Choose the best answer: Chemistry: Quantum Mechanical Model of Atom

Quantum Mechanical Model of Atom

Choose the best answer

1. Electronic configuration of species M2+ is 1s2 2s2 2p6 3s2 3p6 3d6 and its atomic weight is 56. The number of neutrons in the nucleus of species M is

a) 26

b) 22

c) 30

d) 24 

Answer: (c) 30

Solution:

M2+ : 1s2 2s2 2p6 3s2 3p6 3d6

M : 1s2 2s2 2p6 3s2 3p6 3d8

Atomic number = 26

Mass number =56

No. of neutrons = 56 -26 = 30

2. The energy of light of wavelength 45 nm is

a) 6.67 × 1015J

b) 6.67 × 1011J

c) 4.42 × 10–18J

d) 4.42 × 10–15J 

Answer: c) 4.42 × 10–18J

Solution:

E = hν = hc/λ

=6.626 Ã— 10-34 J s Ã— 3 Ã—108 m s-1  / 45 Ã—10-9 m

= 4 .42 Ã—10-18 J

3. The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths ie λ1 and λ2 will be

a) λ1/λ2 = 1

b) λ1 = 2λ2

c) λ1 =  √(25 × 50 λ2)

d) 2λ1 = λ2 

Answer: b) λ1 = 2λ2

Solution:

4.  Splitting of spectral lines in an electric field is called

a) Zeeman effect 

b) Shielding effect

c) Compton effect 

d) Stark effect 

Answer: d) Stark effect

Solution:

Spliting of spectral lines in magnetic field is called Zeeman effect and spliting of spectral lines in electric field, is called Stark effect.

5. Based on equation E = – 2.178 × 10–18 J (z2/n2) certain conclusions are written. Which of them is not correct ?

a) Equation can be used to calculate the change in energy when the electron changes orbit

b) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit

c) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.

d) Larger the value of n, the larger is the orbit radius. 

Answer: b) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit

Solution:

Correct statement: For n=1, the electron has more negative energy than it does for n=6 which means that the electron is strongly bound in the smallest allowed orbit.

6. According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon ?

a) n = 6 to n = 1

b) n = 5 to n = 4

c) n = 5 to n = 3

d) n = 6 to n = 5 

Answer: d) n = 6 to n = 5

Solution:

n = 6 to n=5

E6 = -13.6 / 62 ;

E5 = - 13.6 / 52

E6 - E5 = (-13.6 /62) - (-13.6/52)

= 0.166 eV atom-1

E5 - E4 = (-13.6 /52) - (-13.6/42)

= 0.306 eV atom-1

7. Assertion : The spectrum of He+ is expected to be similar to that of hydrogen

Reason : He+ is also one electron system.

(a) If both assertion and reason are true and reason is the correct explanation of assertion.

(b) If both assertion and reason are true but reason is not the correct explanation of assertion.

(c) If assertion is true but reason is false

(d) If both assertion and reason are false 

Answer: (a) If both assertion and reason are true and reason is the correct explanation of assertion.

8. Which of the following pairs of d-orbitals will have electron density along the axes ?

a) dz2, dxz

b) dxz, dyz

c) dz2, dx2-y2

d) dxy, dx2-y2 

Answer: c) dz2, dx2-y2

9. Two electrons occupying the same orbital are distinguished by

a) azimuthal quantum number

b) spin quantum number

c) magnetic quantum number

d) orbital quantum number 

Answer: b) spin quantum number

Solution:

Spin quantum number

For the first electron ms = +½

For the second electron ms = -½

10. The electronic configuration of Eu (Atomic no. 63) Gd (Atomic no. 64) and Tb (Atomic no. 65) are

a) [Xe] 4f6 5d1 6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f8 5d1 6s2

b) [Xe] 4f7 , 6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2

c) [Xe] 4f7 , 6s2, [Xe] 4f8 6s2 and [Xe] 4f8 5d1 6s2

d) [Xe] 4f6 5d1 6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2 

Answer: b) [Xe] 4f7 , 6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2

Solution:

Eu : [Xe] 4f7, 5d0, 6s2

Gd : [Xe] 4f7, 5d1, 6s2

Tb: [Xe] 4f9, 5d0, 6s2

11. The maximum number of electrons in a sub shell is given by the expression

a) 2n2

b) 2l + 1

c) 4l + 2

d) none of these 

Answer: c) 4l + 2

Solution:

2(2l+1) = 4l+2

12. For d-electron, the orbital angular momentum is

a) √(2h)/2π

b) √(2)h/2π

c) √(2x4) h/2π

d) √(6)h/2Ï€ 

Answer: d) √(6)h/2π

Solution:

Orbital angular momentum = √(l(l+1) h/2π

For d orbital = √(2 x 3) h/2Ï€ = âˆš6 h/2Ï€

13) What is the maximum numbers of electrons that can be associated with the following set of quantum numbers ? n = 3, l = 1 and m = –1

a) 4

b) 6

c) 2

d) = 10 

Answer: c) 2

Solution:

n = 3; l=1; m = -1

either 3px or 3py

i.e., Maximum two electrons can be accomodated either in 3px or in 3py

14. Assertion : Number of radial and angular nodes for 3p orbital are 1, 1 respectively.

Reason : Number of radial and angular nodes depends only on principal quantum num-ber.

a) both assertion and reason are true and reason is the correct explanation of as-sertion.

b) both assertion and reason are true but reason is not the correct explanation of assertion.

c) assertion is true but reason is false

d) both assertion and reason are false 

Answer: c) assertion is true but reason is false

Solution:

No. of radial node = n-l-1

No. of angular node = l

for 3p orbital

No. of angular node = l = 1

No. of radial node = n-l-1 = 3-1-1 = 1

15. The total number of orbitals associated with the principal quantum number n = 3 is

a) 9

b) 8

c) 5

d) 7 

Answer: a) 9

Solution:

n = 3; l = 0; ml = 0 - one s orbital

n = 3; l = 1; ml = -1, 0, 1 - three p orbitals

n = 3; l = 2; ml = -2, -1, 0, 1, 2 - five d Orbitals

Overall nine orbitals are possible.

16. If n = 6, the correct sequence for filling of electrons will be,

a) ns → (n – 2) f → (n – 1)d → np

b) ns → (n – 1) d → (n – 2) f → np

c) ns → (n – 2) f → np → (n – 1) d

d) none of these are correct 

Answer: a) ns → (n – 2) f → (n – 1)d → np

Solution:

n = 6

According Aufbau principle,

6s →4f →5d →6p

ns → (n-1)f →(n-2)d → np

17. Consider the following sets of quantum numbers :

Which of the following sets of quantum number is not possible ?

a) (i), (ii), (iii) and (iv)

b) (ii), (iv) and (v)

c) and (iii)

d) (ii), (iii) and (iv) 

Answer: b) (ii), (iv) and (v)

Solution:

(ii) l can have the values from 0 to n-1

n=2; possible 'l' values are 0, 1 hence l = 2 is not possible.

(iv)for l=0; m=-1 not possible

(v)for n=3 l = 4 and m = 3 not possible

18. How many electrons in an atom with atomic number 105 can have (n + l) = 8 ?

a) 30

b) 17

c) 15

d) unpredictable 

Answer: b) 17

Solution:

n+l = 8

Electronic configuration of atom with atomic number 105 is [Rn] 5f14 6d3 7s2

19. Electron density in the yz plane of 3dx2 âˆ’y2   orbital is

a) zero

b) 0.50

c) 0.75

d) 0.90 

Answer: a) zero

Solution:

Option (a) - Zero (Refer to Figure 2.9)

20. If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is

Answer:  (c) 1/2m √h/Ï€

Solution:

21. A macroscopic particle of mass 100 g and moving at a velocity of 100 cm s–1 will have a de Broglie wavelength of

a) 6.6 × 10–29 cm

b) 6.6 × 10–30 cm

c) 6.6 × 10–31 cm

d) 6.6 × 10–32 cm 

Answer: c) 6.6 × 10–31 cm

Solution:

m = 100 g = 100 x 10-3 kg

v= 100 cm s-1 = 100 x 10-2 m s-1

λ =h/mv

22) The ratio of de Broglie wavelengths of a deuterium atom to that of an α - particle, when the velocity of the former is five times greater than that of later, is

a) 4

b) 0.2

c) 2.5

d) 0.4 

Answer: d) 0.4

23) The energy of an electron in the 3rd orbit of hydrogen atom is –E. The energy of an electron in the first orbit will be

a) –3E

b) −E/3

c) −E/9

d) –9E 

Answer: d) –9E

Solution:

24) Time independent Schnodinger wave equation is

Answer: (a) H∧ Ïˆ = Eψ

25. Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle ?

a) ∆x.∆p ≥h/4π

b) ∆x.∆v ≥h/4πm

c) ∆E.∆t ≥h/4π

d) ∆E.∆x ≥h/4Ï€ 

Answer: d) ∆E.∆x ≥h/4π