Hybridisation and Geometry Questions in English

(C) According to the spectrochemical series,$CN^-$ and $CO$ are strong field ligands,while $Cl^-$ is a weak field ligand.
$Ni^{2+}$ $(3d^8)$ with four strong field ligands forms a square planar,diamagnetic complex like $[Ni(CN)_4]^{2-}$.
$Ni(0)$ $(3d^8 4s^2)$ with four strong field ligands forms a tetrahedral,diamagnetic complex like $[Ni(CO)_4]$.
$Ni^{2+}$ $(3d^8)$ with four weak field ligands $(Cl^-)$ forms a tetrahedral,paramagnetic complex $[Ni(Cl)_4]^{2-}$.
In $[Ni(Cl)_4]^{2-}$,the $Ni^{2+}$ ion has an electronic configuration of $[Ar] 3d^8$. Due to the weak field ligand,no pairing occurs,resulting in two unpaired electrons in the $d$-orbitals,making it paramagnetic.

(D) $[Ni(CO)_4]$: The central metal $Ni$ is in $0$ oxidation state with configuration $3d^8 4s^2$. Due to the strong field ligand $CO$,the electrons pair up to give $3d^{10} 4s^0$. It undergoes $sp^3$ hybridisation,resulting in a tetrahedral geometry.
$[PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d$ series metal ion. For $5d$ elements,the crystal field splitting energy is large,which favors square planar geometry for $d^8$ complexes,even with weak field ligands like $Cl^-$. Thus,it is square planar.
$[Co(NH_3)_6]^{3+}$: $Co^{3+}$ has a $3d^6$ configuration. $NH_3$ acts as a strong field ligand,causing pairing of electrons to give $t_{2g}^6 e_g^0$. It undergoes $d^2sp^3$ hybridisation,resulting in an octahedral geometry.

(A) According to $VBT$:
$1$. $[ZnCl_4]^{2-}$ has $sp^3$ hybridisation and tetrahedral geometry. So,$(A-ii-q)$.
$2$. $[Fe(CN)_6]^{3-}$ has $d^2sp^3$ hybridisation and octahedral geometry. So,$(B-iii-p)$.
$3$. $[Ni(NH_3)_4]^{2+}$ has $dsp^2$ hybridisation and square planar geometry. So,$(C-i-r)$.
Therefore,the correct match is $(A-ii-q), (B-iii-p), (C-i-r)$.

(A) $1$. In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $x + 4(-1) = -2$,so $x = +2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$2$. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
$3$. The $3d^8$ configuration has $2$ unpaired electrons.
$4$. To accommodate $4$ ligands,$Ni^{2+}$ undergoes $sp^3$ hybridisation using one $4s$ and three $4p$ orbitals.
$5$. $sp^3$ hybridisation results in a tetrahedral geometry.
$6$. Therefore,the hybridisation is $sp^3$,the shape is tetrahedral,and the number of unpaired electrons is $2$.

(A) $1$. For $[NiCl_4]^{2-}$,$Ni$ is in the $+2$ oxidation state ($3d^8$ configuration). $Cl^-$ is a weak field ligand,so it does not cause pairing. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with $2$ unpaired electrons,making it paramagnetic. Thus,option $A$ is correct.
$2$. For $[Ni(CO)_4]$,$Ni$ 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 with $0$ unpaired electrons,making it diamagnetic.
$3$. For $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry with $0$ unpaired electrons,making it diamagnetic.

(D) In $[Ni(Cl)_4]^{2-}$,$Ni$ exists as $Ni^{2+}$ ion $(3d^8)$. Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons. Thus,it is paramagnetic and tetrahedral with $sp^3$ hybridization.
$(B)$ In $[Co(C_2O_4)_3]^{3-}$,$(C_2O_4)^{2-}$ is a bidentate ligand,resulting in an octahedral structure.
$(C)$ In $[Ni(CN)_4]^{2-}$,$Ni$ exists as $Ni^{2+}$ ion $(3d^8)$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbital. Thus,it is diamagnetic and square planar with $dsp^2$ hybridization.
$(D)$ In $[Ni(CO)_4]$,$Ni$ has an oxidation state of $0$ $(3d^8 4s^2)$. $CO$ is a strong field ligand that causes the pairing of electrons,promoting the $4s$ electrons into the $3d$ orbital. The complex is tetrahedral with $sp^3$ hybridization and is diamagnetic.
Therefore,$(D)$ is the correct answer.

(A) The hybridization of the given complexes is determined as follows:
$A. sp^3$: $[Ni(CO)_4]$ is a tetrahedral complex with $sp^3$ hybridization. Thus,$A-(ii)$.
$B. dsp^2$: $[Pt(NH_3)_2Cl_2]$ is a square planar complex with $dsp^2$ hybridization. Thus,$B-(iii)$.
$C. sp^3d^2$: $[CoF_6]^{3-}$ is an outer orbital octahedral complex with $sp^3d^2$ hybridization. Thus,$C-(iv)$.
$D. d^2sp^3$: $[Co(NH_3)_6]^{3+}$ is an inner orbital octahedral complex with $d^2sp^3$ hybridization. Thus,$D-(i)$.
Therefore,the correct matching is $A-(ii), B-(iii), C-(iv), D-(i)$.

(D) The correct answer is $[Ni(NH_3)_6]^{2+}$.
$1$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,so all $6$ electrons are paired in $t_{2g}$ orbitals $(t_{2g}^6 e_g^0)$. It is diamagnetic.
$2$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,so $t_{2g}^5 e_g^0$. It has $1$ unpaired electron in $t_{2g}$ orbitals.
$3$. $[Zn(NH_3)_6]^{2+}$: $Zn^{2+}$ is $3d^{10}$. All orbitals are fully filled. It is diamagnetic.
$4$. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand for $Ni^{2+}$. The crystal field splitting results in $t_{2g}^6 e_g^2$ configuration. Here,all $6$ electrons in $t_{2g}$ orbitals are paired,and there are $2$ unpaired electrons in $e_g$ orbitals,making it paramagnetic.

(A) During the extraction of silver,the finely powdered silver ore is treated with a dilute $KCN$ solution to form a cyano complex of silver,which is water-soluble.
$Ag_2S + 4NaCN \rightleftharpoons 2Na[Ag(CN)_2] + Na_2S$
Then,zinc is added to the solution. Being more electropositive,zinc displaces silver from the complex.
$2Na[Ag(CN)_2] + Zn \longrightarrow Na_2[Zn(CN)_4] + 2Ag$
In the complex $[Zn(CN)_4]^{2-}$,the zinc ion $(Zn^{2+})$ undergoes $sp^3$ hybridization,resulting in a tetrahedral coordination geometry.
Hence,the correct option is $(A)$.

(A) The oxidation state of $Co$ in $[Co(NH_3)_6]^{3+}$ is $+3$ because $NH_3$ is a neutral ligand.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Since $NH_3$ is a strong-field ligand,it causes pairing of the $3d$ electrons.
This results in two vacant $3d$ orbitals,which participate in $d^2sp^3$ hybridization.
Because the inner $3d$ orbitals are used,it is an inner-orbital complex.
All $3d$ electrons are paired,so the number of unpaired electrons is $0$.

(A-IV, B-II, C-I, D-V) To find the correct match,we determine the hybridisation of each compound/ion:
$A. sp^3d$: $PCl_5$ $(IV)$ has $5$ bond pairs and $0$ lone pairs,total $5$ hybrid orbitals,corresponding to $sp^3d$ hybridisation.
$B. sp^3d^2$: $SF_6$ $(II)$ has $6$ bond pairs and $0$ lone pairs,total $6$ hybrid orbitals,corresponding to $sp^3d^2$ hybridisation.
$C. dsp^2$: $[PtCl_4]^{2-}$ $(I)$ has $4$ bond pairs and $0$ lone pairs,total $4$ hybrid orbitals,corresponding to $dsp^2$ hybridisation.
$D. dsp^3$: $ClF_3$ $(V)$ has $3$ bond pairs and $2$ lone pairs,total $5$ hybrid orbitals,which can be $sp^3d$ or $dsp^3$ (depending on the model,but $ClF_3$ is commonly associated with $sp^3d$ geometry). However,based on standard matching options provided in such questions,$D$ matches $V$ $(ClF_3)$.
Thus,the correct matching is: $A-IV, B-II, C-I, D-V$.

(C) In $\left[ Ni(CN)_4 \right]^{2-}$,the oxidation state of $Ni$ is $+2$,which has a $3d^8$ configuration.
$CN^{-}$ is a strong field ligand,which causes the pairing of the two unpaired electrons in the $3d$ orbital.
This results in $dsp^2$ hybridization,leading to a square planar geometry.
Since all electrons are paired,the complex is diamagnetic.

(C) $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. Hybridization is $dsp^2$,resulting in a square planar,diamagnetic complex. Matches $III$.
$[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of electrons. Hybridization is $sp^3$,resulting in a tetrahedral,diamagnetic complex. Matches $II$.
$[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,no pairing occurs. Hybridization is $sp^3$,resulting in a tetrahedral,paramagnetic complex. Matches $I$.
Therefore,the correct match is $A-III, B-II, C-I$.

(C) In $[MnCl_6]^{3-}$,the oxidation state of $Mn$ is $+3$. The electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$. Since $Cl^-$ is a weak field ligand,no pairing occurs,resulting in $4$ unpaired electrons.
In $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $CN^-$ is a strong field ligand,pairing occurs,leaving $1$ unpaired electron.
In $[Co(C_2O_4)_3]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$. Since $C_2O_4^{2-}$ acts as a strong field ligand with $Co^{3+}$,all electrons are paired,resulting in $0$ unpaired electrons.

(A) The correct matches are as follows:
$(A)$ $[Ni(CO)_4]$ has $sp^3$ hybridisation (tetrahedral geometry).
$(B)$ $[Pt(NH_3)_2Cl_2]$ has $dsp^2$ hybridisation (square planar geometry).
$(C)$ $[CoF_6]^{3-}$ has $sp^3d^2$ hybridisation (outer orbital complex).
$(D)$ $[Co(NH_3)_6]^{3+}$ has $d^2sp^3$ hybridisation (inner orbital complex).
Therefore,the correct sequence is $(ii), (iii), (iv), (i)$.

(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 5.92 \ BM$,we have $\sqrt{n(n+2)} = 5.92$.
Solving for $n$,we get $n(n+2) \approx 35$,which implies $n = 5$.
For a high spin octahedral complex with $5$ unpaired electrons,the electrons occupy the $t_{2g}$ and $e_g$ orbitals according to Hund's rule.
The configuration is $t_{2g}^3 e_g^2$.

(C) The central metal ion is $Co^{2+}$,which has a $d^7$ electronic configuration.
$(i)$ Since there are $6$ ligands $(H_2O)$,the coordination number is $6$,making the geometry octahedral.
$(ii)$ $H_2O$ is a weak field ligand,so it forms a high-spin (outer orbital) complex using $sp^3d^2$ hybridization.
$(iii)$ The complex has one unpaired electron (as stated in the question),so it is paramagnetic,not diamagnetic.
Therefore,statements $(i)$ and $(ii)$ are true.

(A) In oxyhaemoglobin,$Fe^{2+}$ is in a low-spin state,which makes it diamagnetic. Therefore,the statement that $Fe^{2+}$ is paramagnetic in oxyhaemoglobin is incorrect.

(A) The central metal ion is $Ni^{2+}$,which has an electronic configuration of $[Ar] 3d^8$.
Since the complex $[NiX_4]^{2-}$ is paramagnetic and has a coordination number of $4$,it adopts a tetrahedral geometry.
In a tetrahedral geometry,the hybridisation is $sp^3$.
For $Ni^{2+}$ $(3d^8)$,the $3d$ orbitals have two unpaired electrons.
Therefore,the hybridisation is $sp^3$ and the number of unpaired electrons is $2$.

(C) $1$. The atomic number of $Ni$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.
$2$. In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$,so the configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$3$. $CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
$4$. Due to this pairing,one $3d$ orbital becomes vacant,allowing $dsp^2$ hybridization.
$5$. The $dsp^2$ hybridization results in a square planar geometry.
$6$. Since all electrons are paired,the complex is diamagnetic.

(B) $1$. For $[NiCl_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,so no pairing occurs. It has $2$ unpaired electrons.
$2$. 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 $3d$ orbitals,resulting in $0$ unpaired electrons.
$3$. For $[Cu(NH_{3})_{4}]^{2+}$: $Cu$ is in $+2$ oxidation state $(3d^9)$. It has $1$ unpaired electron.
Thus,the number of unpaired electrons are $2, 0, 1$ respectively. The correct option is $B$.

(A) When excess potassium iodide $(KI)$ is added to mercuric chloride $(HgCl_{2})$,the reaction is as follows:
$HgCl_{2} + 4KI \longrightarrow K_{2}[HgI_{4}] + 2KCl$
In the complex $[HgI_{4}]^{2-}$,the central metal ion is $Hg^{2+}$.
The electronic configuration of $Hg$ $(Z=80)$ is $[Xe] 4f^{14} 5d^{10} 6s^{2}$.
For $Hg^{2+}$,the configuration is $[Xe] 4f^{14} 5d^{10} 6s^{0}$.
The four $I^-$ ligands donate electron pairs to the $6s$ and $6p$ orbitals of $Hg^{2+}$,resulting in $sp^{3}$ hybridization.
This leads to a tetrahedral geometry for the complex $K_{2}[HgI_{4}]$.

(D) In the complex $\left[Cu(NH_3)_4\right]^{2+}$,the central metal ion is $Cu^{2+}$.
$Cu^{2+}$ has the electronic configuration $[Ar] 3d^9$.
During the formation of the complex,one electron from the $3d$ orbital is promoted to the $4p$ orbital to allow for $dsp^2$ hybridization.
This results in a square planar geometry for the complex.

(B) $[Co(NH_{3})_{6}]^{3+}$: $Co^{3+}$ is $3d^{6}$. $NH_{3}$ is a strong field ligand $(SFL)$,causing pairing. Hybridization is $d^{2}sp^{3}$,which is an inner orbital complex. (Correct)
$(B)$ $[MnCl_{6}]^{3-}$: $Mn^{3+}$ is $3d^{4}$. $Cl^{-}$ is a weak field ligand $(WFL)$. Hybridization is $sp^{3}d^{2}$,which is an outer orbital complex. (Correct)
$(C)$ $[CoF_{6}]^{3-}$: $Co^{3+}$ is $3d^{6}$. $F^{-}$ is a $WFL$. Hybridization is $sp^{3}d^{2}$,which is an outer orbital complex. (Incorrect as stated $d^{2}sp^{3}$)
$(D)$ $[FeF_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $F^{-}$ is a $WFL$. Hybridization is $sp^{3}d^{2}$,which is an outer orbital complex. (Correct)
$(E)$ $[Ni(CN)_{4}]^{2-}$: $Ni^{2+}$ is $3d^{8}$. $CN^{-}$ is a $SFL$. Hybridization is $dsp^{2}$,which is an inner orbital complex. (Incorrect as stated $sp^{3}$)
Therefore,the correct statements are $A$,$B$,and $D$.

(C) The complex formed is bis(dimethylglyoximato)nickel$(II)$.
$A$) It is a bright red precipitate,so statement $A$ is correct.
$B$) It is insoluble in water and precipitates in basic medium,so statement $B$ is incorrect.
$C$) $Ni^{2+}$ has a $3d^8$ configuration. In the presence of strong field ligands (dimethylglyoxime),it undergoes $dsp^2$ hybridization,resulting in a square planar geometry with $0$ unpaired electrons. Thus,statement $C$ is incorrect.
$D$) In a square planar geometry,the $N-Ni-N$ bond angle is approximately $90^{\circ}$,so statement $D$ is correct.
$E$) The complex contains $2$ five-membered chelate rings formed by the coordination of $Ni$ with the nitrogen atoms of the two dimethylglyoxime ligands. Statement $E$ is incorrect.
Therefore,the incorrect statements are $B, C,$ and $E$.

(B)

Species	Electrons
$Sc^{3+}$	$18$
$Cr^{2+}$	$22$
$Mn^{3+}$	$22$
$Co^{3+}$	$24$
$Fe^{3+}$	$23$

The isoelectronic species are $Cr^{2+}$ and $Mn^{3+}$,so $n = 2$.
The complex $CoCl_{3}(en)_{2}NH_{3}$ reacts with excess $AgNO_{3}$ to form $2$ moles of $AgCl$,indicating the formula is $[Co(en)_{2}NH_{3}Cl]Cl_{2}$.
In this complex,$Co$ is in the $+3$ oxidation state ($3d^{6}$ configuration).
For an octahedral complex with a strong field ligand like $en$,the $6$ electrons occupy the $t_{2g}$ orbitals as $t_{2g}^{6} e_{g}^{0}$.
Therefore,the number of electrons in the $t_{2g}$ orbital is $6$.

(C) $1$. For $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It forms $dsp^2$ hybridization,making it diamagnetic.
$2$. For $Ni(CO)_4$: $Ni$ is $3d^8 4s^2$ $(3d^{10})$. $CO$ is a strong field ligand. It forms $sp^3$ hybridization,making it diamagnetic.
$3$. For $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,so electrons do not pair. It forms $sp^3$ hybridization with two unpaired electrons,making it paramagnetic.
Therefore,$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic,while $[NiCl_4]^{2-}$ is paramagnetic.

(B) $[Ni(PPh_3)_2Cl_2]$ is a $d^8$ complex. Since it is paramagnetic,it must have a tetrahedral geometry.
$(A)$ Tetrahedral complexes of the type $[MA_2B_2]$ do not show geometrical isomerism. Thus,statement $A$ is incorrect.
$(B)$ The complex is blue in colour,not white. Thus,statement $B$ is incorrect.
$(C)$ For $Ni^{2+}$ $(d^8)$ in a tetrahedral field,the configuration is $e^4 t_2^4$,which has $2$ unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$. Thus,statement $C$ is correct.
$(D)$ The $CFSE$ for a tetrahedral complex is calculated as $CFSE = (-0.6 \times n_e + 0.4 \times n_{t_2}) \Delta_t$. For $d^8$,$CFSE = (-0.6 \times 4 + 0.4 \times 4) \Delta_t = -0.8 \Delta_t$. The statement mentions $-0.8 \Delta_0$,which is incorrect.
$(E)$ $Ni(CO)_4$ is tetrahedral,and $[Ni(PPh_3)_2Cl_2]$ is also tetrahedral. Thus,statement $E$ is correct.
The incorrect statements are $A, B,$ and $D$.

(A) For low spin octahedral complexes,electrons fill the $t_{2g}$ orbitals first.
$Co^{3+} (3d^6): t_{2g}^6 e_g^0$,unpaired electrons $= 0$.
$Fe^{3+} (3d^5): t_{2g}^5 e_g^0$,unpaired electrons $= 1$.
$Mn^{3+} (3d^4): t_{2g}^4 e_g^0$,unpaired electrons $= 2$.
$Cr^{3+} (3d^3): t_{2g}^3 e_g^0$,unpaired electrons $= 3$.
Thus,the decreasing order is $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$.

(B) $Ni^{2+}$ ion has a $3d^8$ electronic configuration.
In $[NiCl_4]^{2-}$,$Cl^-$ is a weak field ligand. It does not cause the pairing of electrons in the $3d$ orbitals. Therefore,the complex undergoes $sp^3$ hybridization,utilizing the $4s$ and $4p$ orbitals,making it an outer orbital complex.
In $[Ni(CN)_4]^{2-}$,$CN^-$ is a strong field ligand. It causes the pairing of electrons in the $3d$ orbitals,leaving one $3d$ orbital vacant. Therefore,the complex undergoes $dsp^2$ hybridization,utilizing the $3d$,$4s$,and $4p$ orbitals,making it an inner orbital complex.

$(D)$ $A: [NiCl_4]^{2-}, Ni^{2+}(3d^8)$,weak field ligand,$sp^3$ hybridization,tetrahedral geometry,paramagnetic (two unpaired electrons). Correct.
$B: [Ni(NH_3)_6]^{2+}, Ni^{2+}(3d^8)$,weak field ligand,$sp^3d^2$ hybridization,octahedral geometry,paramagnetic (two unpaired electrons). Correct.
$C: [Ni(CO)_4], Ni^0(3d^8 4s^2)$,strong field ligand $(CO)$,$sp^3$ hybridization,tetrahedral geometry,diamagnetic (all electrons paired). Incorrect (given as paramagnetic).
$D: [Ni(CN)_4]^{2-}, Ni^{2+}(3d^8)$,strong field ligand $(CN^-)$,$dsp^2$ hybridization,square planar geometry,diamagnetic (all electrons paired). Correct.
Thus,the correct sequences are $A, B$,and $D$.

(C) $[Pt(Cl)_2(NH_3)_2]$ is a $d^8$ complex,which exhibits square planar geometry $(III)$.
$[Co(NH_3)_6]Cl_3$ contains the $[Co(NH_3)_6]^{3+}$ ion,which is an octahedral complex $(I)$.
$[NiCl_4]^{2-}$ is a $d^8$ tetrahedral complex $(IV)$.
$[Fe(CO)_5]$ is a $d^8$ complex with trigonal bipyramidal geometry $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.