# class 11 chemistry solved questions chapter 2

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Question 1:
The atomic number of an element is 5 and mass number is 11. Find the number
of electrons, protons and neutrons present in an atom of it. How can this element
be represented?
Answer 1:
Number of electrons (Z = 5) = 5
Number of protons (Z = 5) = 5
Number of neutrons (A = Z) = 11 – 5 = 6
Since the atomic number is 5, the element is boron and is represented as 5B11
Question 2:
The Kinetic energy of an electron is 4.55 × 10–25 J. The mass of electron is
9.1×10–31 kg. Calculate velocity, momentum and the wavelength of the electron.
Answer 2:
1
Kinetic energy =  2 = 4.55 × 10–25 J
2
2 ×  2 × 4.55 × 10−25
⇒ =
=
=1

9.1 × 10−31
⇒  = 1 /
Hence, velocity is 1 m/s.
Momentum = mv = 9.1×10–31 × 1 = 9.1×10–31 kg m/s
2
Wavelength  =
ℎ

=
6.626×10−34
9.1×10−31
= 0.728 × 10−3 = 7.28 × 10−4 m
Question 3:
Which has higher energy: 3d or 4s? Explain?
Answer 3:
3d because n + l for 3d is 3 + 2 = 5, but for 4s it is 4 + 0 = 5
Question 4:
State and explain Hund’s rule of maximum multiplicity?
Answer 4:
It states that pairing of e in the degenerate orbital is not possible until there is one
e in each orbital.
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Question 5:
Define (i) work function (ii) Threshold frequency.
Answer 5:
(i) Work function is the minimum energy of light radiations required to remove
an electron from the metal surface. (ii) Frequency related to work function is
threshold frequency.
Question 6:
Half-filled and full filled orbitals have extra stability. Why?
Answer 6:
Due to exchange energy. There is extra stability associated with electronic
configurations.
Question 7:
What are isoelectronic series?
Answer 7:
Species having same no of electrons.
Average
Question 1:
The uncertainty in the position of a moving bullet of mass 10g is 10 -5m.Calculate
the uncertainty in the velocity.
Answer 1:
According to uncertainty principle ∆. ∆ =
−34
⇒ ∆ =
ℎ
4
ℎ
6.626 × 10
=
= 5.27 × 10−28 /
−2
−5
4∆ 4 × 3.14 × 10 × 10
Question 2:
Light of wavelength 5000 Å falls on a metal surface of work function 1.9 eV.
Find (a) the energy of photos (b) kinetic energy of photoelectrons?
Answer 2:
Wavelength of light () = 5000 Å = 5000 × 10−10 = 5 × 10−7 m
Work function (ℎ ) = 1.9  = 1.9 × 1.6 × 10−19 J = 3.04 × 10−19 J
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Energy of photons
ℎ 6.626 × 10−34 × 3 × 108
= ℎ =
=
= 3.97 × 10−19 J
−7

5 × 10
Kinetic energy of photoelectrons = ℎ − ℎ = 3.97 × 10−19 − 3.04 × 10−19
= 9.3 × 10−20 J
Question 3:
Which of the following are iso-electronic species?
Na+, K+, Mg2+, Ca2+, S2-, Ar.
Answer 3:
Na+ and Mg2+ are iso-electrons species (have 10 electrons)
K+, Ca2+, S2- are iso-electronic species (have 18 electrons).
Question 4:
(i) Write the electronic configuration of the following ions?
(a) H(b) Na+
(c) O2(d) F-.
(ii) What are the atomic numbers of the elements whose outermost electronic
configurations are represented by
(a) 3s1
(b) 2p3
(c) 3d6?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s1
(b) [Ne] 3s2 3p3
(c) [Ar] 4s2 3d1.
Answer 4:
(i) (a)1s2
(b)1s22s22p6
(c)1s22s22p6
(d)1s22s22p6.
(ii) (a) Na (Z = 11) has outermost electronic Configuration = 3s1
(b) N (Z = 7) has outermost electronic Configuration = 2p3
(c) Fe (Z = 26) has outermost electronic Configuration = 3d6
(iii) (a) Li
(b) P
(c) Sc
Question 5:
The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will
experience more effective nuclear charge from the nucleus?
Answer 5:
Configuration of the two elements are:
Al (Z = 13): [Ne]103s23p1; Si (Z = 14) : [Ne]103s23p2
The unpaired electrons in silicon (Si) will experience more effective nuclear
charge because the atomic number of the element Si is more than that of Al.
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Question 6:
The electronic energy is negative. How will you account for that?
Answer 6:
Electron just out of atom has zero energy. It enters in the atom due to attraction
of atom so it loses energy.
Question 7:
How many unpaired electrons are present in Pd (Z = 46)?
Answer 7:
Zero.
Question 8:
Write electronic configuration of Ca(Z = 24) and Cu( Z = 22).
Answer 8:
Ca = 1s2 2s2 2p6 3s2 3p6 4s2
Cu =1s2 2s2 2p6 3s2 3p6 3d10 4s1
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