Chemistry NCERT Exemplar Solutions Class 11th Chapter Two
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New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: option (iv)
The number of radial nodes is given by n-l-1
where n is principal quantum number, l is azimuthal quantum number
For 3p orbital, n=3 and l=1
Thus, the number of radial nodes
= n-l-1
= 3-1-1
= 1
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option (iv)
sum of the number of protons and neutrons is same but the number of protons is different
Isobars are the atoms which have same mass number (sum of number of neutrons and protons) but different atomic number (i.e. different proton number)
For example, 146C and 147N.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option (i)
Overall neutrality of atom.
According to the J.J Thomson model the positive charge is uniformly distributed and the electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement just as watermelon of positive charge with plums or seeds (electrons) embedded into it. Thus this model is able to explain the overall neutrality of the atom
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option (ii)
The mass of an electron is equal to the mass of a neutron.
The neutron is heavier than the electron as
Mass of the neutron = 1.67 x 10-27 kg
Mass of the electron = 9.11 x 10-31kg
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option (iv)
Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube. As per the result obtained from the cathode ray discharge tube experiment the characteristics of cathode rays (electrons) does not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube. Which concludes that electrons are the basic constituent of all the atoms
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
ANS – Option (iv)
The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases. Electrons at the 1s orbital decreases as we move far from the nucleus, however in case of 2s the probability decreases initially then it increases with the distance and thereafter at a certain point it starts decreasing with the distance.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: (ii) 1s2 2s2 2p6 3s2 3p6 3d9 4s2
As per the Hund's rule the half-filled and fully filled orbital leads to the extra stability due to the symmetry thus fully filled 3d and half-filled 4s is preferred.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option (iii)
Electrons move in a circular path of fixed energy called orbits .
As per Rutherford's α-particle scattering experiment the nucleus is surrounded by electrons that move around the nucleus with a very high-speed in circular paths called orbits. It does not mention the energy or stability of the electrons revolving around the nucleus.
New answer posted
4 months agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar
Ans: The energy of electrons is determined by the value of n in the hydrogen atom and by n + l in the multielectron atom. Thus for a given principal quantum number the electrons of different orbitals would have different energy.
New answer posted
4 months agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar
Ans: The uncertainty principle is significantly only for the microscopic particles and not for the macroscopic particles can be concluded by considering the following example. Let us consider a particle or an object of mass 1 milligram i.e. 10-6 kg Then its uncertainty can be calculated as,
? x ? v = 6.626 10-34 / 4x 3.14 106
= 10-28 m-2 s-1
Thus, the value obtained is negligible and insignificant for the uncertainty principle to be applied to this particle.
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