# Plus One Chemistry Improvement Question Paper 2018

## Kerala Plus One Chemistry Improvement Question Paper 2018

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to Candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to get familiar with questions and to plan your answers.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all questions from question numbers 1 to 7. Each carry one score.

Question 1:
Name the type of smog generally formed during cool and humid climate.

Question 2:
Predict the shape of XeF4 molecule, according to VSEPR theory.

Question 3:
Which is the acidic oxide among the following?

Question 4:
How many angular nodes are present in a 5f-orbital?

Question 5:
The allotrope of carbon with the highest thermodynamic stability is …………

Question 6:
The critical temperatures of some gases are given in the following table:

Question 7:
The number of oxygen atoms present in 5 moles of glucose (C6H12O6) is ……….

Answer any ten from question numbers 8 to 20. Each carries two scores.

Question 8:
By using the concept of hybridization, explain the structure of H2O molecule.

Question 9:
Give the IUPAC names of the following compounds.

Question 10:
Write any two applications of green chemistry in day-to-day life.

Question 11:
Explain the effects of temperature and pressure on the following equilibrium.

Question 12:
Draw the structure of orthoboric acid. Why it is not a protonic acid?

Question 13:
Find the molecular formula of the compound with molar mass, 78g mol-1, and empirical formula CH.

Question 14:
What is meant by entropy of a system? What happens to the entropy during the following changes?

Question 15:
Represent graphically, the variation of prob-ability density ($${ \Psi }^{ 2 }$$ (r )) as a function of distance (r) of the electron from the nucleus for 1s and 2s orbitals.

Question 16:
Derive an equation relating molar mass of an ideal gas with its density.

Question 17:
What is Wurts reaction? Give an example.

Question 18:
Define buffer solutions and write one example for an acidic buffer.

Question 19:
Cycloheptatrienyl Cation is given below :

Is this ion aromatic or not? Justify the answer.

Question 20:
Write the thermochemical equation corresponding to the standard enthalpy of formation of benzene.
Hint: ∆f H$$\Theta$$ of benzene = +49.0 Kjmol-1

Answer any seven from question numbers 21 to 29. Each carries three scores.

Question 21:
Justify the following:
a. ‘Ne’ has positive value for electron gain enthalpy.
b. The electron gain enthalpy of ‘F’ is lower than that of ‘Cl’.
c. The size of ‘Al3+’ is lower than that of F.

Question 22:
Identify X, Y and Z in the following sequence of reactions :

Question 23:
Calculate the mass of oxalic acid dihydrate (H7C2O4.2H2O) required to prepare 0.1M, 250 ml of its aqueous solution.

Question 24:
Balance the following Redox reaction by the ion-electron method or oxidation number method (Acid medium)

Question 25:
Explain any one method of preparation and structure of diborane.

Question 26:
The value of the equilibrium constant is useful to predict the extent of reaction and the direction of the reaction at a given stage. Explain.

Question 27:
Give a reason for the following :
a. H2O2 is stored in wax-lined glass or plastic vessels in dark.
b. Hard water is not suitable for laundry.

Question 28:
Calculate the total pressure in a mixture of 3.5 g of dinitrogen and 16 g of dioxygen confined in a vessel of 2 dm3 at 27°C. (R=0.083 bar dm3K-1 mol-1).

Question 29:
The reaction of cyanamide (NH2CN) with dioxygen was carried out in a bomb calorimeter and AM was found to be -742.7kJ mol-1, at 298 K. Calculate enthalpy change for the reaction at 298 K.

Answer any three from question numbers 30 to 33. Each carries four scores.

Question 30:
Write the molecular orbital electronic configurations of N2 and O2 and calculate their bond orders. Give a comparison of their stability and magnetic behavior.

Question 31:
Briefly describe the principles of the following techniques, taking an example in each case.
a. Crystallization
b. Simple distillation
c. Distillation under reduced pressure
d. Paper chromatography

Question 32:
Give the postulates of Bohr model of hydrogenatom. Also write two merits and two limitations of this model.

Question 33:
Account for the following :
a. Blue coloured solutions are obtained when alkali metals are dissolved in liquid ammonia.
b. ‘Li’and ‘Mg’ show similar properties.
c. Aqueous solution of Na2CO3 is alkaline.
d. BeS04 and MgSO4 are readily soluble in water.

Classical smog

Square planar

a. Cl207

3

Graphite

O2

5 mole glucose contains 30 moles oxygen.
So the number of Oxygen atoms present in 5 moles of glucose = 30 x 6.022 x 1023

The shape should have been tetrahedral if there were all bp but two Ip are present so the shape is distorted tetrahedral or angular. The reason is Ip-Ip repulsion is more than Ip- bp repulsion which is more than bp-bp repulsion. Thus, the angle is reduced to 104.5° from 109.5°

a. 3 – Ethyl – 1,1 – dimethylcyclohexane
b. 3 – Bromo – 3 – chloroheptane

• Dry cleaning of clothes. Tetrachloroethene
(Toxic and contaminates groundwater)
which is used as a solvent for dry cleaning is replaced by liquified CO2.
• Bleaching of paper. Cl2 gas is replaced H2O2.
• Synthesis of chemicals.

Low temperature favours forward reaction.
High pressure favours forward reaction.

Protonic acid are those acids which ionizes in solution and produce protons . But Boric Acid/orthoboric acid is monobasic and a weak acid. It doesn’t donating proton it accepts electrons of OH- which produces due to ionisation of water into H + and OH- hence we can say it act as a lewis acid and proton is remove from water instead of boric acid.

The empirical formula of the compound = CH
Empirical formula mass =12 x 1 + 1 x 1 = 13
Molar mass = 78

Measure of degree of disorders or randomness of a system.
a. Entropy decreases
b. Entropy increases

Ideal gas equation can be rearranged as follows:
$$\frac { n }{ V }$$ = $$\frac { P }{ RT }$$
Replacing n by $$\frac { m }{ M }$$
$$\frac { m }{ MV }$$ = $$\frac { P }{ RT }$$
We get $$\frac { d }{ M }$$ = $$\frac { P }{ RT }$$ ; where d is the density.
On rearranging equation we get the rela-tionship for calculating the molar mass of a gas.
M = $$\frac { dRT }{ P }$$

Alkyl halides on treatment with sodium metal in dry ethereal (free from moisture) solution give higher alkanes. This reaction is known as Wurtz reaction and is used for the preparation of higher alkanes containing even number of carbon atoms.

A buffer solution is one which can resist change in its pH value on the addition of small quantities of acid or base.
Example for acidic buffer is blood.

Yes. It is aromatic. In the case of cycloheptatriene, there are seven carbon atoms. Except one carbon (one is in sp³ hybrid state), all are in sp² hybrid state and each has an unhybridized pure p orbital perpendicular to the molecular plane, in the case of cycloheptatrienyl cation contains 6 π electrons, fulfilling the Huckle rule and the carbocation also in sp² hybrid state leads continuous delocalization. Hence it is aromatic.

6C(s) + 3H2(g) → C6H6 (I); ∆f H$$\Theta$$ = + 49 kJ/mol

a. Noble gases have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very un stable electronic configuration. It may be noted that electron gain enthalpies have large negative values toward the upper right of the periodic table preceding the noble gases.
b. Electron gain enthalpy of F is less negative than that of the succeeding element This is because when an electron is added to F, the added electron goes to the smaller n = 2 quantum level and j suffers significant repulsion from the i other electrons present in this level. For the n = 3 quantum level ( Cl), the ; added electron occupies a larger > region of space and the electron-electron repulsion is much less.
c. For iso electronic species, size of ion decreases with increase in nuclear charge atomic number or cation has a smaller size than anion.

Molarity = $$\frac { No. of moles in the solute }{ Volume of solution in ml }$$ x 1000
Moles of oxalic acid dihydrate in the solution = 0.1 x 0.25 = 0. 025 moles.
Molecular mass of oxalic acid dihydrate is 2+12 x 2 + 16 x 4 + 2(2+16) = 126
Mass of the acid used = 126 x 0.025= 3.15g

In diborane 4 terminal H-atom and the 2 boron atom lies one plane. Above and below this plane, there are 2 bridging H-atom. Each bridged H atom is bonded to 2 boron atom only by sharing of 2 electrons. Such a bond is called 3 centered 2 elec¬tron bond. Thus diborane have two (B-H-B) bonds, ie., two 3-centered 2 electron bond and four (B-H) bonds, ie., four 2 centered 2 electron bond.

Larger the value of K, greater is the extent to which the reactants are converted into the products.

• If K >103, the product predominates over reactants at equilibrium. If K is very large the reaction proceeds nearly to completion.
• If K<103, the product predominates over product. If K is very small the reaction proceeds to very small extend or rarely.
• If K is in the range of 10-3 to 103, appreciable concentrations of both reactants and products are present.

Prediction of direction of a reaction

The reaction quotient in terms of concentration or concentration Quotient (Q) with molar concentrations (Qc) and with partial pressures (Qp) is related to the equilibrium constant (K) for the reaction.
For a general reaction, aA+ bB ⇌ cC + dD

• If Qc > Kc, the reaction will proceed in the direction of reactants (reverse reaction)
• If Qc < Kc, the reaction will proceed in the direction of the products (forward reaction)
• If Qc = K, the reaction is at equilibrium.

a. H2O2 decomposes in the presence of light and alkali content in glass. To prevent decomposition H2O2 is stored in wax – lined glass or plasic vessels indark.
b. Hard water do not form lather easily with soap, due to the presence of dissolved salts in it.

Since the bond order is 3, there is a triple bond in N2 and the bond energy is very high. Since all the electrons are paired. It is diamagnetic.

Oxygen has unpaired electrons and hence I it is paramagnetic in nature.

Crystallization:
This is one of the most commonly used techniques for the purification of solid organic compounds. It is based on the difference in the solubilities of the compound , and the impurities in a suitable solvent.
e.g., sugar and salt.

Simple distilation :
It is the process of eva-poration followed by condensation. It is used for separating a liquid which does not decompose on boiling from other liquids = having diferent boiling points or should contain a non-volatile component,
e.g., Benzene and Toluene. The vapours of a substance formed are condensed and the liquid is collected in the conical flask.

Distillation under reduced pressure: The method is used to separate those liquids which decompose at or below their normal ; boiling points. In this method, the distillation is carried at a lower pressure of 10 – i 20 mm Hg, so that the liquid can be boiled i at lower temperature without undergoing decomposition, e.g., Glycerol can be sep- arated from spentlye in the soap industry.

Paper chromatography: Partition chromaography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases, Paper chromatography is a type of partition chromatography. In paper chromatograph, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase.

1. The electron in an atom revolve around the nucleus only in certain circular path called orbits.
2. The energy of the electron in a particular orbit remains constant.
3. Only those orbits are permitted in which the angalar momentum of the electron is a whole number of multiple of h/27r, where h is plank’s constant.
4. Energy is emitted or absorbed by an atom only when an electron moves one from one orbit to another.

Merits

1.  It could account for the stability of hydrogen atom
2.  It could account for the line spectra of hydrogen atom and hydrogen like ions (for example, He+, Li2+, Be3+, and so on)

Demerits

1. Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect).
2. It could not explain the ability of atoms to form molecules by chemical bonds