NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 1)

Question:1

Which of the following complexes formed by ${\text{Cu}}^{2+}$ ions is most stable?
(i) ${\text{Cu}}^{2+}+{\text{4NH}}_3\rightleftharpoons {\left[\text{Cu}{\left({\text{NH}}_3\right)}_4\right]}^{2+}\text{logK}=11.6$
(ii) ${\text{Cu}}^{2+}+{\text{4CN}}^{-}\rightleftharpoons {\left[\text{Cu}{\left(\text{CN}\right)}_4\right]}^{2-}\text{logK}=27.3$
(iii) ${\text{Cu}}^{2+}+\text{2en}\rightleftharpoons {\left[\text{Cu}{\left(\text{en}\right)}_2\right]}^{2+}\text{logK}=15.4$
(iv) ${\text{Cu}}^{2+}+{\text{4H}}_2\text{O}\rightleftharpoons {\left[\text{Cu}{\left({\text{H}}_2\text{O}\right)}_4\right]}^{2+}\text{logK}=8.9$
Answer:

Option (ii) is the correct answer.
Explanation:
The stability of a compound will be more when the value of log K increases.
For reaction,
${\text{Cu}}^{2+}+{\text{4CN}}^{-}\rightleftharpoons {\left[\text{Cu}{\left(\text{CN}\right)}_4\right]}^{2-}$
$\text{K}=\frac{\left[\text{Cu}{\left({\text{CN}}_4\right)}^{2-}\right]}{\left[{\text{Cu}}^{2+}\right]{\left[{\text{CN}}^{-}\right]}^4}\text{and}\text{log K}=27.3$
Log K has the highest value for this reaction amongst all the four reactions. The value of K will also be higher so, the stability of this complex will be highest.

Question:2

The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, ${\left[\text{Co}{\left({\text{NH}}_3\right)}_6\right]}^{3+}$, ${\left[\text{Co}{\left(\text{CN}\right)}_6\right]}^{3-}$, ${\left[\text{Co}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}$
(i) ${\left[\text{Co}{\left(\text{CN}\right)}_6\right]}^{3-}>{\left[\text{Co}{\left({\text{NH}}_3\right)}_6\right]}^{3+}>{\left[\text{Co}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}$
(ii) ${\left[\text{Co}{\left({\text{NH}}_3\right)}_6\right]}^{3+}>{\left[\text{Co}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}>{\left[\text{Co}{\left(\text{CN}\right)}_6\right]}^{3-}$
(iii) ${\left[\text{Co}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}>{\left[\text{Co}{\left({\text{NH}}_3\right)}_6\right]}^{3+}>{\left[\text{Co}{\left(\text{CN}\right)}_6\right]}^{3-}$
(iv) ${\left[\text{Co}{\left(\text{CN}\right)}_6\right]}^{3-}>{\left[\text{Co}{\left({\text{NH}}_3\right)}_6\right]}^{3+}>{\left[\text{Co}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}$
Answer:

Option (iii) is the correct answer.
Explanation :
${\mathrm{\Delta }}_0$ values follow the order : ${\left[\text{Co}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}<{\left[\text{Co}{\left({\text{NH}}_3\right)}_6\right]}^{3+}<{\left[\text{Co}{\left(\text{CN}\right)}_6\right]}^{3-}$ and therefore,absorption wavelength follows the order:
${\left[\text{Co}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}>{\left[\text{Co}{\left({\text{NH}}_3\right)}_6\right]}^{3+}>{\left[\text{Co}{\left(\text{CN}\right)}_6\right]}^{3-}$

Question:3

When 0.1 mol ${\text{CoCl}}_3{\left({\text{NH}}_3\right)}_5$ is treated with an excess of ${\text{AgNO}}_3$, 0.2 mol of $\text{AgCl}$ are obtained. The conductivity of the solution will correspond to
(i) 1:3 electrolyte
(ii) 1:2 electrolyte
(iii) 1:1 electrolyte
(iv) 3:1 electrolyte
Answer:

Option (ii) is the correct answer.
Explanation:
One mole of chloride ion gets precipitated by one mole of ${\text{AgNO}}_3$. $\text{AgCl}$, is obtained when 0.1 mole of ${\text{CoCl}}_3{\left({\text{NH}}_3\right)}_5$ is treated with ${\text{AgNO}}_3$ in excess. So this leaves two free chloride ions in the solution of electrolyte for every 1 mole reaction.
Hence the molecular formula should be $\left[\text{Co}{\left({\text{NH}}_3\right)}_5\text{Cl}\right]{\text{Cl}}_2$ and the solution of electrolyte should contain ${\left[\text{Co}{\left({\text{NH}}_3\right)}_5\text{Cl}\right]}^{2+}$ and two Cl as their constituent ions. Therefore, it is an 1:2 electrolyte.
$\left[\text{Co}{\left({\text{NH}}_3\right)}_5\text{Cl}\right]{\text{Cl}}_2\to {\left[\text{Co}{\left({\text{NH}}_3\right)}_5\text{Cl}\right]}^{2+}\text{(aq)}+{\text{2Cl}}^{-}\text{(aq)}$

Question:4

When 1 mol ${\text{CrCl}}_3.{\text{6H}}_2\text{O}$ is treated with an excess of ${\text{AgNO}}_3$, 3 mol of $\text{AgCl}$ are obtained. The formula of the complex is:
(i) $\left[{\text{CrCl}}_3{\left({\text{H}}_2\text{O}\right)}_3\right].{\text{3H}}_2\text{O}$
(ii) $\left[{\text{CrCl}}_2{\left({\text{H}}_2\text{O}\right)}_4\right]\text{Cl}.{\text{2H}}_2\text{O}$
(iii) $\left[\text{CrCl}{\left({\text{H}}_2\text{O}\right)}_5\right]{\text{Cl}}_2.{\text{H}}_2\text{O}$
(iv) $\left[\text{Cr}{\left({\text{H}}_2\text{O}\right)}_6\right]{\text{Cl}}_3$
Answer:

Option (iv) is the correct answer.
Explanation: 3 mol of $\text{AgCl}$ indicates that ${\text{3Cl}}^{-}$ ions are given in the solution. Therefore, the formula should be $\left[\text{Cr}{\left({\text{H}}_2\text{O}\right)}_6\right]{\text{Cl}}_3$.

Question:5

The correct $\text{IUPAC}$ name of $\left[\text{Pt}{\left({\text{NH}}_3\right)}_2{\text{Cl}}_2\right]$ is
(i) Diamminedichloridoplatinum (II)
(ii) Diamminedichloridoplatinum (IV)
(iii) Diamminedichloridoplatinum (0)
(iv) Dichloridodiammineplatinum (IV)
Answer:

Option (i) is the correct answer.
Explanation: diamminedichloridoplatinum (II) is $\left[\text{Pt}{\left({\text{NH}}_3\right)}_2{\text{Cl}}_2\right]$.

Question:6

The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?
(i) $\left[\text{Fe}{\left(\text{CO}\right)}_5\right]$
(ii) ${\left[\text{Fe}{\left(\text{CN}\right)}_6\right]}^{3-}$
(iii) ${\left[\text{Fe}{\left({\text{C}}_2{\text{O}}_4\right)}_3\right]}^{3-}$
(iv) ${\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}$
Answer:

Option (iii) is the correct answer.
Explanation: Chelation is the formation of cycle of linkages between the metal ion and ligands is responsible for stabilizing the coordination compounds. A ligand which chelates any metal ion is known as chelating ligand.
Here, oxalate ion is the chelating ligand and ${\left[\text{Fe}{\left({\text{C}}_2{\text{O}}_4\right)}_3\right]}^{3-}$ is the coordination compound. Hence, the oxalate ions stabilize the coordination compound by chelating ${\text{Fe}}^{3+}$ ions.

Question:7

Indicate the complex ion which shows geometrical isomerism.
(i) ${\left[\text{Cr}{\left({\text{H}}_2\text{O}\right)}_4{\text{Cl}}_2\right]}^{+}$
(ii) $\left[\text{Pt}{\left({\text{NH}}_3\right)}_3\text{Cl}\right]$
(iii) ${\left[\text{Co}{\left({\text{NH}}_3\right)}_6\right]}^{3+}$
(iv) ${\left[\text{Co}{\left(\text{CN}\right)}_5\left(\text{NC}\right)\right]}^{3-}$
Answer:

Option (i) is the correct answer.
Explanation :
(a) ${\left[\text{Cr}{\left({\text{H}}_2\text{O}\right)}_4{\text{Cl}}_2\right]}^{+}$

Question:8

The CFSE for octahedral ${\left[{\text{CoCl}}_6\right]}^{4-}$ is 18,000 cm^–1. The CFSE for tetrahedral ${\left[{\text{CoCl}}_4\right]}^{2-}$ will be
(i) 18,000 ${\text{cm}}^{-1}$
(ii) 16,000 ${\text{cm}}^{-1}$
(iii) 8,000 ${\text{cm}}^{-1}$
(iv) 20,000 ${\text{cm}}^{-1}$
Answer:

Option (iii) is the correct answer.
Explanation :
(c) CFSE for tetrahedral complex is ${\mathrm{\Delta }}_t=\left(\frac{4}{9}\right){\mathrm{\Delta }}_0$
${\mathrm{\Delta }}_t=\frac{4}{9}\times 18000=8,000{\text{cm}}^{-1}$

Question:9

Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type $\left[\text{Pd}{\left({\text{C}}_6{\text{H}}_5\right)}_2{\left(\text{SCN}\right)}_2\right]$ and $\left[\text{Pd}{\left({\text{C}}_6{\text{H}}_5\right)}_2{\left(\text{NCS}\right)}_2\right]$ are
(i) linkage isomers
(ii) coordination isomers
(iii) ionisation isomers
(iv) geometrical isomers
Answer:

Option (i) is the correct answer.
Explanation: Ambident ligands are those ligands which have two different bonding sites. For example: ${\text{NCS,NO}}_2$ etc.
In this case, $\text{NCS}$ contains two binding sites at N and S. Therefore, $\text{NCS}$ (thiocyanate) will be bonding to the metal in these two ways; $\text{M}\leftarrow \text{NCS}$ or $\text{M}\to \text{SNC}$
Hence, the coordination compounds containing $\text{NCS}$ as ligands can show linkages isomerism; $\left[\text{Pd}{\left({\text{C}}_6{\text{H}}_5\right)}_2{\left(\text{SCN}\right)}_2\right]$ and $\left[\text{Pd}{\left({\text{C}}_6{\text{H}}_5\right)}_2{\left(\text{NCS}\right)}_2\right]$ are linkage isomers.

Question:10

The compounds $\left[\text{Co}\left({\text{SO}}_4\right){\left({\text{NH}}_3\right)}_5\right]\text{Br}$ and $\left[\text{Co}\left({\text{SO}}_4\right){\left({\text{NH}}_3\right)}_5\right]\text{Cl}$ represent
(i) linkage isomerism
(ii) ionisation isomerism
(iii) coordination isomerism
(iv) no isomerism
Answer:

Option (iv) is the correct answer.
Explanation:$\left[\text{Co}\left({\text{SO}}_4\right){\left({\text{NH}}_3\right)}_5\right]\text{Br}$ and $\left[\text{Co}\left({\text{SO}}_4\right){\left({\text{NH}}_3\right)}_5\right]\text{Cl}$ are different compounds and show no isomerism.

Question:11

A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent?
(i) thiosulphate
(ii) oxalate
(iii) glycinato
(iv) ethane-1,2-diamine
Answer:

Option (i) is the correct answer.
Explanation: Thiosulphate or ${\text{S}}_2{\text{O}}_3^{2-}$ is a monodentate ligand and not a chelating agent.

Question:12

Which of the following species is not expected to be a ligand?
(i) $\text{NO}$
(ii) ${\text{NH}}_4^{+}$
(iii) ${\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2$
(iv) $\text{CO}$
Answer:

Option (ii) is the correct answer.
Explanation: To form M-L bonds, the ligand must donate a pair of electrons or any loosely held electron pair to the metal.
e. g
Among all ${\text{NH}}_4^{+}$ does not have any pair of electrons.
Hence ${\text{NH}}_4^{+}$ is not a ligand.

Question:13

What kind of isomerism exists between $\left[\text{Cr}{\left({\text{H}}_2\text{O}\right)}_6\right]{\text{Cl}}_3\text{(violet)}$ and $\left[\text{Cr}{\left({\text{H}}_2\text{O}\right)}_5\text{Cl}\right]{\text{Cl}}_2.{\text{H}}_2\text{O}\text{(greyish-green)}$ ?
(i) linkage isomerism
(ii) solvate isomerism
(iii) ionisation isomerism
(iv) coordination isomerism
Answer:

Option (ii) is the correct answer.
Explanation: The compound has different water molecules in number inside and outside of the coordination sphere.

Question:14

IUPAC name of $\left[\text{Pt}{\left({\text{NH}}_3\right)}_2\text{Cl}\left({\text{NO}}_2\right)\right]$ is :
(i). Platinum diaminechloronitrite
(ii). Chloronitrito-N-ammineplatinum (II)
(iii). Diamminechloridonitrito-N-platinum (II)
(iv). Diamminechloronitrito-N-platinate (II)
Answer:

Option (iii) is the correct answer.
Explanation: diamminechloridonitrito-N-platinum (II) is $\left[\text{Pt}{\left({\text{NH}}_3\right)}_2\text{Cl}\left({\text{NO}}_2\right)\right]$

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 2)

Question:15

The atomic number of $\text{Mn, Fe and Co}$ are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complexions are diamagnetic?
(i) ${[Co(N{H}_3{)}_6]}^{3+}$
(ii) ${[Mn(CN{)}_6]}^{3-}$
(iii) ${[Fe(CN{)}_6]}^{4-}$
(iv) ${[Fe(CN{)}_6]}^{3-}$
Answer:

Option (i) and (iii) are the correct answers.
Explanation :
(a,c)
(a) Molecular orbital electronic configuration of $C{o}^{3+}$ in ${[Co(N{H}_3{)}_6]}^{3+}$is

Number of unpaired electron = 0
Magnetic property = Diamagnetic
(b) Molecular orbital electronic configuration of $M{n}^{3+}$ in ${[Mn(CN{)}_6]}^{3-}$

Number of unpaired electron = 2
Magnetic property = Paramagnetic
(c) Molecular orbital electronic configuration of $F{e}^{2+}$ in ${[Fe(CN{)}_6]}^{4-}$

Number of unpaired electron = 0
Magnetic property = Diamagnetic
(d) Molecular orbital electronic configuration of $F{e}^{3+}$ in ${[Fe(CN{)}_6]}^{3-}$

Number of unpaired electron = 1
Magnetic property = Paramagnetic

Question:16

The atomic number of $\text{Mn, Fe, Co and Ni}$ are 25, 26 27 and 28 respectively. Which of the following outer orbital octahedral complexes have the same number of unpaired electrons?
(i) ${\left[MnC{l}_6\right]}^{3-}$
(ii) ${\left[Fe{F}_6\right]}^{3-}$
(iii) ${\left[Co{F}_6\right]}^{3-}$
(iv) ${[Ni(N{H}_3{)}_6]}^{2+}$
Answer:

Option (i) and (iii) are the correct answer.
Explanation :
(a,c)
$[MnC{l}_6{]}^{3-}:M{n}^{3+}(3d^4)$
$[Co{F}_6{]}^{3-}:C{o}^{3+}(3d^6)$

Question:17

Which of the following options are correct for $[Fe(CN{)}_6{]}^{3-}$ complex?
(i) $d^{2}s{p}^3$hybridisation
(ii) $s{p}^3{d}^2$hybridisation
(iii) paramagnetic
(iv) diamagnetic
Answer:

Option (i) and (iii) are the correct answers.
Explanation :
(a,c) $[Fe(CN{)}_6{]}^{3-}$

Magnetic nature -paramagnetic.

Question:18

An aqueous pink solution of cobalt(II) chloride changes to deep blue on the addition of an excess of $HCl$. This is because____________.
(i) $[Co(H_{2}O{)}_6{]}^{2+}$ is transformed into $[CoC{l}_6{]}^{4-}$
(ii) $[Co(H_{2}O{)}_6{]}^{2+}$ is transformed into $[CoC{l}_4{]}^{2-}$
(iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes.
(iv) tetrahedral complexes have larger crystal field splitting than octahedral complex.
Answer:

Option (ii) and (iii) are the correct answers.
Explanation: The transition of electrons from t_2g to e_g energy level gives an aqueous pink solution of cobalt (II) chloride in $[Co(H_{2}O{)}_6{]}^{2+}$ complex and when excess $HCl$ is added to the solution:
(i) $[Co(H_{2}O{)}_6{]}^{2+}$ is transformed into $[CoC{l}_4{]}^{2-}$
(ii) Tetrahedral complexes have smaller crystal field spliting than octahedral complexes because ${\mathrm{\Delta }}_r=\frac{4}{9}{\mathrm{\Delta }}_0$
Hence, options (b) and (c) are correct.

Question:19

Which of the following complexes is homoleptic?
(i) ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}$
(ii) ${\left[Co{\left(N{H}_3\right)}_{4}C{l}_2\right]}^{+}$
(iii) ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$
(iv) $\left[Ni{\left(N{H}_3\right)}_{4}C{l}_2\right]$
Answer:

Option (i) and (iii) are the correct answers.
Explanation: Both $\text{Co and Ni}$ are attached the same kind of ligands in the complexes ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}$ and ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$. Hence they are homoleptic.

Question:20

Which of the following complexes are heteroleptic?
(i) ${[Cr(N{H}_3{)}_6]}^{3+}$
(ii) ${[Fe(N{H}_3{)}_{4}C{l}_2]}^{+}$
(iii) ${[Mn(CN{)}_6]}^{4-}$
(iv) $[Co(N{H}_3{)}_{4}C{l}_2]$
Answer:

Option (ii) and (iv) are the correct answers.
Explanation: Metal is bonded to more than one kind of ligands in complexes ${[Fe(N{H}_3{)}_{4}C{l}_2]}^{+}$and $[Co(N{H}_3{)}_{4}C{l}_2]$ therefore they are heteroleptic.

Question:21

Identify the optically active compounds from the following: -
(i) $[Co(en{)}_3{]}^{3+}$
(ii) $trans-[Co(en{)}_{2}C{l}_2{]}^{+}$
(iii) $cis-[Co(en{)}_{2}C{l}_2{]}^{+}$
(iv) $[Cr(N{H}_3{)}_{5}Cl]$

Answer:

Option (i) and (iii) are the correct answers.
(a,c) $[Co(en{)}_3{]}^{3+}$ and $cis-[Co(en{)}_{2}C{l}_2{]}^{+}$ are optically active compounds because their mirror images are non superimposable isomer.

Non-superimposable isomers of $[Co(en{)}_3{]}^{3+}$

Non-superimosable isomers of $[Co(en{)}_{2}C{l}_2{]}^{+}$
Hence, (a) and (c) are correct choices.

Question:22

Identify the correct statements for the behaviour of ethane-1, 2-diamine as a ligand.
(i) It is a neutral ligand.
(ii) It is a didentate ligand.
(iii) It is a chelating ligand.
(iv) It is a unidentate ligand.
Answer:

Option (i), (ii) and (iii) are the correct answers.
Explanation :
(a,b,c) Ethane 1, 2-diamine is a neutral and didentate ligand. It is a chelanting agent.

Question:23

Which of the following complexes show linkage isomerism?
(i) $[Co(N{H}_3{)}_5(N{O}_2){]}^{2+}$
(ii) $[Co(H_{2}O{)}_{5}CO{]}^{3+}$
(iii) $[Cr(N{H}_3{)}_{5}SCN{]}^{2+}$
(iv) $[Fe(en{)}_{2}C{l}_2{]}^{+}$

Answer:

Option (i) and (iii) are the correct answers.
Explanation:$N{O}_2$ and $SCN$ show linkage isomerism because they are ambidentate ligands.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Short Answer Type

Question:24

Arrange the following complexes in the increasing order of conductivity of their solution: $[Co(N{H}_3{)}_{3}C{l}_3],[Co(N{H}_3{)}_{4}C{l}_2]Cl,[Co(N{H}_3{)}_6]C{l}_3,[Cr(N{H}_3{)}_{5}Cl]C{l}_2$
Answer:

The correct order of increase in conductivity is given below:
$[Co(N{H}_3{)}_{3}C{l}_3]<[Co(N{H}_3{)}_{4}C{l}_2]Cl<[Cr(N{H}_3{)}_{5}Cl]C{l}_2<[Co(N{H}_3{)}_6]C{l}_3$

Question:25

A coordination compound $CrC{l}_3.4H_{2}O$ precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write the structural formula of the compound and name it.
Answer:

There should be one free chlorine atom left outside the coordination sphere if silver chloride is formed in the reaction. $\left[Cr{\left(H_{2}O\right)}_{4}C{l}_2\right]Cl$ has to be the formula and it is called tetraaquadichlorido chromium(III) chloride.

Question:26

A complex of the type $[M(AA{)}_2{X}_2{]}^{n+}$is known to be optically active. What does this indicate about the structure of the complex? Give one example of such complex.
Answer:

If the complex is optically active then its structure has to be cis-octahedral. $[Co(en{)}_{2}C{l}_2{]}^{+}$ is one of the prominent examples of an optically stable compound.

Question:27

The magnetic moment of ${\left[MnC{l}_4\right]}^{2-}$ is $5.92BM$. Explain giving reason.
Answer:

A magnetic moment of 5.92 BM indicates that there are at least 5 unpaired electrons. The formula for Magnetic Moment $=\sqrt{n(n+2)}$
As four ligands are attached to $M{n}^{2+}$, therefore the geometry tetrahedral with 5 unpaired electrons gives a magnetic moment of 5.92 BM.

Question:28

Based on crystal field theory explain why $\text{Co(III)}$ forms a paramagnetic octahedral complex with weak field ligands whereas it forms a diamagnetic octahedral complex with strong field ligands.
Answer:

$t_{2g}^4{e}_g^2$ will be the electronic configuration where it indicates that it has 4 unpaired electrons and is paramagnetic, with a weak ligand ${\mathrm{\Delta }}_0<p$. The configuration will be $t_{2g}^6{e}_g^0$ with strong field ligand and there will not be any unpaired electrons. Hence it is diamagnetic.

Question:29

Why are low spin tetrahedral complexes not formed?
Answer:

The crystal field splitting energy is very low for tetrahedral complexes. The crystal field splitting energy is lower than pairing energy, therefore pairing of electrons is not favorable. Therefore, the complexes may not form low spin complexes

Question:30

Give the electronic configuration of the following complexes based on the basis of Crystal Field Splitting theory.
$[Co{F}_6{]}^{3-},[Fe(CN{)}_6{]}^{4-}and[Cu(N{H}_3{)}_6{]}^{2+}.$
Answer:

Question:31

Explain why $[Fe(H_{2}O{)}_6{]}^{3+}$ has a magnetic moment value of 5.92 BM whereas $[Fe(CN{)}_6{]}^{3-}$ has a value of only 1.74 BM.
Answer:

[Fe(CN)₆]^3- involves d²sp³ hybridisation with one unpaired electron and [Fe(H₂O)₆]³⁺ involves sp³d² hybridisation with five unpaired electrons. This difference is due to the presence of strong ligand CN– and weak ligand H₂O in these complexes

Question:32

Arrange following complex ions in increasing order of crystal field splitting energy (${\mathrm{\Delta }}_o$) :
$[Cr(Cl{)}_6{]}^{3-},[Cr(CN{)}_6{]}^{3-},[Cr(N{H}_3{)}_6{]}^{3+}.$

Answer:

The correct increasing order of crystal field is:
$[Cr(Cl{)}_6{]}^{3-}<[Cr(N{H}_3{)}_6{]}^{3+}<[Cr(CN{)}_6{]}^{3-}$
According to the spectrochemical series, this is also the correct increasing of order of the field strength.

Question:33

Why do compounds having similar geometry have a different magnetic moment?
Answer:

The difference lies in the number of paired and unpaired electrons. Strong field ligand can easily cause the pairing of electrons while the weak field ligands are not able to form pairs. The magnetic moment of a compound depends on the number of unpaired or paired electrons. Hence it is different for compounds having similar geometry.

Question:34

${\text{CuSO}}_4.{\text{5H}}_2\text{O}$ is blue while ${\text{CuSO}}_4$ is colourless. Why?
Answer:

In ${\text{CuSO}}_4.{\text{5H}}_2\text{O}$, the extra 5 water act as ligands. They apparently excite the electrons to the higher d orbital which shows the blue color. In ${\text{CuSO}}_4$, there are no water molecule that can act as ligands, so there is no crystal field slitting happening, hence there is no color.

Question:35

Name the type of isomerism when ambidentate ligands are attached to a central metal ion. Give two examples of ambidentate ligands.
Answer:

The ligands which have two binding sites are known as Ambidendate ligands. Some of the examples are: Nitrite-N, Nitrito-O and Isothiocyanato, Thiocyanato.
Linkage Isomerism is a type of isomerism when ambidentate ligands get attached to central metal ions. This happens because it only differs in the atom that they are linked the central metal ion.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Matching Type

Question:36

In the following question match the items given in Columns I and II.
Match the complex ions given in Column I with the colours given in Column II and assign the correct code :

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (2) D (1)
(iii) A (3) B (2) C (4) D (1)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (ii) is the correct answer.

Question:37

In the following question match the items given in Columns I and II.
Match the coordination compounds given in Column I with the central metal atoms given in Column II and assign the correct code :

(i) A (5) B (4) C (1) D (2)
(ii) A (3) B (4) C (5) D (1)
(iii) A (4) B (3) C (2) D (1)
(iv) A (3) B (4) C (1) D (2)

Answer:

Option (i) is the correct answer.

Question:38

In the following question match the items given in Columns I and II.
Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code

(i) A (3) B (1) C (5) D (2)
(ii) A (4) B (3) C (2) D (1)
(iii) A (3) B (2) C (4) D (1)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (i) is the correct answer.
Explanation : (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 $d^{2}s{p}^3$.
According to VBT, hybridisation and number of unpaired electrons of coordination compounds can be calculated as
(a) ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$
MOEC (Molecular orbital electronic configuration) of $C{r}^{3+}$ in ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$ is

Hybridisation = $d^{2}s{p}^3$
n (number of unpaired electrons) = 3
(b) ${\left[Co{\left(CN\right)}_4\right]}^{2-}$ is
MOEC of $C{o}^{2+}$ in ${\left[Co{\left(CN\right)}_4\right]}^{2-}$ is

Hybridisation = $ds{p}^2$
n (number of unpaired electrons) = 1
(c) ${\left[Ni{\left(N{H}_3\right)}_6\right]}^{2+}$
MOEC of $N{i}^{2+}$ in ${\left[Ni{\left(N{H}_3\right)}_6\right]}^{2+}$ is

Hybridisation = $s{p}^3{d}^2$
n (number of unpaired electrons) = 2
(d) ${\left[Mn{F}_6\right]}^{4-}$
MOEC of $M{n}^{2+}$ in ${\left[Mn{F}_6\right]}^{4-}$ is

Hybridisation = $s{p}^3{d}^2$
n (number of unpaired electrons) = 5
Hence , correct choice can be represented by (a).

Question:39

In the following question match the items given in Columns I and II.
Match the complex species given in Column I with the possible isomerism given in Column II and assign the correct code :

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (2) D (1)
(iii) A (4) B (1) C (5) D (3)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (iv) is the correct answer.
Explanation : The types of ligands, arrangement of ligands and the geometry of coordination decide the isomerism in coordination compounds.
(a) $[Co(N{H}_3{)}_{4}C{l}_2{]}^{+}$ shows geometrical isomerism due to presence of two types of ligand whoes $[Co(N{H}_3{)}_{4}C{l}_2{]}^{+}$ arrangement around central metal ion

(b) $cis-[Co(en{)}_{2}C{l}_2{]}^{+}$ shows optical isomer due to its non - superimposable mirror image relationship.

(c) $[Co(N{H}_3{)}_5(N{O}_2)]C{l}_2$ shows ionization isomer due to its interchanging ligand from outside the ionization sphere.
(d) $[Co(N{H}_3{)}_6][Cr(CN{)}_6]$ shows coordination isomerdue to interchanging of ligand in between two metal ions from one coordination sphere to another coordination sphere.
Hence, correct choice is (d)

Question:40

In the following question match the items given in Columns I and II.
Match the compounds given in Column I with the oxidation state of cobalt present in it (given in Column II) and assign the correct code.

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (2) D (1)
(iii) A (5) B (1) C (4) D (2)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (iii) is the correct answer.
Explanation: The oxidation state of the whole molecule is considered to calculate the oxidation state of CMI (central metal ion).
(a) $\left[Co\left(NCS\right){\left(N{H}_3\right)}_5\right]\left(S{O}_3\right)$
Let oxidation state of $Co$ be x.
$x-1+5\times 0=+2$
$x=+2+1=+3$
(b) $\left[Co{\left(N{H}_3\right)}_{4}C{l}_2\right]S{O}_4$
Let oxidation state of $Co$ = x.
$\Rightarrow x+4\times 0+2\times (-1)=+2$
$x-2=+2$
$x=4$
(c) $N{a}_4[Co(S_2{O}_3{)}_3]$
Let oxidation state of $Co$ = x.
$x+3\times (-2)=-4$
$x-6=-4$
$x=-4+6=+2$
(d) $[C{o}_2(CO{)}_8]$
Let oxidation state of $Co$ = x.
$x-8\times 0=0$
$x=0$
Hence, correct choice is (d)

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Assertion and Reason Type

Question:41

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Toxic metal ions are removed by the chelating ligands.
Reason: Chelate complexes tend to be more stable.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (i) is correct answer.
Explanation: Ligands chelate the metal ions by forming stable complex when solution of chelating ligands is added to a solution containing toxic metals.

Question:42

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion:$[Cr(H_{2}O{)}_6]C{l}_2$ and $[Fe(H_{2}O{)}_6]C{l}_2$ are reducing in nature.
Reason: Unpaired electrons are present in their d-orbitals.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (ii) is correct answer.
Explanation: In both the complexes, $\text{Cr and Fe}$ Coexist as ${\text{Cr}}^{2+}$ and ${\text{Fe}}^{2+}$ respectively, these both complexes become stable when oxidation of metal ion happens to ${\text{Cr}}^{3+}$ and ${\text{Fe}}^{3+}$.

Question:43

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Linkage isomerism arises in coordination compounds containing ambidentate ligand.
Reason: Ambidentate ligand has two different donor atoms.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (i) is correct answer.
Explanation: Two different donor atoms in ambidentate ligand gives rise to linkage isomerism.

Question:44

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Complexes of MX₆ and MX₅L type (X and L are unidentate) do not show geometrical isomerism.
Reason: Geometrical isomerism is not shown by complexes of coordination number 6.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (iii) is correct answer.
Explanation: A different arrangement of ligands is not possible for complexes of MX₆ and MX₅L type. The complexes MA₄B₂, M(AA)₂B₂ and MA₃B₃ with coordination number 6 show geometrical isomerism.

Question:45

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion:$[Fe(CN{)}_6{]}^{3-}$ ion shows magnetic moment corresponding to two unpaired electrons.
Reason: Because it has $d^{2}s{p}^3$ type hybridisation.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (iv) is correct answer.
Explanation: Corresponding to one unpaired ion, $[Fe(CN{)}_6{]}^{3-}$ shows magnetic moment.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Long Answer Type

Question:46

Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following :(i) $[Co{F}_6{]}^{3-},[Co(H_{2}O{)}_6{]}^{2+},[Co(CN{)}_6{]}^{3-}$
(ii) $[Fe{F}_6{]}^{3-},[Fe(H_{2}O{)}_6{]}^{2+},[Fe(CN{)}_6{]}^{4-}$
Answer:

(i)
$[Co{F}_6{]}^{3-}$ :
${\text{Co}}^{3+}={\text{3d}}^6$
Number of unpaired electrons =4
Magnetic moment
$=\sqrt{n(n+2)}=\sqrt{4(4+2)}=4.9B.M$
$[Co(H_{2}O{)}_6{]}^{2+}$

${\text{Co}}^{2+}={\text{3d}}^7$
Number of unpaired electrons =3
Magnetic moment =$\sqrt{3(3+2)}=3.87B.M$
$[Co(CN{)}_6{]}^{3-}$

${\text{Co}}^{3+}={\text{3d}}^6$
Number of unpaired electrons =0
Diamagnetic .
(ii)
$[Fe{F}_6{]}^{3-}$

$F{e}^{3+}=3d^5$
Number of unpaired electrons =5
Magnetic moment = $\sqrt{5(5+2)}=5.92B.M$
$[Fe(H_{2}O{)}_6{]}^{2+}$

$F{e}^{2+}=3d^6$
Number of unpaired electrons =4
Magnetic moment = $\sqrt{4(4+2)}=4.9B.M$
$[Fe(CN{)}_6{]}^{4-}:$

$F{e}^{2+}=3d^6$
Diamagnetic.

Question:47

Using valence bond theory, explain the following in relation to the complexes gives below:
${[Mn(CN{)}_6]}^{3-},{[Co(N{H}_3{)}_6]}^{3+},{[Cr(H_{2}O{)}_6]}^{3+},{\left[FeC{l}_6\right]}^{4-}$
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.

Answer:

${[Mn(CN{)}_6]}^{3-}:$
$M{n}^{3+}=3d^4$
(i) Hybridisation - $d^{2}s{p}^3$
(ii) Inner orbital complex
(iii) Paramagnetic
(iv) Magnetic moment $\sqrt{2(2+2)}=2.87B.M$
${[Co(N{H}_3{)}_6]}^{3+}:$
$C{o}^{3+}=3d^6$

(i) Hybridisation - $d^{2}s{p}^3$
(ii) Inner orbital complex
(iii) Diamagnetic
(iv) Magnetic moment=0
${[Cr(H_{2}O{)}_6]}^{3+}$
$C{r}^{3+}=3d^3$

(i) Hybridisation - $d^{2}s{p}^3$
(ii) Inner orbital complex
(iii) Paramagnetic
(iv) Magnetic moment=$\sqrt{3(3+2)}=3.87B.M$
${\left[FeC{l}_6\right]}^{4-}$
$F{e}^{2+}=3d^6$

(i) Hybridisation - $s{p}^3{d}^2$
(ii) Outer orbital complex
(iii) Paramagnetic
(iv) Magnetic moment=$\sqrt{4(4+2)}=4.9B.M$

Question:48

${\text{CoSO}}_4\text{Cl}.{\text{5NH}}_3$ exists in two isomeric forms ‘A’ and ‘B’. Isomer ‘A’ reacts with ${\text{AgNO}}_3$ to give white precipitate, but does not react with ${\text{BaCl}}_2$ . Isomer ‘B’ gives white precipitate with ${\text{BaCl}}_2$ but does not react with ${\text{AgNO}}_3$. Answer the following questions.

(i) Identify ‘A’ and ‘B’ and write their structural formulas.
(ii) Name the type of isomerism involved.
(iii) Give the IUPAC name of ‘A’ and ‘B’.
Answer: