Given below are the Structure Of Atom Class 11 Important Questions
a. Concepts questions
b. Calculation problems
c. Short Answer
d. Quantum number and EC problems
Short Answer Type
Question 1.
Why are the half filled and Fully Orbitals more stable?
Question 2.
Describe the Rutherford's Model of the atom and decribe the drawback of it
Question 3.
Give the electronic configuration of the following ions
(i) H-
(ii) N3-
(iii)O2-
(iv) Li+
(v)Na+
((vi)F- Solution
Question 4.
What are the drawbacks of Bohr's Theory
Question 5.
What are degenerate orbitals.Explain with examples
Question 6.
Differentiate betweem an orbit and orbital
Question 7.
Using s,p,d notations, describe the orbital with the following quantum?
(i) n=2, l=1
(ii) n=4, l=1
(iii) n=4, l=2
(iv) n=3, l=1
(v) n=2 ,l=0
(vi) n=4, l=3 Solution
(i) 2p
(ii) 4p
(iii) 4d
(iv) 3p
(v) 2s
(vi) 4f
Question 8.
State de-broglie relation
Question 9.
What is Ridberg's constant?. What its numerical value
Question 10.
What is Probability Density
Question 11.
State and explain Pauli's exclusion principle
Question 12.
State and explain Heisenberg's uncertainty principle.
Question 13.
Arange the orbitals represented by the following sets of energy
(i)n=4,l=0,ml=0, ms= +1/2
(ii)n=3,l=2,ml=0, ms= -1/2
(iii)n=3,l=1,ml=0, ms= +1/2
(iv)n=3,l=0,ml=0, ms= -1/2 Solution
(i) represent 4s
(ii) represent 3d
(iii) represent 3p
(iv) represent 3s
The decreasing order of energy of these orbitals is
3d > 4s > 3p > 3s
(ii) > (i) > (iii) > (iv)
Question 14.
An electron is one of the 3d orbitals. What are the possible values of n,l,ml for this electron Solution
n=3 and l=2 represent 3d.
Now possible orbitals are
ml=-2,-1,0, +1, +2
Question 15.
State the number of protons, neutrons and electrons in
(i) 12C6
(ii) 14C6 Solution
(i) Number of electrons=6
Number Of protons=6
Mass number =12
Number of Neutrons= 12 -6 =6
(ii) Number of electrons=6
Number Of protons=6
Mass number =14
Number of Neutrons= 14 -6 =8
Numericals Problems
Question 1.
Calculate the Kinetic energy of a moving electron which has a wavelength of 4.9 pm.
Mass of Electron=9.11 × 10-31 kg
Plank's Constant (h)=6.63 × 10-34 J-s Solution
Using
$\lambda = \frac {h}{mV}$
or
$V= \frac {h}{m \lambda}$
Subsituting the values
V=.1516 × 109 m/s
Kinetic energy = $\frac {1}{2}mV^2 $
Subtituting the value of V and m
K.E=1.047 × 10-14 J
Question 2.
Calculate the wavelength and energy of the radiation emitted for the electronic transition from fourth orbit to stationary state 1 of He+ ion.
RH=1 × 107 m-1
Plank's Constant (h)=6.63 × 10-34 J-s Solution
Using
$\frac {1}{\lambda} = R_H Z^2 (\fracc{1}{n_1^2} - \fracc{1}{n_2^2})$
Substituting the values $n_1 =2$ and $n_2=4$, Z=1
$\lambda =.2667 \times 10^{-7}$ m
Now energy is given by
$E= \frac {hc}{\lambda} = 74.47 \times 10^{-20} $ J/atom
Question 3.
what will be the wavelength of a ball of mass .1 kg moving with velocity 10 m/s? Solution
Using de-broglie relation
$\lambda = \frac {h}{p} = \frac {h}{mV}$
Substituting the values
$\lambda = 6.626 \times 10^{-34} $ m
Question 4.
Calculate the wave number for the longest wavelength transition in the balmer series of atomic hydrogen ( Ans $1.525 \times 10^6 m^{-1}$)
Question 5.
An electron is moving with the kinetic energy of 2.275 × 10-25 J. Calculate its De-Broglie wavelength?
Mass of Electron=9.11 × 10-31 kg
Plank's Constant (h)=6.63 × 10-34 J-s Solution
Now
$K= \frac {1}{2}mV^2$
or
V = \sqrt {\frac {2K}{m}}$
Subsituting the values
V=.7 × 103 m/s
Now De-Broglie wavelength is given by
$\lambda = \frac {h}{p} = \frac {h}{mV}$
Substituting the values
$\lambda = 1.036 \times 10^{-6} $ m
Question 6.
A photon ($\lambda = 4 \times 10^{-7} \ m$) strikes a metal surface of Work function(W=2.13 eV).Calculate
(i) Energy of Photon in eV
(ii) K.E of the emission
(iii)Velocity of the photoelectron
1 eV=1.6020 × 10-19 J
Plank's Constant (h)=6.63 × 10-34 J-s
Mass of Electron=9.11 × 10-31 kg Solution
(i) $E= h \nu= \frac {hc}{\lambda}$
Substituting the values
$E=3.1 eV$
(ii) From Photo-electric effect equation
K= E - W= 3.1 - 2.13 =.97 eV
(iii) Now
$\frac {1}{2}mV^2 =.97 \times 1.6020 \times 10^{-19}$
Substituting the values
v=5.84 × 105 m/s
Question 7.
Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length $1 \times 10^{-10}$ metre. Solution
According to Heisenberg's uncertainty principle,
$\Delta x \Delta p \ge \frac {h}{2 \pi}$
Here $\Delta x =1 \times 10^{-10}$ m
Plank's Constant (h)=6.63 × 10-34 J-s
Therefore uncertainty in the momentum of an electron will be
$\Delta p \ge \frac {h}{2 \pi \Delta x}$
or
$\Delta p \ge 5.27 \times 10^{-25} \ kgms^-1$
Therefore uncertainty in the momentum of an electron =$5.27 \times 10^{-25} \ kgms^-1$
Question 8.
Calculate
(i)first excitation energy of the electron in the hydrogen atom
(ii) Ionization energy of the hydrogen atom Solution
The energy of the electron in the nth shell of hydrogen atom is given by
$E_n= - \frac {2 \pi ^2 me^4}{n^2 h^2}= - \frac {1.312 \times 10^6}{n^2} J/mol$
(i) First excitation energy is the amount of energy required to excite the electron from n = 1 (ground state) to n = 2 (first excited state)
$\Delta E = E_2 - E_1 $
$E_2= - \frac {1.312 \times 10^6}{2^2}$
$E_1= - \frac {1.312 \times 10^6}{1^2}$
Therefore
$\Delta E = 9.84 \times 10^5 $ J/mol
(ii) Ionization energy is the amount of energy required to remove the electron from n=1 to $n=\infty$
$\Delta E = E_{\infty} - E_1 $
$E_{\infty}= - \frac {1.312 \times 10^6}{(\infty)^2}=0$
$E_1= - \frac {1.312 \times 10^6}{1^2}$
Therefore
$\Delta E = 1.312 \times 10^6 $ J/mol
Question 9.
Calculate the velocity of an electron placed in the third orbit of the hydrogen atom.Also Calculate the number of revolution per second that this electron makes around the nucleus? Solution
Radius is given by
$r_n= n^2 \times .529 \times 10^{-10} $ m
$r_3= 3^2 \times .529 \times 10^{-10}=4.761 \times 10^{-10} \ m$
Now
$mvr= \frac {nh}{2 \pi}$
or
$v= \frac {nh}{2 \pi m r}$
Substituting the values
$v=7.29 \times 10^5$ m/s
Number of Revolutions per sec
$= \frac {v}{2 \pi r}$
Substituting the values
$=2.4 \times 10^{14} $ rev/sec
Question 10.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the ionisation energy of Sodium in KJ/mol Solution
Ionization energy will be equal to energy possessed by the Electromagnetic radiation
$E= \frac {hc}{\lambda}$
Substituting the values
$E=8.21 \times 10^{-19}$ J
Now This is the energy per atom
Energy per Mole
$E=8.21 \times 10^{-19} \times 6.023 \times 10^{23} = 494000$ J = 494 KJ
Question 11.
Determine the number of unpaired electrons in the ground state of below ions
(i) Ar+
(ii)Sn2+
(iii) F+
(iv)Bi3+ Solution
(i)1
(ii) 0
(iii) 2
(iv) 0
Question 12.
The arrangement of orbitals on the basis of energy is based upon their (n+l ) value. Lower the value of (n+l ), lower is the energy. For orbitals having same
values of (n+l), the orbital with lower value of n will have lower energy.
I. Based upon the above information, arrange the following orbitals in the increasing order of energy.
(a) 1s, 2s, 3s, 2p
(b) 4s, 3s, 3p, 4d
(c) 5p, 4d, 5d, 4f, 6s
(d) 5f, 6d, 7s, 7p
II. Based upon the above information, solve the questions given below :
(a) Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
(b) Which of the following orbitals has the highest energy?
5p, 5d, 5f, 6s, 6p Solution
