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Oxidation is a process in which loss of electrons take place. Reduction is a process in which gain of electrons takes place.

(i) Propyle Benzene 

(ii) 3 Chloro Propenal

Cathode is electrode towards which cations are attracted. Anode is electrode which attracts anions.

(i) C₄H₁₀ + 3/2O₂ → 5H₂O + 4CO₂ 

(ii) C₆H₅CH₃ + 9O₂ → 7CO₂ + 4H₂O

Because of de-localisation of 671 electrons in the benzene ring.

Correct option: (d) All of the above

Correct option: a) C/N

Synthetic detergents are preferred for cleaning clothes. When soaps are dissolved in water
containing calcium ions, these ions form insoluble salts that are of no further use. However, when synthetic detergents are dissolved in water containing calcium ions, these ions form soluble salts that act as cleansing agents.

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

(a)1s² (b)1s²2s²2p⁶ (c)1s²2s²2p6 (d)1s²2s²2p⁶.

(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

9.11 × 10^-31 kg

1. No. 

2. Pauli’s exclusion principle. 3. No two electrons in an atom can have same set of four quantum numbers. 

4. Forthe1st electron: n = 2, l = 0, m = 0, s = +½.

For the 2nd electron: n = 2, l = 0, m = 0, s = -½

(a) (i) The positive charge and most of the mass of the atom is densely concentrated in extremely small region of the atom called nucleus. 

(ii) The nucleus is surrounded by electrons that ’ move around the nucleus with a very high speed in circular paths called orbits. 

(iii) Electrons and the nucleus are held together by electrostatic forces of attraction. 

(b) i) It failed to explain the stability of atom, 

(ii) It says nothing about the electronic structure of atoms.

1. J.J. Thomson proposed that an atom possess a spherical shape in which the positive charge is uniformly distributed. The electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement. The mass of the atom is assumed to be uniformly distributed over the atom. This model explained the overall neutrality of the atom.

2. Thomson’s model of atom can be visualised as a pudding or watermelon of positive charge with electrons embedded into it like the plums or seeds. 

3. Thomson’s model was not consistent with the results of later experiments. It failed to explain the observations of Rutherford’s α-particle scattering experiment.

a) 1. Bohr’s model could explain the stability of an atom. 

2. Bohr’s model could explain the atomic spectrum of hydrogen. 

b) 1. Failed to explain the finer details of hydrogen atom spectrum observed by using sophisticated spectroscopic techniques.

2. It could not explain the ability of atoms to form molecules by chemical bonds.

The orbitals having same energy are called degenerate orbitals.

λ = \(\frac{h}{mv}\) 

= \(\frac{6.626\times 10^{-34}Js}{9.1\times10^{-31}kg \times 2.05 \times 10^7ms^{-1}}\) 

= 3.55 x 10^-11m.

1. Mass of one electron = 9.11 × 10^-31 kg 

∴ Number of electrons in one gram = \(\frac{10^{-3} kg}{9.11\times 10^{-31}kg}\) = 1.098 x 10²⁷

2. Mass of one electron = 9.11 × 10^-31 kg 

∴ Mass of 1 mole of electrons = 9.11 × 10^-31kg × 6.022 × 10²³  

= 5.486 x 10^-7 kg

Charge on one electron = 1.602 × 10^-19 C 

∴ Charge on one mole of electrons = (1.602 × 10^-19 C) × (6.022 × 10²³) 

= 9.65 × 10⁴ C.

(a) (i) 3s¹ - Atomic number is 11 (Na) 

(ii) 3p⁵ - Atomic number is 17 (Cl)

(b) Na⁺, Mg²⁺ (both of them have same no. of electrons, i.e.,10 each)

(c) λ = \(\frac{h}{mv}\) = \(\frac{6.626\times10^{-34}Js}{0.1kg\times10ms^{-1}}\) = 6.626 x 10^-34m

Xe gas, because as size increases ionization energy decrease and relativity increases.

Correct option is (3) 10⁹

f - C/λ - \(\frac{3\times10^8}{6000\times10^{-9}}\) = 5 x 10¹³ Hz

Total band width used for transmission

 = 2% of 5 x 10¹³ = 10¹² Hz

Number of channels

 = \(\frac{10^{12}}{1\times10^3}\) = 10⁹ channels

Concept used:

The number which is not divisible by 2 is an odd number.

Calculation:

The odd number from 1 to 45 which is divisible by 3 is 

3, 9, 15, 21, 27, 33, 39, 45

∴ The number at the 6^th position which is odd and divisible by 3 is 33

Calculation 

Prime factorisation of both numbers 

⇒ 5320 = 2 × 2 × 2 × 5 × 7 × 19 

⇒ 4389 = 3 × 7 × 11 × 19 

LCM = 2 × 2 × 2 × 5 × 7 × 3 × 11 × 19 = 175560 

∴ The required answer is 175560 

The answer is (4) B > C > A

Option "A" represent Guanadine type compound while (C) represent the imidine class of compound and (B) represent aliphatic amines. The conjugate acid of Guanadine is most resonance stabilised followed by imidine. Hence A is more basic than C than B.

KNO₃(aq) + Nacl(aq) \(\rightleftharpoons\) kcl (aq) + NaNO₃ (aq) is a reversible reaction.

Number of oxide in per unit cell o^-2 = 4

Number of M⁺² metal per unit cell = 1/2 x θ = 4

∵ 50% metal M⁺²  oxidized into M⁺³ ions it means , 2 metal al M⁺² is oxidized into M+3 ions applying electrical Meutraility  ⇒ 2 M⁺³ = 3 M⁺²

it means , on oxidation of 50% metal M⁺² , or on oxidation of 2 metal of M⁺² metal per unit cell.

∴ total number of metal M per unit cell = 3

∴ number of vacant tetra hedral oxide = θ - 3 = 5

Answer is (4)

Greenhouse gases – carbondioxide, other green house gases are methane, watervapour, nitrousoxide CFCs and Ozone.

Answer is (4) (a), (b), (c) and (e)

CO₂, H₂O, CFCs and O₃ are green house gases.

From Henry law

P = K_H X\(_{CO_2}\)

0.835 = 1.67 x 10³ x 1.67 x 10³x \(\cfrac{W_{CO_2/44}}{\frac{W_{CO_2}}{44}+\frac{1000}{18}}\) 

\(W_{CO_2}\) = 1.2228g = 1.222.8 x 10^-3 g

Or

P = K_HX\(_{CO_2}\)

0.835 = 1.67 x 10³ x \(\frac{n_{CO_2}}{n_{CO_2}+n_{H_2O}}\)

0.835 = 1.67 x 10³ x \(\cfrac{W_{CO_2}/44}{\frac{1000}{18}}\)

W\(_{CO_2}\) = 1.222g = 1222.2 x 10^-3 g