Answer

(b) Greater the value of $\log K$, greater will be stability of complex compound formed. For reaction,

$$ \begin{aligned} & Cu^{2+}+4 CN^{-} \longrightarrow[Cu(CN)_4]^{2-} \\ & K=\frac{[(Cu(CN)_4)^{2-}]}{[Cu^{2+}][CN^{-}]^{4}} \text { and } \log K=27.3 \end{aligned} $$

For this reaction, $\log K$ has highest value among the given four reactions. Hence, $K$ will also be higher among these four. i.e., stability of the complexes will be highest among these four complexes.

Thinking Process

This problem is based on the concept of crystal field splitting and colour of coordination compounds. Follow the steps to answer this question

(i) Arrange the given complexes in increasing order of their crystal field splitting energy.

(ii) Now arrange them in decreasing order of the wavelength of light.

(iii) As energy and wavelength are related as $\Delta E=h v=\frac{h c}{\lambda}$

Answer

(c) As we know that, strong field ligand split the five degenerate energy levels with more energy separation than weak field ligand, i.e., as strength of ligand increases crystal field splitting energy increases.

Hence, $\quad \Delta E=\frac{h c}{\lambda}$

$\Rightarrow \quad \Delta E \propto \frac{1}{\lambda} \Rightarrow \lambda \propto \frac{1}{\Delta E}$

As energy separation increases, the wavelength decreases.

Thus, the correct order is

$$ [Co(H_2 O)_6]^{3+}>[Co(NH_3)_6]^{3+}>[Co(CN)_6]^{3-} $$

Here, strength of ligand increases, $\Delta E$ increases, CFSE increases and $\lambda$ absored decreases.

Hence, correct choice is (c).

Answer

(b) One mole of $AgNO_3$ precipitates one mole of chloride ion. In the above reaction, when 0.1 mole $CoCl_3(NH_3)_5$ is treated with excess of $AgNO_3, 0.2$ mole of $AgCl$ are obtained thus, there must be two free chloride ions in the solution of electrolyte.

So, molecular formula of complex will be $[Co(NH_3)_5 Cl] Cl_2$ and electrolytic solution must contain $[Co(NH_3)_5 Cl]^{2+}$ and two $Cl^{-}$as constituent ions. Thus, it is $1: 2$ electrolyte.

$$ [Co(NH_3)_5 Cl] Cl_2 \longrightarrow[Co(NH_3)_5 Cl]^{2 \oplus}(aq)+2 Cl^{-}(aq) $$

Hence, option (b) is the correct.

Answer

(d) 1 mole of $AgNO_3$ precipitates one free chloride ion $(Cl^{-})$.

Here, 3 moles of $AgCl$ are precipitated by excess of $AgNO_3$. Hence, there must be three free $Cl^{-}$ions.

So, the formula of the complex can be $[Cr(H_2 O)_6] Cl_3$ and correct choice is (d).

Thinking Process

This problem is based on IUPAC nomenclature of coordination compound. IUPAC nomenclature of any coordination compound can be done as follows

(i) Positively charged ions are named first.

(ii) Negatively charged ions are named in alphabetical order of ligands including their numbers followed by metal ending with -ium and oxidation state in the bracket.

Answer

(a) The complex compound is $[Pt(NH_3)_2 Cl_2]$.

The ligands present in the compound are

(i) $NH_3$ - neutral ligand represented as amine.

(ii) $Cl$ - anion ligand (ending with-o-) represented as chlorido di prefixed to represent two ligands.

The oxidation number of platinum in the compound is 2 . Hence, correct IUPAC name of $[Pt(NH_3)_2 Cl_2]$ is

Diammine dichloridoplatinum (II)

So, (a) option is correct.

Answer

(c) Chelation (formation of cycle by linkage between metal ion and ligand) stabilises the coordination compound. The ligand which chelates the metal ion are known as chelating ligand.

Here, only $[Fe(C_2 O_4)_3]^{3-}$ is a coordination compound which contains oxalate ion as a chelating ligand. Hence, it stabilises coordination compound by chelating $Fe^{3+}$ ion.

Thinking Process

This problem includes concept of isomerism in coordination compound. Complex of $MA_4 B_2$ type show geometrical isomerism.

Answer

(a) $[Cr(H_2 O)_4 Cl_2]^{+}$shows geometrical isomerism because it is a $MA_4 B_2$ type coordination compound which contains two set of equivalent ligands, four $H_2 O$ and $2 Cl$.

Hence, the possible geometrical isomers are

Hence, correct choice is (a).

Answer

(c) CFSE for octahedral and tetrahedral complexes are closely related to each other by formula $\Delta_{t}=\frac{4}{9} \Delta_0$

where, $\Delta_0=$ CFSE for octahedral complex, $\Delta_{t}=$ CFSE for tetrahedral complex According to question, $\Delta_0=18,000 cm^{-1}$

$$ \begin{aligned} \therefore \quad \Delta_{t} & =\frac{4}{9} \Delta_0=\frac{4}{9} \times 18,000 cm^{-1} \\ & =4 \times 2,000 cm^{-1}=8,000 cm^{-1} \end{aligned} $$

Hence, correct choice is (c).

Answer

(a) The ligand(s) which has two different bonding sites are known as ambident ligands e.g., $NCS, NO_2$ etc.

Here, NCS has two binding sites at N and S.

Hence, NCS (thiocyanate) can bind to the metal ion in two ways

$$ M \rightarrow NCS \text { or } M \rightarrow SNC $$

Thus, coordination compounds containing NCS as a ligand can show linkage isomerism i.e., $[Pd(C_6 H_5)_2(SCN)_2]$ and $[Pd(C_6 H_5)_2(NCS)_2]$ are linkage isomers.

Hence, correct choice is (a).

Answer

(d) Compounds having same molecular formula but different structural formula are known as isomers. $[Co(SO_4)_2(NH_3)_5] Br$ and $[Co(SO_4)(NH_3)_5] Cl$ have not same molecular formula. Hence, they are not isomers.

Answer

(a) A chelating ligand has two or more binding donor atoms to a single metal ion e.g.,

Here $(\rightarrow)$ denotes binding site.

thiosulphato $(S_2 O_3^{2-})$ is not a chelating ligand because geometrically it is not favourable for $S_2 O_3^{2-}$ to chelate a metal ion.

Answer

(b) Ligand must donate a pair of electron or loosely held electron pair to metal and form a $M-L$ bond.

e.g.

Among $\stackrel{+}{N} H_4$ does not have any pair of electron.

Hence $\stackrel{+}{N} H_4$ is not a ligand.

Answer

(c) Soluate isomerism to shown when two compounds having same molecular formula differ by whether or solvent molecule is directly bonded to metal ion or is present as free solvent molecules in the crystal lattice.

When water is present as solvent and show this type of isomerism then it is known as hydrate isomerism.

Coordination compound $[Cr(H_2 O)_6] Cl_3$ and $[Cr(H_2 O)_5 Cl_1 H_2 O \cdot Cl_2.$ are solvate isomers, because water is exchanged by chloride ion. This is why both of them show different colour on exposure to sunlight.

Answer

(c) Correct IUPAC name can be written as The ligands present in the given coordination compound are

(i) $(NH_3)$ represented as amine

(ii) $Cl^{3}$ represented as chlorido

(iii) $NO_2^{s}$ represented as nitrito- $N$

According to IUPAC rule, ligands are named in an alphabetical order before central atom. Prefex di-will be used to indicate the number of $NH_3$ ligands present.

Oxidation state of metal is indicated by Roman numeral in parenthesis.

So, IUPAC name will be

diamminechloronitrito-N-platinum (II)

Hence, option (c) is correct.

Thinking Process

This problem is based on magnetic property of coordination compound. Coordination compound containing at least one unpaired electron(s) are paramagnetic and coordination compounds all containing all paired electrons are diamagnetic in nature.

Answer

$(a, c)$

Molecular orbital electronic configuration of $Co^{3+}$ in $[Co(NH_3)_6]^{3+}$ is

$[18 Ar]$

Number of unpaired electron $=0$

Magnetic property = Diamagnetic

Molecular orbital electronic configuration of $Mn^{3+}$ in $[Mn(CN)_6]^{3-}$

Number of unpaired electrons $=2$

Magnetic property $=$ Paramagnetic

Molecular orbital electronic configuration of $Fe^{2+}$ in $[Fe(CN)_6]^{4-}$ is

Number of unpaired electron $=0$

Magnetic property $=$ Diamagnetic

Molecular orbital electronic configuration of $Fe^{3+}$ in $[Fe(CN)_6]^{3-}$

Number of unpaired electron $=1$

Magnetic property $=$ Paramagnetic

Thus, $[Co(NH_3)_6]^{3+}$ and $[Fe(CN)_6]^{4-}$ are diamagnetic.

Hence, correct choices are options (a) and (c).

Answer

(a, c)

Molecular orbital electronic configuration of $Mn^{3+}$ in $[MnCl_6]^{3-}$ is

Number of unpaired electrons $=4$

Magnetic property $=$ Paramagnetic

Molecular orbital electronic configuration of $Co^{3+}$ in $[CoF_6]^{3-}$ is

Number of unpaired electrons $=4$

Magnetic property $=$ Paramagnetic

Molecular orbital electronic configuration of $Fe^{3+}$ in $[FeF_6]^{3-}$ is

Number of unpaired electrons $=5$

Magnetic property $=$ Paramagnetic

Molecular orbital electronic configuration of $Ni^{2+}$ in $[Ni^{2}(NH_3)_6]^{2+}$ is

Number of unpaired electrons $=2$

Magnetic property $=$ Paramagnetic

Thus, $[MnCl_6]^{3-}$ and $[CoF_6]^{3-}$ are paramagnetic having four electrons each.

Hence, correct choices are (a) and (c).

Answer

$(a, c)$

According to VBT, the molecular orbital electronic configuration of $Fe^{3-}$ in $[Fe(CN)_6]^{3-}$ is

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

Number of unpaired electron $=1$

Magnetic property $=$ Paramagnetic

Hence, correct choices are options (a) and (c).

Answer

$(b, c)$

Aqueous pink solution of cobalt (II) chloride is due to electronic transition of electron from $t_{2 g}$ to $e_{g}$ energy level of $[Co(H_2 O)_6]^{2+}$ complex. When excess of $HCl$ is added to this solution

(i) $[Co(H_2 O)_6]^{2+}$ is transformed into $[CoCl_4]^{2-}$.

(ii) Tetrahedral complexes have smaller crystal field splitting than octahedral complexes because $\Delta_{t}=\frac{4}{9} \Delta_0$

Hence, options (b) and (c) are correct choices.

Answer

(a, c)

Homoleptic complex The complex containing only one species or group as ligand is known as homoleptic ligand.

e.g.,

$[Co(NH_3)_6]^{3+},[Ni(CN)_4]^{2-}$

Here, $[Co(NH_3)_6]^{3+}$ contain only $NH_3$ as a ligand and $[Ni(CN)_4]^{2-}$ contain $CN$ as a ligand.

While other two complexes $[Co(NH_3)_4 Cl_2]^{+}$and $[Ni(NH_3)_4 Cl_2]$ contain $NH_3$ and $Cl$ as ligands.

Hence, options (a) and (c) are correct choices.

Answer

( $b, d)$

Heteroleptic complexes Coordination complexes which contain more than one type of ligands are known as heteroleptic complexes.

e.g., $[Fe(NH_3)_4 Cl_2]^{+}$contain $NH_3$ and $Cl$ as a ligand is as heteroleptic complex. Similarly, $[Co(NH_3)_4 Cl_2]$ contain $NH_3$ and $Cl$ as ligand is also a heteroleptic complex.

Hence, optons (b) and (d) are correct choices.

Answer

(a, c)

$[Co(en)_3]^{3+}$ and cis $-[Co(en)_2 Cl_2]^{+}$are optically active compounds because their mirror images are non-superimposable isomer.

Hence, (a) and (c) are correct choices.

Answer

$(a, b, c)$

Molecular formula of ethane-1, 2-diamine is

(a) Ethane-1, 2-diamine is a neutral ligand due to absence of any charge.

(b) It is a didentate ligand due to presence of two donor sites one at each nitrogen atom of amino group.

(c) It is a chelating, ligand due to its ability to chelate with the metal.

Hence, options (a), (b) and (c) are correct choices.

Answer

$(a, c)$

Coordination compounds containing a ligand with more than one non-equivalent binding position (known as ambident ligand) show linkage isomerism.

e.g., $[Co(NH_3)_5(NO_2)^{+}.$contains $NO_2$ which have two donor sites $N$ and $O$ can be shown by arrow $ (\rightarrow)$ as

$$ \rightarrow S-C \equiv N \rightarrow $$

Hence, $[Co(NH_3)_5(NO_2)]^{2+}$ and $[Cr(NH_3)_5 SCN]^{2+}$ show linkage isomerism. While $[Co(H_2 O)_5 CO]^{3+}$ and $[Fe(en)_2 Cl_2]^{+}$has no ambident ligand. So, these two will not show linkage isomerism.

Hence, options (a) and (c) are correct choices.

Thinking Process

This problem is based on the concept of conductivity of coordination compound. Greater the number of ions, greater the conductivity of coordination compound.

Answer

Ions or molecules present outside the coordination sphere are ionisable. A complex which gives more ions on dissolution, is more conducting.

$$ \underset{(1 \text { ion })}{[Co(NH_3)_3 Cl_3]}<\underset{(2 \text { ions })}{[Co(NH_3)_4 Cl_2] Cl}<\underset{(3 \text { ions })}{[Cr(NH_3)_5 Cl] Cl_2}<\underset{(4 \text { ions })}{[Co(NH_3)_6] Cl_3} $$

Here, number of ions increases and conductivity increases.

Answer

Formation of white precipitate with $AgNO_3$ shows that atleast one $Cl$ ion is present outside the coordination sphere. Moreover only two ions are obtained in solution, so only one $Cl^{-}$is present outside the sphere.

Thus, the formula of the complex is $[Co(H_2 O)_4 Cl_2] Cl$ and its IUPAC name is Tetraaquadichloridocobalt (III) chloride.

Answer

An optically active complex of the type $[M(A A)_2 X_2]^{n+}$ indicates cis-octahedral structure, e.g., cis- $[Pt(en)_2 Cl_2]^{2+}$ or cis- $[Cr(en)_2 Cl_2]^{+}$because its mirror image isomers are non-superimposable.

Non-superimposable isomers of $[Pt(en)_2 Cl_2]^{2+}$.

Answer

The magnetic moment $5.92 BM$ shows that there are five unpaired electrons present in the $d$-orbitals of $Mn^{2+}$ ion. As a result, the hybridisation involved is $s p^{3}$ rather than $d s p^{2}$. Thus tetrahedral structure of $[MnCl_4]^{2-}$ complex will show $5.92 BM$ magnetic moment value.

Answer

With weak field ligands; $\Delta_{O}<P$, (pairing energy) so, the electronic configuration of $Co$ (III) will be $t_{2 g}^{4} e_g^{2}$ i.e., it has 4 unpaired electrons and is paramagnetic.

With strong field ligands, $\Delta_0>P$ (pairing energy), so pairing occurs thus, the electronic configuration will be $t_{2 g}^{6} e_g^{0}$. It has no unpaired electrons and is diamagnetic.

Answer

In tetrahedral complex, the $d$-orbital is splitting to small as compared to octahedral. For same metal and same ligand $\Delta_{t}=\frac{4}{9} \Delta_0$.

Hence, the orbital splitting energies are not enough to force pairing. As a result, low spin configurations are rarely observed in tetrahedral complexes.

Answer

According to spectrochemical series, ligands can be arranged in a series in the order of increasing field strength i.e., $F^{-}<NH_3<CN^{-}$.

Hence, $CN^{-}$and $NH_3$ being strong field ligand pair up the $t_{2 g}$ electrons before filling $e_{g}$ set. $[CoF_6]^{3-} ; Co^{3+}=(d^{6}) t_{2 g}^{4} e_g^{2}$

$[Fe(CN) _6^{4-}, Fe^{2+}=(d^{6}) t _{2 g}^{6} e _g^{0}.$

$[Cu(NH _3)c_6]^{2+}, Cu^{2+}=(d^{9}) t _{3 g}^{6} e _g^{3}$

Answer

As we know, where,

$$ \begin{aligned} \mu_{m} & =\sqrt{n(n+2} BM \\ \mu_{m} & =\text { magnetic moment } \\ \mu_{n} & =\text { number of unpaired electrons } \\ It \quad\mu_{m} & =1.74 \text { i.e., } n=1 \\ and \quad \mu_{m} & =5.92 \text { i.e., } n=5 \end{aligned} $$

$[Fe(CN)_6]^{3-}$ involves $d^{2} s p^{3}$ hybridisation with one unpaired electron (as shown by its magnetic moment $1.74 BM$ ) and $[Fe(H_2 O)_6]^{3+}$ involves $s p^{3} d^{2}$ hybridisation with five unpaired electrons (because magnetic moment equal to $5.92 BM$ ).

$CN^{-}$is stronger ligand than $H_2 O$ according to spectrochemical series. $\Delta_0>P$ for $CN^{-}$hence, fourth electron will pair itself. Whereas for water pairing will not happen for $[Fe(CN)_6]^{3-}$ the electronic configuration of $Fe^{3+}$ is

[Ar]

One unpaired electron

For $[Fe(H_2 O)_6]^{3+}$ the electronic configuration of $Fe^{3+}$ is

Five unpaired electron

Hence, $[Fe(CN)_6]^{3-}$ and $[Fe(H_2 O)_6]^{3+}$ are inner orbital and outer orbital complex respectively.

Answer

CFSE is higher when the complex contains strong field ligand. Thus, crystal field splitting energy increases in the order

$$ [Cr(Cl)_6]^{3-}<[Cr(NH_3)_6]^{3+}<[Cr(CN)_6]^{3-} . $$

Because according to spectrochemical series the order of field strength is

$$ Cl^{-}<NH_3<CN^{-} $$

Answer

It is due to the presence of weak and strong field ligands in complexes. If CFSE is high, the complex will show low value of magnetic moment and vice-versa, e.g. $[CoF_6]^{3-}$ and $[Co(NH_3)_6]^{3+}$, the former is paramagnetic, and the latter is diamagnetic because $F^{-}$is a weak field ligand and $NH_3$ is a strong field ligand while both have similar geometry.

Answer

In $CuSO_4 \cdot 5 H_2 O$, water acts as ligand and causes crystal field splitting. Hence, $d-d$ transition is possible thus $CuSO_4 \cdot 5 H_2 O$ is coloured. In the anhydrous $CuSO_4$ due to the absence of water (ligand), crystal field splitting is not possible and hence, it is colourless.

Answer

Ligand having more than one different binding position are known as ambidentate ligand. e.g., SCN has two different binding positions S and N. Coordination compound containing ambidentate ligands are considered to show linkage isomerism due to presence of two different binding positions.

e.g., (i) $[Co(NH_3)_5 SCN]^{3+}$ and

(ii) $[Fe(NH_3)_5(NO_2)]^{3+}$

Answer

A. $\rightarrow$ (4) $\quad$

B. $\rightarrow$ (3) $\quad$

C. $\rightarrow$ (2) $\quad$

D. $\rightarrow$ (1)

Colour of coordination compound is closely related to CFSE of coordination compound. Depending upon the CFSE of given coordination compounds. Correct matching will be as follows

Hence, correct choice is (b).

Answer

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow(1)$

D. $\rightarrow(2)$

Central metal ions present on coordination compounds determine the properties of coordination compound and their biological role.

Hence, correct choice is (a).

Answer

A. $\rightarrow$ (3) $\quad$

B. $\rightarrow(1) \quad$

C. $\rightarrow(4) \quad$

D. $\rightarrow(2)$

Formation of inner orbital complex and outer orbital complex determines hybridisation of molecule which inturn depends upon field strength of ligand and number of vacant $d$ orbitals.

(i) Strong field ligand forms inner orbital complex with hybridisation $d^{2} s p^{3}$.

(ii) Weak field ligand forms outer orbital complex with hybridisation $s p^{3} d^{2}$.

According to VBT, hybridisation and number of unpaired electrons of coordination compounds can be calculated as A. $[Cr(H_2 O)_6]^{3+}$

MOEC (Molecular orbital electronic configuration) of $Cr^{3+}$ in $[Cr(H_2 O)_6]^{3+}$ is

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

$n$ (number of unrpaired electrons) $=3$

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

MOEC of $Co^{2+}$ in $[Co(CN)_4]^{2-}$ is

Hybridisation $=d s p^{2}$

$n=1$

C. $[Ni(NH_3)_6]^{2+}$

MOEC of $Ni^{2+}$ in $[Ni^{-}(NH_3)_6]^{2+}$ is

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

$n=2$

D. $[MnF_6]^{4-}$

MOEC of $Mn^{2+}$ in $[MnF_6]^{4-}$ is

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

$$ n=5 $$

Hence, correct choice can be represented by (a).

Answer

A. $\rightarrow(4)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

D. $\rightarrow(3)$

Isomerism in coordination compound is decided by type of ligands and geometry of coordination and arrangement of ligands.

A. $[Co(NH_3)_4 Cl_2]^{+}$shows geometrical isomerism due to presence of two types of ligand whose $[Co(NH_3)_4 Cl_2]^{+}$arrangement around central metal ion.

B. cis $-[Co(en)_2 Cl_2]^{+}$shows optical isomer due to its non-superimposable mirror image relationship.

C. $[Co(NH_3)_5(NO_2)] Cl_2$ shows ionisation isomer due to its interchanging ligand from outside the ionisation sphere.

D. $[Co(NH_3)_6][Cr(CN)_6]$ shows coordination isomer due to interchanging of ligand in between two metal ions from one coordination sphere to another coordination sphere.

Hence, correct choice is (d).

Answer

A. $\rightarrow(4)$

B. $\rightarrow(1)$

C. $\rightarrow(3)$

D. $\rightarrow(2)$

Oxidation state of $CMI$ (central metal ion) can be calculated by considering the oxidation state of whole molecule is equal to charge present on coordination sphere.

A. $[Co(NCS)(NH_3)_5] SO_3$.

Let oxidation state of $Co$ is $x$.

$$ \begin{aligned} x-1+5 \times 0 & =+2 \\ x & =+2+1=+3 \end{aligned} $$

B. $[Co(NH_3)_4 Cl_2] SO_4$

Let oxidation state of $Co=x$

$$ \begin{aligned} \Rightarrow & & x+4 \times 0+2 \times(-1) & =+2 \\ \Rightarrow & & x-2 & =+2 \\ & & x & =4 \end{aligned} $$

C. $Na_4[Co(S_2 O_3)_3]$

Let oxidation state of $Co=x$

$$ \begin{aligned} x+3 \times(-2) & =-4 \\ x-6 & =-4 \\ x & =-4+6=+2 \end{aligned} $$

D. $[Co(CO)_8]$

Let oxidation state of $Co=x$

$$ \begin{matrix} x-8 \times 0=0 \\ x=0 \end{matrix} $$

Hence, correct choice is (d).

Answer

(a) Assertion and reason both are correct and reason is the correct explanation of assertion.

Toxic metal ions are removed by chelating ligands. When a solution of chelating ligand is added to solution containing toxic metals ligands chelates the metal ions by formation of stable complex.

Answer

(b) Assertion and reason both are true but reason is not correct explanation of assertion. Correct reason is

$[Cr(H_2 O_6)] Cl_2$ and $[Fe(H_2 O)_6] Cl_2$ are reducing in nature due to formation of more stable complex ion after gaining of electron.

Answer

(a) Assertion and reason both are correct and reason is correct explanation of assertion. Linkage isomerism arises in coordination compounds containing ambidentate ligands because ambidentate ligand has two different donor atoms.

e.g., $SCN, NO_2$ etc.

Answer

(b) Assertion and reason both are correct and reason is not correct explanation of assertion.

Complexes of $M X_6$ and $M X_5 L$ type ( $X$ and $L$ are unidentate) do not show geometrical isomerism due to presence of plane of symmetry and necessary condition for showing geometrical isomerism is that complex is must of $M A_4 B_2$ type or $[M(A B)_2 X_2]$ type

Answer

(d) Assertion is false but reason is true.

According to VBT, MOEC of $Fe^{3+}$ in $[Fe(CN)_6]^{3-}$ is

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

$$ n=1 $$

Hence, correct assertion is

$[Fe(CN)_6]^{3-}$ ion shows magnetic moment corresponding to one unpaired electron.

i.e.,

$$ \begin{aligned} \mu & =\sqrt{n(n+2)} \\ & =\sqrt{1(1+2)} \\ & =\sqrt{3}=1.73 BM \end{aligned} $$

Thinking Process

This problem includes conceptual mixing of crystal field theory and magnetic moment $(\mu)$ determination.

$\mu=\sqrt{n(n+2)} B M$ where, $n=$ number of unpaired electrons

Answer

(a) $[CoF_6]^{3-}$.

$F^{-}$is a weak field ligand.

Configuration of $Co^{3+}=3 d^{6}(.$ or $.t_{2 g}^{4} e_g^{2})$

Number of unpaired electrons $(n)=4$

Magnetic moment $(\mu)=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.9 BM$

$[Co(H_2 O)_6]^{2+}$,

$H_2 O$ is a weak field ligand.

Configuration of $Co^{2+}=3 d^{7}(.$ or $.t_{2 g}^{5} e_g^{2})$

Number of unpaired electrons $(n)=3$

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

$[Co(CN)_6]^{3-}$ i.e., $Co^{3+}$

$\because CN$ is strong field ligand.

$Co^{3+}=3 d^{6}(.$ or $.t_{2 g}^{6} e_g^{0})$

There is no unpaired electron, so it is diamagnetic.

$$ \mu=0 $$

(b) $[FeF_6]^{3-}$,

Number of unpaired electrons, $n=5$

$$ \begin{aligned} \mu & =\sqrt{5(5+2)} \\ & =\sqrt{35}=5.92 BM \end{aligned} $$

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

$$ Fe^{2+}=3 d^{6}(\text { or } t_{2 g}^{4} e_g^{2}) $$

Number of unpaired electrons, $n=4$

$$ \begin{aligned} \mu & =\sqrt{4(4+2)} \\ Z & =\sqrt{24} \\ & =4.98 BM \end{aligned} $$

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

Since, $CN^{-}$is a strong field ligand, all the electrons get paired.

$Fe^{2+}=3 d^{6}(.$ or $.t_{2 g}^{6} e_g^{0})$

Because there is no unpaired electron, so it is diamagnetic in nature.

Answer

(a) $[Mn(CN)_6]^{3-}$

(i) $d^{2} s p^{3}$ hybridisation

(ii) Inner orbital complex because $(n-1) d$-orbitals are used.

(iii) Paramagnetic, as two unpaired electrons are present.

(iv) Spin only magnetic moment $(\mu)=\sqrt{2(2+2)}=\sqrt{8}=2.82 BM$

(b) $[Co(NH_3)_6]^{3+}$

$Co^{3+}=3 d^{6} 4 s^{0}$

$(NH_3.$ pair up the unpaired $3 d$ electrons.)

(i) $d^{2} s p^{3}$ hybridisation

(ii) Inner orbital complex because of the involvement of $(n-1) d$-orbital in bonding.

(iii) Diamagnetic, as no unpaired electron is present.

(iv) $\mu=\sqrt{n(n+2)}=\sqrt{0(0+2)}=0$ (Zero)

(c) $[Cr(H_2 O)_6]^{3+}$

(i) $d^{2} s p^{3}$ hybridisation

(ii) Inner orbital complex (as $(n-1) d$-orbital take part.)

(iii) Paramagnetic (as three unpaired electrons are present.)

(iv) $\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15}=3.87 BM$

(d) $[Fe(Cl)_6]^{4-}$

$$ Fe^{2+}=3 d^{6} $$

(i) $s p^{3} d^{2}$ hybridisation

(ii) Outer orbital complex because $n d$-orbitals are involved in hybridisation.

(iii) Paramagnetic (because of the presence of four unpaired electrons).

(iv) $\mu=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.9 BM$

Thinking Process

This problem is based on chemical properties of coordination compounds, ionisation isomerism, and nomenclature of coordination compounds.

Answer

’ $A$ ’ gives precipitate with $AgNO_3$, so in it $Cl$ is present outside the coordination sphere.

’ $B$ ’ gives precipitate with $BaCl_2$, so in it $SO_4^{2-}$ is present outside the coordination sphere.

(a) $So, A$ - $[Co(NH_3)_5 SO_4] Cl$ $B$ - $[Co(NH_3)_5 Cl] SO_4$

(b) Ionisation isomerism (as give different ions when subjected to ionisation.)

(c) $[A]$, Pentaamminesulphatocobalt (III) chloride.

[B], Pentaamminechloridocobalt (III) sulphate.

Answer

When white light falls on the complex, some part of it is absorbed. Higher the crystal field splitting energy, lower will be the wavelength absorbed by the complex. The observed colour of complex is the colour generated from the wavelength left over.

e.g., if green light is absorbed, the complex appears red.

In terms of crystal field theory, suppose there is an octahedral complex with empty $e_{g}$ level and unpaired electrons in the $t_{2 g}$ level in ground level. If the unpaired electron absorbs light corresponding is blue-green region, it will excite to $e_{g}$ level and the complex will appear violet in colour.

In absence of ligand, crystal field splitting does not occur and the substance is colourless.

e.g., anhydrous $CuSO_4$ is while, but $CuSO_4 \cdot 5 H_2 O$ is blue in colour.

Answer

Extent of splitting of $d$-orbitals is different in octahedral and tetrahedral field. CFSE in octahedral and tetra federal field are closely related as.

$$ \Delta_{t}=(\frac{4}{9}) \Delta_{O} $$

where,$\quad \Delta_{t}=$ crystal field splitting energy in tetrahedral field

$\Delta_0=$ crystal field splitting energy in octahedral field

Wavelength of light and CFSE are related to each other by formula

$$ \begin{aligned} \Delta_0=E & =\frac{h c}{\lambda} \\ E & \propto \frac{1}{\lambda} \end{aligned} $$

So, higher wavelength of light is absorbed in octahedral complexes than tetrahedral complexes for same metal and ligands. Thus, different colours are observed.