Atomic Structure MCQs for NEET — Chemistry Questions with Answers

The context states: 'energies of the orbitals in the same subshell decrease with increase in the atomic number (Zeff). For example, energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium and that of lithium is greater than that of sodium and so on, that is, $E_{2s(H)} > E_{2s(Li)} > E_{2s(Na)} > E_{2s(K)}$.' Therefore, the correct order is $E_{2s(H)} > E_{2s(Li)} > E_{2s(Na)}$. (Reference: '$E_{2s(H)} > E_{2s(Li)} > E_{2s(Na)} > E_{2s(K)}$')

The text mentions: 'It may be noted that different subshells of a particular shell have different energies in case of multi-electrons atoms. However, in hydrogen atom, these have the same energy.' (Reference: 'However, in hydrogen atom, these have the same energy.')

The context states: 'if two orbitals have the same value of (n + l), the orbital with lower value of n will have the lower energy.' (Reference: 'if two orbitals have the same value of (n + l), the orbital with lower value of n will have the lower energy.')

The text states: 'The filling of electrons into the orbitals of different atoms takes place according to the aufbau principle which is based on the Pauli’s exclusion principle, the Hund’s rule of maximum multiplicity and the relative energies of the orbitals.' Heisenberg's Uncertainty Principle is not listed as a governing rule for electron filling. (Reference: 'The filling of electrons into the orbitals of different atoms takes place according to the aufbau principle which is based on the Pauli’s exclusion principle, the Hund’s rule of maximum multiplicity and the relative energies of the orbitals.')

The text explains two ways of representing electronic configurations: 'sa pbdc ...... notation' and 'Orbital diagram'. It then states: 'The advantage of second notation over the first is that it represents all the four quantum numbers.' The orbital diagram represents spin using arrows (↑ for positive spin, ↓ for negative spin), along with n, l, and $m_l$ implicitly by the box's position. (Reference: 'The advantage of second notation over the first is that it represents all the four quantum numbers.')

The context mentions that 'It has been observed that half filled and fully filled degenerate set of orbitals acquire extra stability due to their symmetry'. (Reference: 'It has been observed that half filled and fully filled degenerate set of orbitals acquire extra stability due to their symmetry (see Section, 2.6.7).')

For 4s: n=4, l=0, so (n+l) = 4+0 = 4. For 3d: n=3, l=2, so (n+l) = 3+2 = 5. According to the (n+l) rule, the lower the (n+l) value, the lower the energy. Therefore, 4s (n+l=4) has lower energy than 3d (n+l=5). (Reference: 'the lower the value of (n + l) for an orbital, the lower is its energy.')

According to the provided text, 'For a frequency $\nu$ of incident radiation, lower than the cut-off frequency $\nu_0$, no photoelectric emission is possible even if the intensity is large.' This directly contradicts option 4.

The text states: 'For a given photosensitive material and frequency of incident radiation (above the threshold frequency), the photoelectric current is directly proportional to the intensity of incident light (Fig. 11.2)'. This means it increases linearly.

The context mentions, 'the maximum kinetic energy of photoelectrons varies linearly with the frequency of incident radiation, but is independent of its intensity.' Also, from Fig 11.3 description, 'the stopping potential remains the same as that for the incident radiation of intensity $I_1$ ... Thus, for a given frequency of the incident radiation, the stopping potential is independent of its intensity.' Since $K_{max} = eV_0$, $K_{max}$ is also independent of intensity.