BITSAT Chemistry practise questions

2)- I can only think of Hydrogen bonding right now.(Greater the extent of H-Bonding, higher the solubility in water.)

8) 2 & 4 both are Definitely wrong.

2) Perhaps- 1. If we consider the extent of HBonding that is.

3) -3. (Agreeing with sridhar)

4) We can go with -4, i think.

5) -4.

6) -2. (Na- Electrolysis, Ag- Cyanide process, Cu- Self reduction.)

7) -4 (Sridhar's rite).

8) -2,4 both seem to be wrong. (Picric ACid has an -OH group naa...?)

9) -2. (@Sandy, u gave just the opposite,,,, gave fr covalent charac. order i think, in hadbadi ;) )

10) -2.

6. option 2.Na is higher in Electrochemical series is extraxted by electrolysis,Ag by complex formation with NaCN and Cu by self reduction.
5. option 4.zeolites are sodium alumino silicates.
1. option 4.
3. option 3.agree with sridhar's reason in #11
4. option 4.all the given statements are correct.for B the cations have noble gas nature with no unpaired electrons.

4----> 4

all of them are the properties of s-block elements...

2---->2

P.S. though a guess....

9--->3

for a fixed cation .....larger the difference between the size of cation and anion more is ionic character

8) Is it 3? Not sure.

Q2) Correct order of hydrolysis is

(1) CCl4 < AlCl3 < SiCl4 < PCl5
(2) PCl5 < AlCl3 < CCl4 < SiCl4
(3) AlCl3 < PCl5 < CCl4 < SiCl4
(4) CCl4 < AlCl3 < PCl5 < SiCl4
90

Q3)A substance will be deliquescent, if its
vapour pressure is

(1) equal to the atmospheric pressure.
(2) equal to that of water vapour in air.
(3) lesser than that of water vapour in air.
(4) greater than that of water vapour in
air.
93

Q4)Characteristics of s-block elements are:

(A) They are prepared by the electrolysis
of their fused salts.
(B) Their cations are diamagnetic.
(C) Elements have low density, good
conductor of heat and electricity.
(D) Their oxides are basic in property.

(1) Only D
(2) A and B
(3) B and D
(4) A, B, C and D
111

Q5) Zeolites are

(1) double salts of B and Al.
(2) double sulphates of alkali metals with
Cr+3, Al+3, Fe+3.
(3) double halides of K and Mg.
(4) double silicates of Al and Na.
112

Q6) Consider the following metallurgical process:

(A) Heating the sulphide ore in blast
furnace so that a part of it is converted
to oxide and then on further heating
in the absence of air, it undergoes reduction
to get the metal.
(B) Electrolysing the molten electrolyte
with CaCl2 and the metal is got by
electrolytic reduction.
(C) The sulphide ore is mixed with aqueous
NaCN in presence of air, the aqueous
metal complex formed is precipitated
by adding zinc powder.

The process used for obtaining sodium,
silver and copper respectively are

(1) A, B and C
(2) B, C and A
(3) C, B and A
(4) B, A and C
120

Q7)On reduction with LiAlH4, amines can be
prepared by
(1) RCN
(2) RNC
(3) RCONH2
(4) All of these
107

Q8)Which is not true about aniline?

(1) It undergoes bromination at ortho
and para positions.
(2) It will undergo Friedel-CraftÂ’s acylation
and alkylation in presence of
anhydrous AlCl3.
(3) It undergoes bromination at meta
position in presence of strong acid.
(4) It forms picric acid with nitrating
mixture.
109

Q9)Correct order of ionic character of silver
halides is

(1) AgCl > AgF > AgBr > AgI
(2) AgF > AgCl > AgBr > AgI
(3) AgI > AgBr > AgCl > AgF
(4) AgF > AgI > AgBr > AgCl
88

Q10) Which is the correct statement among
the following?

(1) Peroxide ion has a bond order 2
while O2 has a bond order 1.
(2) Peroxide ion has a longer and weaker
bond than O2.
(3) Peroxide ion and O2 are paramagnetic.
(4) Bond length of peroxide ion is stronger
than O2.
89

3) 3

Less than the vapour pressure of water vapour in air....A deliquescent compound has affinity for water vapour molecules....It continues to associate with water vapour molecules until its vapour pressure equal to vapour pressure of water vapour in air.

yeah sridhar correct

5) 4 Fact..

7)4 Cyanide on reduction gives primary amine. Isocyanide on reduction gives secondary amine. Amides also give primary amines on reduction

10)2

o₂^-2 has bond order 1 while o₂ has bond order 2.
we know that bond order is inversely proportional to bond length. therefore peroxide ion has larger bond length.

1) 4
SQPR