Coordination Compounds — Page 4 | JEE Chemistry

Q31

The magnetic moment (spin only) of [NiCl4]2− is

A 5.46 BM

B 2.82 BM

C 1.41 BM

D 1.82 BM

Correct Answer

Option B

Solution

{\left[ {NiC{l_4}} \right]^{2 - }}\,\left( {{d^8}} \right)

i.e. the number of unpaired electrons in

{\left[ {NiCl{}_4} \right]^{2 - }}\,\,

is

2.

n = \sqrt {n\left( {n + 2} \right)} = \sqrt {2\left( 4 \right)}

= 2\sqrt 2 = 2 \times 1.41

= 2.82\,BM

Q32

Which among the following will be named as dibromidobis (ethylene diamine) chromium(III) bromide?

A [Cr (en)3 ] Br3

B [Cr (en)2 Br2 ] Br

C [Cr (en) Br4 ]-

D [Cr (en) Br2 ] Br

Correct Answer

Option B

Solution

\left[ {Cr{{\left( {en} \right)}_2}B{r_2}} \right]Br

dibromidobis (ethylenediamine) chromium

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

Bromide.

Q33

Which of the following complex species is not expected to exhibit optical isomerism?

A [Co (en)2 Cl2]-

B [Co (NH3)3 Cl3]

C [Co (en) (NH3)2 Cl2]+

D [Co (en)3]3+

Correct Answer

Option B

Solution

Octahedral coordination entities of the type

M{a_3}{b_3}

exhibit Geometrical isomerism. The compound exists both as facial and meridional isomers.

Q34

The difference in the number of unpaired electrons of a metal ion in its high spin and low-spin octahedral complexes is two. The metal ion is :

A Mn2+

B Ni2+

C Co2+

D Fe2+

Correct Answer

Option C

Solution

$\therefore$ The difference in the number of unpaired electrons = 3 - 1 = 2

Q35

The complex that has highest crystal field splitting energy (

\Delta

), is :

A [Co(NH3)5(H2O)]Cl3

B K2[CoCl4]

C [Co(NH3)5Cl]Cl2

D K3[Co(CN)6]

Correct Answer

Option D

Solution

As complex K3[Co(CN)6] have CN $-$ ligand which is strongfield ligand amongst the given ligands in other complexes.

Q36

Homoleptic octahedral complexes of a metal ion 'M3+' with three monodenate ligands L1, L2 and L3 adsorb wavelenths in the region of green, blue and red respectively. The increasing order of the ligands strength is :

A L3 < L1 < L2

B L3 < L2 < L1

C L1 < L2 < L3

D L2 < L1 < L3

Correct Answer

Option A

Solution

Stronger the ligand, absorption of light having lower wavelength is more. Order of $\lambda$ : Red

>

Green

>

Blue Hence, ligand strength is L3

<

L1

<

L2

Q37

The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO2)] is

A 16

B 8

C 12

D 4

Correct Answer

Option C

Solution

So, The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO2 )] is 12

Q38

The coordination number of Th in K4[Th(C2O4)4(OH2)2] is : (C2O

{_4^{2 - }}

= Oxalato)

A 14

B 10

C 8

D 6

Correct Answer

Option B

Solution

Oxalato (C2O42–) is a bidentate and H2O is unidentate ligand.

4C2O42– creates 8 covalent bonds.

2H2O creates 2 covalent bonds.

$\therefore$ Around Th 10 coordinate covalent bonds will be present.

Q39

The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2, is -

A V2+ and Fe2+

B V2+ and Co2+

C Co2+ and Fe2+

D Cr2+ and Mn2+

Correct Answer

Option B

Solution

Magnetic moment =

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

BM $\therefore$

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

= 3.9 $\Rightarrow$ n = 3 $\therefore$ no. of unpaired electron = 3 H2O is weak field ligand so no pairing of electrons happens.

Fe2+ = t2g4eg2 Co2+ = t2g5eg2 V2+ = t2g3eg0 $\therefore$ M is V, Co.

Q40

The metal d-orbitals that are directly facing the ligands in K3[Co(CN)6] are -

A dx2

-

y2 and dz2

B dxy, dxz and dyz

C dxz, dyz and dz2

D dxy and dx2

-

y2

Correct Answer

Option A

Solution

Due to presence of strong field ligand (CN–) pairing occurs in which two d-orbitals i.e., dx2–y2 and dz2 directly face the CN- ligand.