NEET CHEMISTRY question bank, MCQ CBSE Notes, Lectures

1 Solid State

The number of atoms in one unit cell of a face centered cubic lattice is :

1) 1
2) 2
3) 3
4) 4

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

The number of atoms per unit cell in a simple cubic, face-centered cubic and body-centred cubic are 1,4,2 respectively.

2 Coordination Compounds

Among the following, which are ambidentate ligands?
(i) SCN⁻
(ii) $$NO^-_3$$
(iii) $$NO^-_2$$
(iv) $$C_2O^{2-}_4$$

1) (i) and (iii)
2) (i) and (iv)
3) (ii) and (iii)
4) (ii) and (iv)

Option 1 Solution:

Solution

Ambidentate ligand has two or more donor atoms, but during complex formation, only one donor atom is attached to central metal. SCN− ion has donor atoms N and S. It links to metal as either M−SCN or M−NCS. NO₂⁻ group has donor atoms N and O. It links to metal as either M−ONO or M−NO2.

Option 2

Option 3

Option 4

3 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

4 Organic Compounds Containing Nitrogen

Amines behave as :

1) Lewis acids
2) Lewis bases
3) Amphoteric compounds
4) Neutral compounds

Option 1

Option 2 Solution:

Option 3

Option 4

5 Redox Reactions

In which of the following pair oxidation number of Fe is same:

1) K₃Fe(CN)₆,Fe₂O₃
2) Fe(CN)₅,Fe₂O₃
3) Fe₂O₃,FeO
4) Fe₂(SO₄)₃,K₄Fe(C)₆

Option 1 Solution:

SOLUTION

Let the oxidation state of Fe in K₃Fe(CN)₆and Fe₂O₃ be x and y respectively.

As we know that the sum of the oxidation states of all the atoms or ions in a neutral compound is zero.

Oxidation state of Fe in K₃Fe(CN)₆-

?CN is a negative ligand.

∴Oxidation number of CN is -1.

Oxidation number of K is +1.

∴3(+1)+x+6(−1)=0

⇒x−3=0

⇒x=3

Oxidation state of Fe in Fe₂O₃^-

Oxidation no. of O is -2.

2(y)+3(−2)=0

⇒2y=6

⇒y=3

Option 2

Option 3

Option 4

6 Chemical Equilibrium

Buffer solution can be obtained by the mixing aqueous solutions of:

1) CH₃COONa and CH₃COOH
2) CH₃COOH and excess NaOH
3) CH3₃COONa and excess HCl
4) NaCl and HCl.

Option 1 Solution:

Solution

A mixture of acetic acid (a weak acid) and sodium acetate ( its salt with strong base sodium hydroxide) acts as acidic buffer.

Option 2

Option 3

Option 4

7 Purification and Characterisation of Organic Compounds

The fragrance of the flower is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in the vapour phase. A suitable method for the extraction of these oil from the flower is:

1) distillation with reduced pressure

2) distillation

3) steam distillation

4) chromatography

Option 1

Option 2

Option 3 Solution:

SOLUTION

As the component is miscible with water vapour in vapour phase so steam distillation will be most suitable method for extraction.

Option 4

8 Haloalkanes and Haloarenes

Liebermann's test is answered by :

1) methylamine
2) aniline
3) phenol
4) ethyl benzoate

Option 1

Option 2

Option 3 Solution:

When phenol is reacted with NaNO₂ and concentrated H₂SO₄,
it provides a deep green or blue colour which changes to red on dilution with water.
Generated substance in presence of NaOH / KOH restores original green or blue colour.
This reaction is termed as Liebermann's nitroso reaction

Option 4

9 Haloalkanes and Haloarenes

X on treatment with sodium hydroxide followed by the addition of silver nitrate gives white precipitate at room temp, which are soluble in NH₄OH. X can be

1) Vinyl chloride

2) Benzyl chloride

3) Ethyl bromide

4) Chloro benzene

Option 1

Option 2 Solution:

C6H5CH2Cl+NaOH(aq)→NaCl+C6H5CH2OH
NaCl+AgNO3→AgCl↓(White ppt.)+NaNO3

Option 3

Option 4

10 Haloalkanes and Haloarenes

What is the final product obtained when chlorine reacts with ethyl alcohol in KOH?

1) CHCl₃

2) CCl₃CHO

3) CH₃Cl

4) None

Option 1

Option 2

Option 3 Solution:

SOLUTION

When chlorine reacts with ethyl alcohol in the presence of KOH, chloroform is obtained as a final product.

Cl₂+CH₃CH₂OH?CH₃CHO+2HCl [oxidation]

CCl₃CHO+KOH --^Hydrolysis?CHCl₃+ HCOOK

Option 4

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