Hybridisation and Geometry Questions in English

(B) In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
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 hybridize to form six equivalent $sp^3 d^2$ hybrid orbitals.
This results in an outer orbital octahedral complex.

(B) An octahedral complex is formed when the central metal ion is surrounded by $6$ ligands in an octahedral geometry. Therefore,the coordination number of the central metal ion in an octahedral complex is $6$.

(B) For the complex $[Ni(CN)_4]^{2-}$:
$1$. Let the oxidation state of $Ni$ be $x$.
$x + 4(-1) = -2 \Rightarrow x = +2$.
Thus,the oxidation state of $Ni$ is $+2$,not $+6$.
$2$. $CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
$3$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$4$. Due to the pairing,the complex undergoes $dsp^2$ hybridization,resulting in a square planar geometry.

(B) In the complex $[Ni(CN)_4]^{2-}$,the central metal ion is $Ni^{2+}$.
The atomic number of $Ni$ is $28$,so the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
This results in one empty $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals being available for hybridisation.
Thus,the hybridisation involved is $dsp^2$,which corresponds to a square planar geometry.

(D) The central metal ion is $Co^{3+}$. The atomic number of $Co$ is $27$,so the electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$NH_3$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
In the presence of $NH_3$,the $6$ electrons in the $3d$ subshell pair up in the first three orbitals ($t_{2g}$ set).
Thus,all $3d$ electrons are paired,and there are no unpaired electrons in the $[Co(NH_3)_6]^{3+}$ complex.

(A) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $3d^6$ configuration remains as $t_{2g}^4 e_g^2$,which corresponds to $4$ unpaired electrons.

(B) In $[Ni(CN)_4]^{2-}$,the oxidation state of nickel is $+2$.
The valence shell electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
Since $CN^-$ is a strong field ligand,the $3d$ electrons are paired prior to hybridization.
This results in one vacant $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals,which undergo $dsp^2$ hybridization to accommodate the four electron pairs donated by the four $CN^-$ ligands.

(A) The atomic number of Cobalt $(Co)$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In the complex $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$. Thus,the configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$NH_3$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
After pairing,two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals become available for hybridization.
Therefore,the hybridization involved is $d^2sp^3$.

(A) The complex is $[Co(NH_3)_6]^{3+}$.
In this complex,the central metal ion is $Co^{3+}$.
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 $3d$ orbitals.
This results in two empty $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals being available for hybridization.
Thus,the hybridization is $d^2sp^3$.
This involves a total of $6$ hybrid orbitals to accommodate the $6$ lone pairs donated by the $6$ $NH_3$ ligands.

(D) When cobalt chloride $(CoCl_2)$ is dissolved in water,it forms the hexaaquacobalt$(II)$ chloride complex.
The chemical reaction is as follows:
$CoCl_2 + 6H_2O \rightarrow [Co(H_2O)_6]Cl_2$
In the complex $[Co(H_2O)_6]^{2+}$,the coordination number of the central metal ion $Co^{2+}$ is $6$,which corresponds to an octahedral geometry.

(A) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In the complex $[CoF_6]^{3-}$,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
The $3d$ orbitals have $4$ unpaired electrons.
The complex undergoes $sp^3d^2$ hybridization,which corresponds to an octahedral geometry.

(C) The atomic number of cobalt $(Co)$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In the $[CoF_6]^{3-}$ complex,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
In the $3d$ subshell,the $6$ electrons are distributed as follows: one orbital is doubly occupied,and the remaining four orbitals are singly occupied.
Therefore,there are $4$ unpaired electrons present in the $Co^{3+}$ ion prior to hybridization.

(C) $1$. The central metal ion is $Ni^{2+}$. The atomic number of $Ni$ is $28$,so its electronic configuration is $[Ar] 3d^8 4s^2$.
$2$. In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$,so the configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$3$. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
$4$. The $Ni^{2+}$ ion undergoes $sp^3$ hybridization using one $4s$ and three $4p$ orbitals.
$5$. This results in a tetrahedral geometry.
$6$. Since there are two unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.

(C) The central metal ion is $Ni^{2+}$,which has an electronic configuration of $[Ar] 3d^8$.
$CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
After pairing,the configuration becomes $d^8$ with no unpaired electrons,leading to $dsp^2$ hybridization.
This results in a square planar geometry and diamagnetic behavior due to the absence of unpaired electrons.

(A) The central metal ion is $Co^{3+}$. The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
For $Co^{3+}$,the configuration is $[Ar] 3d^6$.
In the presence of the strong field ligand $NH_3$,the electrons in the $3d$ orbitals pair up.
The $3d^6$ configuration becomes $t_{2g}^6 e_g^0$,meaning all six electrons are paired in the $3d$ orbitals.
Therefore,the number of unpaired electrons is $0$.

(B) In the hexaamminecobalt$(III)$ complex ion,$[Co(NH_3)_6]^{3+}$,the central metal ion is $Co^{3+}$.
The electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$.
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 $3d$ orbitals.
This results in two empty $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals,which undergo $d^2sp^3$ hybridization to form an octahedral complex.

(B) In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state: $Ni^{2+} \rightarrow [Ar] 3d^8$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of $3d$ electrons.
Thus,the hybridization involved is $sp^3$ using $4s$ and $4p$ orbitals.
In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state: $Ni^{2+} \rightarrow [Ar] 3d^8$.
$CN^-$ is a strong field ligand,which causes pairing of $3d$ electrons.
This leaves one $3d$ orbital vacant,resulting in $dsp^2$ hybridization using one $3d$,one $4s$,and two $4p$ orbitals.

(D) For $[CoF_6]^{3-}$,$Co^{3+}$ is $3d^6$. It undergoes $sp^3d^2$ hybridization,resulting in an octahedral geometry and it is paramagnetic.
For $[Co(NH_3)_6]^{3+}$,$Co^{3+}$ is $3d^6$. It undergoes $d^2sp^3$ hybridization due to the strong field ligand $NH_3$,resulting in an octahedral geometry and it is diamagnetic.
For $[NiCl_4]^{2-}$,$Ni^{2+}$ is $3d^8$. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry and it is paramagnetic.
For $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ is $3d^8$. The strong field ligand $CN^-$ causes pairing of $3d$ electrons,leading to $dsp^2$ hybridization,a square planar geometry,and it is diamagnetic.

(D) For the complex $[Co(NH_3)_6]^{3+}$:
$1$. The oxidation state of $Co$ is $+3$.
$2$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6 4s^0$.
$3$. $NH_3$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
$4$. After pairing,all $6$ electrons occupy the first three $3d$ orbitals,leaving no unpaired electrons $(n = 0)$.
$5$. The complex undergoes $d^2sp^3$ hybridization,resulting in an octahedral geometry.

(B) In $[CoF_{6}]^{3-}$,the oxidation state of $Co$ is $+3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^{6}$.
Since $F^{-}$ is a weak field ligand $(WFL)$,it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $Co^{3+}$ ion uses one $4s$,three $4p$,and two $4d$ orbitals to form six hybrid orbitals.
This results in $sp^{3}d^{2}$ hybridisation.
The geometry associated with $sp^{3}d^{2}$ hybridisation is octahedral.

(D) The chemical formula for Cuproammonium sulphate is $[Cu(NH_{3})_{4}]SO_{4} \cdot H_{2}O$.
In the complex ion $[Cu(NH_{3})_{4}]^{2+}$,the oxidation state of $Cu$ is $+2$.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^{9}$.
Due to the presence of $NH_{3}$ (a strong field ligand),the complex undergoes $dsp^{2}$ hybridization,resulting in a square planar geometry.
Since there is one unpaired electron in the $3d$ orbital,the complex is paramagnetic.

(B) The permanganate ion $(MnO_4^-)$ has a tetrahedral geometry.
In this ion,the manganese atom is in the $+7$ oxidation state,meaning it has a $d^0$ electronic configuration.
The $\pi -$ bonding occurs due to the overlap of $p-$ orbitals of oxygen with $d-$ orbitals of manganese.
Since there are no unpaired electrons in the $d^0$ configuration,the ion is diamagnetic.
The structure is shown below:
$Mn$ is at the center bonded to four oxygen atoms in a tetrahedral arrangement.

(C) Werner’s theory was used to describe the structure and formation of complex compounds or coordination compounds.
According to this theory,the primary valency corresponds to the oxidation number,while the secondary valency corresponds to the coordination number.
The geometry of the complex is determined by the number and position of the secondary valences in space,as the ligands satisfying the secondary valencies are always directed towards fixed positions in space.

(A) The atomic number of $Cu$ is $29$. The electronic configuration is $[Ar] 3d^{10} 4s^1$.
In $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
In the presence of $4$ $NH_3$ ligands,one electron from the $3d$ orbital is promoted to the $4p$ orbital to facilitate $dsp^2$ hybridization.
Thus,the hybridization is $dsp^2$.

(C) In $[Co(NH_3)_6]^{3+}$,the oxidation state of cobalt is $+3$. The valence shell electronic configuration of $Co^{3+}$ $(3d^6)$ is represented as: $3d$ ($4$ unpaired electrons),$4s$ (empty),$4p$ (empty).
Since $NH_3$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
After pairing,the configuration becomes $t_{2g}^6 e_g^0$,meaning all electrons are paired.
This makes the complex diamagnetic and a low spin complex.
Therefore,the statement that it is a high spin complex is incorrect.

(D) The oxidation state of $Ni$ in $[Ni(CN)_4]^{-2}$ is $+2$.
The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
The electronic configuration of $Ni^{+2}$ is $[Ar] 3d^8$.
Since $CN^-$ is a strong field ligand,it causes the pairing of electrons in the $3d$ orbitals.
After pairing,all $8$ electrons in the $3d$ subshell are paired,leaving zero unpaired electrons.

(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $3d^{10} 4s^0 4p^0$. Thus,the hybridization is $sp^3$.
In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,forcing the two unpaired electrons in the $3d$ orbitals to pair up,creating one empty $3d$ orbital. This leads to $dsp^2$ hybridization.

(A) $1$. In $K_4[Ni(CN)_4]$,the oxidation state of $Ni$ is $0$. The configuration is $[Ar] 3d^8 4s^2$. Due to the strong field ligand $CN^-$,the $4s$ electrons pair up in the $3d$ orbital,resulting in a $d^{10}$ configuration. It is tetrahedral and diamagnetic.
$2$. In $K_2[Ni(CN)_4]$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $dsp^2$ hybridization (square planar) and it is diamagnetic.
$3$. In $K_2[NiCl_4]$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $Cl^-$ is a weak field ligand,resulting in $sp^3$ hybridization (tetrahedral) and it is paramagnetic.
$4$. Evaluating the options: Option $A$ is incorrect because $K_4[Ni(CN)_4]$ is tetrahedral,not square planar,and $K_2[Ni(CN)_4]$ is diamagnetic,not paramagnetic. Option $C$ is also incorrect as one is diamagnetic and the other is diamagnetic (both have $\mu = 0$),but the statement implies they are the same,which is true,yet $A$ is fundamentally incorrect regarding geometry and magnetism.

(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $3d^{10}$ configuration. The hybridisation is $sp^3$ (tetrahedral geometry).
In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,which forces the two unpaired electrons in the $3d$ orbitals to pair up. This creates one empty $3d$ orbital,which is used for $dsp^2$ hybridisation (square planar geometry).
Therefore,the hybridisation types are $sp^3$ and $dsp^2$ respectively.

(B) In the complex $K_2[NiCl_4]$,let the oxidation state of $Ni$ be $x$.
$2(+1) + x + 4(-1) = 0$ $\Rightarrow 2 + x - 4 = 0$ $\Rightarrow x = +2$.
The atomic number of $Ni$ is $28$.
The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
Thus,the number of electrons in the $3d$ orbital is $8$.

(A) In $[Fe(Cl)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $3d$ electrons remain unpaired,and the complex utilizes the $4s$,$4p$,and $4d$ orbitals for hybridization,resulting in $sp^3d^2$ hybridization.
This is an outer orbital complex.
In contrast,$[Fe(CN)_6]^{3-}$,$[Fe(CN)_6]^{4-}$,and $[Fe(NH_3)_6]^{3+}$ involve strong field ligands (or specific electronic conditions) that lead to $d^2sp^3$ hybridization (inner orbital complexes).

(C) The complex $[Ni(CO)_4]$ involves $Ni$ in the $0$ oxidation state. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In the presence of strong field ligand $CO$,the $4s$ electrons pair up into the $3d$ orbitals,resulting in a $3d^{10}$ configuration. This leads to $sp^3$ hybridization,which corresponds to a tetrahedral geometry. Since all electrons are paired,it is diamagnetic. The other options like $[NiCl_4]^{2-}$ are tetrahedral but paramagnetic due to the presence of unpaired electrons in the $3d$ orbitals.

(D) The central metal atom is $Ni$ with atomic number $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
In the complex ion $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $x + 4(-1) = -2$,which gives $x = +2$.
Thus,$Ni^{2+}$ has the configuration $[Ar] 3d^8$.
$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
After pairing,all $8$ electrons occupy the $d_{xy}, d_{yz}, d_{xz},$ and $d_{z^2}$ orbitals in pairs,leaving no unpaired electrons.
Therefore,the number of unpaired electrons is $0$.

(B) The manganate ion $(MnO_4^{2-})$ has a central manganese atom in the $+6$ oxidation state. The electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1$. The ion involves $sp^3$ hybridization,which results in a tetrahedral geometry. Therefore,the correct shape is tetrahedral.

(D) When $1$ $s$,$3$ $p$ and $2$ $d$-orbitals belonging to the same shell of an atom mix together to form six new equivalent orbitals,the type of hybridisation is called $d^{2}sp^{3}$ or octahedral hybridisation.
The new orbitals are called $d^{2}sp^{3}$ or octahedral orbitals.
These orbitals are directed towards the corners of an octahedron,resulting in an octahedral geometry.

(C) The geometry of a coordination complex is determined by the hybridization of the central metal atom. The relationship is as follows:

Hybridization	Geometry of Complex
$sp^3$	Tetrahedral
$dsp^2$	Square planar
$d^2sp^3$	Octahedral
$sp^3d^2$	Octahedral

Thus,$d^2sp^3$ hybridization results in an octahedral geometry.

(A) For $[NiCl_4]^{2-}$:
$1$. Oxidation state of $Ni$: $x + 4(-1) = -2 \implies x = +2$.
$2$. Coordination number of $Ni$ is $4$.
$3$. Electronic configuration of $Ni^{2+}$ is $[Ar]3d^8$.
$4$. $Cl^-$ is a weak field ligand,so no pairing of electrons occurs in the $3d$ orbitals.
$5$. The complex undergoes $sp^3$ hybridisation,resulting in a tetrahedral geometry.
$6$. Since there are two unpaired electrons,the complex is paramagnetic.
$7$. It is a high spin complex because the electrons remain unpaired.
Therefore,statements $(a), (c), (d),$ and $(e)$ are true.

(C) $(a): \text{In } [Cu(NH_{3})_{4}]^{2+}, Cu \text{ is present as } Cu^{2+}. \ Cu^{2+} = [Ar] 3d^{9} 4s^{0}. \text{ The } NH_{3} \text{ ligand causes } dsp^{2} \text{ hybridisation, resulting in a square planar geometry.}$
$(b): \text{In } [Ni(CO)_{4}], CO \text{ is a neutral ligand.}$
$(c): \text{In } [Fe(CN)_{6}]^{3-}, Fe \text{ is present as } Fe^{3+}. \ Fe^{3+} = [Ar] 3d^{5} 4s^{0}. \text{ Since } CN^{-} \text{ is a strong field ligand, it causes pairing of electrons. Thus, its hybridisation is } d^{2}sp^{3}, \text{ not } sp^{3}d^{2}. \text{ It is an inner orbital complex.}$
$(d): \text{In } [Co(en)_{3}]^{3+}, \text{ the central metal } Co^{3+} \text{ has } 24 \text{ electrons and } 6 \text{ ligands donate } 12 \text{ electrons, total } = 36 \text{ electrons, which follows the EAN rule.}$

(B) For the complex $[CoF_6]^{3-}$:
$1$. The coordination number is $6$,so it has an octahedral geometry. Statement $I$ is true.
$2$. The oxidation state of $Co$ is $x + 6(-1) = -3$,which gives $x = +3$. The coordination number is $6$. Statement $II$ is false.
$3$. $F^-$ is a weak field ligand,so it forms an outer orbital complex with $sp^3d^2$ hybridization. Statement $III$ is true.
$4$. Since it is an outer orbital complex with weak field ligands,it remains a high spin complex. Statement $IV$ is true.
Therefore,statements $I$,$III$,and $IV$ are correct.

(D) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ levels.
For $d^9$ (high spin): $t_{2g}^6 e_g^3$,number of unpaired electrons $(n)$ = $1$.
For $d^6$ (low spin): $t_{2g}^6 e_g^0$,$n = 0$.
For $d^4$ (low spin): $t_{2g}^4 e_g^0$,$n = 2$.
For $d^7$ (high spin): $t_{2g}^5 e_g^2$,$n = 3$.
Therefore,the $d^7$ (high spin) system has the maximum number of unpaired electrons.

(A) Let us analyze each option:
$A$. $[ZnBr_{4}]^{2-}$: $Br^{-}$ is a weak field ligand and does not cause pairing of electrons. $Zn^{2+}$ has a $d^{10}$ configuration. The hybridization is $sp^3$,resulting in a tetrahedral geometry. This statement is correct.
$B$. $[Ni(NH_{3})_{6}]^{2+}$: $Ni^{2+}$ is $3d^8$. In the presence of $NH_{3}$,it forms an outer orbital complex with $sp^3d^2$ hybridization. It is not an inner orbital complex.
$C$. $[Co(NH_{3})_{6}]^{2+}$: $Co^{2+}$ is $3d^7$. In the presence of $NH_{3}$,it forms a low-spin complex,but it is typically oxidized to $Co^{3+}$ in the presence of air/ligands. However,strictly considering $[Co(NH_{3})_{6}]^{2+}$,it has one unpaired electron and is paramagnetic. Note: The question asks for the correct statement,and $A$ is definitively correct.
$D$. $[CoBr_{2}(en)_{2}]^{-}$: This complex does not contain ambidentate ligands (like $NO_{2}^{-}$ or $SCN^{-}$),so it cannot exhibit linkage isomerism.
Therefore,the most accurate statement is $A$.

(D) When aluminium chloride is dissolved in an acidified aqueous solution,it forms the hexa-aquaaluminium$(III)$ complex ion,$[Al(H_2O)_6]^{3+}$.
In this complex,the central $Al^{3+}$ ion has an electronic configuration of $[Ne] 3s^0 3p^0 3d^0$.
To accommodate six lone pairs from six water molecules,the $Al^{3+}$ ion undergoes $sp^3d^2$ hybridisation,utilizing one $3s$,three $3p$,and two $3d$ orbitals.
The reaction is: $AlCl_3 + 6H_2O \longrightarrow [Al(H_2O)_6]^{3+} + 3Cl^-$.

(C) The atomic number of $Co$ is $27$. Its ground state electronic configuration is $[Ar] 3d^7 4s^2$.
In the complex $[CoF_6]^{3-}$,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Since $F^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $Co^{3+}$ ion uses one $4s$,three $4p$,and two $4d$ orbitals to form six $sp^3d^2$ hybrid orbitals to accommodate six pairs of electrons from six $F^-$ ions.
Thus,the hybridisation is $sp^3d^2$.

(A) Since no precipitate of $AgCl$ is obtained on treating the compound with silver nitrate,it means no $Cl^{-}$ ions are available outside the coordination sphere.
Thus,all $Cl^{-}$ ions must be present inside the coordination sphere.
The correct formula of the coordination compound is $[Co(NH_3)_3 Cl_3]$.
Since the central metal $Co^{3+}$ forms six coordination bonds with the ligands,the secondary valency is $6$.

(C) $A. Ni(CO)_4$: $Ni$ is in $0$ oxidation state. $CO$ is a strong field ligand. $Ni(0) = [Ar] 3d^8 4s^2$. Due to $CO$,electrons pair up to give $3d^{10}$. Hybridization is $sp^3$ $(IV)$.
$B. [Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state. $CN^{-}$ is a strong field ligand. $Ni^{2+} = [Ar] 3d^8$. Electrons pair up,leaving one $3d$ orbital vacant. Hybridization is $dsp^2$ $(III)$.
$C. [Co(NH_3)_6]^{3+}$: $Co$ is in $+3$ oxidation state. $NH_3$ is a strong field ligand. $Co^{3+} = [Ar] 3d^6$. Electrons pair up,leaving two $3d$ orbitals vacant. Hybridization is $d^2sp^3$ $(II)$.
$D. [CoF_6]^{3-}$: $Co$ is in $+3$ oxidation state. $F^{-}$ is a weak field ligand. $Co^{3+} = [Ar] 3d^6$. No pairing occurs. Hybridization is $sp^3d^2$ $(I)$.
Therefore,the correct match is $A-IV, B-III, C-II, D-I$,which corresponds to option $C$ $(I-D, II-C, III-B, IV-A)$.

(B) $i$. In $[Mn(CN)_6]^{3-}$,$Mn^{3+}$ is $3d^4$. $CN^-$ is a strong field ligand,so it is $d^2sp^3$ (inner orbital) with $2$ unpaired electrons (paramagnetic). In $[MnCl_6]^{3-}$,$Mn^{3+}$ is $3d^4$. $Cl^-$ is a weak field ligand,so it is $sp^3d^2$ (outer orbital) with $4$ unpaired electrons (paramagnetic). Thus,$i$ is correct.
$ii$. $[NiCl_4]^{2-}$ is $sp^3$ tetrahedral and paramagnetic ($2$ unpaired electrons). $[Ni(CO)_4]$ is $sp^3$ tetrahedral and diamagnetic. Thus,$ii$ is incorrect.
$iii$. $[Co(NH_3)_6]^{3+}$ involves $d^2sp^3$ hybridization (inner orbital). $[Ni(NH_3)_6]^{2+}$ involves $sp^3d^2$ hybridization (outer orbital). Thus,$iii$ is correct.
Therefore,the correct statements are $i$ and $iii$.

(A) Let us analyze each complex:
$1$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong-field ligand,causing pairing. The configuration becomes $t_{2g}^4 e_g^0$. It is an inner-orbital complex $(d^2sp^3)$ and is paramagnetic ($2$ unpaired electrons).
$2$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong-field ligand. The configuration becomes $t_{2g}^5 e_g^0$. It is an inner-orbital complex $(d^2sp^3)$ and is paramagnetic ($1$ unpaired electron).
$3$. $[Co(C_2O_4)_3]^{3-}$: $Co^{3+}$ is $3d^6$. $C_2O_4^{2-}$ is a weak-field ligand. It forms an outer-orbital complex $(sp^3d^2)$ and is diamagnetic.
$4$. $[MnCl_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $Cl^-$ is a weak-field ligand. It forms an outer-orbital complex $(sp^3d^2)$ and is paramagnetic.
$5$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong-field ligand. The configuration becomes $t_{2g}^6 e_g^0$. It is an inner-orbital complex $(d^2sp^3)$ and is diamagnetic.
$6$. $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak-field ligand. It forms an outer-orbital complex $(sp^3d^2)$ and is paramagnetic.
Inner-orbital complexes are those using $(n-1)d$ orbitals. These are $[Mn(CN)_6]^{3-}, [Fe(CN)_6]^{3-}, [Fe(CN)_6]^{4-}$.
Among these,the paramagnetic ones are $[Mn(CN)_6]^{3-}$ and $[Fe(CN)_6]^{3-}$.
Thus,the total number is $2$.

(D) The spin-only magnetic moment is given by the formula $\mu_{s} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu_{s} = 5.9 \ BM$,we have $\sqrt{n(n+2)} = 5.9$.
Squaring both sides,$n(n+2) \approx 34.81$,which gives $n = 5$.
Since $Mn$ $(Z=25)$ has the electronic configuration $[Ar] 3d^5 4s^2$,to have $5$ unpaired electrons,$Mn$ must be in the $+2$ oxidation state $(Mn^{2+}: [Ar] 3d^5)$.
$Br^-$ is a weak-field ligand,so it does not cause pairing of electrons,resulting in $sp^3$ hybridization.
For the complex $[MnBr_4]^{x-}$,the oxidation state of $Mn$ is $x-4 = -2$,so $x = 2$.
The geometry of $sp^3$ hybridized complexes is tetrahedral.

(D) In $[Mn(CN)_6]^{3-}$,$Mn$ is in $+3$ oxidation state ($3d^4$ configuration).
Since $CN^-$ is a strong field ligand,it causes the pairing of $3d$ electrons,making two $3d$-orbitals vacant. These two $3d$,one $4s$,and three $4p$-orbitals hybridise to give $d^2sp^3$ hybridisation.
In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state ($3d^6$ configuration).
Since $F^-$ is a weak field ligand,it cannot cause the pairing of unpaired $3d$ electrons. Thus,one $4s$,three $4p$,and two $4d$-orbitals hybridise to give $sp^3d^2$ hybridisation.

(B) According to valence bond theory,the central metal atom or ion makes available a number of empty orbitals equal to its coordination number for the formation of coordinate bonds with ligands.
These empty orbitals are hybridized to form a set of equivalent hybrid orbitals of definite geometry.
For transition metals,the orbitals involved in hybridization are $(n-1)d$,$ns$,and $np$ orbitals.