Atomic Structure Questions and Answers

Answer: C

There are (2l + 1) values of \(m_{l}\).

For a given l, \(m_{l}\) takes every integer from \(-l\) to \(+l\), including 0: that is l negative values, l positive values and zero, \(2l + 1\) in all. For example, a p subshell (\(l = 1\)) has three orbitals and a d subshell (\(l = 2\)) has five.

2l leaves out zero, and 2l − 1, l + 1 and l − 1 give the wrong count (for instance 3, 3 and 1 for l = 2).

Answer: C

There are 9 orbitals for n = 3.

The number of orbitals in a shell is \(n^{2}\). For n = 3 the subshells are 3s (1 orbital), 3p (3) and 3d (5), giving \(1 + 3 + 5 = 9\).

3 counts only the subshells, 6 or 10 miscount them, and 18 (not offered) would be the number of electrons the shell can hold, \(2n^{2}\).

Answer: E

The uncertainty in position is infinity.

Heisenberg's uncertainty principle states \(\Delta x \cdot \Delta p \ge \frac{h}{4\pi}\). If the momentum is known exactly, \(\Delta p = 0\), then

\[ \Delta x \ge \frac{h}{4\pi \cdot \Delta p} \to \infty \]

A particle with perfectly defined momentum is completely undetermined in position. h/4π is the lower limit of the product of the two uncertainties, not of Δx alone, and the pure numbers 1, 2 and ½ have no basis in the principle.

Answer: D

The statement that is NOT true is that it only explains about the splitting of spectral lines in the presence of electric field.

Bohr's model cannot explain the splitting of spectral lines in an electric field (the Stark effect) or in a magnetic field (the Zeeman effect); this is one of its recognised failures. So the claim that it explains such splitting is false.

The other statements are true. The model works for one-electron species such as He⁺ and Li²⁺, explaining their stability and line spectra. It fails to account for the fine structure of the hydrogen spectrum seen with better spectrometers, cannot explain the spectrum of any atom or ion with two or more electrons, and cannot explain how atoms combine into molecules by chemical bonds.

Answer: C

The orientation is given by the magnetic quantum number \(m_{l}\).

Each quantum number describes one feature of an orbital:

• \(n\), the principal quantum number: the shell, size and energy.
• \(l\), the azimuthal quantum number: the subshell and shape.
• \(m_{l}\): the orientation in space. For \(l = 1\), \(m_{l} = -1, 0, +1\) are the three p orbitals along different axes.
• \(m_{s}\): the spin of the electron, not a property of the orbital.

So neither \(n\) nor \(l\), alone or together, fixes the orientation.

Answer: B

The order is \(E_{2s}(Li) \gt E_{2s}(Na) \gt E_{2s}(K)\).

The energy of a given orbital (here 2s) depends on the effective nuclear charge it feels. As the atomic number rises from Li (3) to Na (11) to K (19), the nucleus attracts the 2s electrons more strongly, so the same orbital becomes lower (more negative) in energy.

So the 2s energy falls in the order Li, Na, K, and the standard textbook statement is \(E_{2s}(H) \gt E_{2s}(Li) \gt E_{2s}(Na) \gt E_{2s}(K)\). Every other option puts a heavier atom's 2s orbital above a lighter one's.

Answer: C

The correct statements are (i), (iv) and (v).

Half filled and completely filled subshells are extra stable because their electron distribution is symmetrical, giving relatively small shielding and smaller coulombic repulsion energy, and because many electrons of parallel spin can exchange places, giving a larger exchange energy.

Statements (ii) and (iii) are the opposites of (iv) and (v), so any option containing them is wrong. That is why chromium is 3d⁵4s¹ and copper 3d¹⁰4s¹.

Answer: C

The kinetic energy is 5 x 10^-22 J.

From \(\lambda = \frac{h}{mv}\) and \(K = \frac{1}{2}mv^{2}\), \(K = \frac{h^{2}}{2m\lambda^{2}}\).

\[ \begin{aligned} v &= \frac{6.3 \times 10^{-34}}{10^{-31} \times 63 \times 10^{-9}} \\ &= 10^{5}\ \text{m s}^{-1} \\ K &= \tfrac{1}{2} \times 10^{-31} \times (10^{5})^{2} \\ &= 5 \times 10^{-22}\ \text{J} \end{aligned} \]

The option \(1 \times 10^{-21}\) J forgets the \(\frac{1}{2}\) in kinetic energy, and \(5 \times 10^{-21}\) J slips a power of ten.

Answer: B

The ultraviolet lines form the Lyman series.

Each series of hydrogen lines comes from electrons falling to one particular lower level. Transitions ending on n = 1 release the most energy and so the shortest wavelengths, which lie in the ultraviolet; this is the Lyman series.

The Balmer series (ending on n = 2) lies in the visible region. The Paschen (n = 3), Brackett (n = 4) and Pfund (n = 5) series involve smaller energy gaps and lie in the infrared.

Answer: E

The wave number is \(1.724 \times 10^{4}\ \text{cm}^{-1}\).

Wave number is the reciprocal of wavelength. Convert 580 nm to centimetres first:

\[ \begin{aligned} \lambda &= 580 \times 10^{-7}\ \text{cm} \\ \bar{\nu} &= \frac{1}{580 \times 10^{-7}\ \text{cm}} \\ &= 1.724 \times 10^{4}\ \text{cm}^{-1} \end{aligned} \]

Options in plain 'cm' have the wrong unit, since a reciprocal of length is in \(\text{cm}^{-1}\). The \(10^{2}\) and \(10^{3}\) values come from converting nanometres to centimetres with the wrong power of ten (1 nm is \(10^{-7}\) cm).