Chemistry:2018:CBSE:[All India]: Set-I
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Q1
Analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. Give reason.
Marks:1View AnswerAnswer:
Iron existsin nature in +2 and +3 oxidation states. Therefore, the composition of iron oxide is non-stoichiometric.
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Q2
CO(g) and H2(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions?
Marks:1View AnswerAnswer:
The selectivity of a catalyst is its ability to direct a reaction to yield a particular product. This property of enzymes enables CO and H2 to give different products when they react in the presence of different catalysts.
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Q3
Write the coordination number and oxidation state of Platinum in the complex [Pt(en)2Cl2].
Marks:1View AnswerAnswer:
en is a bidentate ligand and chlorine is a monodentate ligand.
Hence, the coordination number of Platinum is 6.
Let x be the oxidation state of Platinum in the complex
So, x + 0+ (–1 x 2) = 0
x = +2Therefore, the oxidation state of Platinum is +2in the complex [Pt(en)2Cl2].
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Q4
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why?
Marks:1View AnswerAnswer:
Benzylchloride readily undergoes hydrolysis with aqueous NaOH as compared to chlorobenzene due to following reasons.
(i)In Chlorobenzene, the electron pair on chlorine atom is in conjugation with pi- electrons of the ring. Therefore, C-Cl bond acquires partial double bond character due to resonance and it does not undergo hydrolysis easily.
(ii) In Chlorobenzene, Cl is attached to sp2 hybridised carbon atom whereas in benzylchloride it is attached to sp3 hybridised carbon atom.Therefore,bond length in chlorobenzene is shorter. Since, it is difficult to break a shorter bond than a longer bond; therefore, chlorobenzene does not undergo hydrolysis easily.
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Q5
Write the IUPAC name of the following:
Marks:1View AnswerAnswer:
The IUPAC name of the following compound is 3,3-dimethylpentan–2–ol.
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Q6
Calculate the freezing point of a solution containing 60g of glucose (Molar Mass = 180g mol-1) in 250g of water.
(Kf of water = 1.86K kg mol-1)
Marks:2View AnswerAnswer:
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Q7
For the reaction
2N2O5(g) → 4NO2(g) + O2(g),
the rate of formation of NO2(g) is 2.8 x 10-3Ms-1. Calculate the rate of disappearance of N2O5(g).
Marks:2View AnswerAnswer:
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Q8
Among the hydrides of Group-15 elements,which have the
(a) lowest boiling point?
(b) maximum basic character?
(c) highest bond angle?
(d) maximum reducing character?
Marks:2View AnswerAnswer:
a) lowest boiling point: PH3
b) maximum basic character : NH3
c) highest bond angle : NH3
d) maximum reducing character: BiH3
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Q9
How do you convert the following?
(a) Ethanal to Propanone
(b) Toluene to Benzoic acid
Marks:2View AnswerAnswer:
(a) Ethanal to propanone
(b) Toluene to Benzoic acid
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Q10
Account for the following:
(a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.
(b) pKa value of 4-Nitrobenzoic acid is lower than that of benzoic acid.Marks:2View AnswerAnswer:
a. Carboxylic acids do not undergo Friedel-Crafts reaction because the catalyst aluminium chloride (Lewis acid) gets bonded to the carboxyl group.
b. 4–Nitrobenzoic acid has nitro group at the para position which is an electron withdrawing group. In general, the electron withdrawing groups increase the acidity of carboxylic acids by stabilising the conjugate base through delocalisation of the negative charge by inductive and/or resonance effects.
