Coordination Compounds

The complexes are $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$, $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2 \mathrm{O}\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+}$ It is based on the ligand strength. In spectrochemical series, ligands are weak or strong stronger ligands will split $d$-orbital energies in transition metal complexes. So, strong ligands cause larger splitting. and result in shorter wavelengths of absorbed light. For weak ligands, the splitting is lesser and wavelength is maximum. For the given ligands, $\mathrm{CN}^{-}, \mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{Cl}^{-}$, the older of increasing field (ligand) strength is $Cl^-<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\mathrm{CN}^{-}$(splitting is inversely. proportional to wavelength): $C N^{-}$is a strong-field ligand. So, it will cause a larger splitting and absorb light at a shorter wavelength (higher energy). So, the complex $\left[C_0(\mathrm{CN})_6\right]^{3^-}$ has shorter wavelength$\mathrm{NH}_3$ ligand strength is lower than $C{N}^{-}$ but higher than $\mathrm{Cl}^{-}$and $\mathrm{H}_2 \mathrm{O}$. The splitting caused by $\mathrm{NH}_3$ is lesser than $\mathrm{CN}^{-}$. So, the wavelength is higher than $\left[\mathrm{CO}(\mathrm{CN})_6\right]^{3-}$ complex.$ \mathrm{Co}\left(\mathrm{NH}_3\right)_6^{3+} \text { has wavelength higher than }\left[\mathrm{CO}(\mathrm{CN})_6\right]^{3-} \text { - } $$\mathrm{H}_2 \mathrm{O}$ is a weaker ligand than $\mathrm{NH}_3$ so, the complex $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2O\right]^{3+}$ has higher wavelength than $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$Cl is the weak field ligand than $^{}$ $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ So, the wavelength is maximum for the complex containing $C^{-}$.$ \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} \text { has highest wavelength. } $ The correct order of wavelength maxima of the absorption band is. $ \left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}<\left[\mathrm{C}_0\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{C}_0\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2 \mathrm{O}\right]^{3+} \leq\left[\mathrm{C}_0\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} $$ \text { Answer: Option (A) } $

1.

$ \begin{aligned} & {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\ & \mathrm{Fe}(26) \longrightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^2 \\\\ & \mathrm{Fe}^{2+} \longrightarrow 3 \mathrm{~d}^6, \mathrm{H}_2 \mathrm{O} \text { is a weak field ligand. } \end{aligned} $

So, the pairing does not take place.


2.

$ \begin{aligned} & {\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\ & \mathrm{Mn}(25) \longrightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^2 \\\\ & \mathrm{Mn}^{2+} \longrightarrow 3 \mathrm{~d}^5, \mathrm{H}_2 \mathrm{O} \text { is a weak field ligand. } \end{aligned} $

So, the pairing does not take place.



3.

$\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

$ \mathrm{CO}^{3+} \longrightarrow[\mathrm{Ar}] 3 \mathrm{~d}^6 $

$\mathrm{NH}_3$ is a strong field ligand. So, the pairing takes place.


4.

$ \begin{aligned} & \begin{aligned} {\left[\mathrm{FeCl}_4\right]^{-} } \end{aligned} \\\\ & x+4(-1)=-1 \\\\ & x=+3 \\\\ & \mathrm{Fe}(26) \longrightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^2 \\\\ & \mathrm{Fe}^{+3} \longrightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5 \end{aligned} $

$\mathrm{FeCl}_4$ is tetrahedral complex.


$\begin{aligned} & \text { 5. }\left[\mathrm{COCl}_4\right]^{2-} \\\\ & \mathrm{CO} \longrightarrow 3 \mathrm{~d}^7 4 \mathrm{~s}^2 \\\\ & \mathrm{CO}^{2+} \longrightarrow 3 \mathrm{~d}^7 \end{aligned}$


$\therefore \mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 2, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 1$

$\left[\operatorname{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Br}_2\right]$

Hybridisation : dsp${ }^2$, geometry : square planar


(A) $\left[\mathrm{Pt}(\mathrm{en})(\mathrm{SCN})_2\right]$ : square planar, cis-trans not possible

(B) $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ : tetrahedral, cis-trans not possible

(C) $\left[\operatorname{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_4\right]$ : octahedral, cis-trans possible

$ \text { (D) }\left[\mathrm{Cr}(\mathrm{en})_2\left(\mathrm{H}_2 \mathrm{O}\right) \mathrm{SO}_4\right]^{+} \text {: Octahedral } $

(I) ${[Cr{(CN)_6}]^{4 - }}$

$C{r^{ + 2}} = [Ar]\,3{d^4}\,4{s^0}$

It is d²sp³ hybridised as CN^$-$ is a strong field ligand.

(II) ${[RuC{l_6}]^{2 - }}$

$R{u^{ + 4}} = [Kr]\,4{d^4}\,5{s^0}$

t_2g set contains 4 electron.

(III) ${[Cr{({H_2}O)_6}]^{2 + }}$

$C{r^{ + 2}} = [Ar]\,3{d^4}\,4{s^0}$

It has 4 unpaired e^$-$ as H₂O is weak field ligand.

So, its $\mu$ = 4.9 B.M.

(IV) ${[Fe{({H_2}O)_6}]^{2 + }}$

$F{e^{ + 2}} = [Ar]\,3{d^6}\,4{s^0}$

$ = t_{2g}^4\,e_g^2$

It has 4 unpaired e^$-$, its $\mu$ = 4.9 B.M.

${[Cr{(N{H_3})_6}]^{3 + }}$

Atomic number of Cr = 24

Electron configuration of Cr = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

Electronic configuration of Cr³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s⁰

NH₃ is a weak field ligand. The splitting of d-orbital occur.


Number of unpaired electrons, n = 3

Magnetic moment (spin only), ${\mu _s} = \sqrt {n(n + 2)} BM$

$ = \sqrt {3(3 + 2)} = \sqrt {3 \times 5} = \sqrt {15} = 3.87BM$

${[Cu{F_6}]^{3 - }}$ Atomic number of Cu = 29

Electronic configuration of Cu = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Electronic configuration of Cu²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s⁰

F^$-$ is a weak field ligand, so splitting of d-orbitals occur as follows:


Number of unpaired electrons, n = 2

Spin only magnetic moment, ${\mu _s} = \sqrt {n(n + 2)} BM$

$ = \sqrt {2(2 + 2)} = \sqrt {2 \times 4} = 2\sqrt 2 = 2.84BM$

(a) ${[FeC{l_4}]^ - }$

Atomic number of Fe = 26

Oxidation number of Fe = x + 4 ($-$1) = $-$1

$\Rightarrow$ x = 3

Electronic configuration of Fe = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

Electronic configuration of Fe³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s⁰

Cl^- is a weak field ligand, so no pairing of electrons occurs.


Hybridisation is sp³

Geometry is tetrahedral.

${[Fe{(CO)_4}]^{2 - }}$ Oxidation number of Fe = x + 4(0) = - 2 $\Rightarrow$ x = - 2

Electronic configuration of Fe^2- = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²



CO is a strong field ligand and cause electrons of 3d orbitals to pair up and shift electrons of 4s to 3d orbital.



One 4s and three 4p orbitals undergoes hybridisation to form four sp³ hybrid orbitals which arrange into a tetrahedral geometry.



Each of the four sp³ hybrid orbitals accept a pair of electrons from carbonyl ligand.



Geometry is tetrahedral.

(b) ${[Co{(CO)_4}]^ - }$ Atomic number of Co = 27

Oxidation number of Co = x + 4(0) = $-$1 $\Rightarrow$ x = $-$1

Electronic configuration of Co = 1s²2s²2p⁶3s²3p⁶3d⁷4s²

Electronic configuration of Co^- = 1s²2s²2p⁶3s²3p⁶3d⁸4s²

CO is a strong field ligand, so pairing of electrons occurs.


Hybridisation is sp³

Geometry is tetrahedral.

${[CoC{l_4}]^{2 - }}$

Oxidation number of Co = x + 4($-$1) = $-$2

$\Rightarrow$ x = + 2

Electronic configuration of Co²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁷4s⁰

Cl^- is a weak field ligand, so pairing of electrons do not occur.


Hybridisation is sp³

Geometry is tetrahedral

(c) $[Ni{(CO)_4}]$ Oxidation number of Ni = x + 4(0) = 0 $\Rightarrow$ x = 0

Atomic number of Ni = 28

Electronic configuration of Ni = 1s²2s²2p⁶3s²3p⁶3d⁸4s²

CO is a strong field ligand, so pairing of electrons occur.


Hybridisation is sp³

Geometry is tetrahedral.

${[Ni{(CN)_4}]^{2 - }}$

Oxidation number of Ni = x + 4($-$1) = $-$2 $\Rightarrow$ x = + 2

Electronic configuration of Ni²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁸4s⁰

CN is a strong field ligand, so pairing of electrons occur.


Hybridisation is dsp².

Geometry is square planar.

(d) ${[Cu{(py)_4}]^ + }$

Oxidation number of Cu = x + 4(0) = + 1 $\Rightarrow$ x = + 1

Atomic number of Cu = 29

Electronic configuration of Cu = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

Electronic configuration of Cu⁺ = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s⁰


Hybridisation is sp³

Geometry is tetrahedral.

${[Cu{(CN)_4}]^{3 - }}$

Oxidation number of Cu = x + 4($-$1) = $-$ 3 $\Rightarrow$ x = + 1


Hybridisation is sp³

Geometry is tetrahedral.

Paramagnetism is a form of magnetism whereby some materials are attracted by an externally applied magnetic field, and form internal, induced magnetic fields in the direction of the applied magnetic field. So, magnetic balance shows downward deflection. While diamagnetic substance shows repulsion in magnetic field and magnetic balance shows upward deflection.

(a) X = H₂O(l)
(Water has no unpaired electrons and is thus diamagnetic). Hence, statement (a) is correct.

(b) X = K₄[Fe(CN)₆](s)
(CN is a strong field ligand which forces the d-orbital electrons to pair up ($t_{2g}^6e_g^0$) and making it diamagnetic. Hence, statement (b) is correct.

(c) X = O₂ (g)
Here, O₂(g) is paramagnetic due to two-unpaired electrons present in $\pi $* (antibonding orbitals). Hence, statement (c) is correct.

(d) X = C₆H₆(l)
(Here, C₆H₆ is diamagnetic due to presence of 0 unpaired electrons. Hence, statement (d) is incorrect.

(a) In [FeCl₄]^-, oxidation number of Fe atom = +3

Electronic configuration of Fe in ground state = 3d⁶4s²

Electronic configuration of

Fe³⁺ = 3d⁵4s⁰4p⁰



Thus, [FeCl₄]^- has tetrahedral geometry.

(b) [Co(en)(NH₃)₂Cl₂]⁺ have three geometrical isomers. Thus, statement (b) is incorrect.



(c) Fe³⁺ in [FeCl₄]^- is sp³-hybridised with 5 unpaired electrons. (higher spin-only magnetic moment

= $\sqrt {n(n + 2)} $ = 5.92 BM). While Co³⁺ in [Co(en)(NH₃)₂Cl₂]⁺ is d²sp³-hybridised with zero unpaired electrons (low spin-only magnetic moment = $\sqrt {n(n + 2)} $ = 0 BM).

Thus, the statement (c) is correct.

(d) Co³⁺[Co(en)(NH₃)₂Cl₂]⁺



Co³⁺ in [Co(en)(NH₃)₂Cl₂]⁺ is d²sp³-hybridised and has octahedral geometry with 0 unpaired electron. Thus, statement (d) is incorrect.

1. $\left[\mathrm{FeF}_6\right]^{4-}$

(A) Let oxidation state of iron (Fe) in $\left[\mathrm{FeF}_6\right]^{4-}$

$ \begin{aligned} x+6 \times(-1) =-4 \\\\ \Rightarrow x =-4+6=+2 \end{aligned} $

$ \therefore $ Iron exist as $\mathrm{Fe}^{2+}$


Electron is $3 d$ orbitals remains unpaired as fluorine is weak ligand. One $4 s$, three $4 p$ and two $4 d$ orbitals undergo hybridisation to form six $s p^3 d^2$ hybrid orbital each of which accumulate a pair of electron from fluorine.


$ \text { Iron }(\mathrm{Fe}) \text { in }\left[\mathrm{FeF}_6\right]^{4-} \text { has } s p^3 d^2 \text { hybridisation. } $

2. $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{Cl}_3\right]$ : The arrangement of electrons in $\mathrm{Ti}^{3+}\left(3 d^1\right)$ will be


Electron in $3 d$ orbitals remains unpaired as water $\left(\mathrm{H}_2 \mathrm{O}\right)$ and chlorine $(\mathrm{Cl})$ is a weak ligands. Two $3 d$, one $4 s$ and three $4 p$ orbitals undergo hybridisation to form six $d^2 s p^3$ hybrid orbitals. These orbitals accept lone pair of electrons from each of the chlorine and water molecules.

3. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ : The arrangement of electrons in $\mathrm{Cr}^{3+}\left(3 d^3\right)$ will be


Electrons in $3 d$ orbitals remains paired as ammonia $\left(\mathrm{NH}_3\right)$ is a strong ligand. Two $3 d$, one $4 s$ and three $4 p$ orbitals undergo hybridisation to form six $d^2 s p^3$ hybrid orbitals. These orbitals accept lone pair of electrons from each of the six ammonia molecules.

4. $ \left[\mathrm{FeCl}_4\right]^{2-}: \text { The arrangement of electrons in } \mathrm{Fe}^{2+}\left(3 d^6\right) \text { will be } $


Electrons in $3 d$ orbitals remains unpaired as chlorine is a weak ligand. One $4 s$ and three $4 p$ orbitals undergo hybridization to form four $s p^3$ hybrid orbitals. These $s p^3$ hybrid orbitals accept lone pair of electrons from each of the chlorine ligand.

5. $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ : The arrangement of electrons in $\mathrm{Ni}^0\left(3 d^8 4 s^2\right)$ will be


Electrons in $3 d$ orbitals remains paired and paired electron from $4 \mathrm{~s}$ orbital move to $3 d$ as carbonyl is a strong ligand. One $4 s$, and three $4 p$ orbitals undergo hybridisation to form four $s p^3$ hybrid orbitals. These $s p^3$ hybrid orbitals accept lone pairs from each of the four carbonyl ligand.

6. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}:$ The arrangement of electrons in $\mathrm{Ni}^{2+}\left(3 d^8\right)$ will be


Electrons in $3 d$ orbitals pair up as cyanide is a strong ligand. One $3 d$, one $4 s$ and two $4 p$ orbitals undergo hybridisation to form four $d s p^2$ hybrid orbitals. The $d s p^2$ hybrid orbitals accept lone pair of electrons from each of the cyanide ligand.

Option (A) : Correct. The complex [Co(en)(NH₃)₃(H₂O)]³⁺ has two following isomers:

Option (B) : Correct. When bidentate 'en' is replaced by two cyanide ligands complex [Co(CN)₂(NH₃)₃(H₂O)]⁺ will have three following isomers:

Option (C) : Incorrect. Incomplex [Co(en)(NH₃)₃(H₂O)]³⁺, Co is in + 3 oxidation, that is, [Ar]d⁶. In the presence given ligands the complex will have low spin, hence, diamagnetic.

Option (D) : Correct. In complex [Co(en)(NH₃)₄]³⁺ there is larger gap between t_2g and e_g than in complex [Co(en)(NH₃)₃(H₂O)]³⁺. Hence [Co(en)(NH₃)₃(H₂O)]³⁺ will absorb light at longer wavelength as compared to [Co(en)(NH₃)₄]³⁺.

(A) [(Fe(CO₅)] & [Ni(CO)₄ complexes have 18-electrons in their valence shell.

(B) Due to strong ligand field, carbonyl complexes are predominantly low spin complexes.

(C) As electron density increases on metals (with lowering oxidation state on metals), the extent of synergic bonding increases. Hence M-C bond strength increases.

(D) While positive charge on metals increases and the extent of synergic bond decreases and hence C-O bond becomes stronger.

The pink coloured aqueous solution of $\mathrm{MCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}$ has central metal atom bonded six water molecule by co-ordinate bonds and there are two primary valences of chloride ions. The formula and structure of the complex $X$ can be represented as :

Formula : $\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2$

$ \left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2 \rightarrow\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{2+}+2 \mathrm{Cl}^{-} $



Here, the six non ionisable water molecules are the secondary valence and two ionisable chloride ions are primary valence. Since, the hexa coordinated aqua complex of cobalt is pink in colour; hence, central metal is cobalt.

(i) Reaction of the complex X with ammonium chloride in presence of air gives hexamminecobalt (III) chloride which is a 1: 3 electrolyte.



$ \begin{aligned} & \text { Option (A) : } \\\\ & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3+3 \mathrm{AgNO}_3 \rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left(\mathrm{NO}_3\right)_3+3 \mathrm{AgCl}} \end{aligned} $

Option (B) : $\mathrm{NH}_3$ is a strong field ligand, therefore, pairing of electrons will take place.



Option (C) :



Option (D) : When X and Z are in equilibrium at 0°C

$ \underset{\text{Pink}}{\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]}\rightarrow \underset{\text{Blue}}{\left[\mathrm{CoCl}_4\right]^{2-}} $

Since, formation of chlorine complex takes place in presence of excess chlorine at room temperature $\left(25^{\circ} \mathrm{C}\right)$. At lower temperature $\left(0^{\circ} \mathrm{C}\right)$, the equilibrium is more shifted towards left favouring the formation of pink coloured hexaaquacobalt (II) chloride complex.

Option (D) is correct.

The hybridisation of central atoms and the geometries of ammonia complex are tabulated as follows:

$ \begin{array}{|l|c|c|c|l|} \hline \text { Metal Ion } & \text { Hybridisation } & \text { Coordination Number } & \text { Complex with } \mathbf{N H}_3 & \text { Geometry } \\ \hline \mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 d^8 4 s^2 & d^2 s p^3 & 6 & {\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}} & \text { Octahedral } \\ \hline \mathrm{Pt}^{2+}:[\mathrm{Xe}] 4 f^4 5 d^9 6 s^1 & d s p^2 & 4 & {\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\right]^{2+}} & \text { Square planar } \\ \hline \mathrm{Zn}^{2+}:[\mathrm{Ar}] 3 d^{10} 4 s^2 & s p^3 & 4 & {\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_4\right]^{2+}} & \text { Tetrahedral } \\ \hline \end{array} $

For the given complexes:

In $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ : $\mathrm{Ni}$ (at. no. 28): $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8 4 s^2$

$\mathrm{CO}$ being a strong field ligand causes the electrons to pair; hence, the complex diamagnetic.


$ \text {In }\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{2+} \text { (at. No. 26): } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8 $

There are two unpaired electrons, so the complex is paramagnetic.


$ \begin{aligned} & \mathrm{In}\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}: \mathrm{Co}^{3+}\text {(at. No. 24): } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6 \end{aligned} $


The ligand $\mathrm{NH}_3$ results in formation of low spin complex, which is diamagnetic.

$ \text {In } \mathrm{Na}_3\left[\mathrm{CoF}_6\right]: \mathrm{Co}^{3+}\text {(at. No. 24): } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6 $


$\mathrm{F}$ is a weak field ligand and results in the formation of high spin complex with four unpaired electrons, which is paramagnetic.

In $\mathrm{Na}_2 \mathrm{O}_2$ : The molecular orbital configuration of $\mathrm{O}_2^{2-}$ is

$ \sigma 1 s^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2,\left\{\begin{array}{l} \pi 2 p_x^2 \\ \pi 2 p_y^2 \end{array},\left\{\begin{array}{l} \pi * 2 p_x^2 \\ \pi * 2 p_y^2 \end{array}\right.\right. $

There is no unpaired electron, so it is diamagnetic.

In $\mathrm{CsO}_2$ : The molecular orbital configuration of $\mathrm{O}_2^{-}$is

$ \sigma 1 s^2, \sigma * 1 \mathrm{~s}^2, \sigma 2 s^2, \sigma * 2 s^2, \sigma 2 p_z^2,\left\{\begin{array}{l} \pi 2 p_x^2 \\ \pi 2 p_y^2 \end{array},\left\{\begin{array}{l} \pi * 2 p_x^2 \\ \pi * 2 p_y^1 \end{array}\right.\right. $

There is one unpaired electron, so it is paramagnetic. Hence, there are three paramagnetic compounds.

For [Cr(NH₃)₄Cl₂]Cl,

For [Ti(H₂O)₅Cl](NO₃)₂,

The complex is paramagnetic due to presence of unpaired electron.

Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Pt(en)(NH₃)Cl]NO₃,

The complex is square planar and all electrons are paired. So it is diamagnetic. Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Co(NH₃)₄(NO₃)₂]NO₃, NH₃ is a strong ligand and causes pairing

The electronic configurations of the central metal ions are as follows:

The number of unpaired electrons is P = 5, Q = 3 and R = 4. From the relation $\mu = \sqrt {n(n + 2)} $, where n is the number of unpaired electrons, we have the order of spin only magnetic moment as Q < R < P.

${[Co{(N{H_3})_4}C{l_2}]^ + }$ and ${[Pt{(N{H_3})_2}({H_2}O)Cl]^ + }$ are octahedral and square planar complexes, which show geometrical isomerism. $[Pt{(N{H_3})_3}(N{O_3})]Cl$ and $[Pt{(N{H_3})_3}Cl]Br$ show ionisation isomerism.

The configuration of Ni = 3d⁸ 4s² and that of Ni²⁺ = 3d⁸.

In paramagnetic state, the hybridisation is sp³ and geometry is tetrahedral.

In diamagnetic state, the hybridisation is dsp² and geometry is square planar.

There are two ligands in the coordination entity, that is, ammine and aqua, but according to the IUPAC rule, alphabetical order is followed, that is, ammine is named first. Now calculating the oxidation state of Co, the net charge on the coordination entity is +3 as Cl has $-$1 and there are three Cl^$-$ outside the square brackets. So the oxidation state of Co will be III as both ammine and aqua are neutral ligands. So, its correct name will be diamminetetraaquacobalt (III) chloride.

(i) Nickel in +2 oxidation state $\left(\mathrm{Ni}^{2+}\right)$ prefers to form tetrahedral complexes with chlorine, e.g., $\left[\mathrm{NiCl}_4\right]^{2-}$.

The oxidation state of nickel in $\left[\mathrm{NiCl}_4\right]^{2-}$

$ \begin{gathered} x+4 \times(-1)=-2 \\\\ x=-2+4=+2 \end{gathered} $

Electronic configuration of $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8 4 s^0$

Since, chlorine is a weak ligand, electrons in $\mathrm{Ni}^{2+}$ do not pair up. One $4 s$ and three $4 p$ orbitals undergo hybridisation to form four $s p^3$ hybrid orbitals. These vacant orbitals accept a pair of electron from each of the chloride ion.

The geometry of $s p^3$ hybridised nickel $\left[\mathrm{Ni}^{2+}\right]$ is tetrahedral

(ii) Nickel in +2 oxidation state $\left(\right.$ as $\left.\mathrm{Ni}^{2+}\right)$ forms square planner complex with cyanide ligands, e.g., $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$.

Oxidation state of $\mathrm{Ni}$ in $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

$ \begin{aligned} x+4 \times(-1) & =-2 \\ x & =-2+4=+2 \end{aligned} $

The oxidation state of nickel in $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is $+2$

Electronic configuration of $\mathrm{Ni}^{2+}=[\operatorname{Ar}] 3 d^8 4 s^0$

Since, cyanide is a strong ligand, electrons in $3 d$-orbitals are paired up.



One $3 d$, one $4 s$ and two $4 p$ orbitals undergoes hybridisation to form four $d s p^2$ hybrid orbitals. These vacant orbitals accept a pair of electrons from each of cyanide ligand.



The geometry of $d s p^2$ hybridised nickel complex $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is square planner.



(iii) Nickel in +2 oxidation state as $\mathrm{Ni}^{2+}$ forms octahedral complex with water as ligands, e.g., $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Oxidation state of nickel in $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is:

$ \begin{aligned} x+6 \times 0 & =+2 \\\\ x & =+2 \end{aligned} $

Electronic configuration of $\mathrm{Ni}^{2+}=$



Since, water is a weak ligand, electrons in $3 d$ orbitals of Nickel remains unpaired. One $4 s$, three $4 p$ and two $4 d$ orbitals undergo hybridisation to form $s p^3 d^2$ hybrid orbitals. These vacant orbitals accept lone pair of electron from each of the water molecule.


The geometry of $s p^3 d^2$ hybridised nickel complex $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is octahedral.

The geometries of different $\mathrm{Ni}^{2+}$ complexes are as follows:

$ \begin{aligned} {\left[\mathrm{NiCl}_4\right]^{2-} } & \rightarrow \text { tetrahedral } \\\\ {\left[\mathrm{NiCN}_4\right]^{2-} } & \rightarrow \text { square planner } \\\\ {\left[\mathrm{Ni}_{(}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} } & \rightarrow \text { octahedral } \end{aligned} $

In K₃Fe(CN)₆ :

CN^$-$ being a strong field ligand, it causes pairing of electrons, so only one unpaired electron. The complex is paramagnetic.

In Co[(NH₃)₆]Cl₃ :

NH₃ being a strong field ligand it causes pairing of spins. There is no unpaired electron, hence the complex is diamagnetic.

In Na₃[Co(oxalate)₃] :

Oxalate being a strong field ligand it causes pairing of electron spins. There is no unpaired electron, hence the complex is diamagnetic.

In [Ni(H₂O)₆]Cl₂ :

Since H₂O is a weak field ligand, no pairing takes place. There are two unpaired electrons, so the complex is paramagnetic.

In K₂[Pt(CN)₄] :

CN^$-$ being a strong field ligand, it causes pairing of electrons spins, so no unpaired electron. Hence, the complex is diamagnetic.

In [Zn(H₂O)₆(NO₃)₂] : Zn²⁺ has 3d¹⁰ configuration. Therefore, it is a diamagnetic complex.

EDTA (ethylenediaminetetracetic acid) is a hexadentate ligand. It is also called the amino poly-carboxylic acid.

It is structure consists of the ethane group which is attached to tertiary amine groups at the terminal positions forming an ethylene diamine. Tertiary amine group is situated by the acetic acid groups $\left(-\mathrm{CH}_2 \mathrm{COOH}\right)$.

The ionisation isomers give different ions in solution. In the complex given in option (B), Cl$^-$ is replaced by NO$_2^ - $ in ionisation sphere.

$\mathrm{[Cr(H_2O)_4Cl(NO_2)]Cl}$ and $[Cr{({H_2}O)_4}C{l_2}](N{O_2})$ have different ions inside and outside the coordinate sphere and they are isomers. Therefore, they are ionisation isomers.

$\begin{aligned} & {\left[\mathrm{NiCl}_4\right]^{2-}} \\\\ & \text { Oxidation state of }\left[\mathrm{NiCl}_4\right]^{2-}=+2 \\\\ & \mathrm{Ni}(28)=[\mathrm{Ar}] 3 d^8 4 s^2 \\\\ & \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8\end{aligned}$

Among the given complexes of Ni , only $\mathrm{Cl}^{-}$is a weak ligand and does not cause pairing of electrons to take place.

Thus, the spin only magnetic moment, with two unpaired electrons $(n=2)$ configuration is

$ \sqrt{n(n+2)}=\sqrt{2(2+2)}=2.82 \mathrm{BM}=2.82 \mathrm{BM} $

In [Cr(CO)$_6$]: Cr(24) = [Ar] 3$d^5$ 4$s^1$

Since CO is a strong field ligand, so pairing of electrons will take place and the configuration will become $3d^6$. There will be no unpaired electrons, so, the spin only magnetic moment is zero.

Among the given compounds, the compounds of the type [MA$_2$X$_2$] and [M(AA)$_2$X$_2$] will exhibit geometrical isomerism.

The electronic configuration of Cu$^{2+}$ is [Ar]3d$^9$. It contains one unpaired electron. Hence, CuF$_2$ is coloured compound.

In all other compounds, Cu$^+$ ion is present. It has electronic configuration of Cu$^{2+}$ is [Ar]3d$^{10}$. All electrons are paired.

Hence, they are colourless.

In [Ni(CO)$_4$], Ni is in 0 oxidation state, so there are 8 electrons in 3d subshell and 2 electrons in 4s. CO is the strong ligand, so it cause pairing of electrons in 3d subshell leaving one d subshell vacant. Now, the electron from s shell shifts to d creating vacant space in s subshell and thus the hybridisation is sp$^3$.

Ni(28) = [Ar] 4s$^2$3d$^8$

CO is a strong ligand, causes coupling.

In [Ni(CN)$_4$]$^{2-}$, Ni is in +2 oxidation state so it has 8 electrons in d subshell and CN is also a strong ligand which pairs with d subshell atom and leaves one d orbital empty and thus its hybridisation is dsp$^2$.

In [Ni(CN)$_4$]$^{2-}$, the oxidation state of Ni is +2

Ni$^{2+}$ = [Ar] 3d$^8$4s$^0$

CN$^-$ is strong ligand, causes coupling.

The IUPAC name for [Ni(NH$_3$)$_4$] [NiCl$_4$] is : Tetraamminenickel (II) - tetrachloronickelate (II)

The name of the complex cation is written first followed by the name of the complex anion. Complex cation contains four ammine ligands and complex anion contains four chloride ligands.

The oxidation state of Fe in the complex, [Fe(H$_2$O)$_5$NO]SO$_4$ is +1 [NO has +1 charge]

Fe$^+$: [Ar] 3d$^+$4s$^1$

When the weak ligand H$_2$O and strong ligand NO attacks, the configuration changes.

NO$^+$ causes pairing of 4s electrons inside. Thus the configuration is 3d$^7$ and number of unpaired electrons = 3.

Fe$^+$ : [Ar] 3d$^7$4s$^0$

$\therefore$ Fe$^+$ has 3 unpaired electrons.

For a complex to be optically active, it should not possess either centre of symmetry or axis of symmetry or plane of symmetry. Here, the given complex is an octahedral complex that possesses plane of symmetry as well as axis of symmetry and thus is optically inactive.

The $cis$ and $trans$ both form of complex [M(NH$_3$)$_4$Cl$_2$] are optically inactive.

(A) $\mathrm{Mn^+=3}d^5 4s^1$; in presence of CO, effective configuration = $3d^64s^0$

Three lone pair of electrons in metal 3d orbitals for back bonding with vacant ($\pi$) pi* orbital of C in CO.

(B) $\mathrm{Fe^0}=3d^6 4s^2$; in presence of CO, effective configuration = $3d^8$

Four lone pair for back bonding with CO.

(C) $\mathrm{Cr^0}=3d^5 4s^1$; effective configuration = $3d^6$

Three lone pair for back bonding with CO.

(D) $\mathrm{V^-}=3d^4 4s^2$; effective configuration = $3d^6$

Three lone pair for back bonding with CO.

Since, Vanadium has a negative charge, it is present in lower oxidation state and can easily back donate the electrons from metal $d$ orbitals to ($\pi$) pi* orbital of carbonyl ligand. Hence, it forms a much stronger bond with carbon or carbonyl thereby weakening the carbon oxygen bond.