Coordination Compounds — Page 9 | JEE Chemistry

Q81

The compound used in the treatment of lead poisoning is :

A desferrioxime B

B Cis-platin

C D-penicillamine

D EDTA

Correct Answer

Option D

Solution

EDTA is used in treatment of lead poisoning.

Q82

The compound that inhibits the growth of tumors is :

A cis-[Pd(Cl)2(NH3)2]

B trans-[Pd(Cl)2(NH3)2]

C cis-[Pt(Cl)2(NH3)2]

D trans-[Pt(Cl)2(NH3)2]

Correct Answer

Option C

Solution

Cis-platin or cis-[Pt(Cl)2(NH3)2] is used as an anti-cancer drug.

Q83

The Crystal Field Stabilization Energy (CFSE) of [CoF3(H2O)3] (

\Delta

0 < P) is :

A -0.8

\Delta

0

B -0.4

\Delta

0

C -0.8

\Delta

0 + 2P

D -0.4

\Delta

0 + P

Correct Answer

Option B

Solution

As

\Delta

0 < P, so all ligands behaves as weak field ligands. For octahedral Crystal field stabilization energy (CFSE) = (-0.4

\Delta

0) $\times$ nt2g + (+0.6

\Delta

0) $\times$ neg + np nt2g = number of electrons in t2g orbital neg = number of electrons in eg orbital n = number of extra pairs p = Pairing energy Here nt2g = 1 neg = 0 n = 0 ( For weak field ligands n always zero.

Here H2O is weak field ligand) $\therefore$ Crystal field stabilization energy (CFSE) = (-0.4

\Delta

0) $\times$ 4 + (+0.6

\Delta

0) $\times$ 2 + 0 $\times$ P = -1.6

\Delta

0 + 1.2

\Delta

0 = -0.4

\Delta

0

Q84

The number of isomers possible for [Pt(en)(NO2)2] is :

A 3

B 1

C 4

D 2

Correct Answer

Option A

Solution

Total 3 geometrical isomers are possible.

Q85

The d-electron configuration of [Ru(en)3 ]Cl2 and [Fe(H2O)6]Cl2 , respectively are :

A

t_{2g}^4e_g^2

and

t_{2g}^6e_g^0

B

t_{2g}^6e_g^0

and

t_{2g}^6e_g^0

C

t_{2g}^6e_g^0

and

t_{2g}^4e_g^2

D

t_{2g}^4e_g^2

and

t_{2g}^4e_g^2

Correct Answer

Option C

Solution

[Ru(en)3 ]Cl2 : Here CN = 6 so octahedral splitting happens. 'en' is strong field ligand so

\Delta

0 > P(pairing energy). That is why pairing of electrons happens.

\Delta

0 = Energy gap between eg and t2g orbital.

[Fe(H2O)6]Cl2 : Here CN = 6 so octahedral splitting happens.

H2O is weak field ligand so

\Delta

0 < P(pairing energy). That is why no pairing of electrons happens.

\Delta

0 = Energy gap between eg and t2g orbital.

Q86

The electronic spectrum of [Ti(H2O)6]3+ shows a single broad peak with a maximum at 20,300 cm-1 . The crystal field stabilization energy (CFSE) of the complex ion, in kJ mol-1, is :

A 83.7

B 242.5

C 145.5

D 97

Correct Answer

Option D

Solution

[Ti(H2O)6]3+ Ti = [Ar]3d24s2 Ti+3 = [Ar]3d1 For octahedral Crystal field stabilization energy (CFSE) = (-0.4

\Delta

0) $\times$ nt2g + (+0.6

\Delta

0) $\times$ neg + np nt2g = number of electrons in t2g orbital neg = number of electrons in eg orbital n = number of extra pairs p = Pairing energy Here nt2g = 1 neg = 0 n = 0 ( For weak field ligands n always zero.

Here H2O is weak field ligand) $\therefore$ Crystal field stabilization energy (CFSE) = (-0.4

\Delta

0) $\times$ 1 + (+0.6

\Delta

0) $\times$ 0 + 0 $\times$ P = -0.4

\Delta

0 = –0.4 × 20300 = –8120 cm–1 CFSE (in kJ) =

{{8120} \over {83.7}}

= 97 kJ/mol

Q87

The complex that can show optical activity is :

A cis-[Fe(NH3)2(CN)4]–

B trans-[Cr(Cl2)(ox)2]3–

C trans-[Fe(NH3)2(CN)4]–

D cis-[CrCl2(ox)2]3– (ox = oxalate)

Correct Answer

Option D

Solution

It does not have symmetry, so, optically active.

Q88

The one that is not expected to show isomerism is :

A [Pt(NH3)2Cl2]

B [Ni(NH3)4(H2O)2]2+

C [Ni(en)3]2+

D [Ni(NH3)2Cl2]

Correct Answer

Option D

Solution

[Pt(NH3)2Cl2] is dsp2 hybridisation and shows geometrical isomerism.

[Ni(NH3)4(H2O)2]2+ is Octahedral, show geometrical isomerism.

[Ni(en)3]2+ is Octahedral and shows optical isomerism.

[Ni(NH3)2Cl2] is sp3 hybridisation and tetrahedral complex, therefore does not show geometrical and optical isomerism and structural isomerism.

Q89

The one that can exhibit highest paramagnetic behaviour among the following is : gly = glycinato; bpy = 2, 2'-bipyridine

A [Fe(en)(bpy)(NH3)2]2+

B [Pd(gly)2]

C [Co(OX)2(OH)2]– (

\Delta

0 > P)

D [Ti(NH3)6]3+

Correct Answer

Option C

Solution

[Co(OX)2(OH)2]– (As here

\Delta

0 > P so those ligands act as strong field ligand and pairing happens.) 27Co+5 = [18Ar] 3d4 4s0(

t_{2g}^{2,1,1},e_g^{0,0}

) It has highest number of unpaired e– s. so it is most paramagnetic.

Q90

The molecule in which hybrid MOs involve only one d-orbital of the central atom is :

A XeF4

B [Ni(CN)4]2–

C [CrF6]3–

D BrF5

Correct Answer

Option B

Solution

XeF4 = sp3d2 [Ni(CN)4]2– = dsp2 [CrF6]3– = d2sp2 BrF5 = sp3d2