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Test: Main Group - 2 - Chemistry MCQ


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20 Questions MCQ Test Inorganic Chemistry - Test: Main Group - 2

Test: Main Group - 2 for Chemistry 2024 is part of Inorganic Chemistry preparation. The Test: Main Group - 2 questions and answers have been prepared according to the Chemistry exam syllabus.The Test: Main Group - 2 MCQs are made for Chemistry 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Main Group - 2 below.
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Test: Main Group - 2 - Question 1

The compound having the highest melting point is:

Test: Main Group - 2 - Question 2

The degree of hydration is expected to be maximum for:

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Test: Main Group - 2 - Question 3

Which of the following order is correct for the first ionization energies of their elements?

Detailed Solution for Test: Main Group - 2 - Question 3

Boron, B is smaller than beryllium, Be atom. Hence we expect increase in ionization energy from Be to B.

However, Be atom has greater ionization energy than B atom. The reason is - in case of Beryllium, the last electron is in the s-orbital and in Boron, the last electron is in the p-orbital. We know that removal of electron from s-orbital requires more energy than the electron from p-orbital.

2s22p3 configuration is more stable than 2s22p4 due to half filled p-sublevel. The electrons in the p-orbital in N atom experience less repulsion, thus more stable and more ionization energy.  Whereas in case of O atom the 4th p-electron atom can be removed more easily as it experiences more repulsion from the electrons in the p-orbitals already present.  Hence nitrogen, Oxygen atom has less ionization energy than Nitrogen atom.

Test: Main Group - 2 - Question 4

The decreasing order of ionic nature of the following compound is:

Test: Main Group - 2 - Question 5

The atomicity and the total number of bonds in the elemental white phosphorous molecule are respectively:

Test: Main Group - 2 - Question 6

The correct order of the ionic radii is:

Test: Main Group - 2 - Question 7

Which of the following reactions does not give H3PO3:

Detailed Solution for Test: Main Group - 2 - Question 7

Explanation : Balanced Chemical Equation 

P4O6 + H2O = H3PO4 + PH3

Test: Main Group - 2 - Question 8

The conductance at infinite dilution follows the order:

Test: Main Group - 2 - Question 9

The compound having as S–S single bond is: 

Test: Main Group - 2 - Question 10

In a reaction, Na2S2O3 is converted to Na2S4O6. The equivalent weight of Na2S2O3 for this reaction is (mol. wt. of Na2S2O3 = M)

Test: Main Group - 2 - Question 11

Among the following, the incorrect statement is:

Detailed Solution for Test: Main Group - 2 - Question 11

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them.

Test: Main Group - 2 - Question 12

The number of P=O bonds present in the tetra basic acid H4P2O7 is:

Test: Main Group - 2 - Question 13

Egyptian blue CaCuSi4O10 is an example of:

Test: Main Group - 2 - Question 14

The formal charges on the nitrogen atom from left to right in the azide anion, [N=N=N]– are: 

Test: Main Group - 2 - Question 15

B2H6 and B4H10, respectively, are examples of:

Test: Main Group - 2 - Question 16

At room temperature, HCl is a gas while HF is a liquid because:

Test: Main Group - 2 - Question 17

P4O10 is the anhydride of​

Detailed Solution for Test: Main Group - 2 - Question 17

Phosphorus Pentoxide, Reagent, ACS is the anhydride of phosphoric acid and a powerful dehydrating and desiccant agent. It is used extensively in organic synthesis as a dehydrating agent.

Test: Main Group - 2 - Question 18

The correct order of acidic character is:

Test: Main Group - 2 - Question 19

In the structure of B4O5 (OH)42-

Test: Main Group - 2 - Question 20

Hydrolysis of (CH3)2 SiCl2 and CH3SiCl3 leads to:

Detailed Solution for Test: Main Group - 2 - Question 20

Correct Answer :- a

Explanation : The silicon containing residual hydroxyl groups will be cross linked using boric acid and this tri functional acid, B(OH)3 groups, forms Si-O-B linkage between siloxane chains.

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