes. However,in many competitive contexts,if a single choice is required,$[Ni(CN)_4]^{2-}$ is the classic example of $dsp^2$ square planar geometry.

(C) The empirical formula is $Ni(NH_3)_4(NO_3)_2 \cdot 2H_2O$.
For the first complex,the magnetic moment is $0 \ BM$,which implies it is diamagnetic.
$Ni^{2+}$ has a $d^8$ configuration. In a square planar geometry,$d^8$ ions form low-spin complexes where electrons pair up,resulting in $0$ unpaired electrons and a magnetic moment of $0 \ BM$.
Thus,the first complex is $[Ni(NH_3)_4](NO_3)_2 \cdot 2H_2O$ with a square planar geometry.
For the second complex,the magnetic moment is $2.84 \ BM$,which corresponds to $n=2$ unpaired electrons $(\mu = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ BM)$.
This corresponds to an octahedral geometry where $Ni^{2+}$ is in a high-spin state,such as $[Ni(NH_3)_4(H_2O)_2](NO_3)_2$.

(D) In both complexes,the central metal ion is $Co^{3+}$,which has a $d^6$ electronic configuration.
For $[CoF_6]^{3-}$,$F^-$ is a weak field ligand,so it forms an outer orbital complex ($sp^3d^2$ hybridization) with $4$ unpaired electrons,making it paramagnetic.
For $[Co(NH_3)_6]^{3+}$,$NH_3$ is a strong field ligand,so it causes pairing of electrons,forming an inner orbital complex ($d^2sp^3$ hybridization) with $0$ unpaired electrons,making it diamagnetic.
Therefore,the statement that $[CoF_6]^{3-}$ is an outer orbital complex and $[Co(NH_3)_6]^{3+}$ is an inner orbital complex is correct.