Coordination Compounds

In the complex [Co(NH$_3$)$_6$]$^{3+}$, Co is in the +3 oxidation state. This means it has a configuration of $d^6$. NH$_3$ is a strong field ligand, which causes the d-electrons to pair up. Therefore, this complex is a low-spin (also referred to as spin-paired) complex.

The term "low-spin" means that the electrons prefer to pair up in the lower energy d-orbitals rather than occupy the higher energy orbitals. When all the electrons are paired, the complex is diamagnetic, which means it's not attracted to an external magnetic field. The hybridization of this complex is $d^2sp^3$, also known as inner orbital complex, because the inner d-orbitals are used in the hybridization.

The complexes [CoF$_6$]$^{3-}$ and [CoCl$_6$]$^{3-}$ are both octahedral, but they do have unpaired electrons because $F^-$ and $Cl^-$ are not strong enough field ligands to cause pairing of all the electrons in the d-orbitals. Therefore, they are not low-spin or diamagnetic.

Lastly, the complex [NiCl$_4$]$^{2-}$ is not octahedral. Because $Cl^-$ is a weak field ligand, the electrons in the d-orbitals do not pair up, and the $Ni^{2+}$ ion, with a $d^8 $ configuration, undergoes sp^3 hybridization, resulting in a tetrahedral complex.

Therefore, the only octahedral, diamagnetic, and low-spin complex in the options provided is [Co(NH$_3$)$_6$]$^{3+}$ (Option A).

For option (A)

$ \begin{aligned} & \mathrm{Cr}^{+3}: 3 \mathrm{~d}^3 \\\\ & \mathrm{CN}^{-} \rightarrow \mathrm{SFL} \end{aligned} $

No. of unpaired electrons $=3$

For option (B)

$ \begin{aligned} & \mathrm{Fe}^{+2}: 3 \mathrm{~d}^6 \\\\ & \mathrm{H}_2 \mathrm{O}: \mathrm{WFL} \end{aligned} $

No. of unpaired electrons $=4$

For option (C)

$ \begin{aligned} & \mathrm{Co}^{+3}: 3 \mathrm{~d}^6 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned} $

No. of unpaired electrons $=0$

For option (D)

$ \begin{aligned} & \mathrm{Ni}^{+2}: 3 \mathrm{~d}^8 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned} $

No. of unpaired electrons $=2$

Option A : $\mathrm{Na}_{3}\left[\mathrm{CoCl}_{6}\right]$

This is an octahedral complex, but it is not diamagnetic. Co in this complex is in +3 oxidation state, and the d-electron configuration is $t_{2g}^6 e_{g}^0$, leading to two unpaired electrons.

Option B : $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$

This is also an octahedral complex, but it is not diamagnetic. The Co ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^5 e_{g}^2$, leading to 3 unpaired electrons.

Option C : $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$

This complex is both octahedral and diamagnetic. Co in this complex is in +3 oxidation state and the d-electron configuration is $t_{2g}^6 e_{g}^0$, which means no unpaired electrons. This complex is more stable due to a strong field ligand (CN^-) which leads to a greater splitting of d-orbitals and more pairing of electrons.

Option D : $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}$

This is an octahedral complex, and it is diamagnetic. The Ni ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^8 e_{g}^0$, leading to no unpaired electrons. However, NH₃ is a weaker field ligand compared to CN^-, so the complex is less stable than Option C.

Hence, the complex that is octahedral, diamagnetic, and the most stable among the given options is $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$.

The spin-only magnetic moment of a complex ion is calculated using the formula $\mu = \sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons in the complex ion. In the case of transition metal complexes, the number of unpaired electrons and hence the magnetic moment can be influenced by the field strength of the ligands (the Ligand Field Theory).

Here, CN$^-$ is a strong-field ligand (or high-spin ligand) that can cause pairing of electrons, while F$^-$ and Br$^-$ are considered weak-field ligands (or low-spin ligands) that do not promote electron pairing.

1. $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$: In this complex, Fe is in +3 oxidation state and has 5 d-electrons. Due to the strong field ligand, CN$^-$, all the 5 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{0(0+2)} = 0$.




2. $\left[\mathrm{CoF}_{6}\right]^{3-}$: Here, Co is in +3 oxidation state and has 6 d-electrons. As F$^-$ is a weak field ligand, no electron pairing occurs, and hence there are 4 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{4(4+2)} = \sqrt{24}$.




3. $\left[\mathrm{MnBr}_{4}\right]^{2-}$: Here, Mn is in +2 oxidation state and has 5 d-electrons. Since Br$^-$ is a weak field ligand, no electron pairing occurs, and hence there are 5 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35}$.




4. $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$: In this complex, Mn is in +3 oxidation state and has 4 d-electrons. Due to the strong field ligand, CN$^-$, all the 4 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{0(0+2)} = 0$.

So the correct order of spin-only magnetic moments is:

$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{CoF}_{6}\right]^{3-}<\left[\mathrm{MnBr}_{4}\right]^{2-}$

To determine the IUPAC name of the compound K₃[Co(C₂O₄)₃], let's follow the nomenclature rules for coordination compounds:

Identify the Cation and Anion: The compound consists of potassium ions (K⁺) and a complex anion [Co(C₂O₄)₃]³⁻.

Name the Cation First: The cation is named first. So, we start with "Potassium."

Determine the Ligand Name: The ligand C₂O₄²⁻ is called "oxalato."

Count the Number of Ligands: There are three oxalato ligands. Since "oxalato" does not contain any numerical prefixes (like di-, tri-), we use the prefixes "di-", "tri-", etc., for simple ligands. So, we use "trioxalato."

Name the Central Metal Atom: Since the complex ion is an anion, the metal name ends with the suffix "-ate." Therefore, "cobalt" becomes "cobaltate."

Specify the Oxidation State: The oxidation state of cobalt in this complex can be calculated:

Let the oxidation state of Co be x.

Each oxalato ligand has a charge of -2.

The overall charge of the complex ion is -3.

So, x + 3(-2) = -3 ⇒ x -6 = -3 ⇒ x = +3.

Therefore, we indicate the oxidation state as (III).

Combining all these, the IUPAC name is Potassium trioxalatocobaltate(III).

Answer: Option C

Potassium trioxalatocobaltate(III)

Unpaired electron $=1$

$ \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1 \times 3}=1.74 \text { B.M. } $

$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ No pairing because $\mathrm{H}_2 \mathrm{O}$ is WFL

Number of unpaired electrons $=5, \mu=5.92 \mathrm{BM}$

Assertion is true, Reason is true but not correct explanation.

Complex $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}$ will have two isomer one linked through N (Nitro) and one through O (Nitrite).

CN^– is strongest field ligand among given ligands. So maximum splitting in d-orbitals take place.

$A=\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ - double salt

B. $\mathrm{CuSO}_{4} \cdot 4 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$ $=\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O} $ - complex salt

C. $\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}$ - double salt

D. $\mathrm{Fe}(\mathrm{CN})_{2} \cdot 4 \mathrm{KCN}$ = $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ - complex salt

Option A, $\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$, is a double salt because it is formed by the combination of two different salts: iron(II) sulfate ($\mathrm{FeSO}_{4}$) and ammonium sulfate ($\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$).

Option B, $\mathrm{CuSO}_{4}\cdot 4 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$, is a complex salt because it contains a central copper ion coordinated to several ammonia molecules.

In this compound, the copper(II) ion (Cu²⁺) acts as the central metal ion, while the four ammonia molecules (NH₃) act as ligands, coordinating to the copper ion through their lone pairs of electrons. The sulfate ion (SO₄^2-) and water molecules (H₂O) are not involved in the coordination sphere and simply crystallize along with the complex.

Option C, $\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}$, is also a double salt, formed by the combination of two different salts: potassium sulfate ($\mathrm{K}_{2} \mathrm{SO}_{4}$) and aluminum sulfate ($\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$).

The 24 water molecules in the formula are not directly involved in the salt formation but instead function to stabilize the crystal lattice by forming hydrogen bonds with the sulfate ions and other water molecules.

Option D, $\mathrm{Fe}(\mathrm{CN})_{2}\cdot4 \mathrm{KCN}$, is not an example of a double salt. It is a complex salt, formed by the coordination of iron(II) ions to cyanide ligands and potassium ions.

$\mathrm{Fe}^{3+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \longrightarrow \mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$(Prussium Blue)

$\mathrm{Ni{(CO)_4}\buildrel {} \over \longrightarrow s{p^3}}$

$\mathrm{{\left[ {Cu{{(N{H_3})}_4}} \right]^{2 + }}\buildrel {} \over \longrightarrow ds{p^2}}$

$\mathrm{{\left[ {Fe{{(N{H_3})}_6}} \right]^{2 + }}\buildrel {} \over \longrightarrow {d^2}s{p^3}}$

$\mathrm{{\left[ {Fe{{({H_2}O)}_6}} \right]^{2 + }}\buildrel {} \over \longrightarrow s{p^3}{d^2}}$

All the Cl$-$Co$-$Cl bond angles are of 90$^\circ$.

Cis-platin is [Pt(NH$_3$)$_2$Cl$_2$]; cis platin and other complexes of pt are used to inhibit the growth of tumours.

Ligand field strength

$\mathrm{{S^{2 - }} < {C_2}O_4^{2 - } < N{H_3} < en < CO}$

${\left[ {Fe{F_6}} \right]^{3 - }}\buildrel {} \over \longrightarrow 5$ unpaired electrons

${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}\buildrel {} \over \longrightarrow 0$ unpaired electron

${\left[ {Co{F_6}} \right]^{3 - }}\buildrel {} \over \longrightarrow 4$ unpaired electrons

So, correct answer is : (3)

This is chiral complex form.

Co-ordination compounds absorb a particular wavelength following certain rules.

$ \begin{aligned} \text{Wavelength of light absorbed} & \propto \frac{1}{\text { Oxidation state of metal ion }} \\\\ & \propto \frac{1}{\text { Strength of ligand }} \end{aligned} $

Ligand field strength : $\mathrm{CN}^{-}>\mathrm{NH}_{3}>\mathrm{H}_{2} \mathrm{O}>\mathrm{Cl}^{-}$

C.	$\left[\mathrm{Co}^{||||}(\mathrm{CN})_{6}\right]^{3-}$	I.	$310 \mathrm{~nm}$
B.	$\left[\mathrm{Co}^{||||}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$	II.	$475 \mathrm{~nm}$
A.	$\left[\mathrm{Co}^{||||}\mathrm{Cl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$	III.	$535 \mathrm{~nm}$
D.	$\left[\mathrm{Co}^{||}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}$	IV.	$600 \mathrm{~nm}$

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is diamagnetic with $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation of $\mathrm{Co}^{+3}$.

This is because $\mathrm{NH}_{3}$ is a strong field ligand and forces electrons to pair up in a $\mathrm{d}^{6}$ configuration.

CFT does not explain the order of spectrochemical series because as per CFT, anionic ligands should exert greatest splitting effect. However, they lie on lower end of the spectrochemical series.

$\mathrm{[Co(NH_3)_5Cl]Cl_2}$

x + 5 (0) – 1 = + 2 $ \Rightarrow $ x = + 3

Secondary valency is the number of ligands in the complex which is equal to 6.

According to John teller any nonlinear molecular system in a degenerate electronic state will be unstable and will undergo some kind of distortion which will lower its symmetry and energy and split the degenerate state.

In case of octahedral $d^{9}$ configuration, the last electron may occupy either $d^{2}$ or $d_{x^{2} y^{2}}$ orbitals of $e_{g}$ set.

If it occupies $\mathrm{dz}^{2}$ orbital most of the electron density will be concentrated between the metal and the two ligands on the $\mathrm{z}$ axis. Thus there will be greater electrostatic repulsion associated with these ligands than with the other four on xy plane.

The Jahn Teller effect is mostly observed in octahedral environments. The considerable distortions are usually observed in high spin $\mathrm{d}^{4}$, low spin $\mathrm{d}^{7}$ and $\mathrm{d}^{9}$ configuration.

Ni(CO)₄ Hybridisation sp³

[Ni(CN)₄]^2– Hybridisation dsp²

[Co(CN)₆]^3– Hybridisation d²sp³

[Co(F)₆]^3– Hybridisation sp³d²

Ligands like : CO, are sigma donor and $\pi$-acceptor and they make stronger bond with lower oxidation state metal ion, in this case back bonding is more effective

$4 \mathrm{Fe}^{3+}+3\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-4} \longrightarrow\underset{\text{Prussian Blue}} {\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3}$

$\Rightarrow \mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]$ is diamagnetic

$\mathrm{Cu}(\mathrm{I}) \Rightarrow \mathrm{d}^{10}$ configuration $\Rightarrow$ No unpaired electrons.

$\Rightarrow \mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right], \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{4}\right]$ and $\mathrm{K}_{4}\left[\mathrm{FeCl}_{6}\right]$ are paramagnetic in nature

Stronger is ligand attached to metal ion, greater will be the splitting between $\mathrm{t}_{2} \mathrm{g}$ and $e_g$ (hence greater will be $\Delta \mathrm{U})$

$ \therefore$ Greater will be absorption of energy.

Hence correct order

$\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}>\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}>\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$

Only the counter ions are replaced easily by other reagents.

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}+\mathrm{BaCl}_{2} \rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{Cl}_{2}+\underset{\text{White ppt.}}{\mathrm{BaSO}_{4}{\downarrow}}$

White ppt. It does not have chloride ion as counter ion so it will not give white ppt with $\mathrm{AgNO}_{3}$.

Statement I is true but statement II is false. Only oxygen atom forms a Co-ordinate bond with Cu⁺² in CuSO₄.5H₂O

[Ni(CN)₄]^2– : d⁸ configuration, SFL, sq. planar splitting (dsp²), diamagnetic.

[Ni(CO)₄] : d¹⁰ config (after excitation), SFL, tetrahedral splitting (sp³), diamagnetic.

[NiCl₄]^2– : d⁸ config, WFL, tetrahedral splitting (sp³), paramagnetic(2 unpaired e^–).

Coordination Compound	Number of unpaired e– (n)	Magnetic moment (µ) (B.M)
A. [FeF6]3$-$ - d5	5	5.91
B. [Fe(CN)6]3$-$ - d5	1	1.73
C. [MnCl6]3$-$ - d4	4	4.89
D. [Mn(CN)6]3$-$ - d4	2	2.82

Hence, correct order of magnetic moment is $2 < 4 < 3 < 1$

The given complexes are:

$\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{-3},\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$

$\mathrm{CN}^{-}$is the strongest ligand among the given complexes CFSE value for the $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{-3}$ complex will be highest as it has $d^{6}$ configuration with a CFSE value of $-2.40 \,\Delta_{0}+2 \mathrm{P}$, where $P$ represents pairing energy and $\Delta_{0}$ represents splitting energy in octahedral field.

The value of $\Delta_{0}$ is high for cyanide complexes.

Brown ring test

$ \begin{aligned} &\mathrm{NO}_{3}^{-}+3 \mathrm{Fe}^{+2}+4 \mathrm{H}^{+} \rightarrow \mathrm{NO}+3 \mathrm{Fe}^{+3}+2 \mathrm{H}_{2} \mathrm{O} \\\\ &{\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \mathrm{NO} \rightarrow \underset{\text{Brown ring}}{\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}]^{2+} \mathrm{H}_{2} \mathrm{O}\right.}} \end{aligned} $

Crystal field splitting $\left(\Delta_{0}\right)$ for octahedral complexes depends on oxidation state of the metal as well as to which transition series the metal belongs. For the same oxidation state, the crystal field splitting $\left(\Delta_{0}\right)$ increases as we move from $3 d \rightarrow 4 d \rightarrow 5 d $

$\mathrm{Cr}^{3+}$ and $\mathrm{Fe}^{3+}$ belong to $3 d$ series, $\mathrm{Mo}^{3+}$ belongs to $4 d$ series and $\mathrm{Os}^{3+}$ belongs to $5 d$ series.

Therefore crystal field splitting $\left(\Delta_{0}\right)$ is highest for $\left[\mathrm{Os}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$.

	List - I		List -II
(A)	${[PtC{l_4}]^{2 - }}$	(III)	$ds{p^2}$
(B)	$Br{F_5}$	(IV)	$s{p^3}d^2$
(C)	$PC{l_5}$	(I)	$s{p^3}d$
(D)	${[Co{(N{H_3})_6}]^{3 + }}$	(II)	$d^2s{p^3}$

Co²⁺ = [Ar] 3d⁷4s^o

in [Co(H₂O)₆]²⁺, H₂O will behave as weak field ligand

Co²⁺ = t_2g⁵ , e_g²


CFSE = (–0.4 × 5 + 2 × 0.6) Δ₀ = –0.8 Δ₀

Magnetic moment (spin only) can be calculated as :

$\mu = \sqrt {n\left( {n + 2} \right)} $

where, $\mu $ = spin only magnetic moment and

n = number of unpaired electrons = 3

= $\sqrt {3\left( {3 + 2} \right)} = \sqrt {15} $ = 3.87 BM

FeCl₃ + K₄[Fe(CN)₆] $\to$ Fe₄[Fe(CN)₆]₃ (Prussian blue)

[Fe(CO)₄(C₂O₄)]⁺



One unpaired electron Spin only magnetic moment = $\sqrt 3 $ B.M. = 1.73 BM

Biuret ligand is

It forms complexes like



The denticity of organic ligand is 2.

(a) [CoCl₂(en)₂] show Cis-trans isomerism

(b) [Co(CN)₅(NC)]^3$-$ can't show G.I.

(3) [Co(NH₃)₃(NO₂)₃] show fac & mer isomerism

(d) [Co(NH₃)₄Cl₂]⁺ show cis & trans isomerism

(i) CFSE $\propto$ charge or oxidation no. of central metal ion.

(ii) CFSE $\propto$ strength of ligand

en > NH₃ > H₂O > F^$-$

$\therefore$ order of CFSE

$\mathop {{{[Co{{(en)}_3}]}^{ + 3}}}\limits^{III} > \mathop {{{[Co{{(N{H_3})}_6}]}^{ + 3}}}\limits^{III} > \mathop {{{[Co{F_6}]}^{ - 3}}}\limits^{III} > \mathop {{{[Co{{({H_2}O)}_6}]}^{ + 2}}}\limits^{III} $

Hybridization of all the given complexes are given in following table.

${[Mn{(CN)_6}]^{3 - }} \Rightarrow M{n^{3 + }}$

CN^$-$ is a strong field ligand.

$\therefore$ Pairing of electron takes place.

${[Fe{(CN)_6}]^{3 - }} \Rightarrow F{e^{3 + }}$

CN^$-$ is a strong ligand.

$\therefore$ Pairing of electron takes place.

${[Co{({C_2}{O_4})_3}]^{3 - }} \Rightarrow C{o^{3 + }}$

${C_2}O_4^{2 - }$ acts as strong field ligand with Co³⁺.

$\therefore$ Pairing of electrons takes place.

${[Fe{F_6}]^{3 - }} \Rightarrow F{e^{3 + }}$

F is a weak field ligand.

$\therefore$ Pairing of electron does not occurs.

${[MnC{l_6}]^{3 - }} \Rightarrow M{n^{3 + }}$

Cl^$-$ is a weak field ligand.

$\therefore$ Pairing of electron does not occurs.

${[Fe{F_6}]^{3 - }}$ have 5 unpaired electrons and ${[MnC{l_6}]^{3 - }}$ have 4 unpaired electrons.

So, both statement I and statement II are true.

[MnCl₆]^3$-$



Paramagnetic and having 4 unpaired electrons.

Complex [Co(en)₃]Cl₂ is most stable complex among the given complex compounds because more number of chelate rings are present in this complex as compare to others.

(1) [Co(en)(NH₃)₄]Cl₂ 1 chelate ring

(2) [Co(en)₃]Cl₂ 3 chelate ring

(3) [Co(en)₂(NH₃)₂]Cl₂ 2 chelate ring

(4) [Co(NH₃)₆]Cl₂ 0 chelate ring

1. ${[\mathop {Fe}\limits^{ + 3} {({H_2}O)_6}]^{3 + }}$

$F{e^{3 + }}:[Ar]3{d^5}$

Hybridisation : $s{p^3}{d^2}$

Magnetic nature : Paramagnetic (so this complex response to external magnetic field)

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

$N{i^{2 + }}:[Ar]3{d^8}$

Hybridisation : dsp²

Magnetic nature : diamagnetic

3. ${[Co{(CN)_6}]^{3 - }}$

$C{o^{3 + }}:[Ar]3{d^6}$

Hybridisation : ${d^2}s{p^3}$

Magnetic nature : diamagnetic

4. $[Ni{(CO)_4}]$

$Ni:[Ar]3{d^8}4{s^2}$

Hybridisation : $s{p^3}$

Magnetic nature : diamagnetic

In presence of SFL $\Delta$₀ > P means pairing occurs therefore

For Fe⁺² $\to$ 3d⁶



$\therefore$ No of unpaired e^-(s) = 0

$\therefore$ $\mu$ = $\sqrt {n(n + 2)} $BM = 0

[n = No of unpaired e^$-$(s)]

In NiCl₂, Ni⁺² is having configuration 3d⁸

$\therefore$ Number of unpaired electron = 2

After formation of oxidised product

[Ni(CN)₆]^$-$2$ \Rightarrow $ Ni⁺⁴ is obtained

Ni⁺⁴ $\Rightarrow$ 3d⁶ and CN^$-$ is strong field ligand

$\therefore$ Number of unpaired electrons = 0

$\therefore$ The charge is 2 $-$ 0 = 2

Species does not have a magnetic moment of 1.7 BM means species must not contain single unpaired electron.

(a) Molecular orbital configuration of O$_2^ + $ (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

Unpaired electron = 1

$\mu = \sqrt {n(n + 2)} = \sqrt {1(1 + 2)} = \sqrt 3 = 1.73$

where n = no. of unpaired e^$-$

$\therefore$ $\mu$ = 1.73 BM

(b) Cul $\Rightarrow$ Cu⁺ $-$[Ar]3d¹⁰, Unpaired electron = 0

I^$-$ $\to$ [Xe], unpaired electron = 0

Therefore, $\mu$ = 0

(c) [Cu(NH₃)₄]Cl₂

Cu²⁺ $\to$ [Ar]3d⁹

Unpaired electron = 1,

Therefore, $\mu$ = 1.73 BM

(d) Molecular orbital configuration of $O_2^ - $ (17 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $

Unpaired electron = 1

Therefore, $\mu$ = 1.73 BM

Therefore, Cul does not have magnetic moment of 1.73 BM.

According to VBT i.e. valence bond theory,

Electronic configuration of Ni = [Ar]3d⁸4s².

(a) NiCl₂ . 6H₂O

NiCl₂ . 6H₂O $\rightarrow$ NiCl₂ + 6H₂O

Oxidation number of Ni(x) = x + 2($-$1) = 0; where x, $-$1 and 2 are the oxidation number of Ni, oxidation number of Cl and number of Cl atoms respectively.

NiCl₂ $\Rightarrow$ x + ($-$2) = 0 $\Rightarrow$ x = 2

Electronic configuration of Ni²⁺ = [Ar]3d⁸4s⁰

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

For C.N. = 6


(b) K₂[Ni(CN)₄]

K₂[Ni(CN)₄] $\rightarrow$ 2K⁺ + [Ni(CN)₄]^2$-$

x + 4($-$1) $-$ ($-$2) = 0; where x, 4, $-$1 and $-$2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[Ni(CN)₄]^2$-$ $\Rightarrow$ x $-$ 4 + 2 = 0 $\Rightarrow$ x = +2

Electronic configuration of Ni²⁺ $\Rightarrow$ [Ar]3d⁸4s⁰

CN^$-$ is a strong field ligand. So, pairing of electrons occur. For C.N. = 4


(c) Ni(CO)₄

CO is neutral and strong field ligand. So, pairing of electrons occur.

Oxidation number of Ni is zero

Electronic configuration of Ni = [Ar]3d¹⁰4s⁰

For C.N. = 4


(d) Na₂[NiCl₄]

Na₂[NiCl₄] $\rightarrow$ 2Na + [NiCl₄]^2$-$

x + 4($-$1) $-$ ($-$2) = 0; where x, 4, $-$1 and $-$2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[NiCl₄]^2$-$ $\Rightarrow$ x $-$ 4 + 2 = 0 $\Rightarrow$ x = +2

Electronic configuration of Ni²⁺ = [Ar]3d⁸4s⁰

For C.N. = 4


Hence, only K₂[Ni(CN)₄] has dsp² hybridization.

Correct order of intensity of colours of the compounds is

[NiCl₄]^2$-$ > [Ni(H₂O)₆]²⁺ > [Ni(CN)₄]^2$-$

Ni is in +2 oxidation state in all complexes. The intensity of colour depends on the strength of the ligand attached with the central metal atom because more strong ligand more splitting energy, less is intensity of colour. Strength of ligand is in the order CN^$-$ > H₂O > Cl^$-$.

Splitting energy order [NiCl₄]^2$-$ < [Ni(H₂O)₆]²⁺ < [Ni(CN)₄]^2$-$

$\therefore$ Intensity of colour of compound

[NiCl₄]^2$-$ > [Ni(H₂O)₆]²⁺ > [Ni(CN)₄]^2$-$