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Correct option is C) 2n² 

Option : a. Distillation under reduced pressure

Option : (B) fractional distillation

i. Thin-layer chromatography (TLC) : 

If components are colourless but have the property of fluorescence then they can be visualized under UV light, or the plate can be kept in a chamber containing a few iodine crystals. The iodine vapours are adsorbed by the components and the spots appear brown. Also, spraying agent like ninhydrin can also be used (for amino acids).

ii. Paper Chromatography : 

The spots of the separated colourless components may be observed either under ultra-violet light or by the use of an appropriate spraying agent.

Correct option is  D) 2l + 1

A) n = 3, l = 2, m_I = -1, m_s = -1/2 

Option : (C) Solvent extraction

i. Chromplate/TLC plate 

ii. Ninhydrin

i. Thin layer chromatography

ii. Based on the developed chromatogram, spot ‘x’ has the highest R_f value while spot ‘z’ the lowest R_f value.

iii. Based on the TLC, spot ‘z’ being strongly adsorbed will elute at the end of a column chromatography.

iv. Applications of TLC are :

• Separation of plant pigments from its mixture.
• Separation of impurities from a given organic compound.
• Separation of different amino acid.

Bohr – Sommerfeld model.

Correct option is  C) parallel

The electrons in ground state i.e. lowest, energy state absorb energy and move into excited state where they are unable to stay for long periods so lose the energy and come back to the ground state. The emitted radiation appears as line in line spectrum.

(a) A and R are correct and R is the correct explanation of A.

(a) A and R are correct and R is the correct explanation of A

(a) If both assertion and reason are true and reason is the correct explanation of assertion.

The quantum numbers for two electrons of Helium are given below as per Pauli’s exclusive principle.

Three of quantum numbers are same. So fourth must be different so the two electrons have anti-parallel spins. 

• It is represented by the letter 7′ and can take integral values from zero to n – 1, where n is the principal quantum number. 
• Each l value represents a subshell (orbital). l = 0, 1, 2, 3 and 4 represents the s, p, d, f and g orbitals respectively. 
• The maximum number of electrons that can be accommodated in a given subshell (orbital) is 2(2l + 1). 

It is used to calculate the orbital angular momentum using the expression Angular momentum = (\(\sqrt{ l(l+1)}\frac{h}{2π}\)

• In a multi-electron atom, in addition to the electrostatic attractive force between the electron and nucleus, there exists a repulsive force among the electrons. 
• These two forces are operating in the opposite direction. This results in the decrease in the nuclear force of attraction on electron. 
• The net charge experienced by the electron is called effective nuclear charge. 
• The effective nuclear charge depends on the shape of the orbitals and it decreases with increase in azimuthal quantum number l. 
• The order of the effective nuclear charge felt by a electron in an orbital within the given shell is s > p > d > f.
• Greater the effective nuclear charge, greater is the stability of the orbital. Hence, within a given energy level, the energy of the orbitals are in the following order s < p < d < f.

Formula for total number of nodes = n – 1 

1. For 2s orbital: 

Number of radial nodes =1. 

2. For 4p orbital: 

Number of radial nodes = n – l – 1. 

= 4 – 1 – 1 = 2 

Number of angular nodes = l 

∴ Number of angular nodes = 1 

So, 4p orbital has 2 radial nodes and 1 angular node.

3. For 5d orbital: 

Total number of nodes = n – 1 = 5 – 1 = 4 nodes 

Number of radial nodes = n – l – 1 = 5 – 2 – 1 = 2 radial nodes. 

Number of angular nodes = l = 2 

∴ 5d orbital have 2 radial nodes and 2 angular nodes.

4. For 4f orbital: 

Total number of nodes = n – 1 = 4 – 1 = 3 nodes 

Number of radial nodes = n – 7 – 1 = 4 – 3 – 1 = 0 node. 

Number of angular nodes = l = 3 nodes 

∴ 4f orbital have 0 radial node and 3 angular nodes.

Answer: (a) 3d⁵ 4s² 

(d) n = 6 to n = 5

n = 6 to n = 5

E₆ = -13.6 / 6² ; E₅ = – 13.6 / 5²

E₆ – E₅ = (-13.6 / 6⁶) – (-13.6 / 5²) 

= 0.166 eV atom^-1 

E₅ – E₄ = (-13.6 / 5²) – (-13.6 / 4²) 

= 0.306 eV atom^-1

1. The rule violated is Aufbau principle. The elements which have half filled or completely filled orbitals have greater stability. So copper can get stability by transferring one electron from 4s to 3d (their energies are close to each other).

2. So the electronic configuration of copper is 1s² 2s² 2p⁶3s²3p⁶3d¹⁰4s¹ not 1s² 2s² 2p 3s²3p⁶3d⁹ 4s²

Answer: (c) 3d

(d) ∆E.∆x ≥ h/4π.

Pauli’s exclusion principle states that “No two electrons in an atom can have the same set of values of all four quantum numbers”.

Illustration: H(Z = 1) 1s¹ .

One electron is present in hydrogen atom, the four quantum numbers are n = 1, l = 0, m = 0 and s = + 1/2. For helium Z = 2. He: 1s². In this one electron has the quantum number same as that of hydrogen, n = 1,l = 0, m = 0 and s = + 1/2 For other electron, fourth quantum number is different, 

i.e. n = 1, l = 0, m = 0 and s = – 1/2 .

Let the number of electrons in an ion = x 

number of neutrons = n = x + 11.1/100 eV = 1.111 x (As the number of neutrons are 11.1% more than the number of electrons) 

In the neutral of atom, number of electron. 

e^- = x – 1 (as the ion carries -1 charge) 

Similarly number of protons = P = x – 1 Number of protons + number of neutrons = mass number = 37 

(x – 1) + 1.111 x = 37 .

 2.111 x = 37 +1 

2.111 x = 38

x = 38/2.111 

= 18.009 = 18 

∴ Number of protons = atomic number – 1 

= 18 -1 

= 17 

∴ The symbol of the ion = . \(_{17}^{37}Cl\).

(c) 30

M²⁺ : 1s² 2s² 2p⁶3s² 3p⁶ 3d⁶

M : 1s² 2s² 2p⁶3s² 3p⁶ 3d⁸

Atomic number = 26 

Mass number = 56 

No. of neutrons = 56 – 26 = 30

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 + 30.4x/100 = x + 0.304 x 

Now, 

mass number of ion = Number of protons + Number of neutrons

= (x + 3) + (x + 0.304 x) 

∴ 56 = (x + 3) + (x + 0.304 x) or 2.304 x = 56 – 3 = 53

x = 53/2.304 = 23

Atomic number of the ion (or element) = 23 + 3 = 26 

The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺ .

In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them. Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals. 

The order of filling of various orbitals as per Aufbau principle is –

1 s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d …

For e.g., K (Z =19) 

The electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ . After filling 4s orbital only we have to fill up 3d orbital.

An element X contains 35 electrons, 45 neutrons 

i. The number of protons must be equal to the number of electrons. So the number of protons = 35. 

ii. Number of electrons = 35. So the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵ . 

iii. The last electron i.e. 5^th electron in 4p orbital has the following quantum numbers. n = 4, l = 1, m = +1, s = +1/2

The electronic configuration of lithium is 1s² 2s¹. So differentiating electron enters into 2s. The values of four quantum numbers are as given below.

Correct option is d. (iv) and (v)

C) n = 3, l = 0, m_I = 0, m_s = +1/2 

1. The electronic configuration of sodium is 1s² 2s² 2p⁶ 3s¹.

2. The valance orbital is 3s.

3. The quantum numbers for this orbital is

Correct option is a. the electron may be in d orbital.

(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

(iii) Copper

Correct option is  D) 0

Mass of NH₃ = 34 mg = 34 x 10^-3 g

Molar mass of NH₃ = 17 g/mol

Amount of ammonia = {34 x 10^-3 g}/{17 g/mol} = 2 x 10^-3 mol

Therefore, No. of NH₃ molecules in 34 mg sample

= 2 x 10^-3 mol x 6.023 x 10²³ mol^-1

= 1.2046 x 10²¹

Each molecule of NH₃ contains 10 protons (7 protons in N⁺ 3 in three H atoms).

Therefore,

(a) No. of protons in 34 mg of ammonia = 10 x 1.2046 x 10²¹

= 1.2046 x 10²²

(b) Mass of proton = 1.675 x 10^-27 kg

Therefore, total mass of protons in 34 mg of NH₃

= 1.2046 x 10²² x 1.675 x 10^-27 kg

= 2.015 x 10^-5 kg

The number of neutrons, protons, molecules and mass does not depend upon temperature and pressure. Therefore there will be no change in the answers, if temperature and pressure are changed.

Correct option is  A) s

Correct Option is : (C) 18, 15, 16

Atomic Number = Number of Protons

Mass Number = Number of Protons + Number of Neutrons

Number of Electron = Atomic Number -(Number of Charges)

Therefore,

Number of Protons = 15

Number of Neutrons = 31 - 15 = 16

Number of Electrons = 15 - (-3) = 15 + 3 = 18

Hence , \(^{31} p^{3-} \) contains , 18 electrons , 15 protons and 16 neutrons.

Option : (C) 18, 15, 16

The representation of the given element is \(_6^{14}X\).

For the given nuclide,

Atomic number, Z = 18, 

Mass number, A = 40 

Number of protons = Number of electrons = Z = 18

Number of neutrons (N) = A – Z = 40 – 18 = 22

Number of protons = 18, 

Number of electrons = 18, 

Number of neutrons = 22

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These are those species which have the same number of electrons is called Isoelectronic species.

Atoms and ions having the same number of electrons are isoelectronic.

Hence,

Ne, O^2-, Na⁺ are isoelectronic species.

Thus, Na⁺ and Mg²⁺ are isoelectronic and K⁺, Ca²⁺, S^2- and Ar are also isoelectronic.

C. decomposition of elements 

Decomposition of elements is not an application of the chemical effect of electric current. Electroplating, purification and decomposition of compounds can be done with the help of chemical effect of electric current.

D. sodium hydroxide 

Sodium hydroxide (NaOH) is manufactured by using the chemical effect of electric current when a sodium salt is electrolysed.