Hybridisation and Geometry Questions in English

(A) The oxidation number of $Ni$ in both $[Ni(PPh_3)_2Cl_2]$ and $[NiCl_4]^{2-}$ is $+2$. The electronic configuration of $Ni(II)$ is $[Ar] 3d^8 4s^0$.
In $[Ni(PPh_3)_2Cl_2]$,$PPh_3$ is a strong field ligand,which forces the electrons in the $3d$ orbital to pair up,leaving one $3d$ orbital vacant. This allows for $dsp^2$ hybridisation,resulting in a square planar geometry.
In $[NiCl_4]^{2-}$,$Cl^-$ is a weak field ligand,which does not cause pairing of electrons in the $3d$ orbital. Thus,the $4s$ and $4p$ orbitals are used for $sp^3$ hybridisation,resulting in a tetrahedral geometry.
Therefore,the hybridisation of $Ni$ in $[Ni(PPh_3)_2Cl_2]$ and $[NiCl_4]^{2-}$ are $dsp^2$ and $sp^3$ respectively.

(B) For $[Fe(H_2O)_6]^{3+}$:
$Fe$ is in $+3$ oxidation state,$Fe^{3+} = [Ar] 3d^5$.
$H_2O$ is a weak field ligand $(WFL)$,so it does not cause pairing of electrons.
Thus,the hybridization is $sp^3d^2$ (outer orbital complex).
For $[Co(NH_3)_6]^{3+}$:
$Co$ is in $+3$ oxidation state,$Co^{3+} = [Ar] 3d^6$.
$NH_3$ is a strong field ligand $(SFL)$,which causes pairing of electrons in the $3d$ orbitals.
Thus,the hybridization is $d^2sp^3$ (inner orbital complex).
Therefore,the hybridizations are $sp^3d^2$ and $d^2sp^3$ respectively.

(C) The central metal ion is $Co^{2+}$,which has a $3d^7$ electronic configuration.
$Cl^-$ is a weak field ligand $(WFL)$,so pairing does not occur.
In a tetrahedral crystal field,the $d$-orbitals split into $e$ (lower energy) and $t_2$ (higher energy) sets.
According to Hund's rule,the $7$ electrons are filled as follows: $e^4 t_2^3$.
Here,$m = 4$ and $n = 3$.
The number of unpaired electrons is $3$.
Sum of '$m$' and the number of unpaired electrons $= 4 + 3 = 7$.

(C) $1$. The oxidation state of $Co$ in $[Co(NH_3)_6]^{3+}$ is $+3$.
$2$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$3$. $NH_3$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
$4$. After pairing,the $3d$ orbitals have two vacant $d$-orbitals,one $4s$ orbital,and three $4p$ orbitals available for hybridization.
$5$. This results in $d^2sp^3$ hybridization,which is inner orbital octahedral geometry.
$6$. Since all electrons are paired in the $3d$ orbitals,the complex is diamagnetic.

(D) $1$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing. Hybridisation is $sp^3$ $(I)$.
$2$. $[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ is $3d^9$. The unpaired electron is in $4p_z$,and the hybridisation is $dsp^2$ $(II)$.
$3$. $[Fe(NH_3)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing. Hybridisation is $d^2sp^3$ $(IV)$.
$4$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,no pairing. Hybridisation is $sp^3d^2$ $(III)$.
Thus,the correct match is $A-I, B-II, C-IV, D-III$.

(B) In a $fac-[Co(NH_3)_3Cl_3]$ complex,the three $Cl^-$ ligands occupy one face of the octahedral geometry.
This means the three $Cl^-$ ligands are at the corners of a triangular face.
In this arrangement,each $Cl^-$ ligand is adjacent to the other two $Cl^-$ ligands.
The bond angle between any two adjacent $Cl^-$ ligands in an octahedral geometry is $90^{\circ}$.
Therefore,the only $Cl-Co-Cl$ bond angle present is $90^{\circ}$.

(A) When $CoCl_2$ is dissolved in water,it forms the pink-colored hexaaquacobalt$(II)$ ion,$[Co(H_2O)_6]^{2+}$,which has an octahedral geometry $(X)$.
Upon adding concentrated $HCl$,the water ligands are replaced by chloride ions to form the deep blue tetrachlorocobaltate$(II)$ ion,$[CoCl_4]^{2-}$ $(Y)$,which has a tetrahedral geometry $(Z)$.

(D) $(A) [NiCl_4]^{2-}$: $Ni^{2+} = [Ar] 3d^8$. $Cl^-$ is a weak field ligand,so it is paramagnetic,high spin,and tetrahedral.
$(B) [CoCl_6]^{3-}$: $Co^{3+} = [Ar] 3d^6$. $Cl^-$ is a weak field ligand,so it is paramagnetic,high spin,and octahedral.
$(C) [CoF_6]^{3-}$: $Co^{3+} = [Ar] 3d^6$. $F^-$ is a weak field ligand,so it is paramagnetic,high spin,and octahedral.
$(D) [Co(NH_3)_6]^{3+}$: $Co^{3+} = [Ar] 3d^6$. $NH_3$ is a strong field ligand,which causes pairing of electrons in $3d$ orbitals. Thus,it is diamagnetic,low spin,and octahedral.

(B) The complex $[NiCl_2Br_2]^{2-}$ involves a $3d$ metal ion with weak field ligands,resulting in a tetrahedral geometry,which is paramagnetic and does not exhibit geometrical isomerism.
The complex $[PtCl_2Br_2]^{2-}$ involves a $5d$ metal ion,which forces pairing of electrons,resulting in a square planar geometry,which is diamagnetic and exhibits geometrical isomerism (cis and trans forms).
Therefore,the properties that change are:
$1.$ Geometry (Tetrahedral to Square Planar)
$2.$ Geometrical isomerism (Absent to Present)
$3.$ Magnetic properties (Paramagnetic to Diamagnetic)
Thus,$A, B,$ and $D$ are expected to change.

(A) $[Co(NH_3)_6]^{3+}$: The central metal ion $Co^{3+}$ has a $d^6$ configuration. In the presence of strong field ligand $NH_3$,electrons pair up,providing two vacant $3d$ orbitals for $d^2 sp^3$ hybridization.
$BrF_5$: The central atom $Br$ undergoes $sp^3 d^2$ hybridization.
$[PtCl_4]^{2-}$: The central metal ion $Pt^{2+}$ undergoes $dsp^2$ hybridization.
$SF_6$: The central atom $S$ undergoes $sp^3 d^2$ hybridization.

(A) In an acidified aqueous solution,$AlCl_3$ undergoes hydrolysis to form the hexaaquaaluminium$(III)$ ion,$[Al(H_2O)_6]^{3+}$.
This complex ion exhibits an octahedral geometry.

(A) In the molecule $Mn_2(CO)_{10}$,each manganese atom is bonded to five carbonyl $(CO)$ ligands and one other manganese atom.
This results in a total of six coordination sites around each $Mn$ atom.
The arrangement of these six ligands around the central $Mn$ atom corresponds to an octahedral geometry.

(D) For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $sp^3$ hybridization with $0$ unpaired electrons,making it diamagnetic.
For $[NiCl_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The electronic configuration is $[Ar] 3d^8$. Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons,resulting in $sp^3$ hybridization with $2$ unpaired electrons,making it paramagnetic.

(B) For $[Co(NH_3)_6]^{3+}$: The central metal ion is $Co^{3+}$ ($3d^6$ configuration). $NH_3$ is a strong field ligand,which causes pairing of electrons in the $d$-orbitals. This results in a $d^2sp^3$ hybridization,forming an inner orbital complex (spin-paired complex).
For $[CoF_6]^{3-}$: The central metal ion is $Co^{3+}$ ($3d^6$ configuration). $F^-$ is a weak field ligand,which does not cause pairing of electrons. This results in an $sp^3d^2$ hybridization,forming an outer orbital complex (spin-free complex).
Therefore,the correct sequence is spin-paired complex and spin-free complex.

(B) The complex is $[Fe(NH_3)_{x}(CN)_{y}]^{-}$.
The oxidation state of $Fe$ is $+3$, so the electronic configuration of $Fe^{3+}$ is $3d^5$.
For the $e_g$ orbital to have no electrons, all $5$ electrons must occupy the $t_{2g}$ orbitals.
This requires a strong field ligand environment to cause large crystal field splitting $(\Delta_o > P)$.
$CN^-$ is a strong field ligand, while $NH_3$ is a moderate field ligand.
For the complex to be low-spin with $t_{2g}^5 e_g^0$, the coordination number must be $6$ to satisfy the geometry of an octahedral complex.
Thus, $x+y = 6$.
For example, in $[Fe(CN)_6]^{3-}$, $x=0$ and $y=6$, which satisfies the condition.

(D) To determine the electronic configuration in tetrahedral complexes,we first find the oxidation state of the central metal ion:
$1$. $TiCl_4$: $Ti$ is in $+4$ state $(3d^0)$. Configuration: $e^0, t_2^0$ $(III)$.
$2$. $[FeO_4]^{2-}$: $Fe$ is in $+6$ state $(3d^2)$. Configuration: $e^2, t_2^0$ $(I)$.
$3$. $[FeCl_4]^{-}$: $Fe$ is in $+3$ state $(3d^5)$. Configuration: $e^2, t_2^3$ $(IV)$.
$4$. $[CoCl_4]^{2-}$: $Co$ is in $+2$ state $(3d^7)$. Configuration: $e^4, t_2^3$ $(II)$.
Thus,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) To determine the hybridization,we analyze the electronic configuration of the central metal ion and the nature of the ligand:
$A. K_2[Ni(CN)_4]$: $Ni^{2+}$ is $[Ar] 3d^8$. $CN^-$ is a strong field ligand $(SFL)$,causing pairing of electrons in $3d$ orbitals,resulting in $dsp^2$ hybridization.
$B. [Ni(CO)_4]$: $Ni$ is $[Ar] 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing,resulting in $sp^3$ hybridization.
$C. [Co(NH_3)_6]Cl_3$: $Co^{3+}$ is $[Ar] 3d^6$. $NH_3$ is a strong field ligand for $Co^{3+}$,causing pairing,resulting in $d^2sp^3$ hybridization.
$D. Na_3[CoF_6]$: $Co^{3+}$ is $[Ar] 3d^6$. $F^-$ is a weak field ligand $(WFL)$,so no pairing occurs,resulting in $sp^3d^2$ hybridization.
Matching the results: $A-III, B-I, C-IV, D-II$.

(B) $EDTA^{4-}$ is a hexadentate ligand,meaning it forms $6$ coordinate bonds with the central metal ion.
In the complex $[Ca(EDTA)]^{2-}$,the $Ca^{2+}$ ion is surrounded by $6$ donor atoms from the $EDTA^{4-}$ ligand.
Therefore,the coordination number is $6$,which corresponds to an octahedral geometry.

(D) The group-$IV$ cation forming a black precipitate with $H_2S$ is $Co^{2+}$.
$Co^{2+} + H_2S \rightarrow CoS \downarrow$ (Black precipitate,$A$)
$CoS + \text{aqua regia} \rightarrow Co^{2+} + NOCl + S + H_2O$ ($B$ is $Co^{2+}$)
$Co^{2+} + KNO_2 + CH_3COOH \rightarrow K_3[Co(NO_2)_6] + NO + H_2O$ ($C$ is $K_3[Co(NO_2)_6]$)
In $K_3[Co(NO_2)_6]$,the oxidation state of $Co$ is $+3$.
Electronic configuration of $Co^{3+}$ is $3d^6$.
Since $NO_2^-$ is a strong field ligand,it causes pairing of electrons,resulting in $d^2sp^3$ hybridization.
Number of unpaired electrons $(n)$ = $0$.
Magnetic moment = $\sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \ BM$.

(C) In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,leading to $sp^3$ hybridization and tetrahedral geometry.
In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The electronic configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,which forces the $3d$ electrons to pair up,resulting in $dsp^2$ hybridization and square planar geometry.

(A) In the complex $[Fe(H_2O)_5NO]SO_4$,the $NO$ ligand exists as $NO^+$.
Thus,the oxidation state of $Fe$ is calculated as: $x + 5(0) + 1 = +2$ (since $SO_4$ is $-2$,the complex ion is $[Fe(H_2O)_5NO]^{2+}$). Wait,for $[Fe(H_2O)_5NO]SO_4$,$x + 0 + 1 = 2$,so $x = +1$.
The electronic configuration of $Fe^+$ $(Z=26)$ is $[Ar] 3d^6 4s^1$.
In the presence of the strong field ligand $NO^+$,the $4s$ electron pairs up into the $3d$ orbital.
The configuration becomes $3d^7$,which has $3$ unpaired electrons as shown in the orbital diagram.
Since it has unpaired electrons,the complex is paramagnetic.
Therefore,$STATEMENT-1$ is True and $STATEMENT-2$ is True,and $STATEMENT-2$ correctly explains $STATEMENT-1$.

(A) The ammonia complexes of the given metal ions are:
$1$. $Ni^{2+}$ forms $[Ni(NH_3)_6]^{2+}$,which has an octahedral geometry.
$2$. $Pt^{2+}$ forms $[Pt(NH_3)_4]^{2+}$,which has a square planar geometry.
$3$. $Zn^{2+}$ forms $[Zn(NH_3)_4]^{2+}$,which has a tetrahedral geometry.
Therefore,the geometries are octahedral,square planar,and tetrahedral,respectively.
Thus,the correct option is $A$.

(C) $1$. Identification of complexes: $X$ is $[Co(H_2O)_6]Cl_2$ (pink,octahedral,$d^7$ high spin,$\mu = 3.87 \ B.M.$).
$2$. Formation of $Y$: $[Co(H_2O)_6]Cl_2 + NH_3(aq) + NH_4Cl + air \rightarrow [Co(NH_3)_6]Cl_3 (Y)$. $Y$ is $[Co(NH_3)_6]Cl_3$,which is a $1:3$ electrolyte. The $Co^{3+}$ ion in $Y$ is $d^6$ low spin,so hybridization is $d^2sp^3$ and magnetic moment is $0 \ B.M$.
$3$. Formation of $Z$: $[Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} (Z) + 6H_2O$. $Z$ is $[CoCl_4]^{2-}$,which is tetrahedral and blue.
$4$. Evaluating options:
$[A]$ Correct: $Co^{3+}$ in $[Co(NH_3)_6]^{3+}$ is $d^2sp^3$ hybridized.
$[B]$ Correct: $[CoCl_4]^{2-}$ is a tetrahedral complex.
$[C]$ Incorrect: $[Co(NH_3)_6]Cl_3$ gives $3$ equivalents of $AgCl$ with $AgNO_3$,not $2$.
$[D]$ Correct: The reaction $[Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O$ is endothermic $(\Delta H > 0)$. At $0^{\circ}C$ (low temperature),the equilibrium shifts to the left (towards $X$),making the solution pink.
Thus,statements $A, B, D$ are correct.

(B) $Ni^{2+}$ has a $3d^8$ electronic configuration.
$1$. For $[NiCl_4]^{2-}$,$Cl^{-}$ is a weak field ligand,so it does not cause pairing of electrons. The hybridization is $sp^3$,resulting in a tetrahedral geometry.
$2$. For $[Ni(CN)_4]^{2-}$,$CN^{-}$ is a strong field ligand,which causes pairing of $3d$ electrons. The hybridization is $dsp^2$,resulting in a square planar geometry.
$3$. For $[Ni(H_2O)_6]^{2+}$,$H_2O$ is a ligand that forms an octahedral complex with $Ni^{2+}$ involving $sp^3d^2$ hybridization.
Thus,the shapes are tetrahedral,square planar,and octahedral,respectively. Therefore,option $B$ is correct.

(C) $P: dsp^2 - (6) \ [Ni(CN)_4]^{2-}$. $Ni^{2+}$ has $3d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. Hybridization is $dsp^2$ (square planar).
$Q: sp^3 - (4), (5)$.
$(4) [FeCl_4]^{2-}: Fe^{2+}$ is $3d^6$. $Cl^-$ is a weak field ligand,no pairing. Hybridization is $sp^3$ (tetrahedral).
$(5) Ni(CO)_4: Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,forcing $4s$ electrons into $3d$. Hybridization is $sp^3$ (tetrahedral).
$R: sp^3d^2 - (1)$.
$(1) [FeF_6]^{4-}: Fe^{2+}$ is $3d^6$. $F^-$ is a weak field ligand,no pairing. Hybridization is $sp^3d^2$ (octahedral).
$S: d^2sp^3 - (2), (3)$.
$(2) [Ti(H_2O)_3Cl_3]: Ti^{3+}$ is $3d^1$. Hybridization is $d^2sp^3$ (octahedral).
$(3) [Cr(NH_3)_6]^{3+}: Cr^{3+}$ is $3d^3$. Hybridization is $d^2sp^3$ (octahedral).
Thus,the correct match is $P$ $\rightarrow 6; Q$ $\rightarrow 4,5; R$ $\rightarrow 1; S$ $\rightarrow 2,3$.

(A) $[FeCl_4]^-$: $Fe^{3+}$ is $d^5$ system. $Cl^-$ is a weak field ligand,so hybridization is $sp^3$ (Tetrahedral).
$[Fe(CO)_4]^{2-}$: $Fe^{2-}$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral).
$(B)$ $[Co(CO)_4]^-$: $Co^-$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral).
$[CoCl_4]^{2-}$: $Co^{2+}$ is $d^7$ system. $Cl^-$ is a weak field ligand,so hybridization is $sp^3$ (Tetrahedral).
$(C)$ $[Ni(CO)_4]$: $Ni^0$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral).
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $d^8$ system. $CN^-$ is a strong field ligand,so hybridization is $dsp^2$ (Square planar).
$(D)$ $[Cu(py)_4]^+$: $Cu^+$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral).
$[Cu(CN)_4]^{3-}$: $Cu^+$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral).
Therefore,pairs $(A), (B),$ and $(D)$ exhibit tetrahedral geometry.

(C) The complex $[NiCl_2\{P(C_2H_5)_2(C_6H_5)\}_2]$ contains $Ni^{2+}$ with the electronic configuration $[Ar] 3d^8$.
In the paramagnetic state,the complex adopts a tetrahedral geometry with $sp^3$ hybridization,where the two unpaired electrons in the $3d$ orbitals result in paramagnetism.
In the diamagnetic state,the complex adopts a square planar geometry with $dsp^2$ hybridization,where all electrons are paired,resulting in diamagnetism.
Therefore,the coordination geometries in the paramagnetic and diamagnetic states are tetrahedral and square planar,respectively.

(B) $EDTA^{4-}$ is a hexadentate ligand that forms a complex with $Co^{3+}$ ion.
In the $[Co(EDTA)]^{-}$ complex,the $Co^{3+}$ ion is octahedrally coordinated by two nitrogen atoms and four oxygen atoms from the $EDTA^{4-}$ ligand.
To find the number of $N-Co-O$ bond angles,we look at the coordination geometry.
Each nitrogen atom is bonded to the cobalt center and is part of three five-membered chelate rings.
Specifically,each $N$ atom is involved in $N-Co-O$ angles with the oxygen atoms of the acetate groups.
By analyzing the structure,each of the two $N$ atoms forms $4$ such $N-Co-O$ angles (where the $O$ atom is one of the four coordinated oxygen atoms).
Therefore,the total number of $N-Co-O$ bond angles is $2 \times 4 = 8$.

(D) To determine the number of square planar species,we analyze the geometry of each:
$1$. $XeF_4$: $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar shape.
$2$. $SF_4$: $sp^3d$ hybridization with $1$ lone pair,resulting in a see-saw shape.
$3$. $SiF_4$: $sp^3$ hybridization,resulting in a tetrahedral shape.
$4$. $BF_4^-$: $sp^3$ hybridization,resulting in a tetrahedral shape.
$5$. $BrF_4^-$: $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar shape.
$6$. $[Cu(NH_3)_4]^{2+}$: $dsp^2$ hybridization,resulting in a square planar shape.
$7$. $[FeCl_4]^{2-}$: $sp^3$ hybridization,resulting in a tetrahedral shape.
$8$. $[CoCl_4]^{2-}$: $sp^3$ hybridization,resulting in a tetrahedral shape.
$9$. $[PtCl_4]^{2-}$: $dsp^2$ hybridization,resulting in a square planar shape.
The species with a square planar shape are $XeF_4, BrF_4^-, [Cu(NH_3)_4]^{2+},$ and $[PtCl_4]^{2-}$.
Therefore,the total number of square planar species is $4$.

(B) For $M1$: $Ni^{2+}$ is known to form tetrahedral complexes with $Cl^-$ (e.g.,$[NiCl_4]^{2-}$) and square planar complexes with $CN^-$ (e.g.,$[Ni(CN)_4]^{2-}$). Thus,$M1 = Ni^{2+}$,$Q = HCl$,and $R = KCN$.
For $M2$: $Zn^{2+}$ is a metal ion that forms tetrahedral complexes and is amphoteric. When $Zn^{2+}$ reacts with $KOH$ $(S)$,it forms a white precipitate of $Zn(OH)_2$,which dissolves in excess $KOH$ to form the soluble complex $[Zn(OH)_4]^{2-}$.
Therefore,$M1 = Ni^{2+}$,$Q = HCl$,$R = KCN$,and $S = KOH$.
Matching with the options:
Question $1$: $M1=Ni^{2+}$,$Q=HCl$,$R=KCN$ corresponds to option $(B)$.
Question $2$: $S=KOH$ corresponds to option $(D)$.
The correct answer is $(B, D)$.

(D) $1$. $[Co(CN)_4]^{4-}$: $Co^{0}$ $(d^9 s^2)$ in $CN^-$ field is tetrahedral.
$2$. $[Co(CO)_3(NO)]$: $Co$ is $d^9 s^2$,$CO$ and $NO$ are strong field ligands,geometry is tetrahedral.
$3$. $XeF_4$: $sp^3d^2$ hybridization,square planar geometry.
$4$. $[PCl_4]^{+}$: $sp^3$ hybridization,tetrahedral geometry.
$5$. $[PdCl_4]^{2-}$: $Pd^{2+}$ $(d^8)$ is square planar.
$6$. $[ICl_4]^{-}$: $sp^3d^2$ hybridization,square planar geometry.
$7$. $[Cu(CN)_4]^{3-}$: $Cu^{+}$ $(d^{10})$ is tetrahedral.
$8$. $P_4$: $P$ atoms are $sp^3$ hybridized,tetrahedral structure.
Thus,the species with tetrahedral geometry are $[Co(CN)_4]^{4-}, [Co(CO)_3(NO)], [PCl_4]^{+}, [Cu(CN)_4]^{3-}$,and $P_4$. The total count is $5$.

(B) For $[NiCl_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The electronic configuration is $[Ar] 3d^8$. Due to the weak field ligand $Cl^-$,it undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with $2$ unpaired electrons,making it paramagnetic.
For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^8 4s^2$. In the presence of the strong field ligand $CO$,the electrons pair up to form $[Ar] 3d^{10} 4s^0$. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with $0$ unpaired electrons,making it diamagnetic.

(C) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 3.95 \ BM$,we have $\sqrt{n(n+2)} \approx 3.95$,which implies $n(n+2) \approx 15.6$,so $n \approx 3$.
$Co(II)$ has a $3d^7$ configuration.
In an octahedral field,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
For a $d^7$ system with $3$ unpaired electrons,the configuration is $t_{2g}^5 e_g^2$ (high spin).
This corresponds to $5$ electrons in $t_{2g}$ (with $1$ unpaired) and $2$ electrons in $e_g$ (with $2$ unpaired),totaling $3$ unpaired electrons.

(C) To find the $d$-configuration of the central metal ion, we determine the oxidation state of the metal in each complex:
$(A)$ In $[FeO_4]^{2-}$, $Fe + 4(-2) = -2 \implies Fe = +6$. $Fe^{6+} = [Ar] 3d^2$.
$(B)$ In $[Mn(CN)_6]^{3-}$, $Mn + 6(-1) = -3 \implies Mn = +3$. $Mn^{3+} = [Ar] 3d^4$.
$(C)$ In $[Fe(CN)_6]^{3-}$, $Fe + 6(-1) = -3 \implies Fe = +3$. $Fe^{3+} = [Ar] 3d^5$.
$(D)$ In $[Cr(H_2O)_6]^{2+}$, $Cr + 6(0) = +2 \implies Cr = +2$. $Cr^{2+} = [Ar] 3d^4$.
$(E)$ In $[NiF_6]^{2-}$, $Ni + 6(-1) = -2 \implies Ni = +4$. $Ni^{4+} = [Ar] 3d^6$.
The complexes with $d^4$ configuration are $(B)$ and $(D)$.

(B) . $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so it forms outer orbital complex with $sp^3d^2$ hybridisation.
$B$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,so it forms tetrahedral complex with $sp^3$ hybridisation.
$C$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,so it causes pairing of electrons,forming inner orbital complex with $d^2sp^3$ hybridisation.
$D$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,so it causes pairing of electrons,forming square planar complex with $dsp^2$ hybridisation.
Therefore,the correct matching is $A-III, B-II, C-I, D-IV$.

(D) $(A) [MnBr_4]^{2-}$: $Mn^{2+}$ is $[Ar] 3d^5$. $Br^-$ is a weak field ligand,so it undergoes $sp^3$ hybridization and is paramagnetic.
$(B) [FeF_6]^{3-}$: $Fe^{3+}$ is $[Ar] 3d^5$. $F^-$ is a weak field ligand,so it undergoes $sp^3d^2$ hybridization and is paramagnetic.
$(C) [Co(C_2O_4)_3]^{3-}$: $Co^{3+}$ is $[Ar] 3d^6$. $C_2O_4^{2-}$ is a strong field ligand,so it undergoes $d^2sp^3$ hybridization and is diamagnetic.
$(D) [Ni(CO)_4]$: $Ni^0$ is $[Ar] 3d^8 4s^2$. $CO$ is a strong field ligand,so it undergoes $sp^3$ hybridization and is diamagnetic.
Therefore,the correct matching is $A-IV, B-II, C-I, D-III$.

(B) In $[MnCl_6]^{3-}$,the oxidation state of $Mn$ is $+3$.
The electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$.
Since $Cl^-$ is a weak field ligand $(WFL)$,it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the hybridization involved is $sp^3d^2$ (outer orbital complex).
The $3d^4$ configuration has $4$ unpaired electrons,making it paramagnetic.

(A) $PF_5$: The central atom $P$ has $5$ bonding pairs and $0$ lone pairs,so steric number is $5$,which corresponds to $sp^3d$ hybridisation.
$SF_6$: The central atom $S$ has $6$ bonding pairs and $0$ lone pairs,so steric number is $6$,which corresponds to $sp^3d^2$ hybridisation.
$Ni(CO)_4$: $Ni$ is in $0$ oxidation state. The $CO$ ligand is a strong field ligand,causing pairing of electrons in $3d$ orbitals. The hybridisation is $sp^3$.
$[PtCl_4]^{2-}$: $Pt$ is in $+2$ oxidation state. The $Cl^-$ ligand is a weak field ligand,but due to the high crystal field splitting energy of $5d$ series elements,it undergoes $dsp^2$ hybridisation.

(A) To determine the number of unpaired electrons,we analyze the electronic configuration of the metal ion in each complex:
$1$. $[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. Since $F^-$ is a weak field ligand,electrons remain unpaired. Number of unpaired electrons = $5$.
$2$. $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. Since $F^-$ is a weak field ligand,it forms a high-spin complex. Number of unpaired electrons = $4$.
$3$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. Number of unpaired electrons = $0$.
$4$. $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of all electrons in $3d$ orbital. Number of unpaired electrons = $0$.
Thus,the order is $[FeF_6]^{3-} (5) > [CoF_6]^{3-} (4) > [Ni(CN)_4]^{2-} (0) = [Ni(CO)_4] (0)$.

(A) To determine the hybridization and magnetic nature,we analyze the oxidation state of the metal and the nature of the ligand:

Complex	Hybridization	Magnetic Nature
$[FeF_6]^{3-}$	$sp^3d^2$	Paramagnetic
$[Fe(CN)_6]^{3-}$	$d^2sp^3$	Paramagnetic
$[Mn(CN)_6]^{3-}$	$d^2sp^3$	Paramagnetic
$[Co(C_2O_4)_3]^{3-}$	$d^2sp^3$	Diamagnetic
$[MnCl_6]^{3-}$	$sp^3d^2$	Paramagnetic
$[CoF_6]^{3-}$	$sp^3d^2$	Paramagnetic

From the table,the complexes that involve $d^2sp^3$ hybridization are $[Fe(CN)_6]^{3-}, [Mn(CN)_6]^{3-}$,and $[Co(C_2O_4)_3]^{3-}$.
Among these,$[Fe(CN)_6]^{3-}$ and $[Mn(CN)_6]^{3-}$ are paramagnetic.
Therefore,the number of such complexes is $2$.

(A) $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,so pairing occurs. $t_{2g}^5, e_g^0$. Unpaired electrons = $1$.
$[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,so no pairing. $t_{2g}^3, e_g^2$. Unpaired electrons = $5$.
$[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand. $t_{2g}^4, e_g^2$. Unpaired electrons = $4$.
$[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong field ligand. $t_{2g}^4, e_g^0$. Unpaired electrons = $2$.

(D) $1$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $[Ar] 3d^6$. $NH_3$ is a Strong Field Ligand $(SFL)$,causing pairing. Hybridization is $d^2 sp^3$.
$2$. $SF_6$: Sulphur has $6$ valence electrons and forms $6$ bonds. Hybridization is $sp^3 d^2$.
$3$. $[CrF_6]^{3-}$: $Cr^{3+}$ is $[Ar] 3d^3$. Hybridization is $d^2 sp^3$.
$4$. $[CoF_6]^{3-}$: $Co^{3+}$ is $[Ar] 3d^6$. $F^-$ is a Weak Field Ligand $(WFL)$,no pairing. Hybridization is $sp^3 d^2$.
$5$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $[Ar] 3d^4$. $CN^-$ is $SFL$. Hybridization is $d^2 sp^3$.
$6$. $[MnCl_6]^{3-}$: $Mn^{3+}$ is $[Ar] 3d^4$. $Cl^-$ is $WFL$. Hybridization is $sp^3 d^2$.
Species with $sp^3 d^2$ hybridization are $SF_6, [CoF_6]^{3-}$,and $[MnCl_6]^{3-}$.
Total count is $3$.

(A) $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state. Configuration: $[Ar] 3d^8 4s^2$. Due to strong field ligand $CO$,electrons pair up. Hybridization: $sp^3$. Shape: Tetrahedral. Magnetic moment: $0 \ BM$ (Diamagnetic). Matches $III$.
$(B)$ $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state. Configuration: $[Ar] 3d^8$. Due to strong field ligand $CN^-$,electrons pair up. Hybridization: $dsp^2$. Shape: Square planar. Magnetic moment: $0 \ BM$ (Diamagnetic). Matches $II$.
$(C)$ $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state. Configuration: $[Ar] 3d^8$. Weak field ligand $Cl^-$. Hybridization: $sp^3$. Shape: Tetrahedral. Two unpaired electrons,$\mu = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \approx 2.8 \ BM$. Matches $I$.
$(D)$ $[MnBr_4]^{2-}$: $Mn$ is in $+2$ oxidation state. Configuration: $[Ar] 3d^5$. Weak field ligand $Br^-$. Hybridization: $sp^3$. Shape: Tetrahedral. Five unpaired electrons,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.9 \ BM$. Matches $IV$.
Therefore,the correct matching is $A-III, B-II, C-I, D-IV$.

(B) The electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$.
In $[Co(NH_3)_6]^{2+}$,the oxidation state of $Co$ is $+2$,so the configuration is $3d^7$.
$NH_3$ is a strong field ligand,but in $Co^{2+}$ complexes,it does not force pairing of all electrons to form an inner orbital complex because the resulting configuration would be highly unstable. Instead,one electron is promoted to the $4d$ orbital to allow for $d^2sp^3$ hybridization,making the complex $[Co(NH_3)_6]^{2+}$ unstable and easily oxidized to the more stable $d^6$ $[Co(NH_3)_6]^{3+}$ complex.
The unpaired electron in $[Co(NH_3)_6]^{2+}$ is promoted to the $4d$ orbital,not the $4p$ orbital. Thus,the Assertion is true,but the Reason is false.

(C) For a metal ion with a $d^6$ configuration in an octahedral complex:
$1$. With a strong field ligand (e.g.,$CN^-$),the electrons pair up in the $t_{2g}$ orbitals,resulting in a low-spin complex $(t_{2g}^6 e_g^0)$ which is diamagnetic.
$2$. With a weak field ligand (e.g.,$F^-$),the electrons occupy all orbitals according to Hund's rule $(t_{2g}^4 e_g^2)$,resulting in a high-spin complex which is paramagnetic.
Therefore,the $d^6$ configuration allows for both possibilities depending on the ligand strength.

(B) The distribution of hybrid orbitals is determined by the hybridisation type:
$1$. $sp^3$ hybridisation corresponds to a tetrahedral geometry.
$2$. $sp^2$ hybridisation corresponds to a trigonal planar geometry.
$3$. $sp^3 d$ hybridisation corresponds to a trigonal bipyramidal geometry.
$4$. $sp^3 d^2$ (or $d^2 sp^3$) hybridisation corresponds to an octahedral geometry.
Therefore,the correct matching is: $(a-iii, b-iv, c-i, d-ii)$.

(A) The hybridization of the given complexes is determined as follows:
$(a)$ $[Ni(NH_3)_6]^{+2}$: $Ni^{+2}$ is $3d^8$. $NH_3$ is a weak field ligand,so it forms an outer orbital complex with $sp^3 d^2$ hybridization.
$(b)$ $[PtCl_4]^{-2}$: $Pt^{+2}$ is a $5d^8$ metal ion. $Cl^-$ is a ligand that forces pairing in $5d$ series,resulting in $dsp^2$ hybridization.
$(c)$ $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,forcing pairing to form $sp^3$ hybridization.
$(d)$ $[Co(OX)_3]^{-3}$: $Co^{+3}$ is $3d^6$. $OX^{-2}$ (oxalate) is a strong field ligand,resulting in $d^2 sp^3$ hybridization.
Thus,the correct matching is $a-i, b-ii, c-iii, d-iv$.

(A) . $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^{-}$ is a weak field ligand,so hybridization is $sp^3d^2$ (outer orbital complex) with $4$ unpaired electrons,making it paramagnetic. Matches $M$.
$B$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^{-}$ is a strong field ligand,so hybridization is $d^2sp^3$ (inner orbital complex) with $2$ unpaired electrons,making it paramagnetic. Matches $O$.
$C$. $[Co(C_2O_4)_3]^{3-}$: $Co^{3+}$ is $3d^6$. $C_2O_4^{2-}$ is a strong field ligand,so hybridization is $d^2sp^3$ (inner orbital complex) with $0$ unpaired electrons,making it diamagnetic. Matches $P$.
$D$. $[Zn(H_2O)_6]^{2+}$: $Zn^{2+}$ is $3d^{10}$. Hybridization is $sp^3d^2$ with $0$ unpaired electrons,making it diamagnetic. Matches $N$.
Therefore,the correct match is $A$ $\rightarrow M, B$ $\rightarrow O, C$ $\rightarrow P, D$ $\rightarrow N$.

(D) In $[Ti(H_2O)_6]^{4+}$,the oxidation state of $Ti$ is $+4$. The electronic configuration of $Ti^{4+}$ is $[Ar]3d^0$. Since there are no $d$-electrons,$d-d$ transition is not possible,and it is colourless. Thus,Statement-$1$ is False.
In $[Sc(H_2O)_6]^{3+}$,the oxidation state of $Sc$ is $+3$. The electronic configuration of $Sc^{3+}$ is $[Ar]3d^0$. Since there are no $d$-electrons,$d-d$ transition is not possible,and it is colourless. Thus,Statement-$2$ is True.
Therefore,Statement-$1$ is False and Statement-$2$ is True.

(D) The complex $[Ni(dmg)_2]$ (bis(dimethylglyoximato)nickel$(II)$) is a square planar complex.
$1$. $Ni^{2+}$ reacts with dimethylglyoxime $(dmgH)$ in the presence of a base to form a rosy red precipitate,which is a standard test for $Ni^{2+}$.
$2$. The hybridization of $Ni$ in this complex is $dsp^2$,resulting in a square planar geometry.
$3$. The structure contains $2$ chelating rings formed by the ligand and the metal,and $2$ additional hydrogen-bonded rings,totaling $4$ rings.
$4$. Each of the $2$ metal-ligand chelating rings is $5$-membered,not $6$-membered. Therefore,the statement that all rings are $6$-membered is incorrect.