Coordination Compounds — Page 11 | JEE Chemistry

Q101

The correct order of the spin-only magnetic moments of the following complexes is : (I) [Cr(H2O)6]Br2 (II) Na4[Fe(CN)6] (III) Na3[Fe(C2O4)3] (

\Delta

0

>

P) (IV) (Et4N)2[CoCl4]

A (III) > (I) > (II) > (IV)

B (II)

\approx

(I) > (IV) > (III)

C (III) > (I) > (IV) > (II)

D (I) > (IV) > (III) > (II)

Correct Answer

Option D

Solution

(I) [Cr(H2O)6]Br2 H2O is weak field ligand so it can not pair up all the electrons. Cr2+ : [Ar] 4s03d4 (

t_{2g}^3e{g^1}

) Unpaired e– = 4 Magnetic moment =

\sqrt {24}

= 4.89 BM (II) Na4[Fe(CN)6] CN- is strong field ligand so it pair up all the electrons. Fe2+ = [Ar] 4s03d6 (

t_{2g}^6e{g^0}

) Unpaired e– = 0 Magnetic moment = 0 BM (III) Na3[Fe(C2O4)3] As

\Delta

0

>

P, so pairing of electrons happens. Fe3+ = [Ar] 4s03d5 (

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

) Unpaired e– = 1 Magnetic moment =

\sqrt 3

= 1.73 BM (IV) (Et4N)2[CoCl4] Cl-is weak field ligand so it can not pair up all the electrons. Co2+ = [Ar] 4s03d7 (

e{g^4}t_{2g}^3

) Unpaired e– = 3 Magnetic moment =

\sqrt {15}

= 3.87 BM Hence order of magnetic moment is I > IV > III > II

Q102

The isomer(s) of [Co(NH3)4Cl2] that has/have a Cl–Co–Cl angle of 90°, is/are :

A cis only

B cis and trans

C meridional and trans

D trans only

Correct Answer

Option A

Solution

[Co(NH3)4Cl2] has 2 geometrical isomers. Among cis and trans isomers, cis isomer has Cl–Co–Cl angle of 90o.

Q103

For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following statements: (I) both the complexes can be high spin. (II) Ni(II) complex can very rarely be low spin. (III) with strong field ligands, Mn(II) complexes can be low spin. (IV)aqueous solution of Mn(II) ions is yellow in colour. The correct statements are :

A (I), (III) and (IV) only

B (I) and (II) only

C (II), (III) and (IV) only

D (I), (II) and (III) only

Correct Answer

Option D

Solution

(I) Under weak field ligand, octahedral Mn(II) and tetrahedral Ni(II) both the complexes are high spin complex.

(II) Tetrahedral Ni(II) complex can very rarely be low spin because square planar (under strong ligand) complexes of Ni(II) are low spin complexes.

(III)With strong field ligands Mn (II) complexes can be low spin because they have less number of unpaired electron (unpaired electron = 1) While with weak field ligands Mn(II) complexes can be high spin because they have more number of unpaired electron (unpaired electron = 5) (IV) Aqueous solution of Mn(II) ions is pink in colour.

Q104

The calculated magnetic moments (spin only value) for species

{[FeC{l_4}]^{2 - }}

,

{[Co{({C_2}{O_4})_3}]^{3 - }}

and

MnO_4^{2 - }

respectively are :

A 5.92, 4.90 and 0 BM

B 4.90, 0 and 1.73 BM

C 5.82, 0 and 0 BM

D 4.90, 0 and 2.83 BM

Correct Answer

Option B

Solution

(i)

{[FeC{l_4}]^{2 - }}\overset{{}}\longrightarrow F{e^{2 + }}\overset{{}}\longrightarrow [Ar]3{d^6}

$\therefore$ Cl is weak field ligand so does not pairing occur So, magnetic moment

(\mu ) = \sqrt {n(n + 2)} BM

= \sqrt {4(4 + 2)} BM

(n = Number of total unpaired e $-$ = 4)

= \sqrt {24} BM = 4.90\,BM

(ii)

{[Co{({C_2}{O_4})_3}]^{3 - }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} C{o^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} [Ar]3{d^6}

C2O4 is strong field ligand so pairing occur. All electrons are paired, n = 0 hence, $\mu$ = 0 (iii)

MnO_4^{2 - }\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} M{n^{ + 6}}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} [Ar]3{d^1}

n = 1

= \sqrt {n(n + 2)} BM

= \sqrt {1(1 + 2)} BM

= \sqrt 3 BM

= 1.73\,BM

Q105

In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment? (i) [FeF6]3

-

(ii) [Co(NH3)6]3+ (iii) [NiCl4]2

-

(iv) [Cu(NH3)4]2+

A (iii) > (iv) > (ii) > (i)

B (ii) > (iii) > (i) > (iv)

C (i) > (iii) > (iv) > (ii)

D (ii) > (i) > (iii) > (iv)

Correct Answer

Option C

Solution

Spin only magnetic moment,

\mu = \sqrt {n(n + 2)} BM

where, n = number of unpaired electrons and $\mu$ $\propto$ n. (i)

{[Fe{F_6}]^{3 - }} \Rightarrow F{e^{3 + }} = (3{d^5}) , {F^ - }

(weak field ligand). Thus, pairing of electron does not take place. (ii)

{[Co{(N{H_3})_6}]^{3 + }} \Rightarrow C{o^{3 + }} = (3{d^6}), N{H_3}

(strong field ligand). Thus, pairing of electron takes place. (iii)

{[NiC{l_4}]^{2 - }} \Rightarrow N{i^{2 + }} = (3{d^8}), C{l^ - }

(weak field ligand). Thus, pairing of electron takes place. (iv)

{[Cu{(N{H_3})_4}]^{2 + }} \Rightarrow C{u^{2 + }}(3{d^9}), N{H_3}

(strong field ligand). Thus, pairing of electron takes place. So, the decreasing order of $\mu$ is

\mathop {{\mu _i}}\limits_{(n = 5)} > \mathop {{\mu _{iii}}}\limits_{(n = 2)} > \mathop {{\mu _{iv}}}\limits_{(n = 1)} > \mathop {{\mu _{ii}}}\limits_{(n = 0)}

Q106

Match with :

List - I		List - II
(a)	$[Co{(N{H_3})_6}][Cr{(CN)_6}]$	(i)	Linkage isomerism
(b)	$[Co{(N{H_3})_3}{(N{O_2})_3}]$	(ii)	Solvate isomerism
(c)	$[Cr{({H_2}O)_6}C{l_3}$	(iii)	Co-ordination isomerism
(d)	$cis - {[CrC{l_2}{(ox)_2}]^{3 - }}$	(iv)	Optical isomerism

A (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

B (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)

C (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)

D (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)

Correct Answer

Option B

Solution

(a) [Co(NH3)6][Cr(CN)6] $\to$ (iii) Coordination isomerism (b) [Co(NH3)3(NO2)3] $\to$ (i) Linkage isomerism (c) [Cr(H2O)6]Cl3 $\to$ (ii) Solvate isomerism (d) cis-[CrCl2(ox)2]3– $\to$ (iv) Optical isomerism

Q107

The secondary valency and the number of hydrogen bonded water molecule(s) in CuSO4 . 5H2O, respectively, are :

A 5 and 1

B 4 and 1

C 6 and 5

D 6 and 4

Correct Answer

Option B

Solution

CuSO4.5H2O $\Rightarrow$ [Cu(H2O)4]SO4.

H2O In CuSO4.5H2O, Cu is co-ordinated with 4 water molecule and two more oxygen atom from sulphate ion and fifth water molecule is hydrogen bonded.

So secondary valency = 4 No. of hydrogen bond per molecule = 1

Q108

Which of the following will have maximum stabilization due to crystal field?

A [Ti(H2O)6]3+

B [Co(H2O)6]2+

C [Co(CN)6]3

-

D [Cu(NH3)4]2+

Correct Answer

Option C

Solution

The given complexes are: $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{-3},\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ $\mathrm{CN}^{-}$ is the strongest ligand among the given complexes CFSE value for the $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{-3}$ complex will be highest as it has $d^{6}$ configuration with a CFSE value of $-2.40 \,\Delta_{0}+2 \mathrm{P}$ , where $P$ represents pairing energy and $\Delta_{0}$ represents splitting energy in octahedral field.

The value of $\Delta_{0}$ is high for cyanide complexes.

Q109

According to the valence bond theory the hybridization of central metal atom is dsp2 for which one of the following compounds?

A

NiC{l_2}.6{H_2}O

B

{K_2}[Ni{(CN)_4}]

C

[Ni{(CO)_4}]

D

N{a_2}[NiC{l_4}]

Correct Answer

Option B

Solution

According to VBT i.e. valence bond theory, Electronic configuration of Ni = [Ar]3d84s2. (a) NiCl2 .

6H2O NiCl2 .

6H2O $\rightarrow$ NiCl2 + 6H2O Oxidation number of Ni(x) = x + 2( $-$ 1) = 0; where x, $-$ 1 and 2 are the oxidation number of Ni, oxidation number of Cl and number of Cl atoms respectively.

NiCl2 $\Rightarrow$ x + ( $-$ 2) = 0 $\Rightarrow$ x = 2 Electronic configuration of Ni2+ = [Ar]3d84s0 Cl $-$ is a weak field ligand.

So, no pairing of electrons occurs.

For C.N. = 6 (b) K2[Ni(CN)4] K2[Ni(CN)4] $\rightarrow$ 2K+ + [Ni(CN)4]2 $-$ x + 4( $-$ 1) $-$ ( $-$ 2) = 0; where x, 4, $-$ 1 and $-$ 2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[Ni(CN)4]2 $-$ $\Rightarrow$ x $-$ 4 + 2 = 0 $\Rightarrow$ x = +2 Electronic configuration of Ni2+ $\Rightarrow$ [Ar]3d84s0 CN $-$ is a strong field ligand.

So, pairing of electrons occur.

For C.N. = 4 (c) Ni(CO)4 CO is neutral and strong field ligand.

So, pairing of electrons occur.

Oxidation number of Ni is zero Electronic configuration of Ni = [Ar]3d104s0 For C.N. = 4 (d) Na2[NiCl4] Na2[NiCl4] $\rightarrow$ 2Na + [NiCl4]2 $-$ x + 4( $-$ 1) $-$ ( $-$ 2) = 0; where x, 4, $-$ 1 and $-$ 2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[NiCl4]2 $-$ $\Rightarrow$ x $-$ 4 + 2 = 0 $\Rightarrow$ x = +2 Electronic configuration of Ni2+ = [Ar]3d84s0 For C.N.

= 4 Hence, only K2[Ni(CN)4] has dsp2 hybridization.

Q110

The correct order of intensity of colors of the compounds is :

A

{[Ni{(CN)_4}]^{2 - }} > {[NiC{l_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }}

B

{[Ni{({H_2}O)_6}]^{2 + }} > {[NiC{l_4}]^{2 - }} > {[Ni{(CN)_4}]^{2 - }}

C

{[NiC{l_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }} > {[Ni{(CN)_4}]^{2 - }}

D

{[NiC{l_4}]^{2 - }} > {[Ni{(CN)_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }}

Correct Answer

Option C

Solution

Correct order of intensity of colours of the compounds is [NiCl4]2 $-$ > [Ni(H2O)6]2+ > [Ni(CN)4]2 $-$ Ni is in +2 oxidation state in all complexes.

The intensity of colour depends on the strength of the ligand attached with the central metal atom because more strong ligand more splitting energy, less is intensity of colour.

Strength of ligand is in the order CN $-$ > H2O > Cl $-$ .

Splitting energy order [NiCl4]2 $-$ 2O)6]2+ 4]2 $-$ $\therefore$ Intensity of colour of compound [NiCl4]2 $-$ > [Ni(H2O)6]2+ > [Ni(CN)4]2 $-$