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Coordination Compounds

JEE Chemistry · 225 questions · Page 5 of 23 · Click an option or "Show Solution" to reveal answer

Q41
Mn2(CO)10 is an organometallic compound due to the presence of -
A C–O bond
B Mn – Mn bond
C Mn – O bond
D Mn – C bond
Correct Answer
Option D
Solution

Compounds that contain at least one carbon-metal bond are called organometallic compounds.

Q42
The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 BM. The suitable ligand for this complex is -
A CN−-
B ethylenediamine
C NCS–
D CO
Correct Answer
Option C
Solution

Given, μ\mu = 5.9 BM ∴\therefore n = number of unpaired electron = 5 For Mn+2 with 5 unpaired electronic configuration only possible when ligand is weak field ligand.

Here weak field ligand is NCS−-.

Q43
Which of the following is correct order of ligand field strength?
A S2−<C2O42−<NH3<\mathrm{S}^{2-}<\mathrm{C}_{2} \mathrm{O}_{4}^{2-}<\mathrm{NH}_{3} < en <CO<\mathrm{CO}
B CO<en<NH3<C2O42<S2−\mathrm{CO}<\mathrm{en}<\mathrm{NH}_{3}<\mathrm{C}_{2} \mathrm{O}_{4}^{2}<\mathrm{S}^{2-}
C S2−<NH3<\mathrm{S}^{2-}<\mathrm{NH}_{3}< en <CO<C2O42<\mathrm{CO}<\mathrm{C}_{2} \mathrm{O}_{4}^{2}
D NH3<en<CO<S2−<C2O42\mathrm{NH}_{3}<\mathrm{en}<\mathrm{CO}<\mathrm{S}^{2-}<\mathrm{C}_{2} \mathrm{O}_{4}^{2}
Correct Answer
Option A
Solution

Ligand field strength $$\mathrm{{S^{2 - }}

Q44
A solution of FeCl3\mathrm{FeCl_3} when treated with K4[Fe(CN)6]\mathrm{K_4[Fe(CN)_6]} gives a prussium blue precipitate due to the formation of :
A Fe[Fe(CN)6]\mathrm{Fe[Fe(CN)_{6}]}
B Fe4[Fe(CN)6]3\mathrm{Fe_{4}[Fe(CN)_{6}]_{3}}
C Fe3[Fe(CN)6]2\mathrm{Fe_{3}[Fe(CN)_{6}]_{2}}
D K[Fe2(CN)6]\mathrm{K[Fe_{2}(CN)_{6}]}
Correct Answer
Option B
Solution

Fe3++[Fe(CN)6]4−⟶Fe4[Fe(CN)6]3\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)

Q45
The type isomerism present in nitropentammine chromium (III)\left( {{\rm I}{\rm I}{\rm I}} \right) chloride is :
A optical
B linkage
C ionization
D polymerisation.
Correct Answer
Option B
Solution

The chemical formula of nitropentammine chromium

(III)\left( {{\rm I}{\rm I}{\rm I}} \right)

chloride is

[Cr(NH3)5 NO2]Cl2\left[ {Cr{{\left( {N{H_3}} \right)}_5}\,N{O_2}} \right]C{l_2}

It can exist in following two structures

[Cr(NH3)5NO2]Cl2  \left[ {Cr{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]C{l_2}\,\,

and nitropentammine chromium

(III)\left( {{\rm I}{\rm I}{\rm I}} \right)

chloride

[Cr(NH3)5ONO]Cl2\left[ {Cr{{\left( {N{H_3}} \right)}_5}ONO} \right]C{l_2}

Nitropentammine chromium

(III)\left( {{\rm I}{\rm I}{\rm I}} \right)

chloride Therefore the type of isomerism found in this compound is linkage isomerism as nitro group is linked through

NN

as

−NO2- N{O_2}

or through

OO

as

−ONO.- ONO.
Q46
One mole of the complex compound Co(NH3)5Cl3, gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of AgNO3 solution to yield two moles of AgCl (s). The structure of the complex is :
A [Co(NH3)3Cl3]. 2NH3
B [Co(NH3)4Cl2] Cl. NH3
C [Co(NH3)4Cl] Cl2. NH3
D [Co(NH3)5Cl] Cl2
Correct Answer
Option D
Solution

The complex compound Co(NH3)5Cl3 dissociates in water to form three moles of ions.

This dissociation can be represented as:

Co(NH3)5Cl3  ⇋  [Co(NH3)5Cl]+2+2Cl−\text{Co(NH}_3\text{)}_5\text{Cl}_3\,\,\leftrightharpoons\,\,\left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]^{+2} + 2\text{Cl}^-

Therefore, the structure of the complex is:

[Co(NH3)5Cl]Cl2\left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2

When this complex reacts with two moles of AgNO3 solution, it produces:

[Co(NH3)5Cl]Cl2+2AgNO3→[Co(NH3)5Cl](NO3)2+2AgCl\left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 + 2\text{AgNO}_3 \to \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right](\text{NO}_3)_2 + 2\text{AgCl}

This reaction yields two moles of AgCl, confirming the complex structure as:

[Co(NH3)5Cl]Cl2\left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2
Q47
In the coordination compound, K4[Ni(CN)4], the oxidation state of nickel is :
A 0
B +1
C +2
D -1
Correct Answer
Option A
Solution

Let the

O.O.

No of

NiNi

in

  K4[Fe(CN)6]  \,\,{K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\,\,
bebe
=x=x

then

4(+1)+x+(−1)×4=04\left( { + 1} \right) + x + \left( { - 1} \right) \times 4 = 0
              ⇒4+x−4=0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow 4 + x - 4 = 0
              \,\,\,\,\,\,\,\,\,\,\,\,\,\,
x=0x=0
Q48
The coordination number of central metal atom in a complex is determined by :
A the number of ligands around a metal ion bonded by sigma bonds
B the number of only anionic ligands bonded to the metal ion
C the number of ligands around a metal ion bonded by sigma and pi- bonds both
D the number of ligands around a metal ion bonded by pi-bonds
Correct Answer
Option A
Solution

The coordination number of central metal atom in a complex is equal to number of monovalent ligands, twice the number of bidentate ligands and so on, around the metal ion bonded by coordinate bonds.

Hence coordination number == no. of σ\sigma bonds formed by metals with ligands

Q49
Among the properties (a) reducing (b) oxidising (c) complexing, the set of properties shown by CN– ion towards metal species is :
A c, a
B b, c
C a, b
D a, b, c
Correct Answer
Option A
Solution
CN−C{N^ - }

ion acts good complexing as well as reducing agent.

Q50
The correct order of magnetic moments (spin only values in B.M.) among is : (Atomic numbers: Mn = 25; Fe = 26, Co =27)
A [MnCl4]2- > [CoCl4]-2 > [Fe(CN)6]-4
B [Fe(CN)6]-4 > [CoCl4]2- > [MnCl4]2-
C [Fe(CN)6]4- > [MnCl4]2- > [CoCl4]2-
D [MnCl4]2- > [Fe(CN)6]4- > [CoCl4]2-
Correct Answer
Option A
Solution

NOTE : The greater the number of unpaired electrons, greater the magnitude of magnetic moment.

Hence the correct order will be

[MnCl4]−−>[CoCl4]−−>[Fe(CN)6]4−{\left[ {MnC{l_4}} \right]^{ - - }} > {\left[ {CoC{l_4}} \right]^{ - - }} > {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}

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