Chemistry

(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) For 1p orbital, n and l are both equal to 1. Since, l should always have a lower value than n. So, 1p ortial does not exist.

For 3f orbital, n=3 and l=3. For the same reason, the existence of 3f orbital is not possible. 1p and 3f orbitals are not possible

Number of electrons in:

(i) H₂⁺ = 1

(ii) H₂ = 2

(iii) O₂⁺ = 15

(i) No. of protons in a neutral atom = No. of electrons = 29

(ii) Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

For electron in 3d orbital, n = 3, l = 2, m_i = -2, -1, 0, +1, +2.

The lowest value of l where 'g' orbital can be present = 4

As for any value 'n' of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).

∴ For l = 4, the minimum of n where 'g' orbital can be present = 4+1=5.

(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) Cl (Z = 17) has outermost electronic configuration = 3p⁵

(iii) (a) Li  (b) P    (c) Sc

Na⁺ and Mg²⁺ are iso-electronic species having 10 electrons each. K⁺, Ca²⁺, S^2- are iso-electronic species having 18 electrons each.

Kinetic energy. K.E. = ½ mv²

=> v² = (2 x K.E.) / m

Given, K.E. = 3 x 10^-25 J = 3 x 10^-25 kg m² s^-2

Therefore, v² = [2 x (3 x 10^-25 kg m² s^-2)] / 9.1 x 10^-31 kg

=> v²= 65.9 x 10⁴ m² s^-2

=>v = 8.12 x 10² m s^-1

To calculate the wavelength of the electron

According to de Broglie's equation,

λ=h/mv  = (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (8.12 x 10² m s^-1)

= 0.08967 x 10^-5 m = 8967 x 10^-10 m = 8967Å

We know that the mass of an electron, m_e = 9.1 x 10^-31 kg,

Velocity of electron, v = 2.05 * 10⁷ m s^-1

We know that Planck's constant, h = 6.626 x 10^-34 kg m² s^-1

As per de Broglie's equation, λ=h/mv

= (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (2.05 x 10⁷ m s^-1) = 3.55 x 10^-11 m

Step I: Calculation of energy required

ΔE = E_∞– E₂

= 0 – (–2.18 * 10^-18 J) / 4 = 5.45 x 10^-19 J

Step II: Calculation of the longest wavelength of light in cm used to cause the transition

λ = hc / ΔE = (6.626 x 10^-34 J s) x (3 x 10⁸ ms^-1) / (5.45 x 10^-19 J)

= 3.644 x 10^-7 x 10² = 3.645 x 10^-5 cm.