Hybridisation and Geometry Questions in English

(N/A) According to this theory,the metal atom or ion under the influence of ligands uses its $(n-1)d, ns, np$ or $ns, np, nd$ orbitals for hybridization to produce a set of equivalent orbitals of definite geometry such as octahedral,square planar,and others (as shown in the table below).

Coordination Number	Type of Hybridization	Distribution of Hybrid Orbitals in Space
$4$	$sp^3$	Tetrahedral
$4$	$dsp^2$	Square planar
$5$	$sp^3d$	Trigonal bipyramidal
$6$	$sp^3d^2$	Octahedral
$6$	$d^2sp^3$	Octahedral

These hybrid orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding.

(N/A) $1$. The central metal ion $Co^{3+}$ has an electronic configuration of $[Ar]3d^6$.
$2$. In the presence of the strong field ligand $NH_3$,the electrons in the $3d$ orbitals pair up,leaving two $3d$ orbitals vacant.
$3$. These two $3d$ orbitals,along with one $4s$ and three $4p$ orbitals,undergo $d^2sp^3$ hybridization to form six equivalent hybrid orbitals.
$4$. Six pairs of electrons from six $NH_3$ ligands occupy these hybrid orbitals,resulting in an octahedral geometry.
$5$. Since there are no unpaired electrons,the complex is diamagnetic.
$6$. Because the inner $3d$ orbitals are involved in hybridization,it is called an inner orbital complex or a low-spin complex.

(N/A) $1$. The central metal ion is $Co^{3+}$,which has an electronic configuration of $[Ar] 3d^6$.
$2$. In the presence of the weak field ligand $F^-$,the $d$-electrons do not pair up.
$3$. The $Co^{3+}$ ion uses one $4s$,three $4p$,and two $4d$ orbitals to undergo $sp^3d^2$ hybridization.
$4$. Six pairs of electrons from six $F^-$ ions occupy these six $sp^3d^2$ hybrid orbitals.
$5$. Since the outer $4d$ orbitals are used,it is an outer orbital complex (high spin complex).
$6$. Due to the presence of four unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.

(N/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. Therefore,the $3d$ electrons remain unpaired.
To accommodate four $Cl^-$ ligands,one $4s$ and three $4p$ orbitals undergo $sp^3$ hybridization,resulting in four equivalent $sp^3$ hybrid orbitals directed towards the corners of a tetrahedron.
Four pairs of electrons from four $Cl^-$ ions are donated into these four $sp^3$ hybrid orbitals. Thus,the complex has a tetrahedral geometry.
Due to the presence of two unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.

(N/A) In the complex $[Ni(CN)_4]^{2-}$,the central metal ion $Ni$ is in the $+2$ oxidation state. The electronic configuration of $Ni^{2+}$ is $3d^8$.
Since $CN^-$ is a strong field ligand,it causes the pairing of electrons in the $3d$ orbitals.
This results in one empty $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals,which undergo $dsp^2$ hybridization to form four equivalent $dsp^2$ hybrid orbitals.
These four hybrid orbitals accept electron pairs from four $CN^-$ ligands.
Due to the presence of all paired electrons,the complex is diamagnetic.
Since it involves the inner $3d$ orbital,it is a low-spin or inner-orbital complex with a square planar geometry.

(N/A) $VBT$ can explain the formation,structure,and magnetic behavior of coordination compounds. However,it has the following limitations:
$(i)$ It involves a number of assumptions.
$(ii)$ It does not give a quantitative interpretation of magnetic data.
$(iii)$ It cannot explain the color exhibited by coordination compounds.
$(iv)$ It does not give a quantitative interpretation of the thermodynamic or kinetic stability of coordination compounds.
$(v)$ It does not make exact predictions regarding the tetrahedral and square planar structures of $4$-coordinate complexes.
$(vi)$ It does not distinguish between weak and strong ligands.

(B) 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.
The $4s$ electrons shift to the $3d$ orbital,resulting in $3d^{10}$ configuration.
The hybridization involved is $sp^3$,which corresponds to a tetrahedral geometry.

(B) $1$. In the complex $[Mn(CN)_6]^{3-}$,the central metal ion is $Mn^{3+}$.
$2$. The atomic number of $Mn$ is $25$,so the electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$.
$3$. $CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
$4$. The $4$ electrons in the $3d$ orbitals pair up to occupy two $3d$ orbitals,leaving two $3d$ orbitals vacant.
$5$. These two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals hybridize to form six $d^2sp^3$ hybrid orbitals.
$6$. Thus,the hybridization of $[Mn(CN)_6]^{3-}$ is $d^2sp^3$.

(A) In the complex $[CoF_6]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$F^-$ is a weak field ligand,which does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the complex utilizes the outer $4d$ orbitals for hybridization,resulting in $sp^3d^2$ hybridization.
Since the outer $d$-orbitals are involved,it is classified as an outer orbital complex.

(C) The central metal atom is $Fe$ in the oxidation state $0$. The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$. In the presence of strong field ligand $CO$,the electrons pair up,and the hybridization involved is $dsp^3$. The geometry corresponding to $dsp^3$ hybridization is trigonal bipyramidal.

(N/A) The central metal atom is Chromium $(Cr)$ with an atomic number of $24$. The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
In the $[Cr(CO)_6]$ complex,the oxidation state of $Cr$ is $0$.
$CO$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
The hybridization involved is $d^2sp^3$,which corresponds to an octahedral geometry.
The structure consists of a central $Cr$ atom surrounded by six $CO$ ligands arranged at the corners of an octahedron.

(N/A) In $[MnCl_{4}]^{2-}$,the oxidation state of $Mn$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^{5}$.
Since $Cl^{-}$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ subshell.
Thus,there are $5$ unpaired electrons $(n=5)$.
The magnetic moment is calculated as $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
This confirms the high spin $sp^{3}$ hybridized tetrahedral complex.

(N/A) $(i)$ $[Mn(CN)_{6}]^{3-}$: $Mn^{3+}$ is $3d^{4}$. $CN^-$ is a strong field ligand,causing pairing. Hybridization: $d^{2}sp^{3}$. Inner orbital complex. Paramagnetic (two unpaired electrons). Magnetic moment,$\mu = \sqrt{2(2+2)} = 2.83 \ B.M.$
$(ii)$ $[Co(NH_{3})_{6}]^{3+}$: $Co^{3+}$ is $3d^{6}$. $NH_{3}$ is a strong field ligand,causing pairing. Hybridization: $d^{2}sp^{3}$. Inner orbital complex. Diamagnetic. Magnetic moment,$\mu = 0 \ B.M.$
$(iii)$ $[Cr(H_{2}O)_{6}]^{3+}$: $Cr^{3+}$ is $3d^{3}$. Hybridization: $d^{2}sp^{3}$. Inner orbital complex. Paramagnetic (three unpaired electrons). Magnetic moment,$\mu = \sqrt{3(3+2)} = 3.87 \ B.M.$
$(iv)$ $[FeCl_{6}]^{4-}$: $Fe^{2+}$ is $3d^{6}$. $Cl^-$ is a weak field ligand,no pairing. Hybridization: $sp^{3}d^{2}$. Outer orbital complex. Paramagnetic (four unpaired electrons). Magnetic moment,$\mu = \sqrt{4(4+2)} = 4.90 \ B.M.$

(N/A) Coordination polyhedron: The spatial arrangement of the ligands which are directly attached to the central ion/atom defines a coordination polyhedron about the central atom.
The most common coordination polyhedra are octahedral,square planar,and tetrahedral.

$Coordination \ Entity$	$Shape \ of \ Polyhedron$
$(i) \ [Co(NH_3)_6]^{3+}$	$Octahedral$
$(ii) \ [Ni(CO)_4]$	$Tetrahedral$
$(iii) \ [PtCl_4]^{2-}$	$Square \ planar$

(B) The hybridization for square planar geometry is $dsp^{2}$.
The hybridization for octahedral geometry is $sp^{3}d^{2}$ or $d^{2}sp^{3}$.
The hybridization for linear geometry is $sp$.
Therefore,the correct sequence is $dsp^{2}, sp^{3}d^{2}, sp$.

(A) The hybridization of the given species is as follows:
$(1)$ $[PtCl_4]^{2-}$ involves $dsp^2$ hybridization (square planar).
$(2)$ $BrF_5$ involves $sp^3d^2$ hybridization (square pyramidal).
$(3)$ $PCl_5$ involves $sp^3d$ hybridization (trigonal bipyramidal).
$(4)$ $[Co(NH_3)_6]^{3+}$ involves $d^2sp^3$ hybridization (octahedral).
Therefore,the correct matching is $1-C, 2-A, 3-D, 4-B$.

(C) $(I)$ Under weak field ligands,octahedral $Mn(II)$ and tetrahedral $Ni(II)$ complexes are both high spin.
$(II)$ Tetrahedral $Ni(II)$ complexes are rarely low spin because $Ni(II)$ prefers square planar geometry with strong field ligands,which are low spin.
$(III)$ With strong field ligands,$Mn(II)$ $(d^5)$ complexes can be low spin (unpaired electrons $= 1$).
$(IV)$ Aqueous solution of $Mn(II)$ ions is pink in color,not yellow. Therefore,statement $(IV)$ is incorrect.

(B) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM.$ For $\mu = 4.90 \ BM,$ $n = 4$ unpaired electrons.
For a $d^{6}$ ion,$4$ unpaired electrons are possible in a high-spin tetrahedral complex $(e^{3}t_{2}^{3})$ or a high-spin octahedral complex $(t_{2g}^{4}e_{g}^{2})$.
In a tetrahedral complex $[M(H_{2}O)_{4}]^{2+}$,the configuration is $e^{3}t_{2}^{3}$.
$CFSE = (3 \times -0.6 \Delta_{t}) + (3 \times 0.4 \Delta_{t}) = -1.8 \Delta_{t} + 1.2 \Delta_{t} = -0.6 \Delta_{t}.$
In an octahedral complex $[M(H_{2}O)_{6}]^{2+}$,the configuration is $t_{2g}^{4}e_{g}^{2}$.
$CFSE = (4 \times -0.4 \Delta_{0}) + (2 \times 0.6 \Delta_{0}) = -1.6 \Delta_{0} + 1.2 \Delta_{0} = -0.4 \Delta_{0}.$
Comparing with the given options,option $B$ matches the tetrahedral geometry and calculated $CFSE$.

(B) To determine the highest paramagnetic behaviour,we calculate the number of unpaired electrons $(n)$ for each complex:
$A) [Pd(gly)_2]$: $Pd^{2+}$ is a $4d^8$ metal ion. $Pd$ is a $4d$ series element,so it always forms low-spin square planar complexes. $n = 0$.
$B) [Ti(NH_3)_6]^{3+}$: $Ti^{3+}$ is a $3d^1$ ion. $n = 1$.
$C) [Co(OX)_2(OH)_2]^{-}$: $Co^{3+}$ is a $3d^6$ ion. Given $\Delta_0 > P$,it forms a low-spin complex. $t_{2g}^6 e_g^0$. $n = 0$.
$D) [Fe(en)(bpy)(NH_3)_2]^{2+}$: $Fe^{2+}$ is a $3d^6$ ion. $en$ and $bpy$ are strong field ligands. This forms a low-spin complex. $t_{2g}^6 e_g^0$. $n = 0$.
Wait,re-evaluating the question options. If we consider $[Fe(en)(bpy)(NH_3)_2]^{2+}$,$Fe^{2+}$ is $3d^6$. With strong field ligands,it is $t_{2g}^6$,$n=0$.
Let's re-check $[Ti(NH_3)_6]^{3+}$. $Ti^{3+}$ is $3d^1$,$n=1$.
Since $1 > 0$,$[Ti(NH_3)_6]^{3+}$ has the highest number of unpaired electrons among the choices provided.

(A) $1$. $[Ni(CN)_{4}]^{2-}$ involves $dsp^{2}$ hybridization,which uses one $d-$orbital $(d_{x^2-y^2})$.
$2$. $[CrF_{6}]^{3-}$ involves $d^{2}sp^{3}$ hybridization,which uses two $d-$orbitals.
$3$. $BrF_{5}$ involves $sp^{3}d^{2}$ hybridization,which uses two $d-$orbitals.
$4$. $XeF_{4}$ involves $sp^{3}d^{2}$ hybridization,which uses two $d-$orbitals.
Therefore,the correct option is $A$.

(A) The electronic configuration of $Ru$ $(Z=44)$ is $[Kr] 4d^{7} 5s^{1}$.
For $Ru^{2+}$,the configuration is $4d^{6}$.
Given that $\Delta_{0} > P$,the complex is a low-spin complex.
In an octahedral field,the $6$ electrons in the $4d$ orbital will occupy the $t_{2g}$ orbitals as $(t_{2g})^{6} (e_{g})^{0}$.
Since all electrons are paired,the number of unpaired electrons $(n)$ is $0$.
The magnetic moment $\mu = \sqrt{n(n+2)} \ BM = \sqrt{0(0+2)} \ BM = 0 \ BM$.

(C) According to Valence Bond Theory:
$1. 4, sp^3$ hybridisation results in a $tetrahedral$ geometry $(a-iii)$.
$2. 4, dsp^2$ hybridisation results in a $square planar$ geometry $(b-iv)$.
$3. 5, sp^3d$ hybridisation results in a $trigonal bipyramidal$ geometry $(c-i)$.
$4. 6, d^2sp^3$ hybridisation results in an $octahedral$ geometry $(d-ii)$.
Therefore,the correct matching is $a-iii, b-iv, c-i, d-ii$.

(A) The coordination number of the $Mn^{2+}$ ion in the complex ion $[MnBr_{4}]^{2-}$ is $4$.
This implies it can have either tetrahedral ($sp^{3}$ hybridization) or square planar ($dsp^{2}$ hybridization) geometry.
The given spin-only magnetic moment is $5.9 \ BM$.
Using the formula $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons:
$5.9 \approx \sqrt{n(n+2)} \implies n \approx 5$.
Since $Mn^{2+}$ has a $d^{5}$ configuration,having $5$ unpaired electrons indicates a high-spin tetrahedral complex where the $d$-orbitals are not forced to pair up.
Therefore,the geometry is tetrahedral.

(A) The reaction of $Ag^{+}$ with excess of sodium thiosulphate is given below:
$Ag^{+} + 2S_{2}O_{3}^{2-} \rightarrow [Ag(S_{2}O_{3})_{2}]^{3-}$.
This complex possesses a $-3$ charge.
The coordination number for $Ag$ is $2$,which corresponds to $sp$ hybridization and linear geometry.

(D) The oxidation state of $Ni$ in the complex $[Ni(CO)_4]$ is calculated as follows:
$x + 4(0) = 0 \implies x = 0$
The ground state electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
In the presence of the strong field ligand $CO$,the two $4s$ electrons pair up with the $3d$ electrons,resulting in a fully filled $3d$ subshell $(3d^{10})$.
The $4s$ and $4p$ orbitals hybridize to form four $sp^3$ hybrid orbitals,which accept electron pairs from four $CO$ ligands.
Since all electrons are paired,the complex is diamagnetic and exhibits $sp^3$ hybridisation.

(A) In the coordination complex $[Co(C_2O_4)_3]^{3-}$,the central metal ion is $Co^{3+}$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$C_2O_4^{2-}$ (oxalate) is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
This results in $d^2sp^3$ hybridization,making it an inner orbital complex.
Since all electrons are paired,it is diamagnetic and a low spin complex.

(C) For $[Mn(CN)_6]^{4-}$:
$Mn$ is in $+2$ oxidation state ($3d^5$ configuration).
$CN^-$ is a strong field ligand $(SFL)$,so it causes pairing of electrons.
Thus,the configuration becomes $t_{2g}^5 e_g^0$,leaving one unpaired electron.
The hybridization is $d^2sp^3$ and it is paramagnetic.
For $[Fe(CN)_6]^{3-}$:
$Fe$ is in $+3$ oxidation state ($3d^5$ configuration).
$CN^-$ is a strong field ligand $(SFL)$,causing pairing of electrons.
Thus,the configuration becomes $t_{2g}^5 e_g^0$,leaving one unpaired electron.
The hybridization is $d^2sp^3$ and it is paramagnetic.
Therefore,both complexes have $d^2sp^3$ hybridization and are paramagnetic.

(C) To determine the hybridization of the central metal atom,we analyze the coordination number and the nature of the ligands:
$1$. In $Na_{2}[NiCl_{4}]$,$Ni$ is in the $+2$ oxidation state ($3d^{8}$ configuration). $Cl^-$ is a weak field ligand,leading to $sp^{3}$ hybridization (tetrahedral geometry).
$2$. In $NiCl_{2} \cdot 6 H_{2}O$,the complex is $[Ni(H_{2}O)_{6}]^{2+}$. $Ni^{2+}$ is in an octahedral environment with $sp^{3}d^{2}$ hybridization.
$3$. In $K_{2}[Ni(CN)_{4}]$,$Ni$ is in the $+2$ oxidation state ($3d^{8}$ configuration). $CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals. This leaves one $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals available for hybridization,resulting in $dsp^{2}$ hybridization (square planar geometry).
$4$. In $[Ni(CO)_{4}]$,$Ni$ is in the $0$ oxidation state ($3d^{8} 4s^{2}$ configuration). $CO$ is a strong field ligand,causing the $4s$ electrons to pair into the $3d$ orbitals,resulting in $sp^{3}$ hybridization (tetrahedral geometry).
Therefore,the correct compound is $K_{2}[Ni(CN)_{4}]$.

(C) In $NiCl_{2}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^{8}$.
In the $3d^{8}$ configuration,there are $2$ unpaired electrons.
When $NiCl_{2}$ is heated with excess $NaCN$ in the presence of a strong oxidizing agent,$Ni^{2+}$ is oxidized to $Ni^{4+}$.
The electronic configuration of $Ni^{4+}$ is $[Ar] 3d^{6}$.
$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
For $d^{6}$ in a strong field,all $6$ electrons pair up in the $t_{2g}$ orbitals,resulting in $0$ unpaired electrons.
The change in the number of unpaired electrons = $|0 - 2| = 2$.

(B) For $[Co(NH_3)_6]Cl_2$:
$Co$ is in $+2$ oxidation state. The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.
Since $NH_3$ is a weak field ligand for $Co^{2+}$,no pairing occurs.
The configuration is $t_{2g}^5 e_g^2$,resulting in $3$ unpaired electrons.
For $[Co(NH_3)_6]Cl_3$:
$Co$ is in $+3$ oxidation state. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Since $NH_3$ is a strong field ligand for $Co^{3+}$,pairing occurs.
The configuration is $t_{2g}^6 e_g^0$,resulting in $0$ unpaired electrons.
Total number of unpaired electrons = $3 + 0 = 3$.

(C) In the complex $[MnCl_6]^{3-}$,the oxidation state of $Mn$ is $+3$.
The electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons.
Therefore,the $3d$ electrons remain unpaired,and the complex uses outer $4d$ orbitals for hybridisation.
The hybridisation is $sp^3d^2$,which corresponds to an outer orbital complex.
Due to the presence of $4$ unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.

(D) Statement $I$:
$1.$ $[Mn(CN)_6]^{3-}: Mn^{3+} = [Ar] 3d^4$. $CN^{-}$ is a strong field ligand. Configuration: $t_{2g}^4 e_g^0$. Hybridization: $d^2sp^3$.
$2.$ $[Fe(CN)_6]^{3-}: Fe^{3+} = [Ar] 3d^5$. $CN^{-}$ is a strong field ligand. Configuration: $t_{2g}^5 e_g^0$. Hybridization: $d^2sp^3$.
$3.$ $[Co(C_2O_4)_3]^{3-}: Co^{3+} = [Ar] 3d^6$. $C_2O_4^{2-}$ is a strong field ligand for $Co^{3+}$. Configuration: $t_{2g}^6 e_g^0$. Hybridization: $d^2sp^3$.
Statement $I$ is correct.
Statement $II$:
$1.$ $[MnCl_6]^{3-}: Mn^{3+} = [Ar] 3d^4$. $Cl^{-}$ is a weak field ligand. Configuration: $t_{2g}^3 e_g^1$. Unpaired electrons = $4$. Paramagnetic.
$2.$ $[FeF_6]^{3-}: Fe^{3+} = [Ar] 3d^5$. $F^{-}$ is a weak field ligand. Configuration: $t_{2g}^3 e_g^2$. Unpaired electrons = $5$. Paramagnetic.
Statement $II$ is correct.

(C) In $[Cu(NH_3)_4]^{2+}$,the $Cu^{2+}$ ion undergoes $dsp^2$ hybridization,which results in a square planar geometry.
In contrast,the ions $NH_4^+$,$BF_4^-$,and $[NiCl_4]^{2-}$ all exhibit $sp^3$ hybridization,which corresponds to a tetrahedral geometry.

(C) The atomic number of $Co$ is $27$. The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.
Since $H_2O$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Thus,the $3d$ orbitals have $3$ unpaired electrons.
The hybridization involves the use of one $4s$,three $4p$,and two $4d$ orbitals to accommodate the six ligands.
Therefore,the hybridization is $sp^3d^2$.

(N/A) The metal atom or ion,under the influence of ligands,makes available $(n-1)d, ns, np$ or $ns, np, nd$ orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral,tetrahedral,square planar,etc. The number of available metal ion orbitals is equal to its coordination number.
Table: Number of Orbitals and Types of Hybridisations
| Coordination number | Type of hybridisation | Distribution of hybrid orbitals in space |
| :--- | :--- | :--- |
| $4$ | $sp^3$ | Tetrahedral |
| $4$ | $dsp^2$ | Square planar |
| $5$ | $sp^3d$ | Trigonal bipyramidal |
| $6$ | $sp^3d^2$ | Octahedral |
| $6$ | $d^2sp^3$ | Octahedral |
The hybrid orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding.
Each hybrid orbital of the metal ion receives an electron pair from a ligand.
The complex is said to be inner orbital or low spin if $(n-1)d, ns, np$ orbitals participate in hybridisation and is said to be outer orbital or high spin if $ns, np, nd$ orbitals participate in hybridisation.
It is possible to predict the geometry of a complex from the knowledge of its magnetic behaviour on the basis of valence bond theory. If all the electrons are paired,the complex is diamagnetic and if electrons are unpaired,it is paramagnetic.
It is important to note that the hybrid orbitals do not actually exist. In fact,hybridisation is a mathematical manipulation of the wave equation for the atomic orbitals involved.

(N/A) $(i)$ $[NiCl_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+}$ has $3d^{8}$ configuration. $Cl^-$ is a weak field ligand,so no pairing occurs. It undergoes $sp^3$ hybridization. Geometry: Tetrahedral. Magnetic property: Paramagnetic ($n=2$,$\mu = 2.82 \ B.M.$).
$(ii)$ $[Ni(CN)_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+}$ has $3d^{8}$ configuration. $CN^-$ is a strong field ligand,causing pairing of $3d$ electrons. It undergoes $dsp^2$ hybridization. Geometry: Square planar. Magnetic property: Diamagnetic ($n=0$,$\mu = 0 \ B.M.$).
$(iii)$ $[Ni(CO)_{4}]$: $Ni$ is in $0$ oxidation state. $Ni$ has $3d^{8} 4s^{2}$ configuration. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$ orbitals. It undergoes $sp^3$ hybridization. Geometry: Tetrahedral. Magnetic property: Diamagnetic ($n=0$,$\mu = 0 \ B.M.$).
$(iv)$ $MnO_{4}^{-}$: $Mn$ is in $+7$ oxidation state. $Mn^{7+}$ has $3d^{0}$ configuration. It undergoes $sd^3$ (or $d^3s$) hybridization. Geometry: Tetrahedral. Magnetic property: Diamagnetic $(n=0)$.

(N/A) The limitations of valence bond theory are:
$1$. It involves a number of assumptions.
$2$. It does not give quantitative interpretation of magnetic data.
$3$. It does not explain the colour exhibited by coordination compounds.
$4$. It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds.
$5$. It does not make exact predictions regarding the tetrahedral and square planar structures of $4$-coordinate complexes.
$6$. It does not distinguish between weak and strong ligands.

(B) The hybridization of the given species is as follows:
$1$. $[PtCl_4]^{2-}$: $Pt^{2+}$ has a $d^8$ configuration. It forms a square planar complex with $dsp^2$ hybridization. $(A-III)$
$2$. $BrF_5$: The central atom $Br$ has $7$ valence electrons. It forms $5$ bonds and has $1$ lone pair,resulting in $sp^3d^2$ hybridization. $(B-IV)$
$3$. $PCl_5$: The central atom $P$ has $5$ valence electrons. It forms $5$ bonds with no lone pairs,resulting in $sp^3d$ hybridization. $(C-I)$
$4$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ has a $d^6$ configuration. $NH_3$ is a strong field ligand,forcing pairing of electrons,resulting in $d^2sp^3$ hybridization. $(D-II)$
Thus,the correct match is $A-III, B-IV, C-I, D-II$.

(B) The reaction of $[Co(H_2O)_6]^{2+}$ with excess $NH_3$ in the presence of oxygen leads to the oxidation of $Co^{2+}$ to $Co^{3+}$,forming the complex $[Co(NH_3)_6]^{3+}$.
$NH_3$ is a strong field ligand,which causes pairing of electrons,resulting in a low-spin octahedral complex.
The electronic configuration of $Co^{3+}$ is $3d^6$.
In an octahedral field,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
For a $d^6$ low-spin complex,all $6$ electrons occupy the $t_{2g}$ orbitals,giving the configuration $t_{2g}^6 e_g^0$.
Thus,the number of electrons in the $t_{2g}$ orbitals is $6$.

(C) $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand $(SFL)$,causing pairing,resulting in $dsp^2$ hybridization (square planar) and diamagnetic behavior.
$[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing $4s$ electrons to pair into $3d$,resulting in $3d^{10}$ configuration,$sp^3$ hybridization (tetrahedral) and diamagnetic behavior.
$[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand $(WFL)$,resulting in $sp^3$ hybridization (tetrahedral) and paramagnetic behavior due to $2$ unpaired electrons.
Thus,Statement $I$ is true and Statement $II$ is false.

(B) $Ni(CO)_4$: The oxidation state of $Ni$ is $0$. The configuration is $3d^8 4s^2$. Due to the strong field ligand $CO$,electrons pair up,resulting in $sp^3$ hybridization.
$[Ni(CN)_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The configuration is $3d^8$. Due to the strong field ligand $CN^-$,the electrons pair up,resulting in $dsp^2$ hybridization.
$[Co(CN)_6]^{3-}$: The oxidation state of $Co$ is $+3$. The configuration is $3d^6$. Due to the strong field ligand $CN^-$,the electrons pair up,resulting in $d^2sp^3$ hybridization.
$[CoF_6]^{3-}$: The oxidation state of $Co$ is $+3$. The configuration is $3d^6$. Due to the weak field ligand $F^-$,the electrons do not pair up,resulting in $sp^3d^2$ hybridization.
Therefore,the correct matching is $A-I, B-IV, C-III, D-II$.

(D) $Cu^{2+}$ has a $d^9$ electronic configuration,which leads to an unsymmetrical filling of the $e_g$ orbitals $(t_{2g}^6 e_g^3)$.
This causes Jahn-Teller distortion in octahedral complexes.
The magnitude of structural distortion is directly proportional to the difference in the field strength of the ligands present in the coordination sphere.
In $[Cu(en)_2Cl_2]$,the ligands are $en$ (a strong field ligand) and $Cl^-$ (a weak field ligand).
Because $en$ and $Cl^-$ have the largest difference in their ligand field strengths compared to the other options,$[Cu(en)_2Cl_2]$ exhibits the maximum structural distortion.
Specifically,for $trans-[Cu(en)_2Cl_2]$,the axial ligands $(Cl^-)$ are significantly different from the equatorial ligands $(en)$,leading to maximum elongation.

(C) The correct option is $(C)$. Among the given options,$PdCl_{4}^{2-}$ exhibits square planar geometry.

Complex	Hybridisation
$CdCl_{4}^{2-}$	$sp^{3}$
$Zn(CN)_{4}^{2-}$	$sp^{3}$
$PdCl_{4}^{2-}$	$dsp^{2}$
$Cu(CN)_{4}^{3-}$	$sp^{3}$

The $Pd^{2+}$ ion has a $4d^{8}$ configuration. In the presence of $Cl^{-}$ ligands,it undergoes $dsp^{2}$ hybridisation,which results in a square planar geometry.

(D) The oxidation state of $Ni$ in $[Ni(CN)_4]^{2-}$ is $+2$.
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$.
Since $CN^-$ is a strong field ligand,it causes the pairing of electrons in the $3d$ orbitals.
This results in the availability of one $3d$,one $4s$,and two $4p$ orbitals for hybridisation,leading to $dsp^2$ hybridisation.
Therefore,the complex $[Ni(CN)_4]^{2-}$ exhibits a square planar geometry.

(A) The oxidation state of $Ni$ in $[NiCl_4]^{2-}$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$.
As $Cl^{-}$ is a weak field ligand,pairing of electrons will not occur.
The $Ni^{2+}$ ion in the complex undergoes $sp^3$ hybridisation using one $4s$ and three $4p$ orbitals.
Thus,the geometry is tetrahedral.
Since there are $2$ unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.

(C) The oxidation state of $Mn$ in $[MnBr_4]^{2-}$ is $+2$.
The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^{5} 4s^{0}$.
Since $Br^{-}$ is a weak field ligand,pairing of electrons will not occur.
As the hybridisation of $Mn$ in $[MnBr_4]^{2-}$ is $sp^{3}$,its geometry is tetrahedral and it contains $5$ unpaired electrons.
$Mn^{2+}$ (Ground state): $3d$ $(1|1|1|1|1)$ $4s$ ( ) $4p$ ( ) ( )
$[MnBr_4]^{2-}$: $3d$ $(1|1|1|1|1)$ $4s$ ( ) $4p$ ( ) ( ) ( ) $\rightarrow sp^{3}$ hybridisation

(C) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$.
Thus,the electronic configuration of $Ni(0)$ is $[Ar] 3d^8 4s^2$.
In the presence of the strong field ligand $CO$,the $4s$ electrons pair up with the $3d$ electrons.
This results in a fully filled $3d$ subshell and the use of one $4s$ and three $4p$ orbitals for $sp^3$ hybridisation.
$Ni(CO)_4$ has tetrahedral geometry ($sp^3$ hybridisation) and is diamagnetic due to the absence of unpaired electrons.

(B) The reaction is: $[Co(H_2O)_6]^{2+} + 6 NH_3 \xrightarrow{O_2} [Co(NH_3)_6]^{3+} + 6 H_2O$.
In complex $X$,$[Co(H_2O)_6]^{2+}$,the oxidation state of $Co$ is $+2$ ($d^7$ configuration).
Since $H_2O$ is a weak field ligand,the electrons do not pair up,resulting in $t_{2g}^5 e_g^2$ configuration,which gives $3$ unpaired electrons.
In complex $Y$,$[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$ ($d^6$ configuration).
Since $NH_3$ is a strong field ligand,it causes pairing of electrons,resulting in $t_{2g}^6 e_g^0$ configuration,which gives $0$ unpaired electrons.
Thus,the number of unpaired electrons in $X$ and $Y$ are $3$ and $0$ respectively.

(C) The correct option is $C$.
In $Ni(CO)_4$,the oxidation state of $Ni$ is $0$. Thus,its electronic configuration is $3d^8 4s^2$.
In the ground state,$Ni$ has two unpaired electrons in the $3d$ orbital.
$CO$ is a strong field ligand,which causes the pairing of electrons. The two $4s$ electrons are promoted to the $3d$ orbital,resulting in a $3d^{10} 4s^0$ configuration.
Since $Ni(CO)_4$ undergoes $sp^3$ hybridization,its geometry is tetrahedral.
Furthermore,all electrons are paired,making it diamagnetic in nature.

(C) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$. Its electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,the $4s$ electrons pair up in the $3d$ orbitals,resulting in $sp^3$ hybridisation.
The oxidation state of $Cr$ in $[Cr(H_2O)_6]^{2+}$ is $+2$. Its electronic configuration is $[Ar] 3d^4$. Since $H_2O$ is a weak field ligand,it does not cause pairing of electrons. However,the complex $[Cr(H_2O)_6]^{2+}$ involves $sp^3d^2$ hybridisation (outer orbital complex).