Question 2.14:
How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your Answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).
Answer:
The expression of energy is given by,
Where,
Z = atomic number of the atom
n = principal quantum number
For ionization from n1 = 5 to ,
Hence, the energy required for ionization from n = 5 to n =
is 8.72 × 10–20 J.
Energy required for n1 = 1 to n =
,
Hence, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.
Question 2.15:
What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?
Answer:
When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:
Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.
The number of spectral lines produced when an electron in the nth level drops down to the ground state is given by .
Given,
n = 6
Number of spectral lines = 15
Question 2.16:
(i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Answer:
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:
E5 = –8.72 × 10–20 J
(ii) Radius of Bohr’s nth orbit for hydrogen atom is given by,
rn = (0.0529 nm) n2
For,
n = 5
r5 = (0.0529 nm) (5)2
r5 = 1.3225 nm
Question 2.17:
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer:
For the Balmer series, ni = 2. Thus, the expression of wavenumber is given by,
Wave number is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, has to be the smallest.
For to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3, we get:
= 1.5236 × 106 m–1
Question 2.18:
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs.
Answer:
Energy (E) of the nth Bohr orbit of an atom is given by,
Where,
Z = atomic number of the atom
Ground state energy = – 2.18 × 10–11 ergs
= –2.18 × 10–11 × 10–7 J
= – 2.18 × 10–18 J
Energy required to shift the electron from n = 1 to n = 5 is given as:
ΔE = E5 – E1
Question 2.19:
The electron energy in hydrogen atom is given by En = (–2.18 × 10–18)/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:
Given,
Energy required for ionization from n = 2 is given by,
= 0.545 × 10–18 J
ΔE = 5.45 × 10–19 J
Here, λ is the longest wavelength causing the transition.
= 3647 × 10–10 m
= 3647 Å
Question 2.20:
Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1.
Answer:
According to de Broglie’s equation,
Where,
λ = wavelength of moving particle
m = mass of particle
v = velocity of particle
h = Planck’s constant
Substituting the values in the expression of λ:
Hence, the wavelength of the electron moving with a velocity of 2.05 × 107 ms–1 is 3.548 × 10–11 m.
Question 2.21:
The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.
Answer:
From de Broglie’s equation,
Given,
Kinetic energy (K.E) of the electron = 3.0 × 10–25 J
Substituting the value in the expression of λ:
Hence, the wavelength of the electron is 8.9625 × 10–7 m.
Question 2.22:
Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na+, K+, Mg2+, Ca2+, S2–, Ar
Answer:
Isoelectronic species have the same number of electrons.
Number of electrons in sodium (Na) = 11
Number of electrons in (Na+) = 10
A positive charge denotes the loss of an electron.
Similarly,
Number of electrons in K+ = 18
Number of electrons in Mg2+ = 10
Number of electrons in Ca2+ = 18
A negative charge denotes the gain of an electron by a species.
Number of electrons in sulphur (S) = 16
∴ Number of electrons in S2- = 18
Number of electrons in argon (Ar) = 18
Hence, the following are isoelectronic species:
1) Na+ and Mg2+ (10 electrons each)
2) K+, Ca2+, S2– and Ar (18 electrons each)
Question 2.23:
(i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2–(d) F–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.
Answer:
(i) (a) H– ion
The electronic configuration of H atom is 1s1.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of H– = 1s2
(b) Na+ ion
The electronic configuration of Na atom is 1s2 2s2 2p6 3s1.
A positive charge on the species indicates the loss of an electron by it.
∴ Electronic configuration of Na+ = 1s2 2s2 2p6 3s0 or 1s2 2s2 2p6
(c) O2– ion
The electronic configuration of 0 atom is 1s2 2s2 2p4.
A dinegative charge on the species indicates that two electrons are gained by it.
∴ Electronic configuration of O2– ion = 1s2 2s2 p6
(d) F– ion
The electronic configuration of F atom is 1s2 2s2 2p5.
A negative charge on the species indicates the gain of an electron by it.
∴ Electron configuration of F– ion = 1s2 2s2 2p6
(ii) (a) 3s1
Completing the electron configuration of the element as
1s2 2s2 2p6 3s1.
∴ Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 = 11
∴ Atomic number of the element = 11
(b) 2p3
Completing the electron configuration of the element as
1s2 2s2 2p3.
∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7
∴ Atomic number of the element = 7
(c) 3p5
Completing the electron configuration of the element as
1s2 2s2 2p6 3s2 3p5∴ Number of electrons present in the atom of the element = 2 + 2 + 6+2+5 = 17
∴ Atomic number of the element = 17
(iii) (a) [He] 2s1
The electronic configuration of the element is [He] 2s1 = 1s2 2s1.
∴ Atomic number of the element = 3
Hence, the element with the electronic configuration [He] 2s1 is lithium (Li).
(b) [Ne] 3s2 3p3
The electronic configuration of the element is [Ne] 3s2 3p3= 1s2 2s2 2p6 3s2 3p3.
∴ Atomic number of the element = 15
Hence, the element with the electronic configuration [Ne] 3s2 3p3 is phosphorus (P).
(c) [Ar] 4s2 3d1
The electronic configuration of the element is [Ar] 4s2 3d1= 1s2 2s2 2p6 3s2 3p6 4s2 3d1.
∴ Atomic number of the element = 21
Hence, the element with the electronic configuration [Ar] 4s2 3d1 is scandium (Sc).
Question 2.24:
What is the lowest value of n that allows g orbitals to exist?
Answer:
For g-orbitals, l = 4.
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
∴ For l = 4, minimum value of n = 5
Question 2.25:
An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
Answer:
For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2
Question 2.26:
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Answer:
(i) For an atom to be neutral, the number of protons is equal to the number of electrons.
∴ Number of protons in the atom of the given element = 29
(ii) The electronic configuration of the atom is
1s2 2s2 2p6 3s2 3p6 4s2 3d10.
Question 2.27:
Give the number of electrons in the species , H2 and
Answer:
:
Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2
∴ Number of electrons in = 2 – 1 = 1
H2:
Number of electrons in H2 = 1 + 1 = 2
:
Number of electrons present in oxygen molecule (O2) = 8 + 8 = 16
∴ Number of electrons in = 16 – 1 = 15
Question 2.28:
(i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f
Answer:
(i) n = 3 (Given)
For a given value of n, l can have values from 0 to (n – 1).
∴ For n = 3
l = 0, 1, 2
For a given value of l, ml can have (2l + 1) values.
For l = 0, m = 0
l = 1, m = – 1, 0, 1
l = 2, m = – 2, – 1, 0, 1, 2
∴ For n = 3
l = 0, 1, 2
m0 = 0
m1 = – 1, 0, 1
m2 = – 2, – 1, 0, 1, 2
(ii) For 3d orbital, l = 2.
For a given value of l, mlcan have (2l + 1) values i.e., 5 values.
∴ For l = 2
m2 = – 2, – 1, 0, 1, 2
(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.
For p-orbital, l = 1.
For a given value of n, l can have values from zero to (n – 1).
∴ For l is equal to 1, the minimum value of n is 2.
Similarly,
For f-orbital, l = 4.
For l = 4, the minimum value of n is 5.
Hence, 1p and 3f do not exist.
Question 2.29:
Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n = 1, l = 0; (b) n = 3; l =1 (c) n = 4; l = 2; (d) n = 4; l =3.
Answer:
(a) n = 1, l = 0 (Given)
The orbital is 1s.
(b) For n = 3 and l = 1
The orbital is 3p.
(c) For n = 4 and l = 2
The orbital is 4d.
(d) For n = 4 and l = 3
The orbital is 4f.
Question 2.30:
Explain, giving reasons, which of the following sets of quantum numbers are not possible.
a |
n = 0 |
l = 0 |
ml = 0 |
|
b |
n = 1 |
l = 0 |
ml = 0 |
|
c |
n = 1 |
l = 1 |
ml = 0 |
|
d |
n = 2 |
l = 1 |
ml = 0 |
|
e |
n = 3 |
l = 3 |
ml = – 3 |
|
f |
n = 3 |
l = 1 |
ml = 0 |
(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.Answer:
(b) The given set of quantum numbers is possible.
(c) The given set of quantum numbers is not possible.
For a given value of n, ‘l’ can have values from zero to (n – 1).
For n = 1, l = 0 and not 1.
(d) The given set of quantum numbers is possible.
(e) The given set of quantum numbers is not possible.
For n = 3,
l = 0 to (3 – 1)
l = 0 to 2 i.e., 0, 1, 2
(f) The given set of quantum numbers is possible.