NEET CHEMISTRY question bank, MCQ CBSE Notes, Lectures

1 Chemical Bonding and Molecular Structure

Amongst LiCl, RbCl, BeCl₂ and MgCl₂ the compounds with greatest and least ionic character respectively are:

1) LiCl and RbCl
2) RbCl and BeCl₂
3) RbCl and MgCl₂
4) MgCl₂ and BeCl₂

Option 1

Option 2 Solution:

SOLUTION

Lesser the Cationic size, lesser the ionic character. Among Li, Be, Be is small. Among Rb and Mg, Rb is large so that Be forms least ionic mostly covalent and Rb form great ionic bonds.

Thus RbCl and BeCl₂ are the compounds with least and greatest ionic character respectively.

Option 3

Option 4

2 Solutions

A storage battery contains a solution of H_2?SO₄?38% by weight. What will be the Van't Hoff factor if the ΔT_f experiment? is 29.08.
[Given Kf?=1.86 mol−1Kg]

1) i=1.25
2) i=2.5
3) i=3
4) i=3.5

Option 1

Option 2 Solution:

Solution

Freezing point depression is a colligative property observed in solutions that result from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute.

ΔTf? = Tf?0−Tf ?= i.Kb?.m.

ΔTf? is freezing point depression.

Tf? is freezing point of solution.

Kb? is depression constant.

Let the total mass be 100 g.

Let the mass of sulphuric acid be 38 g.

Molality =m= (38/98) /(62? /1000)??=6.25

ΔTf? = 29.08K

Kf?=1.86

i= 2.50

Option 3

Option 4

3 Haloalkanes and Haloarenes

The organic chloro compound, which shows complete stereochemical inversion during an S_N 2 reaction is

1) (C₂H5)₂ CHCl
2) (CH₃)₃ CCl
3) (CH₃)₂CHCl
4) CH₃Cl

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

Nucleophilic substitution bimolecular (S_N 2) prefers less stericallv hinciered site to attack. Lesser the steric hindrance faster the S_N 2 reaction. So ease of reaction is 1^o > 2^o > 3^o

S_N 2 involves inversion of configuration stereo, chemically. Since, 1o alkyl halides are prone to S_N2 reactions, therefor CH₃Cl undergoes complete strerochemical inversion

4 Electrochemistry

Calculate the standard free energy change for reaction,
2Ag + 2H⁺ → H₂ + 2Ag⁺
E° for Ag⁺ + e- → Ag is 0.80V

1) +154.4 kJ
2) +308.8 kJ
3)-154.4 kJ
4)-308.8 kJ

Option 1 Solution:

SOLUTION

Given potential of Ag is reduction potential,
Hydrogen electrode is cathode(0Volt) and Ag is Anode(0.8Volt)
Cell potential = E cell=reduction potential of (cathode - Anode)
Cell potential E⁰ = 0-0.8 = -0.8Volts
ΔG=-nFe°
ΔG - -2 ×96500 ×(-0.80)=154.4kJ

Option 2

Option 3

Option 4

5 P � Block Elements

The decrease in stability of higher oxidation states in p-block with increasing atomic number is due to :

1) increase in bond energy on going down the group
2) the reluctance of electrons of s-sub shell to participate in chemical bonding?
3) poor screening effect of d-orbitals
4) none

Option 1

Option 2 Solution:

SOLUTION

The inert pair effect is the tendency of the electrons in the outermost atomic s orbital to remain unionized or unshared in compounds of post-transition metals. Down the group in the p-block, the inert pair effect becomes more pronounced on account of enhanced increase in effective nuclear charge. Therefore, due to reluctance of electrons of s-subshell to participate in chemical bonding, there occurs a decrease in stability of higher oxidation states in p-block with increasing atomic number.

Option 3

Option 4

6 Atomic Structure

The correct order of increasing energy of atomic orbital is

1) 5p < 4f < 6s < 5d
2) 5p < 6s < 4f < 5d
3) 4f < 5p < 5d < 6s
4) 5p < 5d < 4f < 6s

Option 1

Option 2 Solution:

Solution

More the (n + l) value ; more will be the energy
Two (n+l) have same values then higher then higher the value

Option 3

Option 4

7 Chemical Equilibrium

The solubility of K₂SO₄ in water is 16 g per 100 mL at 50^oC. The least weight of water that will dissolve 4 g of this substance at the same temperature is

1) 25 g

2) 10 g

3) 50 g

4) 75 g

Option 1 Solution:

Solution

Solubility is the amount of solute in gram per 100g solvent.
16g K₂?SO₄? is present in =100g water

4g K_2?SO_4? will be present in 100 / 16? × 4=25g

Option 2

Option 3

Option 4

8 D and F Block Elements

If each one of the above ionic species is in turn kept in a magnetic field, which one will get attracted ?

1) $$Co^{+2}$$
2) $$Mn^{+7}$$
3) $$Sc^{+3}$$
4) $$Ti^{+4}$$

Option 1 Solution:

Solution

$$Co^{+2} = [Ar]3d^7\,\,\,\,\,\,\,n = 3$$ attracted by a magnetic field due to presence of 3 unpaired electron.

Option 2

Option 3

Option 4

9 S � Block Elements

What is the correct order of increasing solubility of the following salt?

$$NaHCO_3, KHCO_3, Mg(HCO_3)_2, Ca(HCO_3)_2$$

1) $$KHCO_3<Mg(HCO_3)_2<Ca(HCO_3)_2<NaHCO_3$$
2) $$Mg(HCO_3)_2<Ca(HCO_3)_2<NaHCO_3<KHCO_3$$
3) $$NaHCO_3<KHCO_3<Mg(HCO_3)_2<Ca(HCO_3)_2$$>
4) $$Ca(HCO_3)_2<NaHCO_3<KHCO_3<Mg(HCO_3)_2$$

Option 1

Option 2

Option 3 Solution:

Solution

In the give metal bicarbonates, the anion (bicarbonate ion) is same. Hence the solubility will depend on the size of anion. The size of $$Na^+$$ ion is smaller than that of $$K^+$$ ion. Hence sodium bicarbonate has high lattice energy and lower solubility than potassium bicarbonate. Similarly, the size of $$Mg^+$$ ion is smaller than that of $$Ca^+$$ ion. Hence magnesium bicarbonate has high lattice energy and lower solubility than calcium bicarbonate.

Option 4

10 Stoichiometry

When 0.5g. of sulphur is burnt to SO₂ 4.6 kj of heat is liberated. What is the enthalpy of formation of sulphur dioxide? (S=32 and O=16)

1) -147.2KJ
2) + 147.2KJ
3) + 294.4KJ
4) - 294.4KJ.

Option 1

Option 2

Option 3

Option 4 Solution:

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