Coordination Compounds Questions — JEE Chemistry

Q1

The oxidation state of Cr in [Cr(NH3)4Cl2]+ is :

A +3

B +2

C +1

D 0

Correct Answer

Option A

Solution

Oxidation state of

Cr

in

{\left[ {Cr{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + }.

Let it be

x,\,\,1 \times x + 4 \times 0 + 2 \times \left( { - 1} \right) = 1

Therefore

\,\,\,

x=3.

Q2

The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH3) (NH2OH)]+ is (py = pyridine) :

A 3

B 4

C 6

D 2

Correct Answer

Option A

Solution

Square planar complexes of type

M\left[ {ABCD} \right]\,

form three isomers.

Their position may be obtained by fixing the position of one ligand and placing at the trans position any one of the remaining three ligands one by one.

Q3

Which of the following compounds is not colored yellow?

A K3[Co(NO2)6]

B (NH4)3[As(Mo3O10)4]

C BaCrO4

D Zn2[Fe(CN)6]

Correct Answer

Option D

Solution

Zn2[Fe(CN)6] is white in color as it does not have unpaired electrons.

Q4

The theory that can completely/properly explain the nature of bonding in [Ni(Co)4] is :

A Wemer's theory

B Valence bond theory

C Crystal field theory

D Molecular orbital theory

Correct Answer

Option D

Solution

In complex [Ni(CO)4] decrease in Ni–C bond length and increase in C–O bond length as well as it's magnetic property is explained by MOT.

Q5

The species that has a spin-only magnetic moment of 5.9 BM, is : (Td = tetrahedral)

A [MnBr4]2– (Td)

B [NiCl4]2– (Td)

C Ni(CO)4 (Td)

D [Ni(CN)4]2– (square planar)

Correct Answer

Option A

Solution

$\therefore$ Unpaired electrons = 5 $\therefore$ $\mu$ =

\sqrt {5\left( {5 + 2} \right)}

= 5.9 BM

Q6

Consider that a d6 metal ion (M2+) forms a complex with aqua ligands, and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilization energy of the complex is

A tetrahedral and – 1.6

\Delta

t + 1P

B octahedral and –2.4

\Delta

0 + 2P

C tetrahedral and –0.6

\Delta

t

D octahedral and –1.6

\Delta

0

Correct Answer

Option C

Solution

Spin only magnetic moment = 4.9 =

\sqrt {n\left( {n + 2} \right)}

From this, n = 4 (unpaired electrons) (A) In octahedral complex: [M(H2O)6] 2+ C.F.S.E. = (–0.4

\Delta

0) × 4 + (+0.6

\Delta

0) × 2 + 0 × P = –0.4

\Delta

0 (B) In tetrahedral complex: [M(H2O)4] 2+ C.F.S.E. = (–0.6

\Delta

t) × 3 + (+0.4

\Delta

t) × 3 + 0 × P = –0.6

\Delta

t

Q7

The hybridization and magnetic nature of

{[Mn{(CN)_6}]^{4 - }}

and

{[Fe{(CN)_6}]^{3 - }}

, respectively are :

A sp3d2 and diamagnetic

B d2sp3 and paramagnetic

C sp3d2 and paramagnetic

D d2sp3 and diamagnetic

Correct Answer

Option B

Solution

{[Mn{(CN)_6}]^{4 - }}

Mn2+ = 3d5 4s0 CN– is a strong field ligand. $\therefore$ Pairing will occur.

{[Fe{(CN)_6}]^{3 - }}

Fe3+ = 3d54s0 CN– is a strong field ligand. $\therefore$ Pairing will occur.

Q8

Indicate the complex/complex ion which did not show any geometrical isomerism :

A [CoCl2(en)2]

B [Co(CN)5(NC)]3

-

C [Co(NH3)3(NO2)3]

D [Co(NH3)4Cl2]+

Correct Answer

Option B

Solution

(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

Q9

The denticity of an organic ligand, biuret is :

A 2

B 4

C 3

D 6

Correct Answer

Option A

Solution

Biuret ligand is It forms complexes like The denticity of organic ligand is 2.

Q10

Correct formula of the compound which gives a white precipitate with BaCl2 solution, but not with AgNO3 solution, is :

A [Co(NH3)5Br]SO4

B [Co(NH3)5SO4]Br

C [Pt(NH3)4Cl2]Br2

D [Pt(NH3)4Br2]Cl2

Correct Answer

Option A

Solution

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 $\mathrm{AgNO}_{3}$ .