Coordination Compounds — Page 12 | JEE Chemistry

Q111

Spin only magnetic moment of an octahedral complex of Fe2+ in the presence of a strong field ligand in BM is :

A 4.89

B 2.82

C 0

D 3.46

Correct Answer

Option C

Solution

In presence of SFL

\Delta

0 > P means pairing occurs therefore For Fe+2 $\to$ 3d6 $\therefore$ No of unpaired e-(s) = 0 $\therefore$ $\mu$ =

\sqrt {n(n + 2)}

BM = 0 [n = No of unpaired e $-$ (s)] In NiCl2, Ni+2 is having configuration 3d8 $\therefore$ Number of unpaired electron = 2 After formation of oxidised product [Ni(CN)6] $-$ 2 $\Rightarrow$ Ni+4 is obtained Ni+4 $\Rightarrow$ 3d6 and CN $-$ is strong field ligand $\therefore$ Number of unpaired electrons = 0 $\therefore$ The charge is 2 $-$ 0 = 2

Q112

Which one of the following species doesn't have a magnetic moment of 1.73 BM, (spin only value)?

A O

_2^ +

B CuI

C [Cu(NH3)4]Cl2

D O

_2^ -

Correct Answer

Option B

Solution

Species does not have a magnetic moment of 1.7 BM means species must not contain single unpaired electron. (a) Molecular orbital configuration of O

_2^ +

(15 electrons) is

{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ *

Unpaired electron = 1

\mu = \sqrt {n(n + 2)} = \sqrt {1(1 + 2)} = \sqrt 3 = 1.73

where n = no. of unpaired e $-$ $\therefore$ $\mu$ = 1.73 BM (b) Cul $\Rightarrow$ Cu+ $-$ [Ar]3d10, Unpaired electron = 0 I $-$ $\to$ [Xe], unpaired electron = 0 Therefore, $\mu$ = 0 (c) [Cu(NH3)4]Cl2 Cu2+ $\to$ [Ar]3d9 Unpaired electron = 1, Therefore, $\mu$ = 1.73 BM (d) Molecular orbital configuration of

O_2^ -

(17 electrons) is

{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ *

Unpaired electron = 1 Therefore, $\mu$ = 1.73 BM Therefore, Cul does not have magnetic moment of 1.73 BM.

Q113

Which one of the following species responds to an external magnetic field?

A [Fe(H2O)6]3+

B [Ni(CN)4]2

-

C [Co(CN)6]3

-

D [Ni(CO)4]

Correct Answer

Option A

Solution

1.

{[\mathop {Fe}\limits^{ + 3} {({H_2}O)_6}]^{3 + }}

F{e^{3 + }}:[Ar]3{d^5}

Hybridisation :

s{p^3}{d^2}

Magnetic nature : Paramagnetic (so this complex response to external magnetic field) 2.

{[Ni{(CN)_4}]^{2 - }}

N{i^{2 + }}:[Ar]3{d^8}

Hybridisation : dsp2 Magnetic nature : diamagnetic 3.

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

C{o^{3 + }}:[Ar]3{d^6}

Hybridisation :

{d^2}s{p^3}

Magnetic nature : diamagnetic 4.

[Ni{(CO)_4}]

Ni:[Ar]3{d^8}4{s^2}

Hybridisation :

s{p^3}

Magnetic nature : diamagnetic

Q114

Which one of the following metal complexes is most stable?

A [Co(en)(NH3)4]Cl2

B [Co(en)3]Cl2

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

D [Co(NH3)6]Cl2

Correct Answer

Option B

Solution

Complex [Co(en)3]Cl2 is most stable complex among the given complex compounds because more number of chelate rings are present in this complex as compare to others.

(1) [Co(en)(NH3)4]Cl2 1 chelate ring (2) [Co(en)3]Cl2 3 chelate ring (3) [Co(en)2(NH3)2]Cl2 2 chelate ring (4) [Co(NH3)6]Cl2 0 chelate ring

Q115

The type of hybridisation and magnetic property of the complex [MnCl6]3

-

, respectively, are :

A sp3d2 and diamagnetic

B d2sp3 and diamagnetic

C d2sp3 and paramagnetic

D sp3d2 and paramagnetic

Correct Answer

Option D

Solution

[MnCl6]3 $-$ Paramagnetic and having 4 unpaired electrons.

Q116

Which statement is not true with respect to nitrate ion test?

A A dark brown ring is formed at the junction of two solutions.

B Ring is formed due to nitroferrous sulphate complex.

C The brown complex is [Fe(H2O)5 (NO)]SO4.

D Heating the nitrate salt with conc. H2SO4, light brown fumes are evolved.

Correct Answer

Option B

Solution

Brown ring test

\begin{aligned} &\mathrm{NO}_{3}^{-}+3 \mathrm{Fe}^{+2}+4 \mathrm{H}^{+} \rightarrow \mathrm{NO}+3 \mathrm{Fe}^{+3}+2 \mathrm{H}_{2} \mathrm{O} \\\\ &{\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \mathrm{NO} \rightarrow \underset{\text{Brown ring}}{\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}]^{2+} \mathrm{H}_{2} \mathrm{O}\right.}} \end{aligned}

Q117

Given below are two statements : Statement I :

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

,

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

and

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

are d2sp3 hybridised. Statement II :

{[MnCl)_6}{]^{3 - }}

and

{[Fe{F_6}]^{3 - }}

are paramagnetic and have 4 and 5 unpaired electrons, respectively. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is correct but statement II is false

B Both statement I and Statement II are false

C Statement I is incorrect but statement II is true

D Both statement I and statement II are true

Correct Answer

Option D

Solution

Hybridization of all the given complexes are given in following table.

{[Mn{(CN)_6}]^{3 - }} \Rightarrow M{n^{3 + }}

CN $-$ is a strong field ligand. $\therefore$ Pairing of electron takes place.

{[Fe{(CN)_6}]^{3 - }} \Rightarrow F{e^{3 + }}

CN $-$ is a strong ligand. $\therefore$ Pairing of electron takes place.

{[Co{({C_2}{O_4})_3}]^{3 - }} \Rightarrow C{o^{3 + }}

{C_2}O_4^{2 - }

acts as strong field ligand with Co3+. $\therefore$ Pairing of electrons takes place.

{[Fe{F_6}]^{3 - }} \Rightarrow F{e^{3 + }}

F is a weak field ligand. $\therefore$ Pairing of electron does not occurs.

{[MnC{l_6}]^{3 - }} \Rightarrow M{n^{3 + }}

Cl $-$ is a weak field ligand. $\therefore$ Pairing of electron does not occurs.

{[Fe{F_6}]^{3 - }}

have 5 unpaired electrons and

{[MnC{l_6}]^{3 - }}

have 4 unpaired electrons. So, both statement I and statement II are true.

Q118

Arrange the following Cobalt complexes in the order of increasing Crystal Field Stabilization Energy (CFSE) value. Complexes :

\mathop {{{[Co{F_6}]}^{3 - }}}\limits_A ,\mathop {{{[Co{{({H_2}O)}_6}]}^{2 + }}}\limits_B ,\mathop {{{[Co{{(N{H_3})}_6}]}^{3 + }}}\limits_C and \mathop {{{[Co{{({en})}_3}]}^{3 + }}}\limits_D

Choose the correct option :

A A < B < C < D

B B < A < C < D

C B < C < D < A

D C < D < B < A

Correct Answer

Option B

Solution

(i) CFSE $\propto$ charge or oxidation no. of central metal ion. (ii) CFSE $\propto$ strength of ligand en > NH3 > H2O > F $-$ $\therefore$ order of CFSE

\mathop {{{[Co{{(en)}_3}]}^{ + 3}}}\limits^{III} > \mathop {{{[Co{{(N{H_3})}_6}]}^{ + 3}}}\limits^{III} > \mathop {{{[Co{F_6}]}^{ - 3}}}\limits^{III} > \mathop {{{[Co{{({H_2}O)}_6}]}^{ + 2}}}\limits^{III}

Q119

Spin only magnetic moment in BM of [Fe(CO)4(C2O4)]+ is :

A 5.92

B 0

C 1

D 1.73

Correct Answer

Option D

Solution

[Fe(CO)4(C2O4)]+ One unpaired electron Spin only magnetic moment =

\sqrt 3

B.M. = 1.73 BM

Q120

The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion (Mz+) are

-

0.8

\Delta

0 and 3.87 BM, respectively. Identify (Mz+) :

A V3+

B Cr3+

C Mn4+

D Co2+

Correct Answer

Option D

Solution

Co2+ = [Ar] 3d74so in [Co(H2O)6]2+, H2O will behave as weak field ligand Co2+ = t2g5 , eg2 CFSE = (–0.4 × 5 + 2 × 0.6) Δ0 = –0.8 Δ0 Magnetic moment (spin only) can be calculated as :

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

where, $\mu$ = spin only magnetic moment and n = number of unpaired electrons = 3 =

\sqrt {3\left( {3 + 2} \right)} = \sqrt {15}

= 3.87 BM