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Chemistry : Quantum Mechanical Model of Atom : Book Back Important Questions, Answers, Solutions : Answer briefly, Answer in detail, Exercise Numerical Problems

**Quantum Mechanical Model of Atom | ****Chemistry**

**Answer the following questions**

26. Which quantum number reveal information about the shape, energy, orientation and size of orbitals?

i)
Principal quantum number (n)

ii)
Azimuthal quantum number (*l*)

iii)
Magnetic quantum number (m_{1})

27. How many orbitals are possible for n =4?

Answer:

n = 4 *l* = 0, 1, 2, 3

four sub-shells â‡’ s, p, d, f

*l *= 0m*l* = 0 ; one 4s orbital.

*l *= 1 m*l *= â€“1, 0, +1 ; three 4p* *orbitals.

*l *= 2 m*l *= â€“2, â€“1, 0, +1, +2 ; five* *4d orbitals.

*l *= 3 m*l *= -3, â€“2, â€“1, 0, +1, +2 ,* *+3; seven 4f orbitals

Over all Sixteen orbitals.

4s,
4p, 4d and 4f orbitals = 1 + 3 + 5 + 7

=
**16 Orbitals**

28. How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular nodes?

Answer:

29. The stabilisation of a half filled d - orbital is more pronounced than that of the p-orbital why?

"More
exchange energy, increases the stability of orbitals"

Over
all electron exchange for d - Orbitals = 10

Over
all electron exchange for p - Orbitals = 6

Hence,
Half filled d - orbitals are more stable.

30. Consider the following electronic arrangements for the d5 configuration.

(i) which of these represents the ground state

(ii) which configuration has the maximum exchange energy.

Answer:

i)
(c) represents the ground state among the given three d^{5} electronic
arrangements.

ii)
Also, (c) has the maximum exchange energy.

31. State and explain pauli's exclusion principle.

No
two electrons in an atom can have the same set of values of four quantum
numbers.

32. Define orbital? what are the n and l values for 3px and 4dx2-y2 electron?

Answer:

Orbital is a three dismerisional space in which the probability of finding the electron is maximum

33. Explain briefly the time independent schrodinger wave equation?

Based
on the Heisenberg's principle and the dual nature of the microscopic particles,
Schrodinger expressed the wave nature of electron in terms of a differential
equation.

This
equation determines the change of wave function in space depending on the field
of force in which the electron moves.

This
equation does not contain time (t) as a variable and is reffered to as time
independent Schordinger wave equation.

34. Calculate the uncertainty in position of an electron, if Î”v = 0.1% and Ï… = 2.2 x106 ms-1

Answer:

**Given:** Î” = 0.1%

V
= 2.2 Ã—** **10^{6 }ms^{âˆ’1}.

Î”V
= [0.1 / 100] Ã— 2.2 Ã—** **10^{6 }ms^{âˆ’1}= 2.2 Ã—** **10^{3
}ms^{âˆ’1}

*Solution:*

Î”*x*.m.Î”v
â‰¥ h/4Ï€

Î”
Ã— 9.1Ã—10^{âˆ’31}Kg Ã— 2.2 Ã— 10^{3} ms^{âˆ’1 }â‰¥ [(6.626 Ã— 10 ^{âˆ’34
}) / 4Ï€ ] Kgm^{2} s^{âˆ’1}

Î”x
â‰¥ ( 6.626 Ã— 10^{âˆ’34 }Kgm^{2} s^{âˆ’1 }) / (9.1Ã—10^{âˆ’31}Kg
Ã— 2.2 Ã— 10^{3} ms^{1 }Ã— 4Ï€)

â‰¥
( 6.626 Ã— 10^{âˆ’34} ) / ( 251.45 Ã— 10^{âˆ’28} ) = 0.02635 Ã— 10^{-6}

Î”x
â‰¥ 2.635 Ã— 10^{-8} m

__[Alternative Answer]__

35. Determine the values of all the four quantum numbers of the 8th electron in O- atom and 15th electron in Cl atom and the last electron in chromium

Answer:

The
8^{th} electron of O - atom is present in 2P_{x} orbital

The
15^{th} electron of Cl atom is present in 3P_{z} orbital

__[Alternative Answer]__

36. The quantum mechanical treatment of the hydrogen atom gives the energy value:

i) use this expression to find Î”E between n = 3 and m=4

ii) Calculate the wavelength corresponding to the above transition.

Answer:

(i)
E_{3} = âˆ’ 13.6 / 3^{2} = âˆ’1.51 eV atom^{âˆ’1}

**E _{n} = âˆ’ 13.6 / nÂ² = eV
atom^{âˆ’1} **

E_{4}
= âˆ’ 13.6 / 4Â² = âˆ’0.85 eV atom^{âˆ’1}

Î”E
= (E_{4} â€“ E_{3}) = (âˆ’ 0.85) â€“ (âˆ’1.51) = 0.66 eV atom^{âˆ’1}

(ii)
Energy difference Î”E = 0.66 eVatom^{âˆ’1}

=
0.66 Ã— 1.6 Ã— 10^{âˆ’l9}J

=
1.06 Ã— 10^{âˆ’19}J

=
hÎ³

h
(c/Î») = 1.06 Ã—10^{âˆ’19} J

Î»
= (hc) / (1.06 Ã— 10^{âˆ’19}) J

=
(6.626 Ã— 10^{âˆ’34}JS Ã— 3Ã— 10^{8}ms^{âˆ’1} ) / ( 1.06 Ã— 10^{âˆ’19
}) J

Î»
= 1.875Ã— l0^{âˆ’6 }m

37. How fast must a 54g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light 5400Ã…?

Answer:

Î»
= 5400A^{0} = 5400 Ã— 10^{âˆ’10} m ; m = 54g

Î» = 54 Ã— 10^{âˆ’3} Kg ;

V = ?

**Î» = h / mV ;**

V
= h/mÎ»

=
[ 6.626 Ã—10^{âˆ’34} Kgm^{2}s^{âˆ’1} ] / [ 54Ã— 10^{âˆ’3}
Kg Ã— 5400 Ã— 10^{âˆ’10} m]

**V = 2.27 Ã— 10 ^{âˆ’26} ms^{âˆ’1}**

38. For each of the following, give the sub level designation, the allowable m values and the number of orbitals

i) n = 4, l =2, ii) n =5, l = 3 iii) n=7, l=0

Answer:

39. Give the electronic configuration of Mn2+ and Cr3+

Electronic
configuration of Mn^{2+} and Cr^{3+}

_{25}Mn^{2+}
(23 e^{âˆ’}S): 1s^{2}2s^{2} 2p^{6} 3s^{2}3p^{6}4s^{0}3d^{5}

_{24}Cr^{3+}_{
}(21 e^{âˆ’}S): ls^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^{3}

40. Describe the Aufbau principle

In
the ground state of the atoms, the orbitals are filled in the order of their
increasing energies.

That
is once the lower energy orbitals are completely filled that the electrons
enter the next higher energy orbitals.

41. An atom of an element contains 35 electrons and 45 neutrons. Deduce

i) the number of protons

ii) the electronic configuration for the element

iii) All the four quantum numbers for the last electron

Answer:

An
atom of an element contains 35 electrons and 45 neutrons having

1)
The number of protons = 35

ii)
The electronic configuration

=
1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2}
3d^{10}4p^{5}

iii)
Four quantum numbers of the last electron

(4p^{5})

n
= 4; *l* = 1; m = 0; s = âˆ’1/2

__[Alternative Answer]__

42. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wave length associated with the electron revolving arround the nucleus.

To
Prove

"The
Circumference of the circular orbit is an integral multiple of wave
length"

That
is,

Circumference
2Ï€r = nÎ»

If
n = 5, 2Ï€r = 5Î»

43. Calculate the energy required for the process.

He+ (g) â†’ He2+ (g) + e-

The ionisation energy for the H atom in its ground state is - 13.6 ev atom-1.

Answer:

__[Alternative Answer]__

He^{+}_{(g)}^{
}â†’ He^{2+}_{(g)} + e^{âˆ’} E=
?

(Ionisation
energy of H is = 13.6 eVatom^{âˆ’1})

E_{n}
= âˆ’13.6Z^{2} / n^{2}

E_{1}
= [âˆ’13.6 / 1^{2} ] Ã— (2)^{2} = âˆ’56.4ev

(âˆ´Atomic
number of He = 2)

E_{Î±
}= [âˆ’13.6 Ã— (2)^{2} ] / âˆž^{2 }= 0 ev

The
required energy of the process = E_{âˆž} âˆ’ E_{1}

= 0 âˆ’ (âˆ’56.4)

= 56.4 eV

44. An ion with mass number 37 possesses unit negative charge. If the ion contains 11.1% more neutrons than electrons. Find the symbol of the ion.

Answer:

Let
the mono negative ion with mass number 37 be 37_{z} x^{âˆ’ }

No
of protons + No of neutrons = 37

Let
the no of electrons = x

âˆ´ The no
of protons = xâˆ’1

no
of neutrons = 11. % of the no of electrons

=
x + 11.1 % x

=
x + [11.1 / 100]x

=
11.1 x

=>
(xâˆ’1) + 11.1 x = 37

2.11x
âˆ’ 1 = 37 => 2.11x = 38 => x = 38 / 2.11 = 18

âˆ´ Atomic
number Z = No .of Protons =

= x âˆ’ 1 = 17

Hence,
The given ion is ^{37}_{17}Cl^{âˆ’}

__[Alternative Answer]__^{}

45. The Li2+ ion is a hydrogen like ion that can be described by the Bohr model. Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4th orbit.

Answer:

Bohrâ€™s
radius Î³_{n} = [ 0.529n^{2} / Z ] A^{0}

Energy
of electron E_{n} = [ âˆ’13.6z^{2}/ n^{2} ] eV atom^{âˆ’1}

For
Li^{2+} Atomic number = 3,

and
number of electrons = 1

i)
Bohr radius for the 3rd orbit

r_{3}
= 0529(3)^{2} / 3 = l.587 A^{0}

ii)
Energy of an electron in the 4th Orbit

E_{4}
= (âˆ’13.6(3)^{2} ) / 4^{2} = âˆ’7.65 eV atom^{âˆ’1}

46. Protons can be accelerated in particle accelerators. Calculate the wavelength (in Ã…) of such accelerated proton moving at 2.85 Ã—108 ms-1 ( the mass of proton is 1.673 Ã— 10-27 Kg).

Answer:

V=
2.85 Ã— l0^{8} ms^{âˆ’1}

m_{p}=
1.673 Ã— 10^{âˆ’27 }kg

Î»
= ?

Î»
= h / mv

=
[ 6.626 Ã— 10^{âˆ’34}Kg m^{2}s^{âˆ’1} ] / [1.673 Ã— 10^{âˆ’27}Kg
Ã— 2.85 Ã— 10^{8}ms^{âˆ’1 }]

=
[ ( 6.626 ) / ( 1.673Ã— 2.8 ) ]Ã— 10^{âˆ’15} m

Î»
= 1.389 Ã— 10^{âˆ’15}m (or) 1.389 Ã— 10^{-5} A^{0}

47. What is the de Broglie wavelength (in cm) of a 160g cricket ball travelling at140 Km hr-1.

Answer:

V
= 140 kmhr^{âˆ’1} = [(140 Ã— 1000) / 3600] ms^{âˆ’1 }= 38.88ms^{âˆ’1}

m
= 160 g = 160 Ã— l0^{âˆ’3} Kg;

Solution
: Î» = h / mv

=
[6.626 Ã— 10^{âˆ’34}Kgm^{2}s^{âˆ’1} ] / [ 160 Ã— 10^{âˆ’3}Kg
Ã— 38.88ms^{âˆ’1}]

=
[(6.626) / (160 Ã— 38.88)] Ã— 10^{âˆ’31}m

=
1.065Ã— 10^{âˆ’34 }m

Î»
= 1.065Ã— 10^{âˆ’32}cm

48. Suppose that the uncertainty in determining the position of an electron in an orbit is 0.6 Ã…. What is the uncertainty in its momentum?

Answer:

Î”x
= 0.6A^{0} ;

Î”
p = ?

Î”x
Î”p â‰¥ h / 4Ï€

0.6
Ã— 10^{âˆ’10}m Ã— Î” p â‰¥ (6.626 Ã— 10^{âˆ’34} ) / 4Ï€

Î”p
â‰¥ [ ( 6.626 Ã— 10^{âˆ’34} ) / ( 0.6 Ã— 10^{âˆ’10 }Ã— 4Ï€) ] Kgm^{2}s^{âˆ’1}

Î”p
â‰¥ 0.879 Ã— 10^{âˆ’24} Kgms^{âˆ’1}

49. Show that if the measurement of the uncertainty in the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in its velocity is equal to its velocity / 4Ï€

Answer:

If,
Î”x = Î» Then V = ?

Î”x
Î”p â‰¥ h / 4Ï€

Î»
m Î”V â‰¥ h/4Ï€

â‡’ Î”V â‰¥ h / 4Ï€(mÎ»)

â‡’ Î”V â‰¥ h / [4Ï€m(h/mv)]

(as Î»=h/mv)

â‡’ Î”V â‰¥ V / 4Ï€ (Proved)

__[Alternative Answer]__

50. What is the de Broglie wave length of an electron, which is accelerated from the rest, through a potential difference of 100V?

Answer:

Potential
difference = 100V

=
100 Ã— l.6 Ã— l0^{âˆ’19} J = 1.6 Ã— 10^{âˆ’17}J

Î»
= ?

Î»=
h / âˆš(2meV)

=
[ 6.26 Ã— 10^{âˆ’34 }Kgm^{2}s^{âˆ’1 }] / âˆš[ 2Ã—9.1Ã—10^{âˆ’31}Kg
Ã— 1.6Ã—10^{âˆ’17}J ]

=
[ 6.26 Ã— 10^{âˆ’34}] / âˆš[ 29.12Ã—10^{âˆ’48}]

=
[ 6.26 Ã— 10^{âˆ’34}] / [ 5.4Ã—10^{âˆ’24}]

Î»
= 1.227 Ã— 10^{âˆ’10} m (or) 1.227A^{0}

51. Identify the missing quantum numbers and the sub energy level

Answer:

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