Hybridisation and Geometry Questions in English

(B) The central metal ion in the complex $[Ag(NH_3)_2]^+$ is $Ag^+$.
$Ag^+$ has an electronic configuration of $[Kr] 4d^{10}$.
Since the coordination number of $Ag^+$ in this complex is $2$,it undergoes $sp$ hybridization to form a linear geometry.
Therefore,the hybridization of $Ag^+$ is $sp$.

(A) In the complex $[Cu(NH_3)_4]SO_4$,the central metal ion is $Cu^{2+}$.
The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.
Therefore,the electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
$NH_3$ is a strong field ligand,but in the case of $Cu^{2+}$,the $3d^9$ configuration leaves one unpaired electron in the $3d$ orbital.
To accommodate four $NH_3$ ligands,the $Cu^{2+}$ ion undergoes $dsp^2$ hybridization,utilizing one $3d$,one $4s$,and two $4p$ orbitals.
Thus,the hybridization is $dsp^2$.

(B) The central atom $Sb$ has $5$ valence electrons.
It forms $5$ bonds with $F$ atoms and has $1$ lone pair due to the $2-$ charge.
Total electron pairs = $5 + 1 = 6$.
However,the question specifies $sp^3d$ hybridization for the $5$ bonding pairs.
With $5$ bond pairs and $1$ lone pair,the geometry is square pyramidal.

(B) The hybridization of the central atom in each species is determined as follows:
$1$. In $NH_3$,the central atom $N$ has $3$ bond pairs and $1$ lone pair,resulting in $sp^3$ hybridization.
$2$. In $[PtCl_4]^{2-}$,$Pt^{2+}$ has a $d^8$ configuration,which undergoes $dsp^2$ hybridization to form a square planar geometry.
$3$. In $PCl_5$,the central atom $P$ has $5$ bond pairs,resulting in $sp^3d$ hybridization.
$4$. In $BCl_3$,the central atom $B$ has $3$ bond pairs,resulting in $sp^2$ hybridization.
Thus,the correct order is $sp^3$,$dsp^2$,$sp^3d$,$sp^2$.

(B) The square planar geometry corresponds to $dsp^2$ hybridization.
This involves the mixing of one $s$ orbital,two $p$ orbitals (specifically $p_x$ and $p_y$),and one $d$ orbital (specifically $d_{x^2-y^2}$) to form four equivalent hybrid orbitals directed towards the corners of a square.
Therefore,the correct set of orbitals is $s, p_x, p_y, d_{x^2-y^2}$.

(D) To determine the hybridization,we look at the coordination number and the nature of the central metal ion.
$NiCl_4^{2-}$ involves $Ni^{2+}$ with a weak field ligand $Cl^-$,resulting in $sp^3$ hybridization (tetrahedral).
$SCl_4$ has $sp^3d$ hybridization due to the presence of a lone pair on sulfur.
$NH_4^+$ has $sp^3$ hybridization.
$PtCl_4^{2-}$ involves $Pt^{2+}$ (a $5d$ series metal),which always forms square planar complexes with strong or weak field ligands due to high crystal field splitting energy,resulting in $dsp^2$ hybridization.

(C) The correct answer is $C$.
In the formation of ${d^2}sp^3$ hybrid orbitals,two $(n-1)d$ orbitals belonging to the $e_g$ set,specifically the $(n-1)d_{z^2}$ and $(n-1)d_{x^2-y^2}$ orbitals,combine with one $ns$ orbital and three $np$ orbitals $(np_x, np_y, np_z)$ to form six equivalent ${d^2}sp^3$ hybrid orbitals.

(A) The structure of the $[Cu(H_2O)_4]^{2+}$ ion is square planar.
According to Crystal Field Theory $(CFT)$,the $Cu^{2+}$ ion has a $3d^9$ electronic configuration.
Due to the Jahn-Teller distortion,the complex undergoes elongation along the axial bonds,resulting in a square planar geometry where the four water ligands occupy the equatorial positions.

(B) The correct answer is $(B)$.
In the crystal structure of $CuSO_4 \cdot 5H_2O$,the $Cu^{2+}$ ion is coordinated to four water molecules in a square planar arrangement.
The remaining two coordination sites of the distorted octahedron are occupied by oxygen atoms from two different $SO_4^{2-}$ ions.
The fifth water molecule is held by hydrogen bonding between the coordinated water molecules and the sulphate ions.

(A) In the complex $[Ni(CO)_4]$,the central metal atom is Nickel $(Ni)$.
The atomic number of $Ni$ is $28$. Its ground state electronic configuration is $[Ar] 3d^8 4s^2$.
In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$ because $CO$ is a neutral ligand.
$CO$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbital. The two electrons from the $4s$ orbital shift into the $3d$ orbital,making it fully filled $(3d^{10})$.
This leaves one $4s$ orbital and three $4p$ orbitals vacant,which undergo $sp^3$ hybridization to accommodate the four electron pairs donated by four $CO$ ligands.
Therefore,the hybridization of $[Ni(CO)_4]$ is $sp^3$.

(A) In both complexes,the oxidation state of $Fe$ is $+3$,which corresponds to a $d^5$ electronic configuration.
For $[Fe(CN)_6]^{3-}$,$CN^-$ is a strong field ligand,causing pairing of electrons. The configuration becomes $t_{2g}^5 e_g^0$,resulting in $1$ unpaired electron.
For $[FeF_6]^{3-}$,$F^-$ is a weak field ligand,resulting in no pairing. The configuration is $t_{2g}^3 e_g^2$,resulting in $5$ unpaired electrons.
Since both complexes contain unpaired electrons,both are paramagnetic.

(A) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
Since $CO$ is a strong field ligand,it causes the pairing of electrons in the $3d$ and $4s$ orbitals.
As a result,the $4s$ electrons are promoted to the $3d$ orbital,leading to a fully filled $3d$ subshell $(3d^{10})$ and an empty $4s$ orbital.
Therefore,there are $0$ unpaired electrons in $Ni(CO)_4$.

(A) The central metal ion in $[CoCl_4]^{2-}$ is $Co^{2+}$.
Electronic configuration of $Co$ $(Z=27)$ is $[Ar]3d^74s^2$.
Electronic configuration of $Co^{2+}$ is $[Ar]3d^7$.
Since $Cl^-$ is a weak field ligand,the electrons in the $3d$ orbitals of $Co^{2+}$ remain unpaired according to Hund's rule.
The $3d^7$ configuration is represented as: two pairs and three unpaired electrons.
Therefore,the number of unpaired electrons is $3$.

(B) In the $[Cu(NH_3)_4]^{2+}$ complex,the central metal ion is $Cu^{2+}$ ($3d^9$ configuration).
Due to the presence of one unpaired electron in the $3d$ orbital,the hybridization involved is $dsp^2$.
This hybridization corresponds to a square planar geometry.

(A) The central atom $Sb$ in $[SbF_5]^{2-}$ has $5$ valence electrons.
It forms $5$ bonds with $F$ atoms and has $1$ lone pair of electrons,resulting in a total of $6$ electron pairs.
However,the question specifies $sp^3d$ hybridisation.
For a species with $sp^3d$ hybridisation and $1$ lone pair,the geometry is square pyramidal.

(B) $1$. The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
$2$. In the complex $[Ni(H_2O)_6]^{2+}$,the oxidation state of $Ni$ is $+2$. Thus,the configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$3$. $H_2O$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
$4$. To accommodate six pairs of electrons from six $H_2O$ ligands,the metal ion uses one $4s$,three $4p$,and two $4d$ orbitals.
$5$. This results in $sp^3d^2$ hybridization,which corresponds to an outer orbital octahedral complex.

(B) In $K_4[Fe(CN)_6]$,the central metal ion is $Fe^{2+}$.
The atomic number of $Fe$ is $26$,so the electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
After pairing,two $3d$ orbitals become vacant.
These two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals undergo $d^2sp^3$ hybridisation to form six equivalent octahedral hybrid orbitals.
Thus,the hybridisation involved is $d^2sp^3$.

(C) The hybridization $sp^3d^2$ (or $d^2sp^3$) involves six hybrid orbitals directed towards the corners of an octahedron,resulting in an octahedral geometry for the complex.

(A) An octahedral complex requires a coordination number of $6$,meaning $6$ ligands are attached to the central metal ion.
In the complex $[FeF_6]^{3-}$,there are $6$ $F^{-}$ ligands attached to the $Fe^{3+}$ ion,resulting in an octahedral geometry.
The other complexes listed have a coordination number of $4$,which corresponds to tetrahedral or square planar geometries.

(B) $[Ag(NH_3)_2]^+$ is linear due to $sp$ hybridization.
$[Cu(en)_2]^{2+}$ has a coordination number of $4$ and exhibits square planar geometry due to $dsp^2$ hybridization.
$[MnCl_4]^{2-}$ is tetrahedral due to $sp^3$ hybridization.
$[Ni(CO)_4]$ is tetrahedral due to $sp^3$ hybridization.
Therefore,the correct option is $(b)$.

(C) In the $[Fe(CN)_6]^{4-}$ complex,the central metal ion is $Fe^{2+}$.
The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
$CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
This results in $d^2sp^3$ hybridization,which corresponds to an octahedral geometry.

(D) . $Fe(CO)_5$ undergoes $dsp^3$ hybridization,which results in a trigonal bipyramidal geometry.

(A) In $[Fe(CN)_6]^{3-}$,the central metal ion is $Fe^{3+}$.
The atomic number of $Fe$ is $26$,so the electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $CN^-$ 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 hybridize to form $d^2sp^3$ hybrid orbitals.
Thus,the hybridization is $d^2sp^3$.

(B) In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$ ($3d^8$ configuration).
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
This results in one vacant $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals,which undergo $dsp^2$ hybridisation to form a square planar geometry.

(B) The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^{2+}$,the configuration is $[Ar] 3d^9$.
In the $[Cu(NH_3)_4]^{2+}$ complex,the $Cu^{2+}$ ion undergoes $dsp^2$ hybridisation to form a square planar geometry.
One of the $3d$ electrons is promoted to the $4p$ orbital to allow for the $dsp^2$ hybridisation,which involves one $3d$,one $4s$,and two $4p$ orbitals.

(D) In $Ni(CO)_4$,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,which gives a tetrahedral geometry.
In $Ni(PPh_3)_2Cl_2$,the oxidation state of $Ni$ is $+2$. The configuration is $3d^8$. $PPh_3$ is a bulky ligand,and $Cl^-$ is a weak field ligand. This complex also undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.
Therefore,both complexes are tetrahedral.

(D) The complex $[Cu(NH_3)_4]^{2+}$ exhibits a square planar geometry due to $dsp^2$ hybridization of the central $Cu^{2+}$ ion.

(A) In the complex $[Pt(NH_3)_4]Cl_2$,the central metal ion is $Pt^{2+}$.
$Pt$ $(Z=78)$ has an electronic configuration of $[Xe] 4f^{14} 5d^9 6s^1$. For $Pt^{2+}$,the configuration is $[Xe] 4f^{14} 5d^8$.
Since $NH_3$ is a strong field ligand,it causes pairing of electrons in the $5d$ orbitals.
The four $NH_3$ ligands donate electron pairs to the empty $5d$,$6s$,and $6p$ orbitals,resulting in $dsp^2$ hybridisation.
$A$ $dsp^2$ hybridised complex always exhibits a square planar geometry.

(A) In coordination chemistry,the hybridization of the central metal atom determines the geometry of the complex.
For a complex involving $dsp^2$ hybridization,one $d$-orbital,one $s$-orbital,and two $p$-orbitals are involved.
This specific combination of orbitals results in a square planar geometry,where the four ligands are arranged at the corners of a square around the central metal ion.

(B) In coordination chemistry,the geometry of a complex ion is determined by the hybridization of the central metal atom or ion.
For a coordination number of $4$,the complex can exhibit either tetrahedral or square planar geometry.
Tetrahedral geometry arises from $sp^3$ hybridization,where the four orbitals are directed towards the corners of a regular tetrahedron.
Square planar geometry arises from $dsp^2$ hybridization.
Therefore,a tetrahedral complex ion is formed due to $sp^3$ hybridization.

(C) $d^2sp^3$ hybridisation involves the participation of two $d$-orbitals,one $s$-orbital,and three $p$-orbitals to form six equivalent hybrid orbitals.
These orbitals are directed towards the corners of an octahedron,resulting in an octahedral geometry.

(C) The central metal atom in $Cr(CO)_6$ is Chromium $(Cr)$.
The atomic number of $Cr$ is $24$,and its electronic configuration is $[Ar] 3d^5 4s^1$.
In $Cr(CO)_6$,the oxidation state of $Cr$ is $0$ because $CO$ is a neutral ligand.
$CO$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
The $3d$ orbitals become $t_{2g}^6 e_g^0$,leaving two $3d$ orbitals empty.
These two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals hybridize to form six $d^2sp^3$ hybrid orbitals.
Thus,the hybridization is $d^2sp^3$ and the geometry is octahedral.

(C) 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 are used to form $sp^3d^2$ hybrid orbitals.
Thus,$[CoF_6]^{3-}$ is an outer orbital complex with $sp^3d^2$ hybridization.

(D) The geometry of coordination complexes depends on the nature of the ligand and the central metal ion.
$1$. $[NiCl_4]^{2-}$: $Ni^{2+}$ has a $d^8$ configuration. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons. The hybridization is $sp^3$,resulting in a tetrahedral geometry.
$2$. $[PdCl_4]^{2-}$,$[Ni(CN)_4]^{2-}$,and $[Pd(CN)_4]^{2-}$: These complexes involve $Pd^{2+}$ or strong field ligands like $CN^-$,which cause pairing of electrons. They undergo $dsp^2$ hybridization,resulting in a square planar geometry.

(D) $1$. For $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$,resulting in $sp^3$ hybridization.
$2$. For $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of $3d$ electrons,resulting in $dsp^2$ hybridization.
$3$. For $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,no pairing occurs,resulting in $sp^3$ hybridization.
Therefore,the hybridization states are $sp^3, dsp^2, sp^3$ respectively. The correct option is $D$.

(D) . In $K_3[Fe(CN)_6]$,the oxidation state of $Fe$ is $+3$.
The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
The $5$ electrons in $3d$ orbitals rearrange to occupy $3$ orbitals,leaving $2$ empty $3d$ orbitals.
These $2$ empty $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals hybridize to form $6$ equivalent $d^2sp^3$ hybrid orbitals.
Thus,the hybridization of $Fe$ is $d^2sp^3$.

(A) The complex ion $[MnO_4]^-$ has no $d$-electron in the central metal atom.
The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$. In $[MnO_4]^-$,the oxidation number of $Mn$ is $+7$,meaning all the $3d$ and $4s$ electrons are lost to form the $[MnO_4]^-$ complex.
Thus,the electronic configuration of $Mn(VII)$ is $[Ar] 3d^0 4s^0$,confirming it has no $d$-electrons.
In contrast,the central metal atoms in $[Co(NH_3)_6]^{3+}$,$[Fe(CN)_6]^{3-}$,and $[Cr(H_2O)_6]^{3+}$ have $6$,$5$,and $3$ $d$-electrons respectively.

(C) An octahedral complex is formed when the metal ion is surrounded by six ligands.
$d^6$ (low spin) configuration,such as in $[Co(NH_3)_6]^{3+}$,is a classic example of an octahedral complex where the crystal field splitting energy is high,leading to the pairing of electrons in the $t_{2g}$ orbitals.

(D) The atomic number of $Mn$ is $25$. The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$.
In the complex $[Mn(H_2O)_6]^{2+}$,the oxidation state of $Mn$ is $+2$. Thus,the configuration of $Mn^{2+}$ is $[Ar] 3d^5$.
Since $H_2O$ is a weak field ligand,it does not cause pairing of electrons in the $d$-orbitals.
Therefore,the $5$ electrons in the $3d$ subshell remain unpaired.
Hence,the number of unpaired electrons is $5$.

(D) In $Ni(CO)_4$,the oxidation state of $Ni$ is $0$. The configuration is $[Ar] 3d^8 4s^2$. Due to the strong field ligand $CO$,the electrons pair up,resulting in $sp^3$ hybridization,which gives a tetrahedral geometry.
In $Ni(PPh_3)_2Cl_2$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $PPh_3$ is a bulky ligand,and $Cl^-$ is a weak field ligand. The steric hindrance caused by $PPh_3$ groups forces the complex to adopt a tetrahedral geometry ($sp^3$ hybridization) rather than square planar.

(D) An outer orbital complex uses the outer $d$-orbitals (e.g.,$4d$) for hybridization,typically $sp^3d^2$.
$1$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing. Hybridization is $d^2sp^3$ (inner orbital).
$2$. $[Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. Hybridization is $d^2sp^3$ (inner orbital).
$3$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand for $Co^{3+}$,causing pairing. Hybridization is $d^2sp^3$ (inner orbital).
$4$. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand relative to $Ni^{2+}$. Hybridization is $sp^3d^2$ (outer orbital).
Thus,$[Ni(NH_3)_6]^{2+}$ is an outer orbital complex.

(B) The atomic number of $Ni$ is $28$.
The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
When $Ni$ forms $Ni^{2+}$ ion,it loses two electrons from the $4s$ orbital.
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$.
In the $3d^8$ subshell,the electrons are filled as follows:
$3d^8$: $\uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow$.
There are $2$ unpaired electrons in the $3d$ orbital.
Thus,the correct option is $B$.

(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
When $Ni$ forms $Ni^{2+}$ ion,it loses two electrons from the $4s$ orbital.
Thus,the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
In the $3d$ subshell,there are $5$ orbitals. According to Hund's rule,the $8$ electrons are filled as follows: $\uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow$.
There are $2$ unpaired electrons in the $3d$ orbitals.

(B) In the complex $[Ag(NH_3)_2]^+$,the central metal ion is $Ag^+$.
The electronic configuration of $Ag^+$ is $[Kr] 4d^{10}$.
Since the coordination number is $2$ and the geometry is linear,the $Ag^+$ ion undergoes $sp$ hybridization to accommodate the two lone pairs from the $NH_3$ ligands.

(C) The ion $[Cu(NH_3)_4]^{2+}$ has a square planar geometry. In this complex,the central $Cu^{2+}$ ion undergoes $dsp^2$ hybridization,which results in a square planar shape rather than tetrahedral.

(B) In octahedral hybridization,whether it is inner orbital $({d^2}sp^3)$ or outer orbital $(sp^3d^2)$,the $d$-orbitals involved are always the axial orbitals,which are ${d_{x^2-y^2}}$ and ${d_{z^2}}$.

(B) square planar complex involves $dsp^2$ hybridization.
This hybridization utilizes one $d$-orbital $(d_{x^2-y^2})$,one $s$-orbital,and two $p$-orbitals ($p_x$ and $p_y$).

(D) The complex $[Ni(H_2O)_6]^{2+}$ contains $Ni^{2+}$ ion,which has a $d^8$ electronic configuration.
In the presence of water ligands,which are weak field ligands,the $d-d$ transition occurs,resulting in the absorption of light in the red region.
Consequently,the transmitted light appears green.
Therefore,the aqueous solution of $[Ni(H_2O)_6]^{2+}$ is green in color.

(D) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 3.88 \, BM$,we have $\sqrt{n(n+2)} \approx 3.88$,which implies $n = 3$.
In $[Cr(H_2O)_6]Cl_3$,the oxidation state of $Cr$ is $+3$.
The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.
Since there are $3$ unpaired electrons in the $3d$ subshell,they will occupy three different $d$-orbitals according to Hund's rule.
The $d$-orbitals are $d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2},$ and $d_{z^2}$.
Any combination of three distinct $d$-orbitals is possible for the $3d^3$ configuration,but $d_{xy}, d_{yz},$ and $d_{xz}$ are the standard set of $t_{2g}$ orbitals.

(D) $1$. In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. $CO$ is a strong field ligand,causing pairing of $3d$ electrons. The hybridization is $sp^3$.
$2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of $3d$ electrons. The hybridization is $dsp^2$.
$3$. In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,not causing pairing of $3d$ electrons. The hybridization is $sp^3$.
Thus,the hybridization states are $sp^3, dsp^2, sp^3$ respectively.