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Test: Tin and lead compounds - JEE MCQ


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20 Questions MCQ Test - Test: Tin and lead compounds

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Test: Tin and lead compounds - Question 1

Ge(II)compounds are powerful reducing agents whereas Pb(IV) compounds are strong oxidants. It is because

Detailed Solution for Test: Tin and lead compounds - Question 1

Ge(II) tends to acquire Ge (IV) state by loss of electrons. Hence it is reducing in nature. Pb(IV) tends to acquire Pb (II) O.S. by gain of electrons. Hence it is oxidising in nature. This is due to inert pair effect.

Test: Tin and lead compounds - Question 2

The correct order of melting points of Al,Ga,In is

Detailed Solution for Test: Tin and lead compounds - Question 2

The melting points of group 13 elements follow the order:-
B > Al > Tl > In > Ga
This is due to strong crystalline structure of B and soft metallic nature of the rest of the metals.
Gallium has unusually low melting point due to very high electro positivity.

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Test: Tin and lead compounds - Question 3

PbF4,PbCl4 exist but PbBrand PbI4 do not exist

Detailed Solution for Test: Tin and lead compounds - Question 3

F and Cl are more oxidising in nature and can achieve Pb in (IV) O.S. but Br2 and I2 can not achieve Pb in (IV) O.S. secondly Pb4+ is strong in oxidising nature and in its presence, Br and I can not exist.

Test: Tin and lead compounds - Question 4


incorrect statement for Q is

Detailed Solution for Test: Tin and lead compounds - Question 4

Here only statement in option C is incorrect as it combines with FeSO4 and not it combines with Fe2(SO4)3

paramagnetic, coloured gas

Test: Tin and lead compounds - Question 5

A metallic carbide on reaction with water gives a colourless gas which burns readily in air and gives a precipitate with ammonical silver nitrate solution. What is the gas evolved in reaction ?

Detailed Solution for Test: Tin and lead compounds - Question 5

Test: Tin and lead compounds - Question 6

Lead pipes are not suitable for drinking water because

Detailed Solution for Test: Tin and lead compounds - Question 6

Test: Tin and lead compounds - Question 7

Identify the incorrect statements from the following.
I. Tin in +2 state acts as reducing agent while lead in +4 state acts as strong oxidising agent.
II. Silicon exists as both and forms.
III. The hybridisation of carbon in fullerene is s3.
IV. Among Ge,Sn and Pb lowest melting point is for Sn.

Detailed Solution for Test: Tin and lead compounds - Question 7

(II). is known but does not exist, this is because due to larger size of Cl atom as compared to F atom it is not possible to accommodate six atoms of Cl around Si atom. (III). Fullerens structure have carbon atoms both in sp2 and sp3 hybridised system.
Correct option is (c).

Test: Tin and lead compounds - Question 8

A solid element (symbol Y) conducts electricity and forms two chlorides YCln (colourless volatile liquid) and YCln−2 (a colourless solid). To which one of the following groups of the periodic table does Y belong?

Detailed Solution for Test: Tin and lead compounds - Question 8

SnCl4 is colourless volatile liquid and SnCl2 is colourless solid Sn conducts electricity and it belongs to 14 group

Test: Tin and lead compounds - Question 9

Which of the following conceivable structures for CCl4 will have a zero dipole moment?

Detailed Solution for Test: Tin and lead compounds - Question 9

CCl4 is tetrahedral in nature.

Test: Tin and lead compounds - Question 10

Least thermally stable is -

Detailed Solution for Test: Tin and lead compounds - Question 10

The thermal stability of tetrahalides decreases in order CX4 > SiX4 > GeX4 > SnX4 and in terms ofsame metal with different halides is in order of MF4 > MCl4 > MBr4 > MI4

Test: Tin and lead compounds - Question 11

To a piece of charcoal, sulphuric acid is added. Then:

Detailed Solution for Test: Tin and lead compounds - Question 11

Charcoal is a pure form of carbon, its reaction with hot conc. H2SO4 is as follows:

Test: Tin and lead compounds - Question 12

The correct statement with respect to CO is

Detailed Solution for Test: Tin and lead compounds - Question 12

CO react with haemoglobin, forms carboxy haemoglobin and stopes the supply of O2

Test: Tin and lead compounds - Question 13

With respect to graphite and diamond, which of the statement (s) given below is (are) correct?
1. Graphite has higher electrical conductivity than diamond
2. Graphite is harder than diamond
3. Graphite has higher C - C bond order than diamond
4. Graphite has higher thermal conductivity than diamond

Detailed Solution for Test: Tin and lead compounds - Question 13

Here 1, 3 and 4 are correct as Graphite has higher electrical and thermal conductivities than diamond.
All C- atoms in graphite are sp2 hybridised so due to partial double bond character in graphite, bond order will be more than 1 but in diamond Bond order is one.
Here only statement 2 is incorrect as diamond is more hard than graphite.

Test: Tin and lead compounds - Question 14

Which of the following is similar to graphite?

Detailed Solution for Test: Tin and lead compounds - Question 14

Graphite and boron nitride have similar structure

Test: Tin and lead compounds - Question 15

The correct order of increasing C-O bond length of CO, CO2 and is:

Detailed Solution for Test: Tin and lead compounds - Question 15

Structures of CO,CO2 and CO32− are :



Bond order Hence, the decreasing bond length is:

Test: Tin and lead compounds - Question 16

Which of the following statement(s) is / are incorrect for CO2?
(i) In laboratory CO2 is prepared by the action of dilute HCl on calcium carbonate
(ii) Carbon dioxide is a poisonous gas
(iii) Increase in carbon dioxide content in atmosphere lead to increase in green house effect.
(iv) CO2 as dry ice is used as a refrigerant for ice cream and frozen food.

Detailed Solution for Test: Tin and lead compounds - Question 16

Carbon dioxide is not a poisonous gas.

Test: Tin and lead compounds - Question 17

Certain organic compound on combustion produces three gaseous oxides A, B and C. A and C turned lime water milky, B turned anhydrous CuSO4 blue and C turned K2Cr2O7 solution green. The elements present in organic compounds are

Detailed Solution for Test: Tin and lead compounds - Question 17

The elements present in organic compounds are C, H, S and their oxides are CO2,H2O,SO2 respectively.
CO2 turns lime water milky, water turns anhydrous copper sulphate blue and SO2 being reducing agent can reduce K2Cr2Ointo green coloured Cr3+ compound (CrCl3) etc.

Test: Tin and lead compounds - Question 18

The ions present in Al4C3,CaC2 and Mg2C3 are respectively

Detailed Solution for Test: Tin and lead compounds - Question 18

are present in Al4C3,CaC2 and Mg2C3 are respectively

Test: Tin and lead compounds - Question 19

Consider the following statements:
I. In diamond, each carbon atom is sp3-hybridised.
II. Graphite has planar hexagonal layers of carbon atoms.
III. Silicones being surrounded by non-polar alkyl groups are water repelling in nature
IV. The order of catenation in group 14 elements is Si > C > Sn > Ge > Pb.
The correct statements are

Detailed Solution for Test: Tin and lead compounds - Question 19

(I, II) In diamond, hybridisation of carbon is sp3 and graphite has planar hexagonal layer of C-atoms. The hybridisation of each C-atom in graphite is sp2.
(III) Silicones are organo-silicon polymers formed from inorganic and organic monomer having chemical formula (R2SiO)n, due to non-polar alkyl group, silicones are water repelling in nature.
(IV) The catenation tendency in group 14 elements decreases due to decrease in element bond strength. Thus, the correct order is
C > Si > Ge ≈ Sn
Hence, statement (IV) is incorrect.
Thus, option (a) is correct.

Test: Tin and lead compounds - Question 20

Which of the following statements is false?

Detailed Solution for Test: Tin and lead compounds - Question 20

Water gas is CO + H2

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