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As n increases, the orbitals become larger and the electrons in those orbitals are farther from the nucleus.

Features

1. When two atomic orbitals combine or overlap, they lose their identity and form new orbitals. The new orbitals formed are called molecular orbitals.

2. Only those atomic orbitals can combine to form molecular orbitals which have comparable energies and proper orientations.

3. The number of molecular orbitals formed is equal to the number of combining atomic orbitals.

4. When two atomic orbitals combine, they form two new orbitals called ‘bonding molecular orbital’ and ‘antibonding molecular orbital’.

5. The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.

6. The bonding molecular orbitals are represented by σ,π,δ  and antibonding are represented by σ*,π*,δ* .

7. The shapes of the molecular orbitals formed depend upon the type of the combining atomic orbitals.

8. Filling of MO takes place according to following rules :

• Aufbau principle, Pauli’s exclusion principle,
• Hund’s rule of maximum multiplicity.

Correct option is d. O₂ < O₃ < H₂O₂

Correct option is b. SH₂

Correct option is (D) tritium

The smallest unit of matter is atom.

Antioxidants: Substances that prevent oxidation.

E = hu where, h = planks constant = 6.626 × 10^-34 Js, γ = frequency or radiation.

Emission line spectrum

According to the de-Broglie equation λ = h /m×v ; for same value of velocity λ= 1/ m.

∴ O₂ Molecule has shortest wavelength.

Ground state means the least energy state or the most stable state. Excited state means the higher energy state, in which the electrons are in the higher energy level (unstable state).

(i) Wave Number (\(\bar{v}\)) = 400 cm⁻¹ = 4 m⁻¹

\(\therefore\) wavelength \((\lambda)\) = \(\frac{1}{\bar v}\) = \(\frac{1}{4}\) = 0.25 m

\(\because\) v = \(\frac{c}{\lambda}\) = \(\frac{3.0\times10^8ms^{-1}}{0.25m}\)

= 12 × 10⁸ s⁻¹

= 1.2 × 10⁹ s⁻¹

(ii) de-Broglie equation

\(\lambda=\frac{h}{p}\) = \(\frac{h}{mv}\)

Where \(\lambda\) = Wavelength of the particle

m = mass of the particle

p = momentum of the particle

v = velocity of the particle

h = Planck ′ s Constant

Magnetic moment µ = √(n(n + 2)BM) where n is the number of unpaired electrons.

Mass of wagon = 2000 kg

Uncertainty in position, ∆x = \(\pm\)10 m

According to Heisenberg uncertainty principle

∆x × ∆p = \(\frac{h}{4\pi}\)

∆x × m∆v = \(\frac{h}{4\pi}\)

or ∆x × ∆v = \(\frac{h}{4\pi m}\)

∆v = \(\frac{h}{4\pi m\Delta \mathrm{x}}\)

= \(\frac{6.626\times10^{-34}kgm^2s^{-1}}{4\times3.14\times(2000kg)\times(10m)}\)

= 2.636 × 10⁻³⁹ ms⁻¹

Correct option is d. BF₃

Correct Option is : (D) 3 more neutrons

\(U^{235}\) and \(U^{238}\) are isotopes because they have same number of protons but differ in number of neutrons. \(U^{238}\) contain 3 more neutrons as compared to \(U^{235}\) .

Option : (D) 3 more neutrons

Number of electrons in

H²⁺  is 1

H²  is 2

O²⁺ is 15

So, a positive molecule will contain a higher number of a proton(H⁺ions), a negative particle will have a higher number of electrons(e-ions), and a neutral atom will contain the same number of protons & electrons.

Hence,

(i) For an atom to be neutral, the number of protons= number of electrons.

So, Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰. Hence, it is the electronic configuration of copper "

Kinetic energy of electron,

\(\frac{1}{2}\)mv² = 2.275 x 10^-25 J

or = 2.275 × 10⁻²⁵ kg m² s⁻²

m = 9.1 × 10⁻³¹ kg

v² = \(\frac{2\times2.275\times10^{-25}kg\,m^2s^{-2}}{9.1\times10^{-31}kg}\)

= 0.5 × 10⁶ m² s⁻²

v = 0.7.7 × 10³ ms⁻¹

Now,

\(h=\frac{h}{mv}\)

h = \(\frac{6.6\times10^{-34}kgm^2s^{-1}}{(9.1\times10^{-31kg})\times(0.707\times10^3ms^{-1})}\)

= 1.026 × 10⁻⁶ m = 1026 nm.

n = 3

Therefore, l = 0, 1, 2

for l = 0, m₁ = 0

l = 1, m₁ = -1, 0, +1

l = 2, m₁ = -2, -1, 0, +1, +2

Corrct Option is : (B) \(Be^{3+}\)

Hydrogen like species means elements have only one electron in their shell .

Here , \(Be^{+3}\) has only one electron while , \(He^{+2}\) - has no electron & \(Li^+\) has two electrons in their shell.

Option : (B) Be³⁺

No. of electrons in the atom of an element = 29

No. of neutrons in the atom of an element = 35

(i) Number of protons = 29

(ii) Electronic configuration (At. no = 29)

= 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.

H₂⁺ = 1

H₂ = 2

O₂⁺ = 5

Correct option is b. 1/2

H₂^- ion will have two electrons in bonding orbital and one in antibonding orbital

B.O = 1/2 (electrons is bonding orbitals –electrons in antibonding orbitals) = 1/2 (2–1) = 1/2

Correct option is d. presence of quantized energy levels

Correct Option is : (B) \(6.63 \times 10^{-29} m\)

\(\lambda = \frac{h}{p} = \frac{h}{mv}\)

\(\lambda = \frac{6.63 \times 10^{-34} \ Js }{10^{-6} \times 10}\)

\(\lambda = 6.63 \times 10^{-29} m\)

Option : (B) 6.63 × 10^-29m

Electromagnetic wave is produced when an electric charge vibrates (moves back and fortn).

Characteristics of electromagnetic waves : 

1. Electric field and magnetic fields are perpendicular to each other and at right angles to direction of propagation of wave.

2. It travels with speed of light i.e., 3 × 10⁸ ms^-1. 

3. Electromagnetic energy is characterized by wavelength (λ) and frequency (v). The relation is given by c = vλ.

The number of wave peaks that pass by a given point per unit time is called frequency.

1. Name of the divalent element in the table is Calcium. 

2. Name of the element which belongs to 3rd period and VA Group is Phosphorous.

The distance from one wave peak to the next is called wavelength (λ).

Electronic configuration

(1) Cl^- = 1s², 2s², 2p⁶, 3s², 3p⁶ 

(2) Na⁺ = 1s², 2s², 2p⁶

An orbital can have only two electrons, provided these two electrons have anti parallel spins. This arrangement is represented, as ↓↑

(i) Electronic configuration for outer configuration 4s¹ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹. 

Atomic number = 19 

(ii) Exact configuration for outer configuration 3d3. 

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d³ = 23 

Atomic number = 23

1. What is nl^x method?

2. Where does it use ? 

3. What is meant by ‘n’, 7′ and ‘x’? 

4. How can we use nl^x method in the writing of electronic configuration?

Group of wavelengths is called spectrum (or) A collection of dispersed light giving its wavelength composition is called a spectrum.

It is equal to speed of light, i.e. 3 x 10⁸ ms^-1

Electromagnetic waves can have a wide variety of wavelengths. The entire range of wavelengths is known as the electromagnetic spectrum.

Option : (A) crystallization

• A saturated solution of the crude solid is prepared by boiling it in a small but sufficient quantity of a suitable solvent.
• The main solute from the sample of the crude solid dissolves to form a saturated solution on boiling.

 [Note : The solution is not saturated with respect to the soluble impurities, as they are in small proportion.]

Electronic Configuration is the distribution of electrons in the various available orbitals of an atom of an element. 

The sequenced in which the electron occupy the various orbitals is as follows. Is, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p….

Continuous spectrum of emission.

i. When electron jumps from higher energy level to n = 1, it corresponds to Lyman series. 

ii. When electron jumps from higher energy level to n = 1, it corresponds to Bracket series. 

iii. When electron jumps from n = 4 to n = 1, the second line of Balmer series is observed.

Moseley’s periodic law: The physical and chemical properties of elements are periodic functions of their atomic numbers.

(a) 16 electrons 

(b) 2 electrons.

(a) 1s 

(b) 3p 

(c) 4d 

(d) 4f

Dispersion: 

1. Splitting of white light into colours (VIBGYOR) is called Dispersion of light. 

2. The natural example for dispersion of light is formation of Rainbow. It is caused by dispersion of sunlight by tiny water droplets present in atmosphere which act as small prisms.

Correct option is  C) Ga

D) both B and C

Sodium has a larger atomic radii and lower ionization potential than sulphur. Hence sodium is a metal while sulphur is a non metal.