Third Chapter Solved Exercise MCQs Of Fsc Second Year Chemistry

Third Chapter Solved Exercise MCQs Of Fsc Second Year Chemistry

1: Which metal is used in the thermite process because of its activity?

  1. Iron
  2. Copper
  3. Aluminum
  4. Zinc

Explanation

Aluminum is used in thermite process in powder form. Al powder becomes highly reactive for oxygen and removes the oxygen from the oxides of other metals. The separated metals are then obtained in the molten state which can be used for the welding purpose.


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2: Aluminum oxide is

  1. Acidic oxide
  2. Basic oxide
  3. Amphoteric oxide
  4. None of these

Explanation

In majority of its properties Aluminum acts as a metal. But the oxides of Al are amphoteric in nature. It means these oxides work as both basic and acidic as well in nature.


3: Chemical composition of colemnite is

  1. Ca2B6O11.5H2O
  2. CaB4O7.4H2O
  3. Na2B4O7.4H2O
  4. CaNaB5O9.8H2O

Explanation

Colemnite is known as ore of boron and its formula is Ca2B6O11.5H2O. Hence the option a is correct answer.


4: Which element forms an ion with charge +3?

  1. Beryllium
  2. Aluminum
  3. Carbon
  4. Silicon

Explanation

Among all above mentioned elements Aluminum belongs to group III-A of the periodic table. According to the electronic configuration of Al it has 13 electrons out of which three are present in the outermost shell and the distribution is 1s2, 2s2, 2p6, 3s3, 3p1 .  

The last two orbitals indicates three outermost electrons hence Al can develop +3 charge by losing these three electrons. Therefore option b is the correct answer.


5: Which electronic configuration corresponds to an element of group III-A of the periodic table?

  1. 1s2, 2s2, 2p6, 3s2, sp1
  2. 1s2, 2s2, 2p6, 3s2, sp6, 4s6
  3. 1s2, 2s2, 2p6
  4. 1s2, 2s2, 2p6, 3s2, sp3

Explanation

From the above mentioned electronic configuration, Al has 13 electrons out of which three electrons are present in the outermost shell and the distribution of electrons is 1s2, 2s2, 2p6, 3s3, 3p1 .  Therefore option a is the correct answer.


6: Which element among the following belongs to group IV-A of the periodic table?

  1. Barium
  2. Iodine
  3. Lead
  4. Oxygen

Explanation

By taking a look at periodic table, from the above given elements lead is present at the bottom of the group IV-A. Therefore option c is the correct answer.


7: Boric acid cannot be used

  1. As antiseptic in medicine
  2. For washing eyes
  3. In soda bottles
  4. For enamels and glazes

Explanation

Boric acid cannot be used in soda bottles as it can be dissolved into water. However it can be used as antiseptic, washing of eyes and enamels. Therefore, option c is correct answer.


8: Which of the following elements is not present abundantly in earth’s crust?

  1. Silicon
  2. Aluminum
  3. Sodium
  4. Oxygen

Explanation

From the list of above mentioned elements only sodium is not present abundantly. Because only 2.27% earth’s crust is composed of sodium making it the 7th most abundant element on earth’s crust.


9: Tincal is a mineral of

  1. Aluminum
  2. Boron
  3. Silicon
  4. Carbon

Explanation

If we have a look at the formula of Tincal it becomes clear that Tincal is mineral of Boron.

Na2B4O7.10H2O


10: The chief ore of Aluminum is

  1. Na3AlF6
  2. Al2O3.2H2O
  3. Al2O3
  4. Al2O3.H2O

Explanation

Although all the mentioned ores are of aluminum. But the chief ore of aluminum is bauxite. Therefore, option b is the correct answer.