Coordination Compounds — Page 10 | JEE Chemistry

Q91

The values of the crystal field stabilization energies for a high spin d6 metal ion in octahedral and tetrahedral fields, respectively, are :

A –0.4

\Delta

0 and –0.27

\Delta

t

B –1.6

\Delta

0 and –0.4

\Delta

t

C –0.4

\Delta

0 and –0.6

\Delta

t

D –2.4

\Delta

0 and –0.27

\Delta

t

Correct Answer

Option C

Solution

In octahedral CFSE = [-0.4nt2g + 0.6neg]

\Delta

0 = [-0.4 $\times$ 4 + 0.6 $\times$ 2]

\Delta

0 = -0.4

\Delta

0 In tetrahedral fields CFSE = [-0.6neg + 0.4nt2g]

\Delta

t = (– 0.6 × 3 + 0.4 × 3)

\Delta

t = (–1.8 + 1.2)

\Delta

t = –0.6

\Delta

t

Q92

Consider the complex ions, trans-[Co(en)2Cl2]+ (A) and cis-[Co(en)2Cl2]+ (B). The correct statement regarding them is :

A both (A) and (B) can be optically active.

B both (A) and (B) can not be optically active.

C (A) can not be optically active, but (B) can be optically active.

D (A) can be optically active, but (B) can not be optically active.

Correct Answer

Option C

Solution

trans-[Co(en)2Cl2]+ contain a plane of symmetry.

It cannot be optically active. cis-[Co(en)2Cl2]+ does not contain any plane of symmetry.

Hence the compound (B) is optically active.

Q93

For a d4 metal ion in an octahedral field, the correct electronic configuration is :

A

t_{2g}^4e_g^0

when

\Delta

0 < P

B

t_{2g}^3e_g^1

when

\Delta

0 > P

C

e_g^2t_{2g}^2

when

\Delta

0 < P

D

t_{2g}^3e_g^1

when

\Delta

0 < P

Correct Answer

Option D

Solution

back pairing is not possible because pairing energy >

\Delta

0.

Q94

Complex A has a composition of H12O6Cl3Cr. If the complex on treatment with conc.H2SO4 loses 13.5% of its original mass, the correct molecular formula of A is : [Given: atomic mass of Cr = 52 amu and Cl = 35 amu]

A [Cr(H2O)5Cl]Cl2.H2O

B [Cr(H2O)4Cl2]Cl.2H2O

C [Cr(H2O)3Cl3].3H2O

D [Cr(H2O)6]Cl3

Correct Answer

Option B

Solution

Let x molecule of water are lost then 13.5 =

\left[ {{{x \times 18} \over {6 \times 18 + 3 \times 35 + 52}}} \right] \times 100

$\Rightarrow$ x = 1.99

\simeq

2 $\therefore$ Around two moles of water are lost during heating.

$\therefore$ Formula of complex could be [Cr(H2O)4Cl2]Cl.2H2O

Q95

The number of possible optical isomers for the complexes MA2B2 with sp3 and dsp2 hydridized metal atom. respectively, is : Note : A and B are unidentate netural and unidentate monoanionic ligands, respectively.

A 0 and 2

B 0 and 0

C 0 and 1

D 2 and 2

Correct Answer

Option B

Solution

(a) If the complex MA2B2 is sp3 hybridised then the shape of this complex is tetrahedral this structure is opticaly inactive due to the presence of plane of symmetry. (b) If the complex MA2B2 is dsp2 hybridised then the shape of this complex is square planar.

Both isomers are optically inactive due to the presence of plane of symmetry.

$\therefore$ Total number of optical isomer is zero in both the cases.

Q96

Among the statements(a)-(d) the incorrect ones are : (a) Octahedral CO(III) complexes with strong fields ligands have very high magnetic moments. (b) When

\Delta

0 < P, the d-electron configuration of Co(III) in an octahedral complex is

t_{eg}^4e_g^2

(c) Wavelength of light absorbed by [Co(en)3]3+ is lower than that of [CoF6]3- (d) If the

\Delta

0 for an octahedral complex of CO(III) is 18,000 cm-1, the

\Delta

t for its tetrahedral complex with the same ligand be 16,000 cm-1

A (a) and (b) only

B (b) and (c) only

C (c) and (d) only

D (a) and (d) only

Correct Answer

Option D

Solution

(a) Co3+ with strong field complex forms low magnetic moment complex. (b) If

\Delta

0 < P configuration of Co3+ will be

t_{eg}^4e_g^2

. (c) Splitting power of ethylenediamine (en) is greater than fluoride (F–) ligand therefore more energy absorbed by [Co(en)3]3+ as compared to [CoF6]3–.

So wave length of light absorbed by [Co(en)3]3+ is lower than that of [CoF6]3– (d)

{\Delta _t} = {4 \over 9}{\Delta _0}

=

{4 \over 9} \times 18000

= 8000 cm-1 $\therefore$ Statement (a) and (d) are incorrect.

Q97

Among (a) – (d) the complexes that can display geometrical isomerism are : (a) [Pt(NH3)3Cl]+ (b) [Pt(NH3)Cl5]– (c) [Pt(NH3)2Cl(NO2)] (d) [Pt(NH3)4ClBr]2+

A (a) and (b)

B (c) and (d)

C (d) and (a)

D (b) and (c)

Correct Answer

Option B

Solution

[Pt(NH3)2Cl(NO2)] and [Pt(NH3)4ClBr]2+ can display geometrical isomerism.

Q98

The correct order of the calculated spin-only magnetic moments of complexs (A) to (D) is: (A) Ni(CO)4 (B) [Ni(H2O)6]Cl2 (C) Na2[Ni(CN)4] (D) PdCl2(PPh3)2

A (C) < (D) < (B) < (A)

B (C)

\approx

(D) < (B) < (A)

C (A)

\approx

(C)

\approx

(D) < (B)

D (A)

\approx

(C) < (B)

\approx

(D)

Correct Answer

Option C

Solution

(A) Ni(CO)4 Ni = 3d84s2 CO is strong field ligand.

So pairing of elections happens.

$\therefore$ Number of unpaired electrons = 0 $\therefore$ $\mu$ spin = 0 (B) [Ni(H2O)6]Cl2 Ni+2 = 3d84s0 H2O is weak field ligand.

So no pairing of electrons happens.

Number of unpaired electron = 2 $\therefore$ $\mu$ spin =

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

=

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

=

\sqrt 8

B.M (C) Na2[Ni(CN)4] Ni+2 = 3d84s0 CN- is strong field ligand.

So pairing of electrons happens.

Number of unpaired electron = 0 $\therefore$ $\mu$ spin = 0 (D) PdCl2(PPh3)2 Pd2+ = 4d8 This is dsp2 complex.

And shape is square planar. $\therefore$ $\mu$ spin = 0

Q99

[Pd(F)(Cl)(Br)(I)]2– has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilisation energy [CFSE] of [Fe(CN)6]n–6, respectively, are: [Note : Ignore the pairing energy]

A 1.73 BM and –2.0

\Delta

0

B 5.92 BM and 0

C 2.84 BM and –1.6

\Delta

0

D 0 BM and –2.4

\Delta

0

Correct Answer

Option A

Solution

Complex [Pd(F)(Cl)(Br)(I)]2– (square planar geometry)-has 3 geometrical isomers. $\therefore$ n = 3 [Fe(CN)6]n–6 becomes [Fe(CN)6]3- Here Oxidation state of Fe = +3 $\therefore$ Fe+3 = [Ar]3d54s0 CN- is strong field ligand so it pairing of electrons happens.

E.C. according to CFT = (

t_{2g}^5e{g^0}

) $\therefore$ No. of unpaired e- = 1 Then Magnetic moment =

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

=

\sqrt 3

= 1.73 B.M CFSE = [(–0.4 × 5) + (0.6 × 0)]

\Delta

0 = –2.0

\Delta

0

Q100

Complex X of composition Cr(H2O)6Cln has a spin only magnetic moment of 3.83 BM. It reacts with AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of X is :

A Hexaaqua chromium (III) chloride

B Tetraaquadichlorido chromium(IV) chloride dihydrate

C Tetraaquadichlorido chromium (III) chloride dihydrate

D Dichloridotetraaqua chromium (IV) chloride dihydrate

Correct Answer

Option C

Solution

Given complex Cr(H2O)6Cln As the magnetic mement is 3.83 BM then Spin only magnetic moment =

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

BM = 3.83 $\Rightarrow$ n = 3 Chromium in +3 oxidation state so molecular formula is Cr(H2O)6Cln.

$\therefore$ This formula have following isomers (a) [Cr(H2O)6]Cl3 : react with AgNO3 but does not show geometrical isomerism. (b) [Cr(H2O)5Cl]Cl2.

H2O react with AgNO3 but does not show geometrical isomerism. (c) [Cr(H2O)4Cl2]Cl.2H2O react with AgNO3 & show geometrical isomerism. (d) [Cr(H2O)3Cl3].3H2O does not react with AgNO3 & show geometrical isomerism.

Compound will be [Cr(H2O)4Cl2] Cl.2H2O IUPAC NAME : Tetraaquadichlorido chromium(III) chloride dihydrate