Chemistry: Topic-wise Test- 7 - NEET MCQ

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.

Given m_p = 1.67×10^-27kg
h = 6.63×10^-34Js
v = 1.0×10³ms-1

We know wavelength λ = h/mv
∴λ = 6.63×10^-34/(1.67×10^-27 × 1.0×10³)
= 0.40×10^-10
≈ 0.40nm

Hence, Option (B) is the Correct Answer.

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.

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
 

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

The quantity l/A is called cell constant denoted by the symbol, G*.It depends on the distance between the electrodes and their area of cross-section and has the dimension of length–1 and can be calculated if we know l and A. Measurement of l and A is not only inconvenient but also unreliable. The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations and at different temperatures. The cell constant, G*, is then given by the equation:
G* =l/A= Rκ

The correct answer is Option B 
 
k =G*l/A      (Since, G = 1/R)
   = l/AR 
Therefore, kR = l/A 
So, l/A = kR not k/R

for the given reaction

E⁰_{cell  = 2.73 V}

n(electrons exchanged) = 12

(4 Al → Al³⁺ + 12e^-)

F=96500 C mol^-1

ΔG⁰ (element) = 0

∴ ΔG⁰ (reaction) = -nFE⁰_cell

= -12 x 96500 x 2.73

=3161340 J = 3161.34 kJ

thus,

n-pentane, 2-ethylpropane, and 2-methylbutane are the 3 isomeric alkanes of C₅H₁₂ (pentane).

The compounds with each doubly bonded carbon attached to two different groups (like Cab=Cab, Cab=Ccd) exhibit geometrical isomerism i.e., cis and trans forms. The geometrical isomerism arises due to restricted rotation of double bond.

However, even though there is restricted rotation for triple bond, alkynes do not exhibit geometrical isomerism, since the triply bonded carbons are attached to one group each only.

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.

The number of optically active isomers = 2ⁿ⁻¹−2^{n−1/2 ​}=2³⁻¹−2^3−1/2 ​=2.

The correct answer is Option B.
The optically active hydrocarbon X is 3-methyl-1-pentene CH₂=CH−CH(CH₃)CH₂CH_3​. On catalytic hydrogenation, it forms 3-methyl pentane CH₃CH₂CH(CH₃)CH₂CH₃, which is optically inactive.

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.

A sawhorse projection can reveal staggered and eclipsed conformations.

 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 correct answer is option C
1,4-Dimethylcyclohexane consists of two diastereomeric stereoisomers, although, according to the n2 rule, four stereoisomers should be expected due to the fact that 1,4-dimethylcyclohexane contains two asymmetric carbons.

Correct option A.