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Coordination Compounds

JEE Chemistry · 225 questions · Page 13 of 23 · Click an option or "Show Solution" to reveal answer

Q121
The potassium ferrocyanide solution gives a Prussian blue colour, when added to :
A CoCl3
B FeCl2
C CoCl2
D FeCl3
Correct Answer
Option D
Solution

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

Q122
Given below are two statements. ∙\bullet Statement I : In CuSO4 . 5H2O, Cu-O bonds are present. ∙\bullet Statement II : In CuSO4 . 5H2O, ligands coordinating with Cu(II) ion are O-and S-based ligands. In the light of the above statements, choose the correct answer from the options given below.
A Both Statement I and Statement II are correct.
B Both Statement I and Statement II are incorrect.
C Statement I is correct but Statement II is incorrect.
D Statement I is incorrect but Statement II is correct.
Correct Answer
Option C
Solution

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

Q123
Given below are two statements : Statement I : [Ni(CN)4]2−- is square planar and diamagnetic complex, with dsp2 hybridization for Ni but [Ni(CO)4] is tetrahedral, paramagnetic and with sp3-hybridication for Ni. Statement II : [NiCl4]2−- and [Ni(CO)4] both have same d-electron configuration have same geometry and are paramagnetic. In light the above statements, choose the correct answer from the options given below :
A Both Statement I and Statement II are true.
B Both Statement I and Statement II are false.
C Statement I is correct but Statement II is false.
D Statement I is false but Statement II is correct.
Correct Answer
Option B
Solution

[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–).

Q124
To inhibit the growth of tumours, identify the compounds used from the following : A. EDTA B. Coordination Compounds of Pt C. D - Penicillamine D. Cis - Platin Choose the correct answer from the option given below :
A A and B Only
B C and D Only
C B and D Only
D A and C Only
Correct Answer
Option C
Solution

Cis-platin is [Pt(NH

3_3

)

2_2

Cl

2_2

]; cis platin and other complexes of pt are used to inhibit the growth of tumours.

Q125
Transition metal complex with highest value of crystal field splitting (Δ\Delta0) will be :
A [Cr(H2O)6]3+
B [Mo(H2O)6]3+
C [Fe(H2O)6]3+
D [Os(H2O)6]3+
Correct Answer
Option D
Solution

Crystal field splitting

(Δ0)\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

(Δ0)\left(\Delta_{0}\right)

increases as we move from

3d→4d→5d3 d \rightarrow 4 d \rightarrow 5 d
Cr3+\mathrm{Cr}^{3+}

and

Fe3+\mathrm{Fe}^{3+}

belong to

3d3 d

series,

Mo3+\mathrm{Mo}^{3+}

belongs to

4d4 d

series and

Os3+\mathrm{Os}^{3+}

belongs to

5d5 d

series. Therefore crystal field splitting

(Δ0)\left(\Delta_{0}\right)

is highest for

[Os(H2O)6]3+\left[\mathrm{Os}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}

.

Q126
The correct order of energy of absorption for the following metal complexes is : A : [Ni(en)3]2+ , B : [Ni(NH3)6]2+ , C : [Ni(H2O)6]2+
A C < B < A
B B < C < A
C C < A < B
D A < C < B
Correct Answer
Option A
Solution

Stronger is ligand attached to metal ion, greater will be the splitting between

t2g\mathrm{t}_{2} \mathrm{g}

and ege_g (hence greater will be

ΔU)\Delta \mathrm{U})

∴\therefore Greater will be absorption of energy. Hence correct order

[Ni(en)3]2+>[Ni(NH3)6]2+>[Ni(H2O)6]2+\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+}
Q127
The metal complex that is diamagnetic is (Atomic number: Fe,26;Cu,29)\mathrm{Fe}, 26 ; \mathrm{Cu}, 29)
A K3[Cu(CN)4]\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]
B K2[Cu(CN)4]\mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]
C K3[Fe(CN)4]\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{4}\right]
D K4[FeCl6]\mathrm{K}_{4}\left[\mathrm{FeCl}_{6}\right]
Correct Answer
Option A
Solution
⇒K3[Cu(CN)4]\Rightarrow \mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]

is diamagnetic

Cu(I)⇒d10\mathrm{Cu}(\mathrm{I}) \Rightarrow \mathrm{d}^{10}

configuration ⇒\Rightarrow No unpaired electrons.

⇒K2[Cu(CN)4],K3[Fe(CN)4]\Rightarrow \mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right], \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{4}\right]

and

K4[FeCl6]\mathrm{K}_{4}\left[\mathrm{FeCl}_{6}\right]

are paramagnetic in nature

Q128
Low oxidation state of metals in their complexes are common when ligands :
A have good π\pi-accepting character
B have good σ\sigma-donor character
C are having good π\pi-donating ability
D are having poor σ\sigma-donating ability
Correct Answer
Option A
Solution

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

Q129
Fe3+\mathrm{Fe}^{3+} cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of :
A [Fe(H2O)6]2[Fe(CN)6]\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]
B Fe2[Fe(CN)6]2\mathrm{Fe}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{2}
C Fe3[Fe(OH)2(CN)4]2\mathrm{Fe}_{3}\left[\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{CN})_{4}\right]_{2}
D Fe4[Fe(CN)6]3\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}
Correct Answer
Option D
Solution

4Fe3++3[Fe(CN)6]−4⟶Fe4[Fe(CN)6]3Prussian Blue4 \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}

Q130
Octahedral complexes of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper (II) complexes will show the maximum structural distortion? (en - ethylenediamine; H2 N−CH2−CH2−NH2\mathrm{H}_{2} \mathrm{~N}_{-} \mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2})
A [Cu(H2O)6]SO4\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{SO}_{4}
B [Cu(en)(H2O)4]SO4\left[\mathrm{Cu}(\mathrm{en})\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right] \mathrm{SO}_{4}
C cis-[Cu(en)2Cl2]\left[\mathrm{Cu}\left(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]\right.
D trans-[Cu(en)2Cl2]\left[\mathrm{Cu}\left(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]\right.
Correct Answer
Option A
Solution

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

d9d^{9}

configuration, the last electron may occupy either

d2d^{2}

or

dx2y2d_{x^{2} y^{2}}

orbitals of

ege_{g}

set. If it occupies

dz2\mathrm{dz}^{2}

orbital most of the electron density will be concentrated between the metal and the two ligands on the

z\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

d4\mathrm{d}^{4}

, low spin

d7\mathrm{d}^{7}

and

d9\mathrm{d}^{9}

configuration.

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