Hybridisation and Geometry Questions in English

(D) To determine the shape,we analyze the hybridization of each complex:
$1$. $[Zn(NH_3)_4]^{2+}$: $Zn^{2+}$ has a $d^{10}$ configuration. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.
$2$. $[Ni(CO)_4]$: $Ni$ is in the $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.
$3$. $[Cu(CN)_4]^{3-}$: $Cu^+$ has a $d^{10}$ configuration. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.
$4$. $[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ has a $d^9$ configuration. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry.
Thus,$[Cu(NH_3)_4]^{2+}$ has a square planar shape,while the others are tetrahedral.

(C) To determine the number of unpaired electrons,we analyze the electronic configuration in an octahedral field:
$1$. For $d^4$ (low spin): The configuration is $t_{2g}^4 e_g^0$. The number of unpaired electrons is $2$.
$2$. For $d^8$ (high spin): The configuration is $t_{2g}^6 e_g^2$. The number of unpaired electrons is $2$.
$3$. For $d^6$ (high spin): The configuration is $t_{2g}^4 e_g^2$. The number of unpaired electrons is $4$.
Comparing these,$d^6$ (high spin) has the maximum number of unpaired electrons $(4)$.

(C) In octahedral complexes,the coordination number is $6$.
Low spin complexes are formed when strong field ligands cause pairing of electrons in the inner $d$-orbitals.
This requires the use of $(n-1)d$ orbitals for hybridization,resulting in $d^2sp^3$ hybridization.
$sp^3d^2$ hybridization corresponds to high spin outer orbital complexes.

(D) The geometry of a coordination complex depends on its coordination number and the nature of the ligands.
$Tetracarbonylnickel(0)$ $([Ni(CO)_4])$ has a coordination number of $4$ and exhibits tetrahedral geometry.
$Hexaamminecobalt(II)$ nitrate $([Co(NH_3)_6](NO_3)_2)$ has a coordination number of $6$ and exhibits octahedral geometry.
$Pentacarbonyliron(0)$ $([Fe(CO)_5])$ has a coordination number of $5$ and exhibits trigonal bipyramidal geometry.
$Bis(acetylacetonato)oxovanadium(IV)$ $([VO(acac)_2])$ has a coordination number of $5$ and exhibits square pyramidal geometry,where the oxo group $(O^{2-})$ occupies the apical position.

(D) $1$. $[AgF_4]^-$: $Ag(III)$ is a $d^8$ metal ion,which typically forms square planar complexes.
$2$. $[AuCl_4]^-$: $Au(III)$ is a $d^8$ metal ion,which forms square planar complexes.
$3$. $[RhCl(PPh_3)_3]$: This is a $16$-electron $Rh(I)$ complex,which is square planar (Wilkinson's catalyst).
$4$. $[NiCl_2(PMe_3)_2]$: $Ni(II)$ is a $d^8$ ion. While many $Ni(II)$ complexes are square planar,the geometry depends on the ligand field strength. For $[NiCl_2(PMe_3)_2]$,the complex is tetrahedral due to the steric bulk of the $PMe_3$ ligands and the weak field nature of $Cl^-$ ligands.

(D) To determine the geometry,we look at the hybridization of the central metal ion:
$1$. $[Cu(CN)_4]^{-2}$: $Cu^{+2}$ is $d^9$. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry.
$2$. $[PtCl_4]^{-2}$: $Pt^{+2}$ is a $5d$ series element,which always forms square planar complexes due to high crystal field splitting energy.
$3$. $[Cu(H_2O)_4]^{+2}$: This complex has a distorted octahedral geometry (Jahn-Teller distortion) in the solid state,often described as elongated octahedral,not square planar.
$4$. $[Ni(Cl)_4]^{-2}$: $Ni^{+2}$ is $d^8$. $Cl^-$ is a weak field ligand,so it undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.
Both $[Cu(H_2O)_4]^{+2}$ and $[Ni(Cl)_4]^{-2}$ are not square planar. However,in standard textbook contexts for this specific question,$[Ni(Cl)_4]^{-2}$ is the classic example of a tetrahedral complex. Given the options,$[Ni(Cl)_4]^{-2}$ is the most definitive tetrahedral complex.

(B) $1$. For $NH_3$: The central atom $N$ has $5$ valence electrons. It forms $3$ bonds with $H$ and has $1$ lone pair. Steric number = $3 + 1 = 4$,so hybridization is $sp^3$.
$2$. For $[PtCl_4]^{2-}$: $Pt$ is in $+2$ oxidation state ($d^8$ configuration). It forms a square planar complex,so hybridization is $dsp^2$.
$3$. For $PCl_5$: $P$ has $5$ valence electrons and forms $5$ bonds with $Cl$. Steric number = $5$,so hybridization is $sp^3d$.
$4$. For $BCl_3$: $B$ has $3$ valence electrons and forms $3$ bonds with $Cl$. Steric number = $3$,so hybridization is $sp^2$.
Thus,the sequence is $sp^3, dsp^2, sp^3d, sp^2$.

(B) In $d^2sp^3$ hybridization,the two $d$-orbitals involved are the inner $d$-orbitals,which are the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals.
These orbitals are directed along the axes and are used to form the octahedral complex.

(D) The central metal ion in all these complexes is $Pt^{2+}$,which has a $d^8$ electronic configuration.
For $d^8$ metal ions,the crystal field splitting energy is generally high enough to cause pairing of electrons,leading to $dsp^2$ hybridization.
$dsp^2$ hybridization results in a square planar geometry.
Therefore,all the given complexes,$[PtCl_4]^{2-}$,$[PtClBr(H_2O)_2]$,and $[Pt(NH_3)(Py)(H_2O)I]^{+1}$,exhibit square planar geometry.

(D) In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $+3$ $(3d^6)$. $F^-$ is a weak field ligand,so it forms an outer orbital complex $(sp^3d^2)$ which is paramagnetic.
In $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$ $(3d^6)$. $NH_3$ is a strong field ligand,so it causes pairing of electrons,forming an inner orbital complex $(d^2sp^3)$ which is diamagnetic.
Therefore,the statement that $[CoF_6]^{3-}$ is an outer orbital complex and $[Co(NH_3)_6]^{3+}$ is an inner orbital complex is correct.

(D) In $[Cu(CN)_4]^{3-}$,$Cu$ is in $+1$ oxidation state ($3d^{10}$ configuration). It is colourless,tetrahedral,and diamagnetic. This is correct.
In $[Cu(NH_3)_4]^{2+}$,$Cu$ is in $+2$ oxidation state ($3d^9$ configuration). It is coloured due to $d-d$ transition and has a square planar geometry. This is correct.
In $[Pt(NH_3)_2Cl_2]$,$Pt^{2+}$ is a $5d^8$ metal ion,which forms square planar complexes. It is diamagnetic and exhibits geometrical isomerism (cis-trans). This is correct.
In $[Ni(dmg)_2]$,$Ni^{2+}$ is a $3d^8$ ion. It forms a square planar complex (not tetrahedral) and is diamagnetic. It contains intramolecular hydrogen bonding. Therefore,the statement 'tetrahedral' is incorrect.

(D) The complex $[HgI_3]^{-}$ involves the $Hg^{2+}$ ion. The electronic configuration of $Hg^{2+}$ is $[Xe] 4f^{14} 5d^{10}$.
Since the $d$-orbitals are completely filled,the central metal ion does not participate in $sp^2$ hybridisation in the way a main group element would.
In reality,$[HgI_3]^{-}$ adopts a trigonal planar geometry due to the steric arrangement of the three iodide ligands around the mercury center,but the bonding is primarily ionic/covalent with $d^{10}$ configuration.
However,the statement that it has $sp^2$ hybridisation is chemically incorrect because $Hg^{2+}$ is a $d^{10}$ system and does not typically undergo $sp^2$ hybridisation for this geometry.
Since both $A$ and $B$ are technically incorrect descriptions of the bonding and geometry,option $D$ is the correct answer.

(C) Let us analyze each complex:
$1$. $AuCl_4^-$: $Au^{3+}$ is a $5d^8$ ion. It forms a square planar complex with $dsp^2$ hybridisation.
$2$. $[Co(oxalato)_3]^{3-}$: $Co^{3+}$ is a $3d^6$ ion. Oxalate is a strong field ligand,leading to inner orbital complex formation with $d^2sp^3$ hybridisation.
$3$. $RhCl(PPh_3)_3$: This is Wilkinson's catalyst. $Rh^+$ is a $4d^8$ ion,forming a square planar complex with $dsp^2$ hybridisation.
$4$. $[Fe(NH_3)_6]^{2+}$: $Fe^{2+}$ is a $3d^6$ ion. $NH_3$ acts as a strong field ligand for $Fe^{2+}$,resulting in $d^2sp^3$ hybridisation. However,option $C$ is the most standard representation of $dsp^2$ for $Rh(I)$ square planar complexes.

(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$.
The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
Due to the presence of a strong field ligand $(CO)$,the electrons pair up,resulting in $3d^{10}$ configuration.
The $4s$ and $4p$ orbitals undergo $sp^3$ hybridization.
Therefore,the geometry of $[Ni(CO)_4]$ is tetrahedral.

(C) In the complex $[Fe(H_2O)_5NO]^{2+}$,the $NO$ ligand is present as $NO^+$.
Calculating the oxidation state of $Fe$: $x + 5(0) + 1 = +2$,so $x = +1$.
The coordination number is $5 + 1 = 6$.
Since the coordination number is $6$,the geometry is octahedral,not tetrahedral.
Therefore,the statement that it is a tetrahedral complex is incorrect.

(D) In an outer orbital octahedral complex,the ligands are weak field ligands,which means they do not cause pairing of electrons in the $d$-orbitals.
For a $d^4$ configuration: $t_{2g}^3 e_g^1$,number of unpaired electrons = $4$.
For a $d^9$ configuration: $t_{2g}^6 e_g^3$,number of unpaired electrons = $1$.
For a $d^7$ configuration: $t_{2g}^5 e_g^2$,number of unpaired electrons = $3$.
For a $d^5$ configuration: $t_{2g}^3 e_g^2$,number of unpaired electrons = $5$.
Comparing these,the $d^5$ configuration has the maximum number of unpaired electrons $(5)$.

(A) The coordination number of a metal ion in a complex depends on its electronic configuration and the nature of the ligands.
$Pt^{2+}$ (a $d^8$ metal ion) typically forms square planar complexes with a coordination number of $4$.
$Cr^{3+}$ and $Fe^{3+}$ are $d^3$ and $d^5$ ions respectively,which typically form octahedral complexes with a coordination number of $6$.
$Pt^{4+}$ is a $d^6$ ion which also typically forms octahedral complexes with a coordination number of $6$.
Therefore,$Pt^{2+}$ is the correct answer.

(A) In the crystal structure of $CuSO_4 \cdot 5H_2O$,the copper ion $(Cu^{2+})$ is coordinated to $4$ water molecules in a square planar geometry.
The fifth water molecule is held by hydrogen bonding in the crystal lattice,while the sulfate ion $(SO_4^{2-})$ is also involved in the coordination sphere.

(C) The complex ion is $[Ni(CN)_4]^{2-}$.
In this complex,the oxidation state of $Ni$ is $+2$,which corresponds to the electronic configuration $[Ar] 3d^8$.
$CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
This results in $dsp^2$ hybridization.
$dsp^2$ hybridization corresponds to a square planar geometry.

(C) In the $[Cu(NH_3)_4]^{2+}$ complex,the central metal ion is $Cu^{2+}$.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
Although $NH_3$ is a strong field ligand,the geometry of this complex is $dsp^2$ hybridized,resulting in a square planar structure.
This is due to the promotion of one electron from the $3d$ orbital to the $4p$ orbital to facilitate $dsp^2$ hybridization.

(D) In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $+3$. The atomic number of $Co$ is $27$,so its electronic configuration is $[Ar] 3d^7 4s^2$.
For $Co^{3+}$,the configuration is $[Ar] 3d^6$.
$F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $3d$ electrons remain unpaired,and the hybridization involves the outer $4d$ orbitals.
The hybridization is $sp^3d^2$.

(D) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry.
$2$. $[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ is $3d^9$. The unpaired electron is in the $4p_z$ orbital,and the four ligands form bonds using $dsp^2$ hybrid orbitals,resulting in a square planar geometry.
$3$. $[PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d^8$ ion. $Pt$ is a $5d$ series element,so it always forms square planar complexes with $dsp^2$ hybridization regardless of the ligand strength.
Since all three complexes exhibit $dsp^2$ hybridization and square planar geometry,the correct answer is $D$.

(A) In the $[PtCl_6]^{2-}$ ion,the central metal atom is $Pt^{4+}$.
$Pt$ is a $5d$ series transition metal $(5d^8 6s^2)$.
$Pt^{4+}$ has the electronic configuration $[Xe] 4f^{14} 5d^6$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons in the $5d$ orbitals.
Therefore,the $5d$ orbitals remain occupied by $6$ electrons,and the hybridization involves the outer $6s$,$6p$,and $6d$ orbitals.
The hybridization is $sp^3d^2$,resulting in an octahedral geometry.

(A) $1$. In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$. $F^-$ is a weak field ligand,so it does not cause pairing of electrons. Thus,it undergoes $sp^3d^2$ hybridization (outer orbital complex).
$2$. In $[Co(NH_3)_6]^{3+}$,$Co^{3+}$ is $[Ar] 3d^6$. $NH_3$ is a strong field ligand,causing pairing of electrons in $3d$ orbitals,leading to $d^2sp^3$ hybridization.
$3$. In $[Fe(CN)_6]^{3-}$,$Fe^{3+}$ is $[Ar] 3d^5$. $CN^-$ is a strong field ligand,causing pairing,leading to $d^2sp^3$ hybridization.
$4$. In $[Cr(NH_3)_6]^{3+}$,$Cr^{3+}$ is $[Ar] 3d^3$. It utilizes two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals to form $d^2sp^3$ hybridization.
$5$. Therefore,$[CoF_6]^{3-}$ is the only complex that exhibits $sp^3d^2$ hybridization instead of $d^2sp^3$.

(B) In $[Cu(NH_3)_4]^{2+}$,the oxidation state of $Cu$ is $+2$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
In the presence of $NH_3$ (a strong field ligand),the $3d$ electrons are arranged such that one electron remains unpaired in the $3d$ orbital.
The hybridization involved is $dsp^2$,which corresponds to a square planar geometry.
Thus,the complex has a square planar geometry with one unpaired electron.

(B) $1$. In the complex ion $[Co(C_2O_4)_3]^{3-}$,the central metal atom is $Co^{3+}$.
$2$. The atomic number of $Co$ is $27$,so the electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$3$. The ligand $C_2O_4^{2-}$ (oxalate) is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
$4$. Due to the presence of a strong field ligand,the $6$ electrons in the $3d$ orbitals pair up to occupy $3$ orbitals,leaving $2$ empty $3d$ orbitals.
$5$. These $2$ empty $3d$ orbitals,along with one $4s$ and three $4p$ orbitals,undergo $d^2sp^3$ hybridization to form $6$ equivalent hybrid orbitals to accommodate $6$ pairs of electrons from $3$ bidentate oxalate ligands.

(B) $1$. In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$2$. $CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
$3$. This results in one empty $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals,which undergo $dsp^2$ hybridization to form a square planar geometry.
$4$. $[NiCl_4]^{2-}$ and $[Zn(NH_3)_4]^{2+}$ exhibit $sp^3$ hybridization (tetrahedral),while $[FeF_6]^{3-}$ exhibits $sp^3d^2$ hybridization (octahedral).

(C) The formula for the hexafluoroferrate $(II)$ ion is $[FeF_6]^{4-}$.
In this complex,the oxidation state of $Fe$ is $+2$.
The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
Since $F^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Thus,the $3d$ orbitals contain $5$ electrons in the $t_{2g}$ and $e_g$ sets as follows: $t_{2g}^4 e_g^2$.
The number of unpaired electrons is $4$.

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

(A) $1$. For $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. The hybridization is $sp^3$,resulting in a tetrahedral geometry and diamagnetic nature (all electrons paired).
$2$. For $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing. The hybridization is $dsp^2$,resulting in a square planar geometry and diamagnetic nature.
$3$. For $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,no pairing occurs. The hybridization is $sp^3$,resulting in a tetrahedral geometry and paramagnetic nature (two unpaired electrons).
$4$. Comparing these with the options,statement $A$ is incorrect because $[Ni(CO)_4]$ is diamagnetic,not paramagnetic.

(B) $1$. In $K_3[CoF_6]$,the oxidation state of $Co$ is $+3$.
$2$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$3$. $F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
$4$. Since it is an octahedral complex with a coordination number of $6$,the hybridization involves the outer $4d$ orbitals.
$5$. Thus,the hybridization is $sp^3d^2$.

(A) Outer orbital hybridization involves the use of outer $d$-orbitals (i.e.,$nd$ orbitals) for hybridization,typically $sp^3d^2$.
In $[Zn(NH_3)_6]^{2+}$,the oxidation state of $Zn$ is $+2$. The electronic configuration of $Zn^{2+}$ is $[Ar] 3d^{10}$.
Since the $3d$ subshell is completely filled,the $NH_3$ ligands must occupy the $4s$,$4p$,and $4d$ orbitals to form the complex.
Thus,it undergoes $sp^3d^2$ hybridization,which is an outer orbital complex.
In contrast,$[Co(NH_3)_6]^{3+}$,$[Cr(NH_3)_6]^{3+}$,and $[V(NH_3)_6]^{3+}$ involve inner $d$-orbitals $(3d)$ for hybridization ($d^2sp^3$ hybridization).

(A) $1$. Inner orbital complexes involve the use of $(n-1)d$ orbitals for hybridization (e.g.,$d^2sp^3$).
$2$. Outer orbital complexes involve the use of $nd$ orbitals for hybridization (e.g.,$sp^3d^2$).
$3$. In $[CoF_6]^{3-}$,$F^-$ is a weak field ligand,so it does not cause pairing of electrons. The hybridization is $sp^3d^2$ (outer orbital).
$4$. In $[Co(NH_3)_6]^{3+}$,$[Fe(CN)_6]^{3-}$,and $[Cr(NH_3)_6]^{3+}$,the ligands are strong enough or the electronic configuration allows for $d^2sp^3$ (inner orbital) hybridization.
$5$. Therefore,$[CoF_6]^{3-}$ is the complex that does not involve inner orbital hybridization.

(A) An outer orbital octahedral complex is formed when the metal ion uses outer $d$-orbitals (i.e.,$4d$-orbitals) for hybridization,resulting in $sp^3d^2$ hybridization.
This typically occurs with weak field ligands.
For $d^4, d^5, d^6$ configurations,if the ligand is weak,the electrons do not pair up,leading to high spin complexes which are outer orbital complexes.
Specifically,$d^4, d^5, d^6$ (high spin) configurations result in $sp^3d^2$ hybridization.
Among the given options,$d^6$ (high spin) is a classic example of an outer orbital complex (e.g.,$[FeF_6]^{3-}$).

(B) Iron pentacarbonyl,$Fe(CO)_5$,involves the central metal atom $Fe$ in the $d^8$ configuration.
In this complex,the $Fe$ atom undergoes $dsp^3$ hybridization.
According to $VSEPR$ theory and valence bond theory,$dsp^3$ hybridization corresponds to a trigonal bipyramidal geometry.
Thus,the structure of $Fe(CO)_5$ is trigonal bipyramidal.

(A) To determine the geometry,we analyze the hybridization and lone pairs of the central atom:
$1$. $XeF_4$: $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ and has $2$ lone pairs. Steric number = $4 + 2 = 6$ ($sp^3d^2$ hybridization). Due to $2$ lone pairs,it adopts a square planar geometry.
$2$. $SF_4$: $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ and has $1$ lone pair. Steric number = $4 + 1 = 5$ ($sp^3d$ hybridization). It adopts a see-saw geometry.
$3$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is a $d^8$ ion. With a weak field ligand $(Cl^-)$,it undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.
$4$. $[PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d$ series metal ion. It forms square planar complexes due to high crystal field splitting,resulting in $dsp^2$ hybridization.
Therefore,the species with square planar geometry are $(i)$ $XeF_4$ and $(iv)$ $[PtCl_4]^{2-}$.

(D) In $[FeF_6]^{3-}$,the oxidation state of $Fe$ is $+3$ ($3d^5$ configuration).
$F^{-}$ is a weak field ligand,which does not cause pairing of electrons.
Therefore,it forms a high spin complex with $5$ unpaired electrons.
Thus,the Assertion is incorrect,while the Reason is a correct statement regarding the definition of low spin complexes.

(A) $AlCl_{3}$ in an acidified aqueous solution undergoes hydration to form the octahedral complex $[Al(H_{2}O)_{6}]^{3+}$.
The electronic configuration of $Al^{3+}$ is $[Ne] \ 3s^{0} \ 3p^{0} \ 3d^{0}$.
To accommodate six water ligands,the $Al^{3+}$ ion utilizes one $3s$,three $3p$,and two $3d$ orbitals,resulting in $sp^{3}d^{2}$ hybridisation.
Therefore,$A = [Al(H_{2}O)_{6}]^{3+}$ and $B = sp^{3}d^{2}$.

(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
In the complex $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$.
$CO$ is a strong field ligand $(SFL)$,which causes the pairing of electrons in the $3d$ orbital.
The $4s$ electrons are promoted to the $3d$ orbital,and the $4s$ and $4p$ orbitals undergo $sp^3$ hybridisation.
Since all electrons are paired,the complex is diamagnetic.
The $sp^3$ hybridisation results in a tetrahedral geometry.

(B) $1$. For a square pyramidal geometry $(ML_5)$ (e.g.,$Ni$ complex):
- There are $8$ angles of $90^{\circ}$ (between the axial ligand and the four basal ligands,and between adjacent basal ligands).
- There are $2$ angles of $180^{\circ}$ (between the axial ligand and the opposite basal ligand,and between opposite basal ligands).
- Total angles for square pyramidal = $8 + 2 = 10$.
$2$. For a trigonal bipyramidal geometry $(ML_5)$ (e.g.,$Fe$ complex):
- There are $6$ angles of $90^{\circ}$ (between axial and equatorial ligands).
- There are $3$ angles of $120^{\circ}$ (between equatorial ligands).
- There are $1$ angle of $180^{\circ}$ (between the two axial ligands).
- Total angles for trigonal bipyramidal = $6 + 3 + 1 = 10$.
$3$. Sum of all specified angles = $10 + 10 = 20$.

(A) The oxidation state of $Mn$ in $[MnBr_4]^{2-}$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.
Since the coordination number is $4$,the geometry can be either tetrahedral $(sp^3)$ or square planar $(dsp^2)$.
$A$ magnetic moment of $5.9 \ BM$ corresponds to $n = 5$ unpaired electrons,calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
For $Mn^{2+}$ $(3d^5)$,having $5$ unpaired electrons indicates a high-spin complex.
Therefore,the geometry must be tetrahedral,as square planar geometry would involve pairing of electrons,resulting in a lower magnetic moment.

(N/A) In both complexes,$Ni$ is in the $+2$ oxidation state,which corresponds to a $d^8$ electronic configuration.
For $[Ni(CN)_4]^{2-}$:
$CN^-$ is a strong field ligand. It causes the pairing of the two unpaired $3d$ electrons. This results in $dsp^2$ hybridization,leading to a square planar geometry. Since all electrons are paired,the complex is diamagnetic.
For $[NiCl_4]^{2-}$:
$Cl^-$ is a weak field ligand. It does not cause the pairing of the two unpaired $3d$ electrons. Therefore,the complex undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry. Due to the presence of two unpaired electrons,the complex is paramagnetic.

(N/A) Though both $[NiCl_{4}]^{2-}$ and $[Ni(CO)_{4}]$ are tetrahedral,their magnetic characters are different. This is due to a difference in the nature of ligands.
In $[NiCl_{4}]^{2-}$,$Ni$ is in the $+2$ oxidation state with a $3d^{8}$ configuration. $Cl^{-}$ is a weak field ligand and it does not cause the pairing of unpaired $3d$ electrons. Hence,$[NiCl_{4}]^{2-}$ is paramagnetic.
In $[Ni(CO)_{4}]$,$Ni$ is in the zero oxidation state i.e.,it has a configuration of $3d^{8}4s^{2}$.
$CO$ is a strong field ligand. Therefore,it causes the pairing of unpaired $3d$ electrons. Also,it causes the $4s$ electrons to shift to the $3d$ orbital,thereby giving rise to $sp^{3}$ hybridization. Since no unpaired electrons are present in this case,$[Ni(CO)_{4}]$ is diamagnetic.

(N/A)

$[Co(NH_{3})_{6}]^{3+}$	$[Ni(NH_{3})_{6}]^{2+}$
Oxidation state of cobalt $= +3$	Oxidation state of nickel $= +2$
Electronic configuration of cobalt ion $(Co^{3+})$ $= 3d^{6}$	Electronic configuration of nickel ion $(Ni^{2+})$ $= 3d^{8}$
$NH_{3}$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals. This makes two $3d$ orbitals available for $d^{2}sp^{3}$ hybridization. Hence,it is an inner orbital complex.	Even with $NH_{3}$ as a ligand,the $3d^{8}$ configuration leaves only one $3d$ orbital empty after pairing. Thus,it cannot undergo $d^{2}sp^{3}$ hybridization and instead undergoes $sp^{3}d^{2}$ hybridization using $4d$ orbitals. Hence,it is an outer orbital complex.

(0) In the complex $[Pt(CN)_{4}]^{2-}$,the central metal ion is $Pt^{2+}$.
The atomic number of $Pt$ is $78$. The electronic configuration of $Pt$ is $[Xe] 4f^{14} 5d^{9} 6s^{1}$.
For $Pt^{2+}$,the configuration is $[Xe] 4f^{14} 5d^{8}$.
$CN^{-}$ is a strong field ligand,which forces the pairing of electrons in the $5d$ orbitals to facilitate $dsp^{2}$ hybridization,which is characteristic of square planar geometry.
After pairing,all $8$ electrons in the $5d$ orbitals are paired.
Therefore,the number of unpaired electrons is $0$.

(N/A) $(i)$ $[Fe(CN)_6]^{4-}$: Iron is in $+2$ oxidation state $(Fe^{2+}: 3d^6)$. $CN^-$ is a strong field ligand,causing pairing of $3d$ electrons. It undergoes $d^2sp^3$ hybridization. The complex is octahedral and diamagnetic.
$(ii)$ $[FeF_6]^{3-}$: Iron is in $+3$ oxidation state $(Fe^{3+}: 3d^5)$. $F^-$ is a weak field ligand,so no pairing occurs. It undergoes $sp^3d^2$ hybridization. The complex is octahedral and paramagnetic.
$(iii)$ $[Co(C_2O_4)_3]^{3-}$: Cobalt is in $+3$ oxidation state $(Co^{3+}: 3d^6)$. Oxalate $(C_2O_4^{2-})$ is a bidentate ligand. Although it is often considered a weak field ligand,in this specific complex,it forms a low-spin octahedral complex with $d^2sp^3$ hybridization. The complex is octahedral and diamagnetic.
$(iv)$ $[CoF_6]^{3-}$: Cobalt is in $+3$ oxidation state $(Co^{3+}: 3d^6)$. $F^-$ is a weak field ligand,so no pairing occurs. It undergoes $sp^3d^2$ hybridization. The complex is octahedral and paramagnetic.

(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbital. The $4s$ electrons shift to the $3d$ orbital,resulting in a $3d^{10}$ configuration. The hybridization is $sp^3$,which corresponds to a tetrahedral geometry.

(A) The complex $[Co(H_2O)_6]Cl_3$ contains the $[Co(H_2O)_6]^{3+}$ ion.
In this complex,the oxidation state of $Co$ is $+3$.
The electronic configuration of $Co^{3+}$ is $3d^6$.
Due to the presence of $H_2O$ as a weak field ligand,it forms a high-spin octahedral complex.
This complex is known to be pink in color.