Coordination Compounds — Page 21 | JEE Chemistry

Q201

Which of the following complexes will shows geometrical isomerism ?

A aquachlorobis (ethylenediamine) cobalt (II) chloride

B pentaaquachlorohromium (III) chloride

C potassium amminetrichloropltinate (II)

D potassium tris(oxalato) chromate (III)

Correct Answer

Option A

Solution

(a) [Co(H2O) Cl(en)2]Cl $\to$ it shows gemetrical isomerism. (b) [Cr(H2O)5Cl]Cl2 : This complex is [MA5B] type.

It does not show geometrical isomerism. (c) K[Pt(NH3)Cl3] : Thiscomplex is [MAB3] type.

It does not show geometrical isomerism. (d) K3[Cr(OX)3] : This complex is [M(AA)3] type.

It does not show geometrical isomerism.

Q202

The correct order of spin-only magnetic moments among the following is : (Atomic number : Mn = 25, Co = 27, Ni = 28, Zn = 30)

A [ZnCl4]2- > [NiCl4]2- > [CoCl4]2- > [MnCl4]2-

B [CoCl4]2- > [MnCl4]2- > [NiCl4]2- > [ZnCl4]2-

C [NiCl4]2- > [CoCl4]2- > [MnCl4]2- > [ZnCl4]2-

D [MnCl4]2- > [CoCl4]2- > [NiCl4]2- > [ZnCl4]2-

Correct Answer

Option D

Solution

We know, Spin only magnetic moment ( $\mu$ ) =

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

B.M Where, n is the number of unpaired electrons.

So, the complex having higher number of unpaired electrons will have higher value of spin only magnetic moment.

(1) Zn+2 : [Ar] 3d10 Here 0 unpaired electrons present (2) Ni+2 : [Ar] 3d8 Here 2 unpaired electrons present.

(3) Co+2 : [Ar] 3d7 Here 3 unpaired electrons present.

(4) Mn+2 : [Ar] 3d5 Here 5 unpaired electrons present.

So, correct order is $\to$ [MnCl4]2 $-$ > [CoCl4]2 $-$ > [NiCl4]2 $-$ > [ZnCl4]2 $-$

Q203

A reaction of cobalt (III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two isomeric products A (violet coloured) and B (green coloured). A can show optical activity, but B is optically inactive. What type of isomers does A and B represcent?

A Ionisation isomers

B Linkage isomers

C Coordination isomers

D Geometrical isomers

Correct Answer

Option D

Solution

CoCl3 + en $\to$ [Co(en)2Cl2]Cl 1 : 2 A and B are Geometrical isomers.

Q204

Two complexes [Cr(H2O)6]Cl3 (A) and [Cr(NH3)6]Cl3 (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is :

A

\Delta

0 values of (A) and (B) are calculated from the energies of violet and yellow light, respectively

B both are paramagnetic with three unpaired electrons

C both absorb energies corresponding to their complementry colors.

D

\Delta

0 value for (A) is less than that of (B).

Correct Answer

Option A

Solution

NH3 is a strong field ligand. H2O is weak field ligand. Let

\Delta

violet is the crystal field splitting energy of complex A and

\Delta

yellow is for complex B. $\therefore$

\Delta

yellow >

\Delta

violet. We know,

\Delta

=

{{hc} \over {{\lambda _{absorb}}}}

Violet color will absorb yellow color and yellow color will absorb violet color. $\therefore$

\Delta

yellow =

{{hc} \over {{\lambda _{violet}}}}

and

\Delta

violet =

{{hc} \over {{\lambda _{yellow}}}}

So,

\Delta

o of A is calculated from energy of yellow light and

\Delta

o of B is calculated from energy of violet light.

Q205

Match the metals (column I) with the coordination compound(s)/enzyme(s) (column II) .tg .tg (Column I) Metals (Column II) Coordination compounds(s) enzyme(s)

List - I		List - II
(A)	Co	(i)	Wilkinson catalyst
(B)	Zn	(ii)	Chlorophyl
(C)	Rh	(iii)	Vitamin B12
(D)	Mg	(iv)	Carbonic anhydrase

A (A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)

B (A)-(iv); (B)-(iii); (C)-(i); (D)-(ii)

C (A)-(i); (B)-(ii); (C)-(iii); (D)-(iv)

D (A)-(ii); (B)-(i); (C)-(iv); (D)-(iii)

Correct Answer

Option A

Solution

Co $\to$ Vitamin B12 Zn $\to$ Carbonic anhydrase Rh $\to$ Wilkinson catalyst Mg $\to$ Chlorophyll

Q206

Match with $$

List - I		List - II
(B)	${[Ni{(CN)_4}]^{2 - }}$	(I)	$s{p^3}$
(C)	${[Co{(CN)_6}]^{3 - }}$	(II)	$s{p^3}{d^2}$
(D)	${[Co{F_6}]^{3 - }}$	(III)	${d^2}s{p^3}$
()		(IV)	$ds{p^2}$

A A-IV, B-I, C-III, D-II

B A-I, B-IV, C-III, D-II

C A-I, B-IV, C-II, D-III

D A-IV, B-I, C-II, D-III

Correct Answer

Option B

Solution

Ni(CO)4 Hybridisation sp3 [Ni(CN)4]2– Hybridisation dsp2 [Co(CN)6]3– Hybridisation d2sp3 [Co(F)6]3– Hybridisation sp3d2

Q207

Match List I with List II: .tg .tg List I (Complexes) List II (Hybridisation) A.

\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]

I.

\mathrm{sp}^{3}

B.

\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}

II. dsp

^{2}

C.

\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}

III.

\mathrm{sp}^{3}\mathrm{d}^{2}

D.

\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}

IV.

\mathrm{d}^{2} \mathrm{sp}^{3}

A A-I, B-II, C-III, D-IV

B A-II, B-I, C-III, D-IV

C A-I, B-II, C-IV, D-III

D A-II, B-I, C-IV, D-III

Correct Answer

Option C

Solution

\mathrm{Ni{(CO)_4}\overset{{}}\longrightarrow s{p^3}}

\mathrm{{\left[ {Cu{{(N{H_3})}_4}} \right]^{2 + }}\overset{{}}\longrightarrow ds{p^2}}

\mathrm{{\left[ {Fe{{(N{H_3})}_6}} \right]^{2 + }}\overset{{}}\longrightarrow {d^2}s{p^3}}

\mathrm{{\left[ {Fe{{({H_2}O)}_6}} \right]^{2 + }}\overset{{}}\longrightarrow s{p^3}{d^2}}

Q208

Match List I with List II .tg .tg List - I Complex List - II Crystal Field splitting energy (

\Delta_0

) A.

{[Ti{({H_2}O)_6}]^{2 + }}

I.

-1.2

B.

{[V{({H_2}O)_6}]^{2 + }}

II.

-0.6

C.

{[Mn{({H_2}O)_6}]^{3 + }}

III. 0 D.

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

IV.

-0.8

Choose the correct answer from the options given below:

A A-IV, B-I, C-II, D-III

B A-II, B-IV, C-III, D-I

C A-II, B-IV, C-I, D-III

D A-IV, B-I, C-III, D-II

Correct Answer

Option A

Solution

\begin{aligned} & \text{(A)} ~~ {\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\ & \mathrm{Ti}^{2+} \Rightarrow 3 \mathrm{~d}^2 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=2 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=0 \\\\ & \mathrm{CFSE}=[-0.4 \times 2+0.6 \times 0] \Delta_0 \\\\ & \quad=-0.8 \Delta \end{aligned}

\begin{aligned} & \text{(B)} ~~ {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\ & \mathrm{V}^{2+} \Rightarrow 3 \mathrm{~d}^3 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=0 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 0] \Delta_0 \\\\ & =-1.2 \Delta_0 \end{aligned}

\begin{aligned} & \text{(C)} ~~{\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\ & \mathrm{Mn}^{3+} \Rightarrow 3 \mathrm{~d}^4 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=1 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 1] \Delta_0 \\\\ & =-0.6 \Delta_0 \end{aligned}

\begin{aligned} & \text{(D)} ~~{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\ & \mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \quad \mathrm{e}_{\mathrm{g}}=2 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 2] \Delta_0 \\\\ & \quad=0 \Delta_0 \end{aligned}

Q209

Match List I with List II .tg .tg List - I (Substances) List - II (Element Present) (A) Ziegler catalyst (I) Rhodium (B) Blood Pigment (II) Cobalt (C) Wilkinson catalyst (III) Iron (D) Vitamin

\mathrm{B_{12}}

(IV) Titanium Choose the correct answer from the options given below:

A A-IV, B-III, C-I, D-II

B A-II, B-IV, C-I, D-III

C A-III, B-II, C-IV, D-I

D A-II, B-III, C-IV, D-I

Correct Answer

Option A

Solution

Ziegler catalyst $\rightarrow$ Titanium Blood pigment $\rightarrow$ Iron Wilkinson catalyst $\rightarrow$ Rhodium Vitamin

\mathrm{B}_{12} \rightarrow

Cobalt

Q210

\text{ Match the LIST-I with LIST-II }

.tg .tg LIST-I (Molecules/ion) LIST-II (Hybridisation of central atom) A.

\mathrm{PF}_5

I

\mathrm{dsp}^2

B

\mathrm{SF}_6

II

\mathrm{sp}^3 \mathrm{~d}

C

\mathrm{Ni}(\mathrm{CO})_4

III

\mathrm{sp}^3 \mathrm{~d}^2

D

\left[\mathrm{PtCl}_4\right]^{2-}

IV

\mathrm{sp}^3

\text{ Choose the correct answer from the options given below: }

A A-IV, B-I, C-II, D-III

B A-III, B-I, C-IV, D-II

C A-II, B-III, C-IV, D-I

D A-I, B-II, C-III, D-IV

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{PF}_5: 5 \sigma+0 \ell \mathrm{p} \rightarrow \mathrm{sp}^3 \mathrm{~d} \\ & \mathrm{SF}_6: 6 \sigma+0 \ell \mathrm{p} \rightarrow \mathrm{sp}^3 \mathrm{~d}^2 \\ & \mathrm{Ni}(\mathrm{CO})_4: \mathrm{Ni} \rightarrow 0 \end{aligned}

In presence of ligand field :-

\begin{aligned} &\left[\mathrm{PtCl}_4\right]^{2-}: \mathrm{Pt} \rightarrow+2\\ &\text{ In presence of ligand field :- } \end{aligned}