1 Chemical Bonding and Molecular Structure
Which of the following arrangement of molecules is correct on the basis of their dipole moments? 1) BF3 > NF3 > NH3 2) NH3 > BF3 > NH3 3) NH3=NF3 > BF3 4) NH3 > NF3 > BF3
SOLUTION
NH3 > NF3 are pyramidal and have lone pairs. But NH3 have more dipole moment. BF3 have sp2 hybridization, and planer trigonal due to symmetry dipole moment is zero
2 Purification and Characterisation of Organic Compounds
Coal tar is obtained as a by product during :
1) Destructive distillation of wood
2) Destructive distillation of coal
3) Destructive distillation of bones
4) Steam distillation of light oil
Destructive distillation is the decomposition of feedstock by heating to a high temperature; process breaks up large molecules. Coke, coal gas, gas carbon, coal tar, etc are obtained.
3 Chemical Equilibrium
35 g of NH4OHis dissolved in 10L of H2O. The percentage dissociation of NH4OH is- (M.wt NH4OH - 35; pK, NH4OH - 4.75) 1) 13.4 2) 1.34 3) 26.8 4) 2.68
Given that
pKb=4.75⇒Kb=1.8×10−5
NH4OH=35g
Concentration CC of NH4OH=35 /(35×10)
=0.1M
4 Hydrocarbons
An alkyl bromide, RBr of molecular weight 151 is the exclusive product of bromination of which hydrocarbon?
1) Dodecane 2) 2,2 - dimethylpropane 3) 2,2 - dimethylhexane 4) 2,2,3 - trimethylheptane
Molecular weight of RBr=151
Molecular weight of Br=80
∴ Molecular weight of −R=151−80=71
If we consider 2,2−dimethylpropane the molecular formula is C5H12 .
After bromination, C5H11Br
∴ Molecular weight of C5H11=60+11=71
5 Classification of Elements and Periodicity
The highest ionization potential in a period is shown by
1) halogens
2) noble gases
3) alkaline earth metals
4) alkali metals
6 Solid State
The number of atoms in 100 g of a fcc crystal with density of 10.0 g cm−3 and cell edge equal to 200 pm is equal to : 1) 5 × 1024 2) 5 × 1025 3) 6 × 1023 4) 2 × 1025
density $$ = \dfrac{zM}{N_a \times (a \times 10^{-10})^3}$$ $$10 = \dfrac{4 \times M}{6 \times 10^{23} \times (200\times 10^{-10})^3}$$ $$M= 12 \text{ g/mol}$$ Number of atoms is 100 g $$= \dfrac{100}{12} \times 6 \times 10^{23}= 5 \times 10^{24}$$
7 P – Block Elements
What is the oxidation state of nitrogen in N2?O5?? 1) 5 2) 10 3) 4 4) 3
Let the oxidation state of N be x. Oxidation state of O is −2 as it is in oxide form.
⇒2x+5(−2)=0
⇒x=10 /2?=+5
8 Chemical Equilibrium
Buffer solution can be obtained by the mixing aqueous solutions of:
1) CH3COONa and CH3COOH 2) CH3COOH and excess NaOH 3) CH33COONa and excess HCl 4) NaCl and HCl.
Solution
A mixture of acetic acid (a weak acid) and sodium acetate ( its salt with strong base sodium hydroxide) acts as acidic buffer.
9 Chemistry in Everyday Life
The compound used as a preservative for food products such as tomato ketchup and fruit juices is
1) Sodium benzoate 2) Formic acid 3) Sodium Salicylate 4) Sodium acetate.
10 S – Block Elements
The pair of amphoteric hydroxides is:
1) Al(OH)3 and LiOH 2) B(OH)3 and Be(OH)2 3) Be(OH)2 and Zn(OH)2 4) Be(OH)2 and Mg(OH)2
Beryllium hydroxide and zinc hydroxide are amphoteric in nature because they react with both, acid and base. Zinc hydroxide as acid and base:
Zn(OH)2 + 4H+ → Zn2+ + 2H2O
Zn(OH)2 + 2OH− → [Zn(OH)4]2−
Beryllium hydroxide as acid and base:
Be(OH)2 + 2HCl → BeCl2+2H2O
Be(OH)2 + 2NaOH → Na2[Be(OH)4]
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