Ionisation energy of He⁺ is 19.6 × 10^-18 J atom^-1. The energy of the first stationary state (n = 1) of Li²⁺ is

4.41 × 10^-16J atom^-1
–4.41 × 10^-17J atom^-1
–2.2 × 10^-15J atom^-1
8.82 × 10^-17J atom^-1

Solution

The energy required to break one mole of Cl – Cl bonds in Cl₂ is 242 kJ mol^-1. The longest wavelength of light capable of breaking a single Cl – Cl bond is(c = 3 × 10⁸ ms^-1 and N_A = 6.02 × 10²³ mol^-1).

594 nm
640 nm
700 nm
494 nm

Solution

In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6 × 10^–34 kg m²s^-1, mass of electron, e_m = 9.1 × 10^-31kg)

5.10 × 10^–3 m
1.92 × 10^–3 m
3.84 × 10^–3 m
1.52 × 10^–4 m

Solution

The ionization enthalpy of hydrogen atom is 1.312 × 10⁶ J mol^-1.The energy required to excite the electron in the atom from n = 1 to n = 2 is

8.51 × 10⁵ J mol^-1
6.56 × 10⁵ J mol^-1
7.56 × 10⁵ J mol^-1
9.84 × 10⁵ J mol^-1

Solution

Which one of the following constitutes a group of the isoelectronic species?

\(C_{2}^{2-}, O_{2}^{-}, CO, NO\)
\(NO^{+}, C_{2}^{2-}, CN^{-}, N_{2}\)
\(CN^{-}, N_{2}, O_{2}^{2-}, C_{2}^{2-}\)
\(N_{2}, O_{2}^{-}, NO^{+}, CO\)

Solution

Species having same number of electrons are isoelectronic. On calculating the number of electrons in each given species, we get.

CN^-(6 + 7 + 1 = 14); N₂(7 + 7 = 14);

O₂^2- (8 + 8 + 2 = 18); C₂^2- (6 + 6 + 2 = 14);

O₂^-(8 + 8 + 1 = 17); NO⁺(7 + 8 - 1 = 14)

CO(6 + 8 = 14); NO(7 + 8 = 15)

From the above calculation we find that all the species listed in choice (b) have 14 electrons each so it is the correct answer.

Which of the following sets of quantum numbers represents the highest energy of an atom?

n = 3, l = 0, m = 0, s = +1/2
n = 3, l = 1, m = 1, s = +1/2
n = 3, l = 2, m = 1, s = +1/2
n = 4, l = 0, m = 0, s = +1/2

Solution

(a) n = 3, l = 0 means 3s-orbital and n + l = 3
(b) n = 3, l = 1 means 3p-orbital n + l = 4
(c) n = 3, l = 2 means 3d-orbital n + l = 5
(d) n = 4, l = 0 means 4s-orbital n + l = 4

Increasing order of energy among these orbitals is 3s < 3p < 4s < 3d ∴ 3d has highest energy.

According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon ?

n = 6 to n = 1
n = 5 to n = 4
n = 6 to n = 5
n = 5 to n = 3

Solution

Energy of photon obtained from the transition n = 6 to n = 5 will have least energy.

\(\Delta E = 13.6 Z^{2} \left ( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right )\)

If n = 6,the correct sequence for filling of electrons will be :

ns → (n – 2)f → (n – 1)d → np
ns → (n – 1)d → (n– 2) f → np
ns → (n – 2)f → np → (n – 1)d
ns → np → (n – 1)d → (n – 2)f

Solution

ns → (n – 2)f → (n – 1)d → np

The energies E₁ and E₂ of two radiations are 25 eV and 50 eV,respectively. The relation between their wavelengths i.e., ?₁ and ?₂ will be :

?₁ = ?₂
?₁ = 2?₂
?₁ = 4?₂
?₁ = ¹⁄₂ ?₂

Solution

A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be (h = 6.6 × 10^-34 Js):

1.0 × 10^-32m
6.6 × 10^-32m
6.6 × 10^-34m
1.0 × 10^-35m

Solution

\(\lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{0.66 \times 100} = 1 \times 10^{-35} m\)

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete