JEE Advanced (Subjective Type Questions): The d & f-Block Elements & Coordination Compounds- 2 | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.16. Compare qualitatively the first and secon d ion isation potentials of copper and zinc. Explain the observation.   (1996 - 2 Marks)

Solution. 

On the basis of configuration of Cu and Zn, first ionisation potential of Zn is greater than that of copper because in zinc the electron is removed from 4s2 configuration while in copper it is removed from 4s¹ configuration. So more amount of energy is required for the removal of electron of 4s² (completely filled orbital) than that of 4s1 while the second ionisation potential of Cu is higher than that of zinc because Cu⁺ has 3d¹⁰ (stable configuration) in comparison to Zn⁺ (4s¹ configuration).

Q.17. Write the formulae of the following complexes :
 (i) Pentamminechlorocobalt(III)            (1997 - 1 Mark)
 (ii) Lithium tetrahydroaluminate(III).               (1997 - 1 Mark)

Ans. (i) [CoCl(NH₃)₅]⁺², (ii) LiAlH₄

Solution. (i) [CoCl(NH₃)₅]²⁺ Formula of pentamminechlorocobalt (III)
(ii) LiAlH₄ Formula of lithium tetrahydroaluminate (III)

Q.18. When the ore haematite is burnt in air with coke around 2000°C along with lime, the process not only produces steel but also produces a silicate slag that is useful in making building materials such as cement. Discuss the same and show through balanced chemical equations.           (1998 -  4 Marks)

Solution. Haematite(Fe₂O₃) on burning with coke and lime at 2000°C results in the following reactions.

Q.19. Work out the following using chemical equations                   (1998 -  2 Marks) 

In moist air copper corrodes to produce a green layer on the surface.

Solution. 

Q. 20. A, B, and C are three complexes of chromium (III) with the empirical formula H₁₂O₆Cl₃Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H₂SO₄, whereas complexes B and C lose 6.75% and 13.5% of their original mass, respectively, on treatment with concentrated H₂SO₄. Identify A, B and C.                (1999 - 6 Marks)

Ans. [Cr(H₂O)₆]Cl₃, [Cr(H₂O)₅Cl](H₂O)Cl₂, [Cr(H₂O)₄Cl₂](H₂O)₂Cl

Solution. The complex A does not react with concentrated H₂SO₄ implying that all water molecules are coordinated with Cr3+ ion. Hence, its structure would be [Cr(H₂O)₆]Cl₃.
The compound B loses 6.75% of its original mass when treated with concentrated H₂SO₄. This loss is due to the removal of water molecules which is/are not directly coordinated to Cr³⁺ ion.
The mass of water molecules removed per mole of the complex

 molar mass of the complex  

= 17.98 g This corresponds to one mole of water. Hence, the structure of the compound B will be [Cr(H₂O)₅Cl](H₂O)Cl₂
NOTE : The compound C loses 13.5% of its mass when treated with concentrated H2SO4 which is twice of the mass lost by the compound B. Hence, the structure of the compound C will be [Cr(H₂O)₄Cl₂](H₂O)₂ Cl.

Q. 21. Write the chemical reaction associated with the 'brown ring test'. (2000 - 2 Marks)

Solution. NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃

6FeSO₄ + 2HNO₃ + 3H₂SO₄ → 3Fe₂ (SO₄)₃ + 4H₂O + 2NO

Q.22. Draw the structures of [Co(NH₃)₆]³⁺ , [Ni(CN)₄]^2- and [Ni(CO)₄]. Write the hybridisation of atomic orbitals of the transition metal in each case.    (2000 - 4 Marks)

Solution.

Octa hedral compl ex, d²sp³ hybridisation

Q.23. (i) Write the chemical reactions involved in the extraction of metallic silver from argentite.
 (ii) Write the balanced chemical equation for developing photographic films.           (2000 - 4 Marks)

Solution.(i) Argentite is Ag₂S. Silver is extracted from its ore argentite (silver glance, Ag₂S) as follows :
(1) Silver glance is concentrated by froth flotation.
(2) Leaching : The concentrated ore is ground to fine powder and dissolved in dilute solution of sodium cyanide.

Ag₂S + 4 NaCN → 2 NaAg(CN)₂ + Na₂S

Oxygen of air converts Na₂S to Na₂SO₄ thereby preventing reaction to take place in the reversible direction.

(3) Recovery of silver.

Silver is precipitated out by adding electropositive metal, Zn.

2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄]+ 2Ag

(ii) For development, activated grains are preferentially reduced by mild reducing agents like hydroquinone

(Reduction of activated AgBr to elemental silver.) The photographic film is permanently fixed by immediately washing out any non activated AgBr grains in hypo emulsion.

Q.24. A metal complex having composition Cr(NH₃)₄Cl₂Br has been isolated in two forms (A) and (B). The form (A) reacts with AgNO₃ to give a white precipitate readily soluble in dilute aqueous ammonia, whereas (B) gives a pale yellow precipitate soluble in concentrated ammonia. Write the formula of (A) and (B) and state the hybridization of chromium in each. Calculate their magnetic moments (spin– only value).              (2001 - 5 Marks)

Ans. [Cr(NH₃)₄BrCl]Cl : (d²sp³); [Cr(NH₃)₄Cl₂]Br : (d²sp³); 3.87 BM, 3.87 BM

Solution. Compound (A) on treatmen t with AgNO₃ gives wh ite precipitate of AgCl, which is readily soluble in dil.aq.
NH₃.Therefore it has at least one Cl– ion in the ionization sphere furthermore chromium has coordination number equal to 6. So its formula is [Cr(NH₃)₄BrCl]Cl.
Compound (B) on treatment with AgNO₃ gives pale yellow precipitate of AgBr soluble in conc. NH₃. Therefore it has Br^– in the ionization sphere. So its formula is [Cr(NH₃)₄Cl₂]Br

State of hybridization of chromium in both (A) and (B) is d²sp³.

Spin magnetic moment of (A) or (B),

Q.25. Deduce the structure of [NiCl₄]^2- and [Ni (CN)₄]^2- considering the hybridization of the metal ion. Calculate the magnetic moment (spin only) of the species. (2002 - 5 Marks)

Solution. Cl^– is a weak ligand which is unable to pair the electrons of Ni²⁺. Therefore, here hybridisation is sp³ and shape will be tetrahedral.
Electronic configuration of Ni⁺² (No. of electrons = 26) in presence of Cl^– ion, a weak ligand.

Magnetic moment of BM

On the other hand, CN^– is a stong ligand which pairs up the electrons of Ni²⁺. Therefore, here hybridisation is dsp² and shape will be square planar.
Electronic configuration of Ni²⁺ in presence of CN^– ion, a strong ligand.

For structure of [Ni(CN)₄]^2–, refer question 24 in Section (E).
Magnetic moment of 

Q.26. Write the IUPAC nomenclature of the given complex along with its hybridisation and structure.

K₂ [Cr(NO)(NH₃)(CN)₄], μ = 1.73 BM                                      (2003 - 4 Marks)

Ans. Potassium amminotetracyanonitrosoniumchromate (I), (d²sp³), octahedral shape.

Solution. The spin magnetic moment, μ of the complex is 1.73 BM.

It means that nucleus of the complex, chromium ion has one unpaired electron. So the ligand NO is unit positively charged.

IUPAC name :

Potassium amminetetracyanonitrosochr (I)

(a) Electronic configuration of Cr⁺ :

(b) Electronic configuration of Cr⁺ under the influence of strong field ligand CN^–

So, Hybridization : d²sp³; Shape : Octahedral

Q.27. Nickel chloride, when treated with dimethylgyloxime in presence of ammonium hydroxide, a bright red precipitate is obtained. Answer the following. (2004 - 4 Marks) 

(a) Draw the structure of the complex showing H-bonds
 (b) Give oxidation state of nickel and its hybridisation
 (c) Predict the magnetic behaviour of the complex

Ans. (b)⁺², dsp²; (c) diamagnetic

Solution. 

(b) Charge on Ni in the complex is +2 and it is dsp² hybridised
(c) Since number of unpaired electrons in Ni²⁺ is zero, the complex is diamagnetic.

Q.28. Some reactions of two ores, A₁ and A₂ of the metal M are given below.                 (2004 - 4 Marks)

 Green solution

Identify A₁, A₂, M, C, D, and G, and explain using the required chemical reactions.

Ans. 

Solution. Calcination of th e ore A₁ to form CO₂ in dicates that A₁ should be a carbonate. Further, reaction of A₁ with HCl and KI to evolve I₂ indicates that A₁  would also be hydroxide.
So the possible formula for the ore, should be CuCO₃.Cu(OH)₂ which explains all the given reactions

Roasting of A₂ gives gas G whose nature is identified as SO₂ as it gives green colour with acidified K₂Cr₂O₇. So A₂ should be sulphide of copper.

Q.29. 

What are (A) and (B)? Give IUPAC name of (A). Find the spin only magnetic moment of (B).               (2005 - 4 Marks)

Ans. 

Solution. 

IUPAC name of A is pentaaquathiocyanatoferrate (III) ion
IUPAC name of B is hexafluoroferrate (III)
In [FeF₆]^3–  coordination no. of Fe = 6
In [FeF⁶]^3– oxidation state of Fe = + 3
∴ It has 5 unpaired electrons, n = 5, Fe³⁺ is 3d⁵

Q.30. Write the chemical reaction involved in developing of a black and white photographic film. An aqueous Na₂S₂O₃ solution is acidified to give a milky white turbitity. Identify the product and write the balanced half chemical reaction for it.         (2005 - 4 Marks)

Solution. Reaction in volved in developing of a black and white photographic film.

Q.31.  compound; M Transition metal

Identify (A), (B) and MCl₄. Also explain colour difference between MCl₄ and (A).

Ans. 

Solution. [A] = [Ti(H₂O)₆]Cl₃ [B] = HCl

TiCl⁴ is colourless since Ti⁴⁺ has no d electrons, hence d-d transition is impossible. On the other hand, Ti³⁺ is coloured due to d-d transition. Ti³⁺ absorbs greenish yellow compound of white light, hence its aqueous solution is purple which is complementary colour of greenish yellow in white light.