CBSE Solutions for Class 11 English

જવાબ : All have equal energy.

જવાબ : Similarities:

• Both have spherical shape.
• Both have same angular momentum.

• 1s has no node while 2s has one node.
• Energy of 2s is greater than that of 1s.

જવાબ : Electron has particle nature

જવાબ : The electronic configuration of the element palladium (Z = 46) is [Kr]³⁶ 4d¹⁰ 5S^°.
This means that it has no impaired electron.

જવાબ : A quantum is a bundle of energy of a definite magnitude (E = hv) and it may be from any source. However, a photon is quantum of energy associated with light only.

જવાબ : The energy of the electron is regarded as zero when it is at infinite distance from the nucleus.
At that point force of attraction between the electron and the nucleus is almost nil. Therefore, its energy is regarded as zero.

જવાબ : The metals with low ionisation enthalpies are used in photoelectric cells. Caesium (Cs), an alkali metal belonging to group 1 is the most commonly used metal.

જવાબ : I = 0,1, 2
m =- 2, — 1, 0, + 1, + 2 and S = +1/2 and-1/2
for each value of m.

જવાબ : Both have identical shape, consisting of four lobes.

જવાબ : Ground state means the lowest energy state. When the electrons absorb energy and jump to outer orbits, this state is called excited state.

જવાબ : Balmer series.

જવાબ : ‘l’ signifies the secondary quantum number.
‘L’ signifies second energy level (n = 2).

જવાબ : Orbitals having same energy belonging to the same subshell.

જવાબ : In the construction of electron microscope used for the measurement of objects of very small size.

જવાબ : As Fe³⁺ contains 5 impaired electrons while Fe²⁺ contains only 4 unpaired electrons. Fe³⁺ is more paramagnetic.

જવાબ : Hydrogen.

જવાબ : 6.62 x 10³⁴ Js.

જવાબ : Cosmic rays > X-rays > radio waves.

જવાબ : Electron has wave nature.

જવાબ : Z = 35
A = 80
Atomic no. = 35 No. of protons = 35
No. of protons No. of electrons No. of neutrons
= 80 – 35 = 45

જવાબ : α < p < e.

જવાબ : 15

જવાબ : 18

જવાબ : I represents the subshell and L represents shell.

જવાબ : s-orbital

જવાબ : h/4π

જવાબ : Fe (Z = 26) : [Ar]¹⁸ 3d⁶4s²
Fe²⁺ion : [Ar]¹⁸ 3d⁶
No. of unpaired electrons = 4

જવાબ :
The relation is known as de Broglie’s relationship.

જવાબ :

જવાબ :
(ii) For hydrogen atom ; r_n = 0.529 x n² Å
r₅ = 0.529 x (5)² = 13.225 Å = 1.3225 nm.

જવાબ : According to uncertainty principle,

જવાબ : According to uncertainty principle,

જવાબ :

જવાબ : (i)Boron (Z = 5) ; 1s² 2s² 1p¹
(ii)Neon (Z = 10) ; 1s² 2s² 2p⁶
(iii)Aluminium (Z = 13) ; 1s² 2s² 2p⁶ 3s² 3p¹
(iv)Chlorine(Z = 17) ; 1s² 2s² 2p⁶ 3s² 3p⁵
(v)Calcium (Z = 20) ; 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
(vi)Rubidium (Z = 37) ; 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰4s² 4p⁶ 5s¹.

જવાબ :

જવાબ : (i) Mass of an electron = 9.1 × 10^-28 g
9.1 × 10^-28 g is the mass of = 1 electron

(ii) One mole of electrons = 6.022 × 10²³ electrons
Mass of 1 electron = 9.1 × 10^-31 kg
Mass of 6.022 × 10²³ electrons = (9.1 × 10.31kg) × (6.022 × 10²³) = 5.48 × 10^-7 kg
Charge on one electron = 1.602 × 10^-19 coulomb
Charge on one mole electrons = 1.602 × 10^-19 × 6.022 × 10²³ = 9.65 × 10⁴ coulombs

જવાબ : (i) One mole of methane (CH₄) has molecules = 6.022 × 10²³
No. of electrons present in one molecule of CH₄ = 6 + 4 = 10
No. of electrons present in 6.022 × 10²³ molecules of CH₄ = 6.022 × 10²³ × 10
= 6.022 × 10²⁴ electrons (ii) Step I. Calculation of total number of carbon atoms
Gram atomic mass of carbon (C-14) = 14 g = 14 × 10³ mg
14 × 10³ mg of carbon (C-14) have atoms = 6.022 × 10²³

Step II. Calculation of total number and tatal mass of neutrons
No. of neutrons present in one atom (C-14) of carbon = 14 – 6 = 8
No. of neutrons present in 3-011 × 10²⁰ atoms (C-14) of carbon = 3.011 × 10²⁰ × 8
= 2.408 × 10²¹ neutrons
Mass of one neutron = 1.675 × 10^-27 kg
Mass of 2.408 × 10²¹ neutrons = (1.675X10^-27 kg) × 2.408 × 10²¹
= 4.033 × 10^-6 kg. (iii) Step I. Calculation of total number ofNH₃ molecules
Gram molecular mass of ammonia (NH₃) = 17 g = 17 × 10³ mg
17 × 10³ mg of NH₃ have molecules = 6.022 × 10²³

Step II. Calculation of total number and mass of protons
No. of protons present in one molecule of NH₃ = 7 + 3 = 10 . No. of protons present in 12.044 × 10²⁰ molecules of NH₃ = 12.044 × 10²⁰ × 10
= 1.2044 × 10²² protons
Mass of one proton = 1.67 × 10^-27 kg
Mass of 1.2044 × 10²² protons = (1.67 × 10^-27 kg) × 1.2044 × 10²²
= 2.01 × 10^-5 kg.
No, the answer will not change upon changing the temperature and pressure because only the number of protons and mass of protons are involved.

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જવાબ : (i) Energy of photon (E) = hv
h = 6.626 × 10^-34 J s ; v = 3 × 10¹⁵ Hz = 3 × 10¹⁵s^-1
∴ E = (6.626 × 10^-34 J s) × (3 × 10¹⁵ s^-1) = 1.986 × 10¹⁸ J
Energy of photon (E) = hv = hcλ
h = 6.626 × 10 34 J s; c = 3 × 10⁸ m s^-1 ;
λ= 0.50 Å = 0.5 × 10^-10 m.
 

જવાબ :

જવાબ : Energy of photon (E) = hcλ
h = 6.626 × 10^-34 Js, c = 3 × 10⁸ m s^-1, λ = 4000 pm = 4000 × 10^-12 = 4 × 10^-9 m

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જવાબ : The maximum no. of emission lines = n(n–1)2 = 6(6–1)2 =3 × 5 = 15

જવાબ : Cosmic rays < X-rays < amber colour < microwave < FM

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જવાબ : Time duration (t) = 2 ns = 2 x 10^-9 s

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જવાબ : Let the no. of electrons in the ion = x
∴ the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + x×31.7100 = xc + 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons
= (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x) or 2.304x = 56 – 3 = 53
x = 532.304 = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

જવાબ : Let the no. of electron in the ion = x
∴ the no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x + x×11.1100 = 1.111 x
Mass no. or mass of the ion = No. of protons + No. of neutrons
(x – 1 + 1.111 x)
Given mass of the ion = 37
∴ x- 1 + 1.111 x = 37 or 2.111 x = 37 + 1 = 38
x = 382.111 = 18
No. of electrons = 18 ; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17 ; Atom corresponding to ion = Cl
Symbol of the ion = 3717Cl^–

જવાબ : An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + [x×31.7100 = (x × 0.317x)
Now, Mass no. of element = no. of protons =no. neutrons
81 = x + x + 0-317 x = 2.317 x or x = 812.317 = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine 8135Br

જવાબ : In the symbol BAX of an element :
A denotes the atomic number of the element
B denotes the mass number of the element.
The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus,
Symbols 7935Br and ⁷⁹Br are accepted because atomic number of Br will remain 35 even if not mentioned. Symbol 3579Br is not accepted because atomic number of Br cannot be 79 (more than the mass number = 35). Similarly, symbol 35Br cannot be accepted because mass number has to be mentioned. This is needed to differentiate the isotopes of an element.

જવાબ : We have studied that in Rutherford’s experiment by using heavy metals like gold and platinum, a large number of a-particles sufferred deflection while a very few had to retrace their path.

જવાબ : (a) Radius of zinc atom =Undefined control sequence \AA= 1.3 Å = 1.3 × 10^-10m = 130 × 10^-12m = 130 pm
(b) Length of the scale = 1.6 cm = 1.6 × 10¹⁰ pm
Diameter of zinc atom = 260 pm

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જવાબ : The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 ×10⁸

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જવાબ :
Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.

જવાબ : (a) For n = 4
Total number of electrons = 2n² = 2 × 16 = 32
Half out of these will have ms = —1/2
∴ Total electrons with ms (-1/2) = 16
(b) For n = 3
l= 0 ; m_l = 0, m_s +1/2, -1/2 (two e^–)

જવાબ : (i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(ii) The set of quantum numbers is possible.
(iii) The set of quantum numbers is not possible because, for n = 1, l can not be equal to 1. It can have 0 value.
(iv) The set of quantum numbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantum numbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantum numbers is possible.

જવાબ : (a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital

જવાબ : (i) For n = 3; l = 0, 1 and 2.
For l = 0 ; m_l = 0
For l = 1; m_l = +1, 0, -1
For l = 2 ; m_l = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; m_l can have any of the values -2, -1, 0,
+ 1, +2.
(iii) 1p and 3f orbitals are not possible.

જવાબ : (i) (a) 1s²
(b) 1s² 2s² 2p⁶
(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

જવાબ : Na⁺ and Mg²⁺ are iso-electronic species (have 10 electrons) K⁺, Ca²⁺ , S^2- are iso-electronic species (have 18 electrons)

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