Hybridisation and Geometry Questions in English

(D) $1$. The oxidation state of $Ni$ in $[Ni(CN)_4]^{2-}$ is calculated as: $x + 4(-1) = -2$,so $x = +2$.
$2$. The electronic configuration of $Ni^{2+}$ is $[Ar]^{18} 3d^8 4s^0$.
$3$. $CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
$4$. After pairing,one $3d$,one $4s$,and two $4p$ orbitals are available for hybridization.
$5$. Thus,the hybridization is $dsp^2$.

(C) In $[Co(CN)_6]^{3-}$,the oxidation state of $Co$ is $+3$.
The electronic configuration of $Co^{3+}$ is $3d^6$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
This results in all $6$ electrons being paired in the $t_{2g}$ orbitals.
Therefore,the complex has no unpaired electrons and is a low-spin complex.

(A) For $[Ni(NH_3)_6]^{2+}$,the oxidation state of $Ni$ is $+2$ ($3d^8$ configuration). It uses $4s, 4p,$ and $4d$ orbitals to form $sp^3d^2$ hybrid orbitals,making it an outer orbital complex. It has two unpaired electrons in the $3d$ orbitals,so it is paramagnetic.
For $[Zn(NH_3)_6]^{2+}$,the oxidation state of $Zn$ is $+2$ ($3d^{10}$ configuration). It is an outer orbital complex $(sp^3d^2)$,but it is diamagnetic due to no unpaired electrons.
For $[Cr(NH_3)_6]^{3+}$,the oxidation state of $Cr$ is $+3$ ($3d^3$ configuration). It uses $3d, 4s,$ and $4p$ orbitals to form $d^2sp^3$ hybrid orbitals,making it an inner orbital complex. It is paramagnetic.
For $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$ ($3d^6$ configuration). It forms $d^2sp^3$ hybrid orbitals,making it an inner orbital complex. It is diamagnetic.

(C) The reaction of $Ni^{2+}$ with dimethylglyoxime $(DMG)$ in an ammoniacal medium produces a red-coloured complex,$[Ni(DMG)_2]$.
This complex has a square planar geometry,not a tetrahedral one.
Dimethylglyoxime acts as a bidentate ligand,coordinating through the nitrogen atoms.
The complex is highly stable due to the presence of strong intramolecular symmetrical hydrogen bonding between the oxime groups of the two ligands.

(D) Magnetic moment $(\mu) = \sqrt{n(n+2)} \text{ } B.M.$
Given $\mu = 3.83 \text{ } B.M.$
$3.83 = \sqrt{n(n+2)} \implies n \approx 3$
This indicates there are $3$ unpaired electrons.
For the complex $[Cr(H_2O)_6]Cl_3$,the oxidation state of $Cr$ is $+3$.
The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
Thus,$Cr^{3+}$ is $[Ar] 3d^3$.
According to Hund's rule,the $3$ electrons in the $3d$ orbitals occupy the $t_{2g}$ set $(d_{xy}, d_{yz}, d_{xz})$ singly.
Therefore,the distribution is $3d_{xy}^1, 3d_{yz}^1, 3d_{xz}^1$.

(D) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.
In the complex $[NiX_4]^{-2}$,let the oxidation state of $Ni$ be $y$.
$y + 4(-1) = -2 \implies y = +2$.
So,$Ni^{+2}$ has the configuration $[Ar] 3d^8$.
In the $3d$ orbital,there are $8$ electrons. According to Hund's rule,these are distributed as $3$ paired and $2$ unpaired electrons.
Since the complex is paramagnetic and the ligand $X^-$ is a weak field ligand (as it forms a tetrahedral complex),no pairing of electrons occurs in the $3d$ orbitals.
Thus,there are $2$ unpaired electrons.
The hybridization involved is $sp^3$,which corresponds to a tetrahedral geometry.

(A) In $[PtCl_4]^{2-}$,the central metal ion is $Pt^{2 }$ ($5d^8$ configuration).
$Pt$ is a $5d$ series element,and for $5d$ elements,the crystal field splitting energy $(\Delta_o)$ is very high,even with weak ligands like $Cl^-$.
This high splitting energy forces the pairing of electrons in the $5d$ orbitals,resulting in $dsp^2$ hybridization.
Therefore,$[PtCl_4]^{2-}$ exhibits a square planar geometry.
In contrast,$[NiCl_4]^{2-}$,$[CoCl_4]^{2-}$,and $[FeCl_4]^{2-}$ involve $3d$ metals with weak ligands,leading to $sp^3$ hybridization and tetrahedral geometry.

(C) In the complex $[Cr(NH_3)_6]Cl_3$,the oxidation state of $Cr$ is $+3$. The electronic configuration of $Cr^{3+}$ is $[Ar]3d^3$.
Since $NH_3$ is a ligand,it forms an octahedral complex using $d^2sp^3$ hybridization.
Because it uses $(n-1)d$ orbitals ($3d$ orbitals),it is classified as an inner orbital complex,not an outer orbital complex.
Therefore,the statement that it is an outer orbital complex is wrong.

(A) The central metal ion is $Pt^{2+}$,which has a $5d^8$ electronic configuration.
For $5d$ series elements,the crystal field splitting energy $(CFSE)$ is very high,which forces the pairing of electrons.
This leads to $dsp^2$ hybridization,resulting in a square planar geometry.
The complex is diamagnetic because all electrons are paired.

(A) $1$. $[Ni(CO)_4]$: $Ni(0)$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing. Hybridization is $sp^3$,geometry is tetrahedral,and it is diamagnetic $(n=0)$.
$2$. $[Ni(CN)_4]^{2-}$: $Ni(II)$ is $3d^8$. $CN^-$ is a strong field ligand. Hybridization is $dsp^2$,geometry is square planar,and it is diamagnetic $(n=0)$.
$3$. Since both are diamagnetic but have different geometries (tetrahedral vs square planar),option $A$ is correct.

(C) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing. It forms $dsp^2$ hybridization,which is square planar and diamagnetic.
$2$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand. It forms $sp^3$ hybridization,which is tetrahedral and paramagnetic.
$3$. $[Ni(CO)_4]$: $Ni$ is $d^8s^2$. $CO$ is a strong field ligand,forcing electrons to pair. It forms $sp^3$ hybridization,which is tetrahedral and diamagnetic.
$4$. $[Fe(CO)_5]$: $Fe$ is $d^6s^2$. It forms $dsp^3$ hybridization,which is trigonal bipyramidal and diamagnetic.
Therefore,the complex that is both tetrahedral and diamagnetic is $[Ni(CO)_4]$.

(B) The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.
In the complex $[FeF_6]^{3-}$,the oxidation state of $Fe$ is $+3$,resulting in an electronic configuration of $[Ar] 3d^5$.
Since $F^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $4s$,$4p$,and $4d$ orbitals participate in hybridization,resulting in $sp^3d^2$ hybridization.
This hybridization corresponds to an octahedral geometry.
Hence,option $B$ is the correct answer.

(B) $1$. For $[Cu(NH_3)_4]^{2+}$,$Cu^{2+}$ is $3d^9$. It undergoes $dsp^2$ hybridization and is paramagnetic due to one unpaired electron.
$2$. For $K_2[Ni(CN)_4]$,$Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $dsp^2$ hybridization and is diamagnetic.
$3$. $[Co(NH_3)_4(H_2O)_2]Cl_3$ involves $Co^{3+}$ $(3d^6)$. With weak field ligands,it undergoes $sp^3d^2$ hybridization and is paramagnetic.
$4$. $[Pt(NH_3)_2Cl_2]$ is a square planar complex ($dsp^2$ hybridization) and is diamagnetic.
Therefore,option $B$ is correctly matched.

(A) $1$. For $[PdCl_4]^{2-}$,$Pd^{2+}$ is a $4d^8$ ion. Since $Pd$ is a $4d$ series metal,it always forms square planar complexes with $dsp^2$ hybridization. This is correct.
$2$. For $[PtCl_4]^{2-}$,$Pt^{2+}$ is a $5d^8$ ion. Like $Pd^{2+}$,it forms square planar complexes with $dsp^2$ hybridization,not $sp^3$. This is incorrect.
$3$. For $[Fe(NH_3)_6]^{2+}$,$Fe^{2+}$ is a $3d^6$ ion. $NH_3$ is a strong field ligand,but for $Fe^{2+}$,it is not strong enough to force pairing in the $3d$ orbital to form $d^2sp^3$. It forms an outer orbital complex $sp^3d^2$. This is incorrect.
$4$. For $[Co(H_2O)_6]^{3+}$,$Co^{3+}$ is a $3d^6$ ion. $H_2O$ is a weak field ligand,so it forms an outer orbital complex $sp^3d^2$. This is correct.
Note: Both $A$ and $D$ are technically correct based on standard coordination chemistry principles. However,$A$ is a classic example of $4d$ metal behavior.

(C) $1$. In $[Fe(CO)_5]$,the geometry is trigonal bipyramidal ($sp^3d$ hybridization).
$2$. In $[Co(NH_3)_6]^{2+}$,the geometry is octahedral ($d^2sp^3$ or $sp^3d^2$ hybridization).
$3$. In $[NiCl_4]^{2-}$,$Ni^{2+}$ has a $d^8$ configuration. $Cl^-$ is a weak field ligand,so it does not cause pairing. The hybridization is $sp^3$,resulting in a tetrahedral geometry.
$4$. In $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ has a $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing. The hybridization is $dsp^2$,resulting in a square planar geometry.
Therefore,the correct option is $C$.

(A) The central metal ion is $Fe^{2+}$,which has an electronic configuration of $[Ar] 3d^6$.
$H_2O$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbital.
Thus,the $3d$ electrons are arranged as $t_{2g}^4 e_g^2$,resulting in $4$ unpaired electrons.
The hybridization involves the outer $4d$ orbitals,leading to $sp^3d^2$ hybridization (outer orbital complex).

(D) The central metal ion is $Co^{3+}$,which has an electronic configuration of $[Ar] 3d^6$.
$F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the complex uses the outer $4d$ orbitals for hybridisation.
For a coordination number of $6$,the hybridisation is $sp^3d^2$,resulting in an outer orbital high-spin complex.

(B) $Fe(CO)_5$ adopts a trigonal bipyramidal structure with the $Fe$ atom surrounded by five $CO$ ligands: three in equatorial positions and two axially bound. The $Fe-C-O$ linkages are each linear.

(D) The coordination number of a metal ion in a complex depends on the nature of the ligands and the geometry of the complex.
For the $Cu^{2+}$ ion,it commonly forms complexes with a coordination number of $4$ (e.g.,$[Cu(NH_3)_4]^{2+}$ with square planar geometry) and sometimes $6$ (e.g.,$[Cu(H_2O)_6]^{2+}$ with distorted octahedral geometry).
Therefore,$Cu^{2+}$ can exhibit a coordination number of $4$ or $6$ depending on the ligands attached.

(C) $Ni(CO)_4$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^{10} 4s^0$. Due to the strong field ligand $CO$,the electrons are paired,resulting in $sp^3$ hybridization. Thus,the geometry is tetrahedral.
$[Ni(PPh_3)_2Cl_2]$: The oxidation state of $Ni$ is $+2$. The electronic configuration is $[Ar] 3d^8$. $PPh_3$ and $Cl^-$ are ligands that do not cause pairing of $d$-electrons in this specific complex,leading to $sp^3$ hybridization. Thus,the geometry is tetrahedral.

(D) In $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
The $3d^9$ configuration has one unpaired electron.
The magnetic moment $\mu$ is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
For $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$
Due to the square planar geometry of this complex,the hybridisation is $dsp^2$.

(C) In $[FeF_6]^{3-}$,the oxidation state of $Fe$ is $+3$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $F^-$ is a weak field ligand,it does not cause pairing of electrons.
Therefore,the $3d$ electrons remain unpaired,resulting in $5$ unpaired electrons.
The hybridization involves the outer $d$-orbitals,leading to $sp^3d^2$ hybridization.

(C) In $Fe(CO)_5$,the central metal atom is $Fe$ in $0$ oxidation state.
The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
$CO$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
After pairing,all electrons are paired,making the complex diamagnetic.
Since all electrons are paired,it is a low spin complex.

(D) The correct answer is $D$.
$A$. $MnCl_4^{2-}$: $Mn$ is in $+2$ oxidation state $(d^5)$. $Cl^-$ is a weak field ligand,leading to $sp^3$ hybridization and tetrahedral geometry. It has $5$ unpaired electrons,so it is paramagnetic. This statement is true.
$B$. $[Mn(CN)_6]^{2-}$: $Mn$ is in $+4$ oxidation state $(d^3)$. It forms an octahedral complex ($d^2sp^3$ or $sp^3d^2$). It has $3$ unpaired electrons,so it is paramagnetic. This statement is true.
$C$. $[CuCl_4]^{2-}$: $Cu$ is in $+2$ oxidation state $(d^9)$. It is a tetrahedral complex ($sp^3$ hybridization). However,the statement claims square planar geometry,which is incorrect for $[CuCl_4]^{2-}$. Note: $[CuCl_4]^{2-}$ is actually tetrahedral. Wait,checking the options: $[Ni(PPh_3)_2Br_2]$ is a $4$-coordinate complex,which is tetrahedral,not trigonal bipyramidal. Thus,$D$ is definitely false.
$D$. $[Ni(PPh_3)_2Br_2]$: $Ni$ is in $+2$ oxidation state $(d^8)$. It is a $4$-coordinate complex,which typically adopts tetrahedral geometry,not trigonal bipyramidal. Therefore,this statement is false.

(B) The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
In an octahedral geometry,the coordination number is $6$.
For $d^8$ configuration,the $d$-orbitals are $d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}, d_{z^2}$.
Regardless of whether the ligand is strong field or weak field,the $d^8$ configuration will always have two unpaired electrons in the $e_g$ set ($d_{x^2-y^2}$ and $d_{z^2}$).
Since the $3d$ orbitals are not fully available for hybridization,the complex must use the $4d$ orbitals for $sp^3d^2$ hybridization.
Therefore,$Ni(II)$ octahedral complexes are always outer orbital complexes.

(B) The magnetic moment of $5.92 \ B.M.$ corresponds to $n = 5$ unpaired electrons,indicating a high-spin $d^5$ configuration for $Mn^{2+}$.
Since $Br^{-}$ is a weak field ligand,it does not cause pairing of electrons.
For a coordination number of $4$,the $sp^3$ hybridization leads to a tetrahedral geometry.
Square planar geometry for a $d^5$ system would require $dsp^2$ hybridization,which is not possible for a high-spin $Mn^{2+}$ complex as it would require pairing of electrons.
Therefore,$[MnBr_4]^{2-}$ with square planar geometry is not possible.

(A, C, D) $1$. $[Cu(CN)_4]^{3-}$: $Cu^+$ is $3d^{10}$. It forms a tetrahedral complex with $sp^3$ hybridization. This statement is correct.
$2$. $[Ni(CN)_6]^{4-}$: $Ni^{2+}$ is $3d^8$. It cannot form a $d^2sp^3$ complex as it lacks sufficient empty $d$-orbitals. This is incorrect.
$3$. $[ZnBr_4]^{2-}$: $Zn^{2+}$ is $3d^{10}$. It forms a tetrahedral complex with $sp^3$ hybridization and is diamagnetic. This statement is correct.
$4$. $[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. It forms an octahedral complex with $d^2sp^3$ hybridization. This statement is correct.
Note: Multiple options ($A$,$C$,and $D$) are chemically correct based on standard coordination chemistry principles.

(B) An inner orbital complex is one that uses $(n-1)d$ orbitals for hybridization.
In $[Ni(CN)_2(PPh_3)_2]$,the central metal ion is $Ni^{2+}$ ($3d^8$ configuration).
$CN^{-}$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
This results in $dsp^2$ hybridization,which is characteristic of inner orbital complexes in square planar geometry.

(D) $CN^-$ is a strong field ligand,while $Br^-$ and $F^-$ are weak field ligands.
For $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ has $d^8$ configuration. $CN^-$ causes pairing,leading to $dsp^2$ hybridization (square planar) with $0$ unpaired electrons and $\mu = 0 \ BM$.
For $[MnBr_4]^{2-}$,$Mn^{2+}$ has $d^5$ configuration. $Br^-$ is a weak ligand,leading to $sp^3$ hybridization (tetrahedral) with $5$ unpaired electrons and $\mu = \sqrt{5(5+2)} = 5.9 \ BM$.
For $[FeF_6]^{4-}$,$Fe^{2+}$ has $d^6$ configuration. $F^-$ is a weak ligand,leading to $sp^3d^2$ hybridization (octahedral) with $4$ unpaired electrons and $\mu = \sqrt{4(4+2)} = 4.9 \ BM$.

(A) The reaction is $K_2[Ni(CN)_4] + 4K + 4NH_3 \to K_4[Ni(NH_3)_4] + 4KCN$.
Complex $A$ is $K_4[Ni(NH_3)_4]$.
In $K_4[Ni(NH_3)_4]$,the oxidation state of $Ni$ is $0$ ($d^{10}$ configuration).
Since $Ni(0)$ has a $d^{10}$ configuration,all $d$-orbitals are filled.
Therefore,it undergoes $sp^3$ hybridisation to form a tetrahedral geometry.
Since all electrons are paired in the $d^{10}$ configuration,the complex is diamagnetic.
Thus,the complex is $sp^3$ hybridised and diamagnetic.

(D) The reaction proceeds as follows:
$1$. $2CuSO_4 + 4KCN \to 2CuCN + (CN)_2 + 2K_2SO_4$. Here,$Y$ is $(CN)_2$ (cyanogen).
$2$. Cyanogen $(CN)_2$ has a linear structure,not an open book-like structure.
$3$. When $CuCN$ reacts with excess $KCN$,it forms the complex $K_3[Cu(CN)_4]$.
$4$. In $K_3[Cu(CN)_4]$,the copper is in the $+1$ oxidation state $(Cu^+)$,which has a $d^{10}$ configuration.
$5$. The complex $[Cu(CN)_4]^{3-}$ is tetrahedral with $sp^3$ hybridization.
$6$. Since all $d$-orbitals are filled $(d^{10})$,there are no unpaired electrons,making it a diamagnetic complex.
$7$. Therefore,statement $D$ is correct.

(B) The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$. For $Co^{2+}$,the configuration is $3d^7$.
In a tetrahedral field,the $d-$ orbitals split into $e_g$ (lower energy,doubly degenerate) and $t_{2g}$ (higher energy,triply degenerate) sets.
The $7$ electrons fill as follows: $e_g^4 t_{2g}^3$.
The $t_{2g}$ orbitals are triply degenerate and contain $3$ electrons,meaning they are exactly half-filled ($3/6$ capacity).
Thus,$Co^{2+}$ in a tetrahedral field satisfies the condition.

(C) In a square planar complex,the $d$-orbitals split into four energy levels: $d_{x^2-y^2}$ (highest energy),$d_{xy}$,$d_{z^2}$,and the degenerate pair $d_{zx}$ and $d_{yz}$ (lowest energy).
For a $Co^{2+}$ ion ($d^7$ configuration) in a square planar field,the electrons fill these orbitals starting from the lowest energy level.
The filling order is: $(d_{zx}, d_{yz})^4, (d_{z^2})^2, (d_{xy})^1, (d_{x^2-y^2})^0$.
Thus,the single unpaired electron is present in the $d_{xy}$ orbital.

(B) The structure of the peroxoborate ion $[B_2(O_2)_2(OH)_4]^{2-}$ consists of two boron atoms linked by two peroxo bridges. Each boron atom is bonded to two hydroxyl groups and two oxygen atoms of the peroxo bridges. By observing the structure,we can identify three distinct types of $O-B-O$ bond angles:
$1$. The angle between the two hydroxyl oxygen atoms $(HO-B-OH)$.
$2$. The angle between a hydroxyl oxygen and a peroxo oxygen $(HO-B-O)$.
$3$. The angle between the two peroxo oxygen atoms ($O-B-O$ within the ring).
Therefore,there are $3$ different types of $O-B-O$ angles.

(A) In the complex $[Fe(H_2O)_5(NO)]^{2+}$,the $NO$ ligand acts as $NO^+$. The oxidation state of $Fe$ is $+1$. The electronic configuration of $Fe^+$ is $[Ar] 3d^6 4s^1$. Due to the strong field of $NO^+$,the electron from the $4s$ orbital is promoted to the $3d$ orbital to facilitate $d^2sp^3$ hybridization. This involves the transition of an electron from the $4s$ shell to the $3d$ shell of the central metal atom.

(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
If $Hund's$ rule is violated,the $8$ electrons in the $3d$ orbitals would pair up in the lower energy orbitals first.
Specifically,the $3d$ orbitals would be filled as: $(d_{xy})^2 (d_{yz})^2 (d_{zx})^2 (d_{x^2-y^2})^2 (d_{z^2})^0$.
This leaves two inner $d$-orbitals ($d_{x^2-y^2}$ and $d_{z^2}$) vacant.
These two vacant $d$-orbitals,along with one $4s$ and three $4p$ orbitals,undergo $d^2sp^3$ hybridization.
Since all electrons are paired in this configuration,the complex becomes diamagnetic.
Therefore,the correct statement is $d^2sp^3$ and diamagnetic.

(A) The ammine complexes are as follows:
$1$. For $Cu^{2+}$,the complex is $[Cu(NH_3)_4]^{2+}$,which has a square planar geometry due to $dsp^2$ hybridization.
$2$. For $Ni^{2+}$,the complex is $[Ni(NH_3)_6]^{2+}$,which has an octahedral geometry due to $sp^3d^2$ hybridization.
$3$. For $Zn^{2+}$,the complex is $[Zn(NH_3)_4]^{2+}$,which has a tetrahedral geometry due to $sp^3$ hybridization.
Therefore,the shapes are square planar,octahedral,and tetrahedral respectively.

(B) $1$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ $(3d^6)$,$CN^-$ is a strong field ligand. It forms $d^2sp^3$ (inner orbital) and is diamagnetic ($0$ unpaired electrons).
$2$. $[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ $(3d^3)$,$NH_3$ is a ligand. It forms $d^2sp^3$ (inner orbital) and is paramagnetic ($3$ unpaired electrons).
$3$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ $(3d^6)$,$NH_3$ is a strong field ligand. It forms $d^2sp^3$ (inner orbital) and is diamagnetic ($0$ unpaired electrons).
$4$. $[Fe(NO_2)_6]^{4-}$: $Fe^{2+}$ $(3d^6)$,$NO_2^-$ is a strong field ligand. It forms $d^2sp^3$ (inner orbital) and is diamagnetic ($0$ unpaired electrons).
Thus,$[Cr(NH_3)_6]^{3+}$ is both paramagnetic and an inner orbital complex.

(C) The correct answer is $[PtCl_4]^{2-}$.
$Pt$ is a $5d$ series element. For $5d$ series elements,the crystal field splitting energy is high,so even weak field ligands like $Cl^-$ can cause pairing of electrons.
$Pt^{2+}$ has a $5d^8$ configuration. Due to the pairing of electrons,it undergoes $dsp^2$ hybridisation,resulting in a square planar geometry.
In contrast,$[FeCl_4]^{2-}$,$[NiCl_4]^{2-}$,and $[CoCl_4]^{2-}$ involve $3d$ series metals where $Cl^-$ acts as a weak field ligand and cannot cause pairing. Thus,they undergo $sp^3$ hybridisation and exhibit tetrahedral geometry.

(D) $1$. $Ni(CO)_4$ is a tetrahedral complex where $Ni$ is in $0$ oxidation state.
$2$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In the presence of strong field ligand $CO$,electrons pair up,resulting in $sp^3$ hybridization. It is diamagnetic.
$3$. Due to back-bonding from $Ni$ to $CO$,the $C-O$ bond order decreases compared to free $CO$ molecule,while the $Ni-C$ bond order increases (becomes greater than $1$).
$4$. According to the $EAN$ rule,$Ni(CO)_4$ has $28 + 4 \times 2 = 36$ electrons,which is the atomic number of $Kr$. This makes it stable,but it can still act as a reducing agent in certain chemical reactions. Thus,the statement that it cannot act as an oxidizing or reducing agent is incorrect.

(A) In chromyl chloride $(CrO_2Cl_2)$,the oxidation state of Chromium $(Cr)$ is $+6$.
The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
For $Cr^{6+}$,the configuration is $[Ar] 3d^0 4s^0$.
Chromium forms two double bonds with oxygen atoms and two single bonds with chlorine atoms.
There are $4$ sigma bonds and $2$ pi bonds around the central $Cr$ atom.
Since there are $4$ sigma bonds and no lone pairs,the hybridisation is $sp^3$.
The shape corresponding to $sp^3$ hybridisation with $4$ bonding pairs is Tetrahedral.

(D) In the complex $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
According to $C.F.T.$ and valence bond theory,the $Cu^{2+}$ ion undergoes $dsp^2$ hybridisation to form a square planar geometry.
Due to the presence of one unpaired electron in the $3d$ orbital,the complex is paramagnetic.
Therefore,all the given statements are correct.

(D) $1$. The central metal ion is $Co^{3+}$. The electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$. Thus,$Co^{3+}$ is $[Ar] 3d^6$.
$2$. The ligand $C_2O_4^{2-}$ (oxalate) is a strong field ligand $(SFL)$ for $Co^{3+}$,causing pairing of electrons in the $3d$ orbitals.
$3$. The $6$ electrons in $3d$ orbitals pair up to occupy $3$ orbitals,leaving $2$ $d$-orbitals empty for hybridisation.
$4$. The hybridisation is $d^2sp^3$,which corresponds to an inner orbital octahedral complex.
$5$. Since all electrons are paired,the complex is diamagnetic.
$6$. The spin-only magnetic moment $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons. Here $n=0$,so $\mu = 0 \ B.M.$

(D) $1$. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$,outer orbital complex $(sp^3d^2)$,paramagnetic,coloured.
$2$. $[Co(H_2O)_6]^{2+}$: $Co^{2+}$ is $3d^7$,outer orbital complex $(sp^3d^2)$,paramagnetic,coloured.
$3$. $[Ti(H_2O)_6]^{3+}$: $Ti^{3+}$ is $3d^1$,outer orbital complex $(sp^3d^2)$,paramagnetic,coloured.
$4$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing of electrons. It forms an inner orbital complex $(d^2sp^3)$. Since it has no unpaired electrons,it is diamagnetic and colourless.
Wait,re-evaluating the question criteria: Paramagnetic,inner orbital,and coloured. None of the options perfectly fit all three criteria simultaneously based on standard ligand field theory. However,if we consider $[Co(NH_3)_6]^{3+}$ as the target for inner orbital,it is diamagnetic. Given the options,there might be a typo in the question. If we look for paramagnetic and coloured,$A, B, C$ are candidates. If we look for inner orbital,only $D$ is a candidate. Assuming the question implies a complex like $[Fe(CN)_6]^{3-}$,but among these,$[Co(NH_3)_6]^{3+}$ is the only inner orbital complex.

(A) $1$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing,resulting in $d^2sp^3$ hybridisation (inner orbital complex).
$2$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ causes pairing,resulting in $d^2sp^3$ hybridisation.
$3$. Both are octahedral and $d^2sp^3$ hybridised. Thus,option $A$ is correct.
$4$. $[Ni(CO)_4]$ is tetrahedral ($sp^3$,diamagnetic),but $[Ni(CN)_4]^{2-}$ is square planar ($dsp^2$,diamagnetic). Option $B$ is incorrect.
$5$. $[Ni(CO)_4]$ and $[Co(CO)_4]^{-}$ are both isoelectronic and tetrahedral ($sp^3$,diamagnetic). Option $C$ is also technically correct in terms of geometry and hybridisation,but $A$ is the standard textbook example for $d^2sp^3$ comparison.
$6$. $[Co(H_2O)_6]^{3+}$ is $d^2sp^3$ (low spin),while $[Cr(H_2O)_6]^{3+}$ is $d^2sp^3$ (inner orbital). However,$A$ is the most definitive match.

(D) In the $[Ni(NH_3)_6]^{2+}$ complex,the oxidation state of $Ni$ is $+2$.
$Ni$ has an outer electronic configuration of $3d^8, 4s^2$,so $Ni^{2+}$ has an electronic configuration of $3d^8, 4s^0$.
$NH_3$ is a weak field ligand,so it does not cause pairing of the $3d$ electrons.
Therefore,the complex utilizes the $4d$ orbitals for hybridization,resulting in $sp^3d^2$ hybridization.
Since the outer $d$-orbitals are used,it is classified as an outer orbital complex.

(C) In the complex $[MnO_4]^-$,the oxidation state of $Mn$ is calculated as $x + 4(-2) = -1$,which gives $x = +7$.
The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$.
For $Mn^{7+}$,the configuration is $[Ar] 3d^0$.
Since there are no electrons in the $d$-orbitals,the number of electrons in the $e_g$ orbitals is $0$.

(B) In $[Ni(CN)_4]^{2-}$,the central metal ion is $Ni^{2+}$ ($3d^8$ configuration).
$CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
This leaves one $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals vacant,which undergo $dsp^2$ hybridisation to form a square planar geometry.
Conversely,$[Ni(CO)_4]$ and $[NiCl_4]^{2-}$ involve $sp^3$ hybridisation,resulting in a tetrahedral geometry.

(D) When excess $NH_3$ is added to $CuSO_4$ solution,the complex formed is $[Cu(NH_3)_4]^{2+}$.
In this complex,the oxidation state of $Cu$ is $+2$,which corresponds to the electronic configuration $[Ar] 3d^9$.
Due to the presence of one unpaired electron in the $3d$ orbital,the complex is paramagnetic.
Although the geometry is often described as distorted octahedral in aqueous solution,in the context of coordination chemistry theory regarding this specific complex,it is frequently categorized as square planar or distorted square planar with $dsp^2$ hybridization character,exhibiting paramagnetism due to the unpaired electron.

(A) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing. Hybridization is $dsp^2$,geometry is square planar.
$2$. $[Cu(CN)_4]^{3-}$: $Cu^+$ is $3d^{10}$. Hybridization is $sp^3$,geometry is tetrahedral.
$3$. $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing. Hybridization is $sp^3$,geometry is tetrahedral.
$4$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,no pairing. Hybridization is $sp^3$,geometry is tetrahedral.
Thus,$[Ni(CN)_4]^{2-}$ has a square planar geometry,while the others are tetrahedral.