Chemistry 2009 Set 3 Sol

Chemistry 2009 Set 3 Close

Subjective Test

These are additional set 3 questions.

Question 7 ( 1.0 marks)

Give the IUPAC name of the following compound:

Solution:

Nylon is a polymer of 1, 6 hexanedioic acid and hexamethylene diamine, both having 6 carbon atoms.Hence, it is called nylon 6, 6.


Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na2SO4. If this

solution actually freezes at −0.320°C, what would be the value of Van’t Hoff factor? (Kf for water is

1.86 °C mol−1)

Solution:

Molality, m = 0.0711 m

Kf = 1.86ºC mol-1

∴ Depression in freezing point = Kf × m

= 1.86 × 0.0711

= 0.132oC

Freezing point = 0ºC − 0.132ºC

= −0.132ºC

Now, Van’t Hoff factor,

[Note − Theoretically the value of Vant Hoff factor should be 3 but according to the values given inthe question, the value of Vant Hoff factor is coming out to be 2.42. ]


Compare the following complexes with respect to their shape, magnetic behaviour and the hybridorbitals involved:

(i) [CoF4]2−

(ii) [Cr(H2O)2 (C2O4)2]−

(iii) [Ni(CO)4]

(Atomic number: Co = 27, Cr = 24, Ni = 28)

Solution:

(i) [ CoF ]2−

2/4/2011 Subjective Test Paper - Chemistry - Meri…

…meritnation.com/…/OaWmjlXI15m2q6I… 1/6

(i) [ CoF4]2−

Let the oxidation state of Co in [CoF4]2− be x.

x − 4 = −2

∴x = +2

Due to the unpaired electrons, it is paramagnetic in nature.

(ii) [Cr(H2O)2 (C2O4)2]−

Let the oxidation state of Cr in [Cr(H2O)2 (C2O4)2]− be x.

x − 2 × 2 = −1

∴x = +3

Thus, it has d2sp3 hybridisation with octahedral shape. Due to the presence of unpaired electrons, itis paramagnetic in nature.

(iii) [Ni(CO)4]

Let the oxidation state of Ni in [Ni(CO)4] be x.

x + 4(0)= 0

∴x = 0


What are the following substances? Give one example of each type.

(i) Antacid

(ii) Non-ionic detergents

(iii) Antiseptics

Solution:

(i) Antacids are stomach acid neutralisers. They raise the pH to reduce acidity in the stomach.

Example − aluminium hydroxide

(ii) Non-ionic detergents produce electrically neutral colloidal particles in solution.

Example − esters of high molecular mass formed by reaction between stearic acid and polyethyleneglycol



(iii) Antiseptics are used for destroying microorganisms off the skin of humans and animals.

Example − boric acid


(a) Complete the following chemical reaction equations:

(i)

(ii)

(b) Explain the following observations:

(i) In general the atomic radii of transition elements decrease with atomic number in a given series.

(ii) The for copper is positive (+ 0.34 V). It is the only metal in the first series of transition

elements showing this type of behaviour.

(iii) The E0 value for Mn3+ | Mn2+ couple is much more positive than for Cr3+ | Cr2+ or Fe3+ | Fe2+

couple.

OR

(a) What is meant by the term lanthanoid contraction? What is it due to and what consequencesdoes it have on the chemistry of elements following lanthanoids in the periodic table?

(b) Explain the following observations:

(i) Cu+ ion is unstable in aqueous solutions.

(ii) Although Co2+ ion appears to be stable, it is easily oxidised to Co3+ ion in the presence of astrong ligand.

(iii) The value for manganese is much more than expected from the trend for other

elements in the series.

Solution:

(a)

(i)

(ii)

(b)

(i) Among the elements of a particular transition series, as atomic number increases, atomic radiusdecreases. This is due to the fact that on moving across the period, electrons are added to the samed orbital. Thus, effective nuclear charge increases, which is only partly cancelled by the increasedscreening effect of electrons. As a result, atomic radius decreases.

(ii) Copper shows a unique behaviour in its Eº value. The electrode potentials are the measure ofthe value of enthalpy change. Furthermore, enthalpy change in a reaction is the sum of ionisationenergy, enthalpy of hydration of metal ion and sublimation of metal. Cu has very low enthalpy ofhydration & high ∆Hº value, thereby resulting in a positive electrode potential value.

(iii) The change from Mn3+ to Mn2+ results in the formation of half-filled d5 configuration, which has

extra stability. EºM3+/M

2+ has a high positive value since the third ionisation energy is higher. On

the other hand, the low value of Fe is due to the extra stability of Fe3+, i.e., d5 configuration. Cr2+ is

reducing as its configuration changes from d4 to d3. In d3 configuration, t2g level is half-filled. So,

Cr3+/ Cr2+ has a low value.

OR

(a) The steady decrease in the size of lanthanoid ions (M3+) with the increase in atomic number isknown as lanthanoid contraction.

It is due to the filling of the 4f orbitals before the 5d orbitals. So, the effective nuclear chargeincreases. However, the f orbital is rather ineffective in screening the nuclear charge. Due tolanthanoid contraction, the chemistry of elements differs. The early members of lanthanoids are quitereactive, but with increase in atomic number, their reactivity decreases. Lanthanoid compounds areionic and generally coloured. Due to lanthanoid contraction, these elements exhibit slight variation intheir property, and thus, they can be separated easily.

(b)

(i) Cu+ ion disproportionate in aqueous solution to form an oxidised and a reduced species.



It is energetically favourable for these changes to occur.

So, Cu+ ion is unstable in aqueous solution.

(ii) The electronic configuration of Co3+ is 3d6 4sº. So, pairing occurs in the presence of a strongligand. Thus, there are no unpaired electrons and it is stable.

Co − 3d6

In the presence of a strong ligand:

However, in Co2+, whose electronic configuration is 3d7, there is one unpaired electron even after

pairing occurs in the presence of a strong ligand. Hence, Co2+ is oxidised to Co3+.

(iii) Mn2+ has the electronic configuration 3d5 4s2. Hence, due to a half-filled orbital, Mn2+ is stable.Now, d electrons have a poor shielding effect. So, the effective nuclear charge increases and the size

decreases in Mn2+. As such, it will be difficult for Mn2+ to accept two electrons and then get reducedto Mn.


(a) Define the term molar conductivity. How is it related to conductivity of the related solution?

(b) One half cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution ofunknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO3)2. A voltage of 1.48 V is measured for this cell. Use this information to calculate the

concentration of silver nitrate solution used.

OR

(a) Corrosion is essentially an electrochemical phenomenon. Explain the reactions occurring duringcorrosion of iron kept in an open atmosphere.

(b) Calculate the equilibrium constant for the equilibrium reaction

Solution:

(a) Molar conductivity (^m) is defined as the conductance of a solution containing 1 g molecule or 1mol of electrolyte such that the entire solution is placed between two electrodes, 1 cm apart.

= Conductivity

V = Volume of solution

∴ ^m =

(b) At anode:

Zn → Zn2+ + 2e-

At cathode:

Ag+1+ e- → Ag] ×2

Net cell reaction:

Zn + 2Ag+ → Zn2+ + 2 Ag

Eºcell = Eºcathode − Eºanode

= 0.80 − (-0.76)

= +1.56V



OR

(a) Corrosion is defined as the process of slow conversion of metals into their undesirablecompounds due to the reaction with moisture and other gases present in the atmosphere

Due to this process, impure iron surface behaves like a small electrochemical cell in the presence ofwater containing dissolved oxygen or carbon dioxide.

At the anode, oxidation of Fe atom takes place.

At the cathode, electrons are picked up by the Fe+ ions, which are produced by either water orcarbonic acid (formed by CO2 and H2O).

Thus, the net cell reaction is

Eºcell = 1.67 V

The formed ferrous ions move through water. These are oxidised to ferrous state by atmosphericoxygen to form rust.

(b) Eºcell = Eºcathode − Eºanode

At cathode:

At anode:

Now, Eº cell =

∴n = 2

Kc = antilog (1.356)

Kc = 22.695





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Chemistry 2009 Set 3 Sol