Hybridisation and Geometry Questions in English

(A) $1$. The complex $[Cr(NH_3)_6]Cl_3$ dissociates as $[Cr(NH_3)_6]^{3+} + 3Cl^-$.
$2$. $Cr^{3+}$ has an electronic configuration of $[Ar] 3d^3$. It uses two $3d$,one $4s$,and three $4p$ orbitals to form $d^2sp^3$ hybrid orbitals,resulting in an inner orbital octahedral complex.
$3$. Since it involves $d^2sp^3$ hybridization,it is an inner orbital complex,not an outer orbital complex. Thus,statement $A$ is incorrect.
$4$. It reacts with $AgNO_3$ to give $3$ moles of $AgCl$ (white precipitate),so statement $B$ is correct.
$5$. It has $3$ unpaired electrons in the $3d$ subshell,making it paramagnetic,so statement $D$ is correct.

(NONE) The hybridization $sp^{3}d$ corresponds to a trigonal bipyramidal geometry,which is typically observed in $5$-coordinate complexes.
In the given options,all complexes involve $Co$ in a coordination environment that typically leads to octahedral geometry ($sp^{3}d^{2}$ or $d^{2}sp^{3}$).
However,if we analyze the coordination number,all listed complexes have a coordination number of $6$ ($en$ is bidentate,$NH_3$ and $Cl$ are monodentate).
None of the provided options exhibit $sp^{3}d$ hybridization as they are all $6$-coordinate octahedral complexes.
If the question implies a $5$-coordinate species,none of the options fit. Given the standard curriculum,this question appears to have no correct option among the choices provided.

(D) $1$. For $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing. Hybridization is $sp^3$. (Correct)
$2$. For $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing. Hybridization is $dsp^2$. (Incorrect,given as $sp^3$)
$3$. For $[CoF_6]^{3-}$: $Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak field ligand,no pairing. Hybridization is $sp^3d^2$. (Incorrect,given as $d^2sp^3$)
$4$. For $[Fe(CN)_6]^{3-}$: $Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ is a strong field ligand,causing pairing. Hybridization is $d^2sp^3$. (Incorrect,given as $sp^3d^2$)
Thus,$(ii), (iii),$ and $(iv)$ do not match.

(A) In the $[NiCl_4]^{2-}$ complex,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals. Thus,the hybridization is $sp^3$,which corresponds to a tetrahedral geometry.

(C) The complex is $[Cr(en)_2Cl_2]^+$. Let the oxidation state of $Cr$ be $x$.
$x + 2(0) + 2(-1) = +1$
$x - 2 = +1 \Rightarrow x = +3$.
$Cr$ is a $3d$ transition metal,so its valence shell involves $3d$ orbitals.
The ligand $en$ (ethylenediamine) is bidentate (coordination number $2$) and $Cl^-$ is monodentate (coordination number $1$).
Coordination number $= 2(2) + 2(1) = 4 + 2 = 6$.
Thus,the values are $+3, 3d$ and $6$.

(A) $1$. The coordination number of $Co$ is determined by the ligands: $5$ from $NH_3$ (monodentate) and $1$ from $CO_3^{2-}$ (bidentate),so $5 + 2 = 7$ is incorrect; wait,$CO_3^{2-}$ acts as a bidentate ligand,so $5 + 2 = 7$. However,in this specific complex $[Co(NH_3)_5CO_3]^+$,the $CO_3^{2-}$ is often treated as monodentate in some contexts,but standard $IUPAC$ considers it bidentate. Let's re-evaluate: $5(NH_3) + 1(CO_3) = 6$ coordination sites.
$2$. Oxidation state: $x + 5(0) + 1(-2) = +1$ (since $ClO_4$ is $-1$),so $x - 2 = +1$,$x = +3$.
$3$. Electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$. For $Co^{3+}$,it is $[Ar] 3d^6$.
$4$. Number of electrons in $d$-orbitals is $6$.
$5$. Since $NH_3$ is a strong field ligand,it causes pairing. For $d^6$ in an octahedral field,all electrons are paired,so unpaired electrons = $0$.

(D) In the complex $[PtCl_4]^{2-}$,the central metal ion is $Pt^{2+}$,which belongs to the $5d$ series.
For $5d$ series elements,the crystal field splitting energy is high,which forces the pairing of electrons even with weak field ligands like $Cl^-$.
This results in $dsp^2$ hybridization,leading to a square planar geometry.

(A) $1$. In $[MnO_4]^-$,the oxidation state of $Mn$ is $x + 4(-2) = -1$,so $x = +7$. The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. Thus,$Mn^{7+}$ is $[Ar] 3d^0 4s^0$. It has no $d$-electrons.
$2$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ state. $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$,so $Co^{3+}$ is $[Ar] 3d^6$. It has $d$-electrons.
$3$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ state. $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$,so $Fe^{3+}$ is $[Ar] 3d^5$. It has $d$-electrons.
$4$. In $[Cr(H_2O)_6]^{3+}$,$Cr$ is in $+3$ state. $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$,so $Cr^{3+}$ is $[Ar] 3d^3$. It has $d$-electrons.
Therefore,the correct option is $A$.

(B) In coordination chemistry,when $(n-1)d$ orbitals are used for hybridization (e.g.,$d^2sp^3$ hybridization),the complex is referred to as an inner orbital complex.
These complexes are typically formed by strong field ligands that cause pairing of electrons in the $(n-1)d$ orbitals.
Due to this pairing,the number of unpaired electrons decreases,resulting in a lower magnetic moment and a lower total spin value.
Therefore,these are known as low spin complexes.

(C) $1$. In $Ni(CO)_4$,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,resulting in $3d^{10}$ configuration. Thus,it is diamagnetic.
$2$. In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons in $3d$ orbitals,resulting in $d^8$ configuration with no unpaired electrons. Thus,it is diamagnetic.
$3$. In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $Cl^-$ is a weak field ligand,so it does not cause pairing. There are $2$ unpaired electrons in the $3d$ orbitals. Thus,it is paramagnetic.

(B) square planar complex is formed by $dsp^2$ hybridization.
This hybridization involves one $s$ orbital,two $p$ orbitals ($p_x$ and $p_y$),and one $d$ orbital.
The specific $d$ orbital involved is the $d_{x^2-y^2}$ orbital,which lies in the same plane as the $p_x$ and $p_y$ orbitals.

(C) In the complex $[Cu(NH_3)_4]^{2+}$,the central metal ion is $Cu^{2+}$.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
Due to the presence of $NH_3$ (a strong field ligand),the unpaired electron in the $3d$ orbital is promoted to the $4p$ orbital.
This results in $dsp^2$ hybridization,which corresponds to a square planar geometry.
Since there is one unpaired electron in the $4p$ orbital,the complex is paramagnetic.

(D) The complex $[Ag(NH_3)_2]^{+}$ involves $sp$ hybridization of the central $Ag^{+}$ ion.
Due to $sp$ hybridization,the geometry of the complex is linear.

(C) In $K_2[NiCl_4]$,the oxidation state of $Ni$ is calculated as: $2(+1) + x + 4(-1) = 0$,which gives $x = +2$.
The atomic number of $Ni$ is $28$,so its electronic configuration is $[Ar] \ 3d^8 \ 4s^2$.
For $Ni^{2+}$,the configuration is $[Ar] \ 3d^8$.
Thus,there are $8$ electrons in the $3d$ orbital.

(D) $(1) [Ag(NH_3)_2]^+$: $Ag^+$ has $d^{10}$ configuration. It undergoes $sp$ hybridization,resulting in a linear geometry.
$(2) [MnCl_4]^{2-}$: $Mn^{2+}$ has $d^5$ configuration. $Cl^-$ is a weak field ligand,leading to $sp^3$ hybridization,resulting in a tetrahedral geometry.
$(3) [Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ has $d^9$ configuration. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry.
$(4) [Ni(CN)_4]^{2-}$: $Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,leading to $dsp^2$ hybridization,resulting in a square planar geometry.
All the given geometries are correct. Therefore,the correct option is $D$.

(C) In $K_3[CoF_6]$,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.
$Co^{3+}$ has the electronic configuration $[Ar] 3d^6$.
$F^-$ is a weak field ligand,so it does not cause pairing of electrons.
Thus,$Co^{3+}$ has $4$ unpaired electrons,and the magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \, BM$.
This confirms it is a high spin complex with $sp^3d^2$ hybridization.
The primary valency refers to the oxidation state,which is $3$,not $6$. The coordination number is $6$ (secondary valency). Therefore,statement $C$ is incorrect.

(C) $1$. In $[Cr(NH_3)_6]Cl_3$,the oxidation state of $Cr$ is $+3$. The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.
$2$. Since $NH_3$ is a strong field ligand,it causes pairing,but for $d^3$ configuration,there are already $3$ unpaired electrons in $3d$ orbitals. The hybridization is $d^2sp^3$,which corresponds to an inner orbital complex,not an outer orbital complex.
$3$. The geometry is octahedral. The complex is paramagnetic due to the presence of $3$ unpaired electrons.
$4$. It reacts with $AgNO_3$ to give $3$ moles of $AgCl$ (white precipitate) because there are $3$ chloride ions outside the coordination sphere.
$5$. Therefore,the statement that it is an outer orbital complex is incorrect.

(A) To find the number of $d$-electrons,we determine the oxidation state of the central metal atom:
$1$. For $[MnO_4]^-$,$Mn + 4(-2) = -1 \implies Mn = +7$. Electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. $Mn^{7+}$ is $[Ar] 3d^0 4s^0$. It has $0$ $d$-electrons.
$2$. For $[Co(NH_3)_6]^{3+}$,$Co + 6(0) = +3 \implies Co = +3$. $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$. $Co^{3+}$ is $[Ar] 3d^6$. It has $6$ $d$-electrons.
$3$. For $[Fe(CN)_6]^{3-}$,$Fe + 6(-1) = -3 \implies Fe = +3$. $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$. $Fe^{3+}$ is $[Ar] 3d^5$. It has $5$ $d$-electrons.
$4$. For $[Cr(H_2O)_6]^{3+}$,$Cr + 6(0) = +3 \implies Cr = +3$. $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$. $Cr^{3+}$ is $[Ar] 3d^3$. It has $3$ $d$-electrons.
Thus,$[MnO_4]^-$ has $0$ $d$-electrons.

(D) In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state ($3d^6$ configuration). $NH_3$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons.
In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state ($3d^6$ configuration). $F^-$ is a weak field ligand,not causing pairing,resulting in $4$ unpaired electrons.

(A) For $Ni(CO)_4$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
In the presence of a strong field ligand like $CO$,the $4s$ electrons pair up into the $3d$ orbital,resulting in $sp^3$ hybridization.
Since all electrons are paired,$Ni(CO)_4$ is tetrahedral and diamagnetic.
Therefore,the statement '$Ni(CO)_4$ $-$ Tetrahedral,Paramagnetic' is incorrect.

(D) In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$ ($3d^8$ configuration).
$Cl^-$ is a weak field ligand,so it does not cause pairing of electrons.
Thus,the hybridization is $sp^3$,which results in a tetrahedral geometry.
In contrast,$[PdCl_4]^{2-}$,$[Ni(CN)_4]^{2-}$,and $[Pd(CN)_4]^{2-}$ involve strong field ligands or $4d/5d$ metals,leading to $dsp^2$ hybridization and square planar geometry.

(A) In the $MnO_4^-$ complex ion,the oxidation state of $Mn$ is $+7$.
The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$.
For $Mn^{7+}$,the configuration becomes $[Ar] 3d^0 4s^0$.
Thus,the $d$-orbitals of the central metal atom are empty.
In $MnO_4^-$,the hybridization is $sd^3$ (or $d^3s$ depending on the bonding model used for tetrahedral geometry).

(B) The central metal atom in $Fe(CO)_5$ is $Fe$.
$Fe$ has an atomic number of $26$ and its electronic configuration is $[Ar] 3d^6 4s^2$.
In $Fe(CO)_5$,$Fe$ is in the $0$ oxidation state.
Due to the strong field ligand $CO$,the electrons in $3d$ and $4s$ orbitals pair up,resulting in $dsp^3$ hybridization.
This hybridization corresponds to a trigonal bipyramidal geometry.

(D) $1$. The atomic number of $Co$ is $27$. The electronic configuration is $[Ar] 3d^7 4s^2$.
$2$. In the complex $[CoF_6]^{3-}$,let the oxidation state of $Co$ be $x$. Then $x + 6(-1) = -3$,which gives $x = +3$.
$3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$4$. $F^-$ is a weak field ligand,so it does not cause pairing of electrons.
$5$. In the $3d^6$ configuration,the electrons are arranged as $t_{2g}^4 e_g^2$. This results in $4$ unpaired electrons.

(B) The chemical formula for the hexafluoroferrate$(III)$ ion is $[FeF_6]^{3-}$.
In this complex,the oxidation state of $Fe$ is $x + 6(-1) = -3$,so $x = +3$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $F^-$ is a weak field ligand,it forms an outer orbital complex ($sp^3d^2$ hybridization).
In the $3d$ orbitals,the $5$ electrons remain unpaired according to Hund's rule.
Therefore,the number of unpaired electrons is $5$.

(C) The chemical formula $FeCl_3 \cdot 6H_2O$ represents a coordination compound where the iron center is coordinated by water molecules and chloride ions.
Experimental studies and structural analysis confirm that it exists as the complex $[Fe(H_2O)_4Cl_2]Cl \cdot 2H_2O$.
In this structure,four water molecules and two chloride ions are directly bonded to the $Fe^{3+}$ ion in a trans-configuration,while two water molecules are present as lattice water (water of crystallization).

(C) The chemical formula for the hexachloridocobaltate $(III)$ ion is $[CoCl_6]^{3-}$.
In this complex,the oxidation state of $Co$ is $+3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $3d$ electrons remain unpaired,resulting in a high-spin complex.
To accommodate six ligands,the $Co^{3+}$ ion uses one $4s$,three $4p$,and two $4d$ orbitals,leading to $sp^3d^2$ hybridisation.
Thus,the geometry is octahedral and the hybridisation is $sp^3d^2$.

(A) An outer orbital complex uses $4d$ orbitals for hybridization $(sp^3d^2)$.
In $[Ni(NH_3)_6]^{2+}$,the oxidation state of $Ni$ is $+2$,with a $d^8$ configuration.
Since $NH_3$ is a weak field ligand,it does not cause pairing of electrons,resulting in $sp^3d^2$ hybridization.
Therefore,$[Ni(NH_3)_6]^{2+}$ is an outer orbital complex.

(C) $F^-$ is a weak field ligand.
In the complex $[FeF_6]^{3-}$,the central metal ion $Fe^{3+}$ undergoes $sp^3d^2$ hybridization.
Since the outer $d$-orbitals are involved in hybridization,it is classified as an outer orbital complex.

(D) To determine the geometry,we look at the central metal ion and the nature of the ligand.
$1$. $[CoCl_4]^{2-}$: $Co^{2+}$ is $d^7$,$Cl^-$ is a weak field ligand,resulting in $sp^3$ hybridization (tetrahedral).
$2$. $[FeCl_4]^{2-}$: $Fe^{2+}$ is $d^6$,$Cl^-$ is a weak field ligand,resulting in $sp^3$ hybridization (tetrahedral).
$3$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $d^8$,$Cl^-$ is a weak field ligand,resulting in $sp^3$ hybridization (tetrahedral).
$4$. $[PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d$ series metal. For $5d$ series metals,even weak field ligands like $Cl^-$ cause pairing of electrons due to high crystal field splitting energy,resulting in $dsp^2$ hybridization,which corresponds to square planar geometry.

(A) $1$. For $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. The configuration is $[Ar] 3d^8 4s^2$. Due to the strong field ligand $CO$,electrons pair up,resulting in $sp^3$ hybridization. It is tetrahedral and diamagnetic. Thus,statement $A$ is incorrect.
$2$. For $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,causing pairing,resulting in $dsp^2$ hybridization. It is square planar and diamagnetic.
$3$. For $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The configuration is $[Ar] 3d^8$. $Cl^-$ is a weak field ligand,no pairing occurs,resulting in $sp^3$ hybridization. It is tetrahedral and paramagnetic.

(D) $1$. The central metal ion is $Cu^{2+}$,which has an electronic configuration of $[Ar] 3d^9$.
$2$. In the presence of $NH_3$ ligands,which are strong field ligands,the $Cu^{2+}$ ion undergoes $dsp^2$ hybridization.
$3$. This results in a square planar geometry.
$4$. The $3d^9$ configuration leaves one unpaired electron in the $3d_{x^2-y^2}$ orbital (or rather,the hole is in the $d_{x^2-y^2}$ orbital),making the complex paramagnetic with one unpaired electron.

(D) $1$. $Ni(CO)_4$ involves $sp^3$ hybridization,resulting in a tetrahedral geometry.
$2$. $[NiCl_4]^{2-}$ involves $sp^3$ hybridization,resulting in a tetrahedral geometry.
$3$. $[Ni(H_2O)_6]^{2+}$ involves $sp^3d^2$ hybridization,resulting in an octahedral geometry.
$4$. $[Cu(NH_3)_4]^{2+}$ involves $dsp^2$ hybridization,which corresponds to a square planar geometry.

(C) In $Ni(CO)_4$,the $Ni$ atom undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.
In $[Ni(CN)_4]^{2-}$,the $Ni^{2+}$ ion undergoes $dsp^2$ hybridization,resulting in a square planar geometry.

(C) $1$. For $[Cu(NH_3)_4]^{2+}$,the geometry is square planar ($dsp^2$ hybridization).
$2$. In $[Ni(CO)_4]$,$CO$ is a neutral ligand.
$3$. For $[Fe(CN)_6]^{3-}$,$Fe^{3+}$ is $3d^5$. Since $CN^-$ is a strong field ligand,it causes pairing,resulting in $d^2sp^3$ hybridization (inner orbital complex),not $sp^3d^2$.
$4$. For $[Co(en)_3]^{3+}$,$Co^{3+}$ has $Z=27$,so $Co^{3+} = 24$ electrons. $en$ donates $2 \times 3 = 6$ electrons. Total $EAN = 24 + 12 = 36$,which follows the $EAN$ rule (Krypton configuration).

(C) $Ni(CO)_4$ has a $sp^3$ hybridization,making it tetrahedral and diamagnetic because all electrons are paired. Therefore,the statement that it is octahedral is incorrect. $[Ni(CN)_4]^{2-}$ is $dsp^2$ (square planar,diamagnetic) and $[NiCl_4]^{2-}$ is $sp^3$ (tetrahedral,paramagnetic).

(C) $1$. The geometry of a coordination complex depends on the hybridization of the central metal ion.
$2$. $[NiCl_4]^{2-}$ involves $sp^3$ hybridization,resulting in a tetrahedral geometry.
$3$. $[Zn(en)_2]^{2+}$ involves $sp^3$ hybridization,resulting in a tetrahedral geometry.
$4$. $[Pt(NH_3)_2Cl_2]$ (Cisplatin) involves $dsp^2$ hybridization,which corresponds to a square planar geometry.
$5$. $[Cr(NH_3)_2Cl_2]^{2+}$ typically exhibits octahedral geometry.
$6$. Therefore,the correct option is $C$.

(C) 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 causes pairing of electrons in the $3d$ orbitals.
This results in $dsp^2$ hybridization,leading to a square-planar geometry.
Since all electrons are paired,the complex is diamagnetic.

(A) $1$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The hybridization is $sp^3d^2$,resulting in a square planar geometry.
$2$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The hybridization is $sp^3d$,resulting in a see-saw geometry.
$3$. $[NiCl_4]^{2-}$: $Ni$ is in the $+2$ oxidation state ($d^8$ configuration). $Cl^-$ is a weak field ligand,so it forms a tetrahedral complex ($sp^3$ hybridization).
$4$. $[PtCl_4]^{2-}$: $Pt$ is in the $+2$ oxidation state ($5d^8$ configuration). $Cl^-$ is a ligand,but $Pt^{2+}$ is a $5d$ series metal,which always forms square planar complexes ($dsp^2$ hybridization) due to high crystal field splitting.
Therefore,$(i)$ and $(iv)$ have square planar geometry.

(C) In the complex $[NiX_4]^{2-}$,the oxidation state of $Ni$ is $+2$.
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
Since the complex is paramagnetic,the ligand $X^-$ is a weak field ligand that does not cause pairing of electrons.
Thus,the $3d^8$ configuration remains as $t_{2g}^6 e_g^2$,resulting in $2$ unpaired electrons.
Due to the $sp^3$ hybridization,the geometry of the complex is tetrahedral.

(C) $1$. $[Cu(NH_3)_4]^{2+}$ is square planar due to $dsp^2$ hybridization.
$2$. $Ni(CO)_4$ contains $CO$ which is a neutral ligand.
$3$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. Since $CN^-$ is a strong field ligand,it causes pairing,resulting in $d^2sp^3$ hybridization (inner orbital complex),not $sp^3d^2$.
$4$. For $[Co(en)_3]^{3+}$,$EAN = Z - \text{oxidation state} + 2 \times \text{coordination number} = 27 - 3 + 2(6) = 24 + 12 = 36$,which is the atomic number of $Kr$. Thus,it follows the $EAN$ rule.

(B) $1$. For $[Cu(NH_3)_2]^+$,$Cu^+$ is $3d^{10}$,hybridization is $sp$,involving $4s$ and $4p$ orbitals.
$2$. For $[Cu(NH_3)_4]^{2+}$,$Cu^{2+}$ is $3d^9$,hybridization is $dsp^2$ (square planar),involving $3d$,$4s$,and $4p$ orbitals.
$3$. For $[Cu(CN)_4]^{3-}$,$Cu^+$ is $3d^{10}$,hybridization is $sp^3$ (tetrahedral),involving $4s$ and $4p$ orbitals.
$4$. For $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ is $3d^8$,hybridization is $dsp^2$,involving $3d$,$4s$,and $4p$ orbitals.
$5$. All the given complexes involve $4p$ orbitals in their hybridization. However,in the context of standard competitive chemistry questions,$[Cu(NH_3)_4]^{2+}$ is the most common example cited for $dsp^2$ hybridization involving $4p$ orbitals.

(D) An outer orbital complex is formed when the outer $d$-orbitals $(4d)$ are used in hybridization,typically involving $sp^3d^2$ hybridization.
$1$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing. It forms $d^2sp^3$ (inner orbital).
$2$. $[Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. It forms $d^2sp^3$ (inner orbital).
$3$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing. It forms $d^2sp^3$ (inner orbital).
$4$. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a moderate field ligand. It forms $sp^3d^2$ hybridization using $4d$ orbitals,making it an outer orbital complex.

(B) The oxidation state of $Ni$ in $[NiCl_4]^{2-}$ is calculated as $x + 4(-1) = -2$,which gives $x = +2$.
$Ni$ has an atomic number of $28$,so its electronic configuration is $[Ar] 3d^8 4s^2$.
For $Ni^{2+}$,the configuration is $[Ar] 3d^8$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
The $3d^8$ configuration has two unpaired electrons in the $d$-orbitals.
Therefore,the number of unpaired electrons in $[NiCl_4]^{2-}$ is $2$.

(C) The correct option is $C$.
$1$. For $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$2$. $NH_3$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
$3$. The configuration becomes $t_{2g}^6 e_g^0$,utilizing $d^2sp^3$ hybridization,which makes it an inner orbital complex.
$4$. Since all electrons are paired,the complex is diamagnetic.

(B) $1$. The oxidation state of $Mn$ in $[Mn(CN)_6]^{3-}$ is calculated as: $x + 6(-1) = -3 \implies x = +3$.
$2$. The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. Thus,$Mn^{3+}$ is $[Ar] 3d^4$.
$3$. $CN^-$ is a Strong Field Ligand $(SFL)$,which causes pairing of electrons in the $3d$ orbitals.
$4$. After pairing,two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals are available for hybridization,resulting in $d^2sp^3$ hybridization.
$5$. Since the coordination number is $6$,the geometry is octahedral.

(B) The Jahn-Teller effect is a geometric distortion that occurs in non-linear molecules with unsymmetrical electronic configurations in the $e_g$ or $t_{2g}$ orbitals of octahedral complexes.
For high spin complexes:
- $d^4$ configuration is $t_{2g}^3 e_g^1$ (unsymmetrical).
- $d^7$ configuration is $t_{2g}^5 e_g^2$ (unsymmetrical).
- $d^9$ configuration is $t_{2g}^6 e_g^3$ (unsymmetrical).
- $d^8$ configuration is $t_{2g}^6 e_g^2$ (symmetrical).
Since the $d^8$ high spin complex has a symmetrical electronic configuration,it does not show Jahn-Teller distortion.