Coordination Compounds — Page 22 | JEE Chemistry

Q211

Match List I with List II .tg .tg LIST I Complex LIST II CFSE (

\Delta_0

) A.

\mathrm{[Cu(NH_3)_6]^{2+}}

I.

-0.6

B.

\mathrm{[Ti(H_2O)_6]^{3+}}

II.

-2.0

C.

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

III.

-1.2

D.

\mathrm{[NiF_6]^{4-}}

IV.

-0.4

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option A

Solution

In order to match these complexes with their respective crystal field splitting energies (CFSE), we need to understand the coordination number, the geometry of the complex, and the nature of the ligand.

The crystal field splitting energy, often denoted as Δ₀ or Δ, is the energy difference between the higher-energy and lower-energy sets of d-orbitals in a transition metal ion when in a crystal field created by a set of ligands.

The sign of CFSE is often taken as negative because it represents a lowering in energy due to the field of the ligands.

The magnitude of CFSE depends on the nature of the ligand, with ligands causing larger splitting being known as strong-field ligands and those causing smaller splitting being known as weak-field ligands.

Based on spectrochemical series, we know the strength of ligands as follows: F⁻ < H₂O < NH₃ < CN⁻ The crystal field splitting energy also depends on the configuration of the d electrons in the transition metal ion.

For octahedral complexes, the d orbitals split into two sets with an energy difference Δ₀: a lower-energy set (dxy, dyz, dxz) called t₂g, and a higher-energy set (dz², dx²-y²) called eg.

Now let's analyze each complex: [Cu(NH₃)₆]²⁺: Cu²⁺ ion has a d⁹ configuration.

It is an octahedral complex with a strong field ligand (NH₃), hence the splitting will be large.

[Ti(H₂O)₆]³⁺: Ti³⁺ ion has a d¹ configuration.

It is an octahedral complex with a weak field ligand (H₂O), hence the splitting will be small.

[Fe(CN)₆]³⁻: Fe³⁺ ion has a d⁵ configuration.

It is an octahedral complex with a very strong field ligand (CN⁻), hence the splitting will be very large.

[NiF₆]⁴⁻: Ni²⁺ ion has a d⁸ configuration.

It is an octahedral complex with a very weak field ligand (F⁻), hence the splitting will be very small.

So, the correct order of CFSE values for these complexes should be: [NiF₆]⁴⁻ < [Ti(H₂O)₆]³⁺ < [Cu(NH₃)₆]²⁺ < [Fe(CN)₆]³⁻ Therefore, the correct answer is: A-I, B-IV, C-II, D-III

Q212

Match List I with List II .tg .tg LIST I Complex LIST II Colour A.

Mg(N{H_4})P{O_4}

I. brown B.

{K_3}[Co{(N{O_2})_6}]

II. white C.

MnO{(OH)_2}

III. yellow D.

F{e_4}{[Fe{(CN)_6}]_3}

IV. blue Choose the correct answer from the options given below :

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

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

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

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

Correct Answer

Option B

Solution

In this problem, we are matching chemical compounds with their corresponding color. Let's go over each one : A.

Mg(NH_4)PO_4

- Magnesium ammonium phosphate is white in color, so A matches with II. B.

K_3[Co(NO_2)_6]

- Potassium hexanitritocobaltate(III) is a yellow complex, so B matches with III. C.

MnO(OH)_2

- Manganese hydroxide is brown in color, so C matches with I. D.

Fe_4[Fe(CN)_6]_3

- Prussian blue or Ferric ferrocyanide is deep blue in color, so D matches with IV.

So, the correct option is Option B: A-II, B-III, C-I, D-IV.

Q213

Match List I with List II .tg .tg LIST I Coordination Complex LIST II Number of unpaired electrons A.

\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}

I. 0 B.

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

II. 3 C.

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

III. 2 D.

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

IV. 4 Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option C

Solution

For option (A)

\begin{aligned} & \mathrm{Cr}^{+3}: 3 \mathrm{~d}^3 \\\\ & \mathrm{CN}^{-} \rightarrow \mathrm{SFL} \end{aligned}

No. of unpaired electrons $=3$ For option (B)

\begin{aligned} & \mathrm{Fe}^{+2}: 3 \mathrm{~d}^6 \\\\ & \mathrm{H}_2 \mathrm{O}: \mathrm{WFL} \end{aligned}

No. of unpaired electrons $=4$ For option (C)

\begin{aligned} & \mathrm{Co}^{+3}: 3 \mathrm{~d}^6 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned}

No. of unpaired electrons $=0$ For option (D)

\begin{aligned} & \mathrm{Ni}^{+2}: 3 \mathrm{~d}^8 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned}

No. of unpaired electrons $=2$

Q214

Match with }$$

List - I		List - II
(B)	$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$	(I)	$t_{2 g}{ }^2 e_g^0$
(C)	$\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$	(II)	$t_{2 g}{ }^3 e_g{ }^0$
(D)	$\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$	(III)	$t_{2 g}{ }^3 e_g{ }^2$
()		(IV)	$t_{2 g}{ }^6 e_g^2$

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

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

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

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

Correct Answer

Option C

Solution

\begin{aligned} & {\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text{ Contains } \mathrm{Cr}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^3: \mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^{\mathrm{o}}} \\ & {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text{ Contains } \mathrm{Fe}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^5: \mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^2} \\ & {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \text{ Contains } \mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^8: \mathrm{t}_{2 \mathrm{~g}}^6 \mathrm{e}_{\mathrm{g}}^2} \\ & {\left[\mathrm{~V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text{ Contains } \mathrm{V}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^2: \mathrm{t}_{2 \mathrm{~g}}^2 \mathrm{eg}^{\mathrm{o}}} \end{aligned}

Q215

Match List I with List II .tg .tg LIST I LIST II A.

\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]

I.

sp^3

B.

\left[\mathrm{Ni}(\mathrm{CO})_4\right]

II.

sp^3d^2

C.

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3

III.

dsp^2

D.

\mathrm{Na}_3\left[\mathrm{CoF}_6\right]

IV.

d^2sp^3

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option D

Solution

\begin{aligned} & \stackrel{+2}{\text{ (A) } \mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]} \\ & \mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^{\circ},\left(\mathrm{CN}^{-} \text{is } \mathrm{S} . \mathrm{F} . \mathrm{L}\right) \\ & \text{ Pre hybridization state of } \mathrm{Ni}^{+2} \end{aligned}

(B)

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

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

\mathrm{CO}

is S.F.L , so pairing occur Pre hybridization state of

\mathrm{Ni}

(C)

[\mathrm{Co}(\mathrm{NH}_3)_6] \mathrm{Cl}_3

\mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^0

With

\mathrm{Co}^{3+}, \mathrm{NH}_3

act as S.F.L (D)

\mathrm{Na}_3[\mathrm{CoF}_6]

\mathrm{Co}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^6\left(\mathrm{~F}^{\ominus}: \text{ W.F.L }\right)

Q216

Match List I with List II .tg .tg LIST I (Compound) LIST II (Colour] A.

\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH_2O}

I. Violet B.

\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}

II. Blood Red C.

[\mathrm{Fe}(\mathrm{SCN})]^{2+}

III. Prussian Blue D.

\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\cdot12 \mathrm{MoO}_3

IV. Yellow Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option D

Solution

To match List I with List II correctly, we need to identify the colors associated with each compound.

Let's analyze each compound: 1.

\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH_2O}

- This is known as Prussian Blue. 2.

\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}

- This complex is known to have a violet color. 3.

[\mathrm{Fe}(\mathrm{SCN})]^{2+}

- This ion is known for its blood red color. 4.

\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\cdot12 \mathrm{MoO}_3

- This is commonly known as Ammonium phosphomolybdate, which is yellow.

Matching each item, we get: LIST I (Compound) LIST II (Colour) A.

\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH_2O}

III. Prussian Blue B.

\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}

I. Violet C.

[\mathrm{Fe}(\mathrm{SCN})]^{2+}

II. Blood Red D.

\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\cdot12 \mathrm{MoO}_3

IV. Yellow Based on the analysis, the correct matching is: Option D: A-III, B-I, C-II, D-IV

Q217

Match List I with List II. .tg .tg LIST I Tetrahedral Complex LIST II Electronic configuration A.

\mathrm{TiCl}_4

I.

\mathrm{e}^2, \mathrm{t}_2^0

B.

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

II.

\mathrm{e^4, t_2^3}

C.

\left[\mathrm{FeCl}_4\right]^{-}

III.

\mathrm{e}^0, \mathrm{t}_2^0

D.

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

IV.

\mathrm{e}^2, \mathrm{t}_2^3

Choose the correct answer from the options given below :

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

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

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

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

Correct Answer

Option B

Solution

(A)

\mathrm{TiCl}_4: \mathrm{Ti}^{4+}: 3 d^0 4 s^0

or

\mathrm{e}^0 \mathrm{t}_2^0

(B)

\left[\mathrm{FeO}_4\right]^{2-}: \mathrm{Fe}^{6+}: 3 d^2 4 s^0

or

\mathrm{e}^2 \mathrm{t}_2^0

(C)

\left[\mathrm{FeCl}_4\right]^{-}: \mathrm{Fe}^{3+}: 3 d^5 4 s^0

or

\mathrm{e}^2 \mathrm{t}_2^3

(D)

\left[\mathrm{CoCl}_4\right]^{2-}: \mathrm{Co}^{2+}: 3 d^7 4 s^0

or

\mathrm{e}^4 \mathrm{t}_2^3

$\therefore$ Correct matching of tetrahedral complexes given in List-I with their electronic configuration given in List-II is (A)-(III); (B)-(I); (C)-(IV); (D)-(II)

Q218

Match List I with List II .tg .tg LIST I (Hybridization) LIST II (Orientation in Shape) A. sp

^3

I. Trigonal bipyramidal B. dsp

^2

II. Octahedral C. sp

^3

d III. Tetrahedral D. sp

^3

d

^2

IV. Square planar Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option B

Solution

A $\rightarrow$ Tetrahedral (III) B $\rightarrow$ Square planar (IV) C $\rightarrow$ Trigonal bipyramidal (I) D $\rightarrow$ Octahedral (II)

Q219

The pair in which both the species have the same magnetic moment (spin only) is :

A [Cr(H2O)6]2+ and [CoCl4]2–

B [Co(OH)4]2– and [Fe(NH3)6]2+

C [Mn(H2O)6]2+ and [Cr(H2O)]2+

D [Cr(H2O)6]2+ and [Fe(H2O)6]2+

Correct Answer

Option D

Solution

Species with same number of unpaired electrons have equal magnetic moment. .tg .tg Complex Ligand Ligand Type Number of unpaired electrons [Mn(H2O)6]2+ H2O Weak Field Ligand 5 [Cr(H2O)]2+ H2O Weak Field Ligand 4 [CoCl4]2– Cl Weak Field Ligand 3 [Fe(H2O)6]2+ H2O Weak Field Ligand 4 [Co(OH)4]2– OH Weak Field Ligand 3 [Fe(NH3)6]2+ NH3 Weak Field Ligand 4

Q220

Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers : Mn = 25; Fe = 26) A. [FeF6]3

-

B. [Fe(CN)6]3

-

C. [MnCl6]3

-

(high spin) D. [Mn(CN)6]3

-

Choose the correct answer from the options given below :

A A < B < D < C

B B < D < C < A

C A < C < D < B

D B < D < A < C

Correct Answer

Option B

Solution

.tg .tg Coordination Compound Number of unpaired e– (n) Magnetic moment (µ) (B.M) A.

[FeF6]3 $-$ - d5 5 5.91 B.

[Fe(CN)6]3 $-$ - d5 1 1.73 C.

[MnCl6]3 $-$ - d4 4 4.89 D.

[Mn(CN)6]3 $-$ - d4 2 2.82 Hence, correct order of magnetic moment is $2 < 4 < 3 < 1$