Coordination Compounds — Page 23 | JEE Chemistry

Q221

The correct order of the number of unpaired electrons in the given complexes is A.

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

B.

\left[\mathrm{Fe} \mathrm{F}_{6}\right]^{3-}

C.

\left[\mathrm{CoF}_{6}\right]^{3-}

D.

\left.[\mathrm{Cr} \text{ (oxalate})_{3}\right]^{3-}

E.

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

Choose the correct answer from the options given below:

A E < A < B < D < C

B A < E < D < C < B

C A < E < C < B < D

D E < A < D < C < B

Correct Answer

Option D

Solution

The number of unpaired electrons in a complex can be determined by the oxidation state of the central metal atom and the ligand field splitting energy.

In general, a complex with a high-spin configuration will have more unpaired electrons than a complex with a low-spin configuration.

In the given complexes, the oxidation states of the central metal atoms are: Fe(II) in [Fe(CN)6]3- Fe(III) in [FeF6]3- Co(III) in [CoF6]3- Cr(III) in [Cr(ox)3]3- Ni(0) in [Ni(CO)4] The ligand field splitting energy of a complex depends on the type of ligand.

In general, ligands that are strong field ligands will split the d orbitals more than ligands that are weak field ligands.

The following table shows the number of unpaired electrons in each complex : .tg .tg Complex Oxidation state Ligand field splitting energy Number of unpaired electrons [Fe(CN)6]3- Fe(II) Low-spin 1 [FeF6]3- Fe(III) High-spin 5 [CoF6]3- Co(III) High-spin 4 [Cr(ox)3]3- Cr(III) High-spin 3 [Ni(CO)4] Ni(0) None 0 Therefore, the correct order of the number of unpaired electrons in the given complexes is : E

Q222

Match with . (Complex) (Primary valency and Secondary valency)

List - I		List - II
(A)	[Co(en)2Cl2]Cl	(I)	3, 6
(B)	[Pt(NH3)2Cl(NO2)]	(II)	3, 4
(C)	Hg [Co(SCN)4]	(III)	2, 6
(D)	[Mg (EDTA)]2−	(IV)	2, 4

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

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

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

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

Correct Answer

Option C

Solution

Primary valency $=$ Oxidation state Secondary valency $=$ Co-ordination number .tg .tg Complex Primary valency Secondary (A) $\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Cl}$ 3 6 (B) $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right]$ 2 4 (C) $\mathrm{Hg}\left[\mathrm{Co}(\mathrm{SCN})_4\right]$ 3 4 (D) $[\mathrm{Mg}(\mathrm{EDTA})]^{2-}$ 2 6

Q223

Match List I with List II .tg .tg List I Coordination entity List II Wavelength of light absorbed in nm A.

\mathrm{[CoCl(NH_3)_5]^{2+}}

I. 310 B.

\mathrm{[Co(NH_3)_6]^{3+}}

II. 475 C.

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

III. 535 D.

\mathrm{[Cu(H_2O)_4]^{2+}}

IV. 600 Choose the correct answer from the options given below :

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

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

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

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

Correct Answer

Option B

Solution

Co-ordination compounds absorb a particular wavelength following certain rules.

\begin{aligned} \text{Wavelength of light absorbed} & \propto \frac{1}{\text{ Oxidation state of metal ion }} \\\\ & \propto \frac{1}{\text{ Strength of ligand }} \end{aligned}

Ligand field strength : $\mathrm{CN}^{-}>\mathrm{NH}_{3}>\mathrm{H}_{2} \mathrm{O}>\mathrm{Cl}^{-}$ .tg .tg C.

$\left[\mathrm{Co}^{||||}(\mathrm{CN})_{6}\right]^{3-}$ I.

$310 \mathrm{~nm}$ B.

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

$475 \mathrm{~nm}$ A.

$\left[\mathrm{Co}^{||||}\mathrm{Cl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$ III.

$535 \mathrm{~nm}$ D.

$\left[\mathrm{Co}^{||}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}$ IV.

$600 \mathrm{~nm}$

Q224

The IUPAC name of complex [Pt(NH3)2Cl(NH2CH3)]Cl is:

A Diamminechlorido (methanamine) platinum(II)chloride.

B Biasmmine (methanamine)chlorido platinum(II)chloride.

C Diamminechlorido (amminomethane) platinum(II) chloride.

D Diammine (methanamine) chlorido platinum(II)Chloride.

Correct Answer

Option A

Solution

[Pt(NH3)2Cl(NH2CH3)]Cl x + 0 – 1 + 0 = +1 x = +2 So Pt present in Pt+2 form.

$\therefore$ Correct IUPAC form Diamminechlorido (methanamine) platinum(II)chloride.

Q225

The complex that can show fac- and mer-isomers is :

A [CoCl2(en)2]

B [Pt(NH3)2Cl2]

C [Co(NH3)3(NO2)3]

D [Co(NH3)4Cl2]+

Correct Answer

Option C

Solution

[Ma3b3] type complex shows fac and mer isomerism. So [Co(NH3)3(NO2)3] is correct answer.