Solved Problems on Transition Elements

Problem 1: 
The chemical reactivity of lanthanides resemble to which other elements of the periodic table?

Solution: 
The chemical reactivity of the starting lanthanides resemble calcium due to similar first and second ionization energy. But latter lanthanides resemble Al due to ability of showing +3 oxidation state and similarity in I.E.

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Problem 2: 
Enthalpies of atomization of transition elements are higher than those of alkali and alkaline earth metals. Explain.

Solution: 
The number of unpaired electrons in transition elements are more when compared to these in alkali and alkaline earth metals. As a result, the metallic bonds in transition metals are stronger and enthalpies of atomization are higher than those of alkali and alkaline earth metals.

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Problem 3: 
Explain the following:

(a) Chromium is a typical metal while mercury is a liquid metal.

(b) Zinc readily liberates H₂ from cold dil. H₂SO₄ but not form cold conc. H₂SO₄.

Solution:

(a) Chromium has five unpaired electrons in its d-orbitals which make its metallic bond very strong, whereas in mercury there is no unpaired d-electrons so its metallic bond is very weak, hence it is a liquid.

(b) Since, conc. H₂SO₄ act as an oxidizing agent hence does not evolve H₂ when it reacts with zinc.

          Zn + 2H₂SO₄ ————→ ZnSO₄ + SO₂ + H₂O

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Solution: 
ZnO > CdO > HgO

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Problem 5: 
Cu⁺ ion has 3d¹⁰4s⁰ configuration and colourless but Cu₂O is red and Cu₂S is black. Explain.

Solution: 
Cu⁺ ion has 3d¹⁰4s⁰ configuration, i.e. it has no unpaired electron hence there is no d-d transition possible and it is colourless. But Cu₂O and Cu₂S are coloured due to charge transfer of electrons from O^2- or S^2- to the vacant orbital of Cu⁺ ion.

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Problem 6: 
While Cu, Ag and Au are considered as transition elements but Zn, Cd and Hg are not considered as transition elements although all the mentioned elements have complete d-orbitals. Explain.

Solution: 
Although Cu, Ag and Au have their d – orbitals complete in the elemental state. They do have incomplete d orbitals in their compound state. So they are included in the transition elements.

Cu⁺² = 3d⁹

Au⁺³ = 5d⁸

Zn, Cd and Ag have their d-orbitals complete in their elemental state as well as compound state. So they are not included in the transition elements.

Zn⁺² = 3d¹⁰

Hg⁺² = 5d¹⁰ 

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Problem 7:

(i) CrO₃ is an acid anhydride. Explain.

(ii) Between Na⁺ and Ag⁺ which is a stronger Lewis acid and why?

Solution:

(i) CrO₃ + H₂O ————→ H₂CrO₄, i.e CrO₃ is formed by loss of one H₂O molecule from chromic acid.

(ii) Between Na⁺ and Ag⁺, Ag⁺ is stronger Lewis acid. Because Ag⁺ has pseudo noble gas configuration which makes it more polarizing.

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Problem 8: 

It is well known that alkali and alkaline earth metals displace hydrogen from dilute acids. But most of the transition elements do not behave so. Explain.

Solution: 
Alkali and alkaline earth metals have positive oxidation potential. But most of the transition elements have negative oxidation potentials. So they are not as good oxidizing agents as the alkali and alkaline earth metal are.

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Problem 9: 
In the melting point curves of transition metals, one observes a dip in the curves at the end i.e. Cu, Ag & Au and Zn, Cd & Hg have lower melting points when compared to other transition metals. Explain.

Solution: 
In the last two groups of transition elements i.e. Cu, Ag, Au, Zn, Cd and Hg all the electrons are paired which can not take part in metallic bonding. As a result, metallic bond in these elements is weak resulting in the lower melting points of these metals.

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Problem 10: 
Enthalpies of atomization of transition elements are higher than those of alkali and alkaline earth metals. Explain.

Solution: 
The number of unpaired electrons in transition elements are more when compared to those in alkali and alkaline earth metals. As a result, the metallic bonds in transition metals are stronger and enthalpies of atomization are higher than those of alkali and alkaline earth metals.

Problem 11:  
Explain the following:

(a) Scandium forms no coloured ions, yet it is regarded as a transition element.

(b) Transition elements have many irregularities in electronic configurations.

Solution:

(a) Scandium in the ground state has one d electron. Hence it is regarded as transition element.

(b) In the transition elements, the (n – 1)d subshell and ns subshell have very small difference in energy. The incoming electron may enter into either ns or (n-1)d subshell. Hence they show irregularities in their electronic configurations.

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Problem 12: 
Explain the following

(a) Chromium is a typical metal while mercury is a liquid metal.

(b) Cobalt (II) is stable in aqueous solution but in the presence of strong ligands, it is a easily oxidised to cobalt (III).

Solution:

(a) chromium has 5 unpaired electrons in its d – orbitals which make its metallic bond very stronger. Whereas in mercury there are no unpaired d electrons, so its metallic bond is very weak.

(b) CO(III) has greater tendency to form complex than CO(II) hence in the presence of ligands CO(II) changes to CO(III).

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Problem 13: 

Write down the products of the following reactions.

(a) CuSO₄ solution is treated with KI solution.

(b) AgNO₃ solution is added to Na₂S₂O₃ solution.

Solution:

(a) Free iodine is liberated along with the formation of a white precipitate of cupric iodide.

CuSO₄ + 2Kl ————→ Cul₂ + K₂SO₄

2Cul₂ ————→ 2Cul + l₂ 

(b) A white precipitate of Ag₂S₂O₃ is obtained which turns yellow, brown and finally black on keeping.

2AgNO₃ + Na₂S₂O₃ ————→ Ag₂S₂O₃ + 2NaNO₃

Ag₂S₂O₃ + H₂O ————→ Ag₂S + H₂SO₄

                                                         ^{black ppt.}

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Problem 14: 
Explain the following

(a) Zinc readily liberates H₂ form cold dil.H₂SO₄ but not from cold conc. H₂SO₄.

(b) Blue colour of the CuSO₄ solution is discharged slowly when an iron rod is dipped into it.

Solution:

(a) Conc. H₂SO₄ is a covalent compound. Hence does not contain H⁺ ions. Dilute H₂SO₄ contains H₃O⁺ which reacts with Zn and liberates H₂.

H₂SO₄ + H₂O ————→ 2H₃O⁺ + SO₄^2–

Zn + 2H₃O⁺ ————→ Zn²⁺ + 2H₂O + H₂O ↑

(b) Fe is more electropositive than Cu, hence it displaces copper form CuSO₄solution.

Fe_(s) + CuSO_(aq) ————→ FeSO_4(aq) + Cu_(s) 

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Problem 15: 
An aqueous solution containing one mole of HgI₂ and two moles of NaI is orange in colour. On addition of excess NaI the solution becomes colourless. The orange colour reappears on subsequent addition of NaOCl. Explain with equations.

Solution:

Hgl₂ + 2Nal ————→           Na₂[Hgl₄]

^{(orange)                           coloured due to residual Hgl2)}

Hgl₂ + Nal(excess) ————→                   Na₂[Hgl₄]

^{orange)                                        (colourless because there is no residual Hgl2)}

2Na₂[Hgl₄] + 2NaOCl + H₂O ————→ 2Hgl₂ + NaCl + 4NaOH + 2Nal₃

                                                                              ^(orange)

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Problem 16:
Write down the IUPAC name of the complex   K₄[Fe(CN)₆].

Solution:
Firstly the +ve part should be named followed by the negative part in which the name of the ligand should be given in alphabetical order with the metal in the negative part ending in - ate (oxidation state in parenthesis) Thus the name is Potassiumhexacyano - C-ferrate (II). Hyphen C is shown to indicate CN^– bonded via carbon.

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Problem 17: 
Write down the IUPAC name  of K₂[Fe(NC)₃Cl₂(NH₃)₂].

Solution:

The positive part is named first followed by the negative part. In the negative part the names are written in alphabetical order followed by metal. So the name is Potassiumdiamminedichlorotricyano-N-
ferrate (III)

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Problem 18: 

Write the IUPAC name of [Co(NH₃)₄(NO₂)₂]Cl

Solution:
In the earlier two examples the negative part was the complex part while in this case the positive part is the complex. So it is named first with ligands in alphabetical order followed by metal (but not ending in –ate as the metal belong to the positive part of the complex). This is followed by the negative part. So the name is Tetraamminedinitrocobalt (III) chloride.

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Problem 19:
Write the formula of
a)   Tetrachlorocuprate(II)ion

b)   Dichlorotetraaquochromium(III)chloride

c)   Bromo Chlorotetra ammine Cobalt(III) sulphate

d)   Diammine Silver (I) hexacyano ferrate (II)

e)   Dichlorobis (ethylenediammine) chromium (III) tetrachloro palladate (II)

f)    Aluminium tetrachloro aurate (III)

Solution:

a) [CuCl₄]^2–

b) [Cr(H₂O)₄Cl₂]Cl

c) [Co(NH₃)4BrCl₂]SO₄

d) [Ag(NH₃)₂]₄[Fe(CN)₆]

e) [Cr(en)₂Cl₂]₂[PdCl₄]

f) Al[AuCl₄]₃

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Problem 20:

Arrange the following compounds in order of increasing molar conductivity:

a) K[Co(NH₃)₂(NO₂)₄]

b) [Cr(NH₃)₃(NO₂)₃]

c) [Cr(NH₃)₅ (NO₂]₃ [Co(NO₂)₆]₂

d) Mg[Cr(NH₃) (NO₂)₅]

Solution

The larger the number of ions and the larger the charge on each, the larger the conductivity. The compounds from lowest conductivity to highest conductivity are:

b < a < d < c

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