JEE Chemistry Structure Of Atom MCQs Set 16

Answer: (c) IV < II < III < I

Explanation:
1. Identify the sub-shells: I is \( 4d (n + l = 6) \), II is \( 3d (5) \), III is \( 4p (5) \), IV is \( 3p (4) \).
2. A larger \( n + l \) means higher energy; when \( n + l \) is equal, the lower \( n \) has lower energy.
3. Hence IV (4) < II \( (5, n = 3) \) < III \( (5, n = 4) \) < I (6).

Teacher's Note:
1. Use the \( (n + l) \) rule; \( m_l \) and \( m_s \) do not affect the energy of a many-electron atom.
2. For equal \( (n + l) \), compare the \( n \) values.

Answer: (c) 9 : 4

Explanation:
1. Lyman: ν̄_max = \( R(1 - 0) = R \), ν̄_min = \( R(1 - 1/4) = 3R/4 \), so Δν̄ = \( R/4 \).
2. Balmer: ν̄_max = \( R/4 \), ν̄_min = \( R(1/4 - 1/9) = 5R/36 \), so Δν̄ = \( R/4 - 5R/36 = R/9 \).
3. Ratio = \( (R/4)/(R/9) = 9/4 \).

Teacher's Note:
1. The maximum wave number is the series limit \( (n_{2} = \infty) \); the minimum is the first line \( (n_{2} = n_{1} + 1) \).
2. Take the differences separately for each series.

Answer: (d) 2s-orbital

Explanation:
1. Number of radial nodes = \( n - l - 1 \).
2. The curve has exactly one radial node and a non-zero value at the nucleus, so \( l = 0 \) and \( n - 1 = 1 \), giving \( n = 2 \).
3. The orbital is 2s (1s has no node, \( 3s \) has two, and \( 2p \) is zero at the nucleus).

Teacher's Note:
1. Only \( s \) orbitals have a non-zero density at the nucleus.
2. Count the nodes: \( n - l - 1 \).

Answer: (a) in the regions a and c

Explanation:
1. The probability of finding an electron is proportional to |ψ|², so the sign of ψ does not matter.
2. |ψ|² is large at the crest \( (a) \) and the trough \( (c) \) and zero at \( b \), which is a node.

Teacher's Note:
1. A node is where ψ = 0.
2. Negative values of ψ are as probable as positive values.

Answer: (c) 5d

Explanation:
1. Element 71 is lutetium, the last lanthanoid, with the configuration [Xe] \( 4f^{14} 5d^{1} 6s^{2} \).
2. The \( 4f \) sub-shell is completed by the 70th electron (Yb: \( 4f^{14} 6s^{2} \)), so the 71st electron enters \( 5d \).

Teacher's Note:
1. Lu is placed at the end of the \( 4f \) series.
2. The \( 5d^{1} \) configuration is general in the (n−2)f¹⁴(n−1)d¹ns² pattern of Lu.

Answer: (b) Paschen, ∞ → 3

Explanation:
1. \( 1/\lambda = R_H(1/n_{1}^{2} - 1/n_{2}^{2}) \), with \( R_H = 10^{7} m^{-1} \) and \( \lambda = 900 \times 10^{-9} m \).
2. \( 1/n_{1}^{2} - 1/n_{2}^{2} = 1/(10^{7} \times 900 \times 10^{-9}) = 1/9 \).
3. This is satisfied by \( n_{1} = 3 \) and \( n_{2} = \infty \), the Paschen series limit.

Teacher's Note:
1. Convert \( R_H \) to \( m^{-1} \) before substituting.
2. Wavelengths near 900 nm lie in the infrared, the Paschen region.

Answer: (d) \( \lambda ∝ 1/(\nu - \nu _{0})^{1/2} \)

Explanation:
1. λ = h/√(2m·KE) and KE = h(ν − ν₀).
2. So λ = h/√(2mh(ν − ν₀)), which is proportional to \( (\nu - \nu _{0})^{-1/2} \).

Teacher's Note:
1. Substitute the photoelectric equation into the de Broglie relation.
2. Wavelength falls as the kinetic energy rises.

Answer: (b) 0.75

Explanation:
1. The orbit circumference must hold \( n \) wavelengths: \( 2\pir = n\lambda \).
2. \( r = a_{0}n^{2}/Z \), so \( 2\pia_{0}n^{2}/Z = n(1.5\pia_{0}) \).
3. \( n/Z = 1.5/2 = 0.75 \).

Teacher's Note:
1. Bohr's condition is equivalent to a standing electron wave.
2. Use \( r_n = a_{0}n^{2}/Z \).

Answer: (c) √(2meV)

Explanation:
1. Kinetic energy gained = eV = \( p^{2}/2m \), so \( p \) = √(2meV).
2. de Broglie: \( p = h/\lambda \), so \( h/\lambda \) = √(2meV).

Teacher's Note:
1. Equate the work done by the potential with the kinetic energy.
2. \( h/\lambda \) is the momentum of the electron.

Answer: (a) 5, 0, 0, +½

Explanation:
1. Rb = [Kr] \( 5s^{1} \), so the valence electron is in the \( 5s \) orbital.
2. \( n = 5, l = 0 (s), m_l = 0 \) (the only allowed value), \( m_s = +\frac{1}{2} \) (or −½).

Teacher's Note:
1. For \( l = 0 \) the only magnetic quantum number is 0.
2. The valence electron of an alkali metal is in an ns orbital.

Answer: (b) (4) < (2) < (3) < (1)

Explanation:
1. The sub-shells are \( (1) 4p, (2) 4s, (3) 3d, (4) 3p \), with \( n + l = 5, 4, 5, 4 \) respectively.
2. For equal \( n + l \) the lower \( n \) is lower in energy: \( 3p \) < \( 4s \) and \( 3d \) < \( 4p \).
3. Order: \( 3p \) < \( 4s \) < \( 3d \) < \( 4p \), i.e. (4) < (2) < (3) < (1).

Teacher's Note:
1. Apply the \( (n + l) \) rule.
2. \( 4s \) fills before \( 3d \).

Answer: (b) n = 2 to n = 1

Explanation:
1. For He⁺ \( (Z = 2) \): ν \( ∝ 4(1/2^{2} - 1/4^{2}) = 4 \times 3/16 = 3/4 \).
2. For \( H \): \( n = 2 \) to 1 gives \( (1 - 1/4) = 3/4 \), the same value.
3. Other options give \( 8/9, 5/36 \) and 7/144, which do not match.

Teacher's Note:
1. Frequency scales as \( Z^{2}(1/n_{1}^{2} - 1/n_{2}^{2}) \).
2. Compare the bracket values for each hydrogen transition.