NEET Chemistry: Topic-wise Test- 7 Free Online Test 2026

The wavelength λ is inversely proportional to the square root of kinetic energy. So if KE is increased 4 times, the wavelength becomes half.

λ∝1/√KE

Hence Option A is the answer.

Correct Answer is A.

Number of orbitals in a shell = n² = (5)² = 25

Bohr Model of atom:

• An atom is made up of three particles: Electrons, neutrons and protons.
• The protons and neutrons are located in a small nucleus at the centre of the atom.
• The electrons revolve rapidly around the nucleus at the centre of the atom.
• There is a limit to the number of electrons that each energy level can hold.
• Each energy level is associated with a fixed amount of energy.
• There is no change in the energy of electrons as long as they keep revolving in the same energy level.

Bohr explained the stability through the concept of revolution of electrons in different energy levels.

The change in the energy of an electron occurs when it jumps from lower to higher energy levels. When it gains energy, it excites from lower to higher and vice versa.
Thus energy is not lost and the atom remains stable.

• Number of radial nodes = n - l – 1

where n = principal quantum number, l = azimuthal quantum number

• For 3p orbital, n = 3 – 1 – 1 = 1
• Number of radial nodes = 3 – 1 – 1 = 1. 

n = 6, l = 2 means 6d → will have 5 orbitals. 

∴ max 10 electrons can be accommodate as each orbital can have maximum of 2 electrons. 

• An atom consists of a positively charged sphere with electrons filled into it. The negative and positive charges present inside an atom are equal and as a whole, an atom is electrically neutral.
• Thomson’s model of the atom as compared to plum pudding and watermelon.
• He compared the red edible part of the watermelon to a positively charged sphere whereas the seeds of watermelon to negatively charged particles.

• Concept of electrons moving in a circular path of fixed energy called orbits was put forward by Bohr and not derived from Rutherford's scattering experiment.
• Out of a large number of circular orbits theoretically possible around the nucleus.
• The electron revolves only in those orbits which have a tired value of energy Hence, these orbits are called energy level or stationary states.

Anodic

Cathodic

Reaction quotient (Q) 

Anodic

Cathodic

Reaction quotient (Q) 

By Kohlrausch’s law,

= 1.50 x 10^-2 + 2.48 x 10^-2 - 1.26 x 10^-2
= 2.72 x 10^-2 Sm²mol^-1
 

By Kohlrausch’s law,

= 1.50 x 10^-2 + 2.48 x 10^-2 - 1.26 x 10^-2
= 2.72 x 10^-2 Sm²mol^-1
 

Dibasic acid is H₂A   2H⁺ + A^2- 
∴ 0.01M = 0.02N
Equivalent conductance at 0.02 N

Equivalent conductance at 0.004 N 

The internationally recommended unit for conductance is Siemens (S). 1 siemen = 1 ohm^-1

In all cases, we use





Equal number of electrons are involved. 

For equilibrium, AgCl(s)  Ag⁺  + Cl^-    E⁰ = -0.59V

-0.59 = 0.0591logK_sp

∴   log K_sp = -10

∴ K_sp = 10^-10

C₅ H₁₀

 Two times double bond can’t appear as it has asked alkene and not alkadiene. So, these are the only types of alkenes that can be formed which are structural isomers.

According to option A chiral compound at least have one chiral carbon but compounds other than carbon and also chiral like NCl Br I is chiral. According to option b a compound containing only one ka kalakaron is always is always correct that is statement is true it is the minimum requirement for a carbon atom to become chiral. Efficiency is not necessary to be true all the time compound will not be superimposable on its Mirror image. And for option d a plane of symmetry is sufficient but not a mandatory to give achiral compound.

Correct answer: Option C.

A molecule that has a plane of symmetry (σ) or a centre of symmetry (i) is achiral because the optical activity of one part of the molecule is exactly cancelled by the mirror-related part; a classical example is meso-tartaric acid.

The presence of a proper rotation axis (C_n) does not necessarily destroy chirality; many chiral molecules possess a C_n axis. By contrast, an improper rotation axis (S_n) - a rotation followed by reflection - renders a molecule achiral; in particular S₁ ≡ σ and S₂ ≡ i.

Therefore, only statement 3 is correct. Statements 1, 2 and 4 are incorrect: statement 1 is false because a C_n axis alone need not destroy chirality; statement 2 is false because a centre of symmetry (i) does make a molecule achiral; statement 4 is false because an axis of symmetry does not imply the existence of a centre of symmetry.

The given compound is superimposable on its mirror image, so it is achiral. Achiral molecules which have chirality centers are called meso compounds. Meso compounds are possible if two (or more) chirality centers in a molecule have the same set of four substituents.

The compound shown is glyceraldehyde, which contains 3 chiral centers (n = 3). The general formula to calculate the number of stereoisomers is 2ⁿ = 2³ = 8. However, due to the presence of a meso form (an internal plane of symmetry), some stereoisomers are optically inactive.

The number of optically active isomers = 2^n-1 = 2^3-1 = 4. But since one of these is a meso compound (optically inactive), the total number of optically active isomers = 4 - 2 = 2.

Alternatively, using the formula: Number of optically active isomers = 2^n-1 - 2^(n-1)/2 = 2² - 2¹ = 4 - 2 = 2.

Thus, the compound has 2 optically active isomers.

The number of stereoisomers for CH₃ – CHOH – CH = CH – CH₃ is four. This is calculated by the formula 2ⁿ⁺¹. where n is number of chiral centres, which is 1 in this case. So, 2¹⁺¹ = 2²= 4

The correct option is D Molecule in meso form have a plane of symmetry
Meso forms of isomers are single compound and their molecules are achiral and hence they cannot be separated into optically active enantiometric pairs.
Molecule in meso form have a plane of symmetry due to which the optical rotations of upper and lower parts are equal and in the opposite direction which balanced internally and compound become optically inactive. this property is called internal compensation.

 2-bromomethyl-2-methyl-1-butanol: The bromine substitution occurs at the carbon bonded to oxygen but not at a chiral carbon. Thus option b satisfies the condition.

Option b: Decrease concentration of sample four times and measure the optical rotation. Explanation:

• Optical rotation is directly proportional to concentration (specific rotation = observed rotation/concentration).
• If you decrease the concentration of the chiral compound in the sample, the observed optical rotation will decrease proportionally.
• If the initial observed rotation is indeed +60° and not -300°, decreasing the concentration will still result in a positive rotation, but it will be lower.

The other options are less likely to provide useful information in this context.

•  The given compound has a total of six stereoisomers 

• It's meso form upon ozonolysis followed by Zn-hydrolysis gives racemic mixture 

• It's optically active isomers, each upon ozonolysis followed by Zn-hydrolysis gives a single enantiomer.