7/31/2019 d-blcok

1/5

p. 1

Queens College

S.7 Chemistry

Class Exercise (d-block elements)

1. a. Given the following electronic configurations:

Fe2+

1s22s

22p

63s

23p

63d

6Fe

3+1s

22s

22p

63s

23p

63d

5

Mn2+

1s22s

22p

63s

23p

63d

5Mn

3+1s

22s

22p

63s

23p

63d

4

explain why

i. Fe2+(aq) ion is readily oxidized to Fe3+(aq) ion.ii. Mn3+(aq) ion is readily reduced to Mn2+(aq) ion

b. Give the outer valence shell electronic configuration of each of the following ions:Ti

3+, V

4+, Mn

3+, Sc

3+and Cu

+.

Which of the above hydrated ions would you expect to be colourless? Give reasons

for your answer.

c. Write down the ground-state electronic configurations of chromium (Z = 24) andzinc (Z = 30).

2. The analysis of a compound Co(NH3)xCly yielded the following composition by mass:

Co, 25.59%; NH3, 29.00%; Cl, 45.41%

a. Calculate the values of x and y.

b. Draw and name all possible structures for this compound.

(Relative atomic masses: H, 1.0; N, 14.0; Cl, 35.5; Co, 58.9)

3. A quantity of cobalt(II) chloride hexahydrate was dissolved in 20 cm3

of water, giving a

pink solution A (0.035 M). To this pink solution, 10 cm3

of an aqueous solution of

ammonium carbonate (0.2 M) and 50 cm3

of concentrated ammonia were added. The

solution immediately turned blue in colour. This was solution B.

Air was drawn through solution B for 3 hours during which the colour changed from

deep blue to red. The resulting solution was concentrated to about 60 cm3

and excess

ammonium carbonate was added continually at the rate of about 1 g per 15 minutes

during evaporation. The solution was filtered to remove the dark brown by-product C.The filtrate was then further concentrated down to about 30 cm

3. It was then set aside to

crystallize out the desired product D.

D is known to be a monomer and it contains cobalt, ammonia, carbonate and chloride.

Elemental analysis of D gave N: 25.2%; C: 5.4% and Cl: 16.0%.

a. Deduce the molecular formula of D and suggest possible structure(s) for this

complex.

b. Briefly explain the chemical reactions involved in the above preparation. Identify

the chemical species which are responsible for the pink colour in solution A and the

blue colour in solution B. Also identify the solid species C.

7/31/2019 d-blcok

2/5

p. 2

The metal content of the cobalt complex D was determined by removing the organic

matter with concentrated sulphuric(VI) acid and then quantitatively converting the

cobalt ion into cobalt(III). The cobalt(III) solution was acidified with dilute sulphuric(VI)

acid and then an excess of solid potassium iodide was added to it, reducing cobalt(II)

quantitatively to cobalt(II). The amount of iodine liberated was titrated with standard

sodium thiosulphate solution. The final solution was pink in colour.

c. Calculate the percentage by mass of cobalt in D if 0.556 g of D requires 25.00 cm3

of sodium thiosulphate solution (0.100 M) for complete reaction.

A blue solution E was also formed when a large excess of concentrated hydrochloric

acid was added to the pink aqueous solution A of cobalt(II) chloride hexahydrate.

d. What Is the chemical species which is responsible for the blue colour in solution E?

Suggest its structure.

(Relative atomic masses: H=1.0; C=12.0; N=14.0; O=16.0;

Cl=35.5; Co=58.9)

4. When Cu(OH)2.xH2O is shaken with pure water, the liquid portion of the mixture

remains practically colourless; however, if the water contains dissolved ammonium

chloride, a blue solution results.

5. a. Give the formula for hexaaquacobalt(II) chloride.

b. When concentrated hydrochloric acid is added to hexaaquacobalt(II) chloride the

colour of the solution changes from pink to blue. Explain why.

6. a. State THREE characteristic properties of d-block elements, apart from complex ion

formation. In each case, illustrate your answer with an example involving copper or

vanadium.

b. You are provided with the following standard reduction potentials:

E

/ V

Cu2+

(aq) + e-q e Cu

+(aq) +0.15

Cu

2+

(aq) + 2e

-

q e Cu(s) +0.34Cu

+(aq) + e

-q e Cu(s) +0.52

I2(s) + 2e-q e 2I-(aq) +0.54

Cu2+

(aq) + I-(aq) + e

-q e CuI(s) +0.92

Using the above information,

i. explain why copper(I) compounds are unstable in aqueous solutions; andii. predict what will be observed when a potassium iodide solution is added to a

copper(II) suphate(VI) solution. Explain your predication and write a balanced

equation for the reaction involved.

7/31/2019 d-blcok

3/5

Suggested solutions:

1. a. i. Fe2+

(aq) ion is readily oxidized to Fe3+

(aq) ion because in undergoing

oxidation, Fe2+

will become Fe3+

whose electronic configuration has a

half-filled 3d subshell which is preferable.

ii. Mn3+

(aq) ion is readily reduced to Mn2+

(aq) ion because in undergoing

reduction, Mn3+

will become Mn2+

whose electronic configuration has the

preferred half-filled 3d subshell.

b. Sc3+

3d0

Cu+

3d10

Mn3+ 3d4

V4+

3d1

Ti3+

3d1

Sc3+

and Cu+

are colourless

Because (1) absence of d electrons in the case of Sc3+

(2) d orbitals are completely filled in Cu+

The colour of transitional metal ion is due to the presence of partially occupied d

orbitals.

c. Cr: [Ar] 3d54s

1Zn: [Ar] 3d

104s

2

2. a.Co NH3 Cl

% mass 25.59 29.00 45.41Relative no. of moles 25.59/58.9 29.00/17 45.41/35.5

Ratio 0.4345 1.7059 1.2792

Simplest ratio 1 4 3

x = 4, y = 3

b. trans-dichlorotetraamminecobalt(III) chloride

cis- dichlorotetraamminecobalt(III) chloride

p. 3

7/31/2019 d-blcok

4/5

3. a. Mole ratio of N : C : Cl

=25.2/14.0 : 5.4/12.0 : 16.0/35.5

= 4 : 1 : 1

or NH3 : CO3 : Cl = 4 : 1 : 1

molecular formula : Co(NH3)4CO3Cl

b. CoCl2.6H2O in water gives a pink [Co(H2O)6]2+

(A) which in basic medium

rendered by ammonia immediately turned to a blue Co(OH)2 (B).{or [Co(NH3)6]2+

,

[Co(NH3)4(OH)2]}

When air was drawn through the solution, oxidation occurred oxidizing thecobalt(II) species to the cobalt(III) state in the form of either [Co(NH3)4CO3]

+or

[Co(NH3)4CO3]Cl.

Continual addition of ammonium carbonate is to maintain a sufficient carbonate

concentration to keep the cobalt as its carbonate complex.

C is the unreacted Co2O3 (or Co(OH)3). On concentration, the desired product of

[Co(NH3)4CO3]Cl crystallizes Out.

c. 1 Co(III) 1 Co(II) 1 I 1 S2O32-

mass of Co = [25(0.1)/1000](58.9) = 0.147% of Co = 0.147 / 0.556 x 100% = 26.5%

d. E: [CoCl4]2-

or [CoCl6]4-

tetrahedral octahedral

4. Cu(OH)2.xH2O is practically insoluble in water, hence [Cu]2+

is very low.

Cu(OH)2q w e Cu2+

+ 2OH-or low solubility of Cu(OH)2 in water

NH4+

+ OH-q w e NH3 + H2O

4NH3 + Cu2+

q w e [Cu(NH3)4]2+

blue colour is due to [Cu(NH3)4]2+

complex ion

p. 4

7/31/2019 d-blcok

5/5

p. 5

5. a. [Co(H2O)6]2+2Cl-

b. The chloride ion (from the conc. hydrochloric acid) replaces water to form a new

complex.

[Co(H2O)6]2+

+ 4Cl- [CoCl4]

2-+ 6H2O

The main cause for the blue colour is that the Co changes from octahedral to

tetrahedral structure.

6. a. exhibition of variable oxidation states e.g. Cu(I) & Cu(II) / V2+

. V3+

, VO2+

and

VO2+

formation of coloured compounds e.g. Cu2+(aq) is blue / VO2+ is yellow.

Exhibition of catalytic properties e.g. V2O5 in contact process

Formation of non-stoichiometric compounds e.g. sulphide of vanadium

b. i. In aqueous solution, Cu+ disproportionates to give Cu(s) and Cu2+(aq)2Cu+(aq) Cu(s) + Cu2+(aq)

because the above reaction has a +ve E

of +0.52 (+0.15) = +0.37 V

ii. A brown solution and a white precipitate are formed because the reaction

2Cu2+

+ 4I- 2CuI + I2

has a E

value of (0.92 0.54) = +0.38 V