501. A neutral atom has the configuration \([\mathrm{Ar}]\,3d^5\,4s^1\). It forms a \(2+\) ion. The configuration of the ion and the number of unpaired \(d\)-electrons are respectively
ⓐ. \([\mathrm{Ar}]\,3d^5\), \(5\)
ⓑ. \([\mathrm{Ar}]\,3d^3\,4s^1\), \(3\)
ⓒ. \([\mathrm{Ar}]\,3d^4\), \(4\)
ⓓ. \([\mathrm{Ar}]\,3d^5\,4s^2\), \(5\)
Correct Answer: \([\mathrm{Ar}]\,3d^4\), \(4\)
Explanation: \( \textbf{Neutral configuration:} \) \([\mathrm{Ar}]\,3d^5\,4s^1\).
\( \textbf{Ion formed:} \) A \(2+\) ion means loss of \(2\) electrons.
\( \textbf{First electron removed:} \) The \(4s\) electron is removed first.
\[
[\mathrm{Ar}]\,3d^5\,4s^1 \rightarrow [\mathrm{Ar}]\,3d^5
\]
\( \textbf{Second electron removed:} \) The next electron is removed from \(3d\).
\[
[\mathrm{Ar}]\,3d^5 \rightarrow [\mathrm{Ar}]\,3d^4
\]
\( \textbf{Unpaired count:} \) A \(d\) subshell has \(5\) orbitals, and \(d^4\) places four electrons singly before pairing.
\( \textbf{Final answer:} \) The ion has configuration \([\mathrm{Ar}]\,3d^4\) and \(4\) unpaired \(d\)-electrons. The \(4s\) electron is removed before reducing the \(3d\) count.
502. A hydrogen-like ion has one electron and nuclear charge \(Z=3\). If its electron moves from \(n=2\) to \(n=1\), the energy emitted is how many times the energy emitted for the same transition in hydrogen?
ⓐ. \(9\) times
ⓑ. \(3\) times
ⓒ. \(6\) times
ⓓ. \(\frac{1}{9}\) times
Correct Answer: \(9\) times
Explanation: For a hydrogen-like species, the energy of each level is \(E_n=-\frac{13.6Z^2}{n^2}\,\text{eV}\). For the same transition, the energy difference is proportional to \(Z^2\). Hydrogen has \(Z=1\), while the given ion has \(Z=3\). The ratio of emitted energies is \(\frac{3^2}{1^2}=9\). The values of \(n\) are the same in both transitions, so the change comes only from the \(Z^2\) factor. A larger nuclear charge gives wider energy spacing for one-electron species.
503. A line spectrum record for hydrogen is shown below.
| Line | Transition | Claimed series |
|---|
| P | \(n=3\rightarrow n=1\) | Lyman |
| Q | \(n=5\rightarrow n=2\) | Balmer |
| R | \(n=6\rightarrow n=3\) | Paschen |
| S | \(n=4\rightarrow n=1\) | Balmer |
The row with the wrong series claim is
ⓐ. S
ⓑ. P
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: The series name of a hydrogen emission line depends on the final lower level. A transition ending at \(n=1\) belongs to the Lyman series, so row \(P\) is correct. A transition ending at \(n=2\) belongs to the Balmer series, so row \(Q\) is correct. A transition ending at \(n=3\) belongs to the Paschen series, so row \(R\) is correct. Row \(S\) ends at \(n=1\), so it should be Lyman, not Balmer. The starting level changes the line within a series, but the final level names the series.
504. A photon has wavelength \(250\,\text{nm}\), and another photon has wavelength \(500\,\text{nm}\). If both photons fall on the same metal surface, the difference in their maximum photoelectron kinetic energies is
ⓐ. \(\frac{hc}{500\,\text{nm}}-\frac{hc}{250\,\text{nm}}\)
ⓑ. \(\frac{hc}{250\,\text{nm}}+\frac{hc}{500\,\text{nm}}\)
ⓒ. \(\frac{hc}{250\,\text{nm}}-\frac{hc}{500\,\text{nm}}\)
ⓓ. zero, because the metal surface is the same
Correct Answer: \(\frac{hc}{250\,\text{nm}}-\frac{hc}{500\,\text{nm}}\)
Explanation: For the same metal, the work function \(\phi\) is the same in both cases. The maximum kinetic energy is \(K_{\max}=E-\phi\). Photon energy is \(E=\frac{hc}{\lambda}\), so the shorter-wavelength photon has greater energy. The difference in maximum kinetic energies is \(\left(\frac{hc}{250\,\text{nm}}-\phi\right)-\left(\frac{hc}{500\,\text{nm}}-\phi\right)\). The work function terms cancel because the metal is unchanged. Therefore, the difference is \(\frac{hc}{250\,\text{nm}}-\frac{hc}{500\,\text{nm}}\).
505. A particle has momentum \(p\), and its position uncertainty is \(\Delta x\). Another particle has momentum \(2p\), and its position uncertainty is \(\frac{\Delta x}{2}\). The correct comparison is
ⓐ. the second has twice the de Broglie wavelength and half the minimum momentum uncertainty
ⓑ. both have the same de Broglie wavelength and same uncertainty limit
ⓒ. the second has zero uncertainty in momentum
ⓓ. half the wavelength and twice the minimum momentum uncertainty
Correct Answer: half the wavelength and twice the minimum momentum uncertainty
Explanation: De Broglie's relation is \(\lambda=\frac{h}{p}\). If momentum changes from \(p\) to \(2p\), the wavelength becomes \(\frac{h}{2p}\), which is half the original wavelength. The uncertainty principle is \(\Delta x\Delta p\ge\frac{h}{4\pi}\). If \(\Delta x\) is reduced to \(\frac{\Delta x}{2}\), the minimum allowed \(\Delta p\) becomes twice as large. Both parts involve inverse relationships, but the quantities involved are different. Momentum controls de Broglie wavelength, while position uncertainty controls the lower bound on momentum uncertainty.
506. A proposed set of quantum numbers for an electron is \(n=4,\ l=0,\ m_l=0,\ m_s=-\frac{1}{2}\). The orbital type and number of angular nodes are respectively
ⓐ. \(4p\), \(1\)
ⓑ. \(4d\), \(2\)
ⓒ. \(4s\), \(0\)
ⓓ. \(4f\), \(3\)
Correct Answer: \(4s\), \(0\)
Explanation: The principal quantum number \(n=4\) places the electron in the fourth shell. The value \(l=0\) corresponds to an \(s\) subshell. For \(l=0\), the only possible \(m_l\) value is \(0\), so the given \(m_l\) is valid. The number of angular nodes is equal to \(l\). Therefore, the orbital is \(4s\), and it has \(0\) angular nodes. The spin value \(-\frac{1}{2}\) is allowed but does not determine the orbital shape.
507. A neutral atom has the configuration \([\mathrm{Ne}]\,3s^2\,3p^4\). A student says it has \(4\) unpaired electrons because the last superscript is \(4\). The correct number of unpaired electrons is
ⓐ. \(0\)
ⓑ. \(2\)
ⓒ. \(1\)
ⓓ. \(4\)
Correct Answer: \(2\)
Explanation: The unpaired electrons must be counted from the orbital arrangement, not directly from the superscript. A \(p\) subshell has three degenerate orbitals. For \(3p^4\), the first three electrons occupy the three \(p\) orbitals singly according to Hund's rule. The fourth electron then pairs in one of these orbitals. This leaves two orbitals singly occupied. The \(3p^4\) subshell therefore has \(2\) unpaired electrons, not \(4\).
508. A graph of \(K_{\max}\) versus frequency \(\nu\) for two different metals gives two parallel straight lines with different x-intercepts. The best interpretation is that the two metals have
ⓐ. different values of Planck's constant but the same work function
ⓑ. the same threshold frequency and different slopes
ⓒ. no threshold frequency at all
ⓓ. the same Planck constant but different work functions
Correct Answer: the same Planck constant but different work functions
Explanation: The photoelectric equation is \(K_{\max}=h\nu-\phi\). In a graph of \(K_{\max}\) against \(\nu\), the slope is \(h\). Planck's constant is universal, so different metals give lines with the same slope if the same units are used. The x-intercept is the threshold frequency \(\nu_0\), and it depends on the work function of the metal. Different x-intercepts therefore show different work functions. Parallel lines show the same slope, not the same threshold frequency.
509. An orbital is described as having \(2\) radial nodes and \(1\) angular node. The orbital could be
ⓐ. \(3p\)
ⓑ. \(4d\)
ⓒ. \(4p\)
ⓓ. \(5f\)
Correct Answer: \(4p\)
Explanation: \( \textbf{Given radial nodes:} \) \(2\).
\( \textbf{Given angular nodes:} \) \(1\).
\( \textbf{Angular node relation:} \)
\[
\text{angular nodes}=l
\]
\( \textbf{So:} \)
\[
l=1
\]
\( \textbf{Subshell type:} \) \(l=1\) means a \(p\) orbital.
\( \textbf{Radial node relation:} \)
\[
\text{radial nodes}=n-l-1
\]
\( \textbf{Substitution:} \)
\[
2=n-1-1
\]
\( \textbf{Solve:} \)
\[
n=4
\]
\( \textbf{Final answer:} \) The orbital is \(4p\). The angular node count first fixes the subshell type, and the radial node count then fixes the shell number.
510. A sample contains \(60\%\) of isotope \({}^{20}X\), \(25\%\) of isotope \({}^{21}X\), and \(15\%\) of isotope \({}^{22}X\). Taking isotopic masses as \(20\), \(21\), and \(22\), the average atomic mass is
ⓐ. \(20.55\)
ⓑ. \(21.00\)
ⓒ. \(21.40\)
ⓓ. \(22.00\)
Correct Answer: \(20.55\)
Explanation: \( \textbf{Isotopic abundances:} \) \(60\%=0.60\), \(25\%=0.25\), and \(15\%=0.15\).
\( \textbf{Weighted average relation:} \)
\[
\bar{m}=m_1f_1+m_2f_2+m_3f_3
\]
\( \textbf{Substitution:} \)
\[
\bar{m}=20(0.60)+21(0.25)+22(0.15)
\]
\( \textbf{First contribution:} \)
\[
20(0.60)=12.00
\]
\( \textbf{Second contribution:} \)
\[
21(0.25)=5.25
\]
\( \textbf{Third contribution:} \)
\[
22(0.15)=3.30
\]
\( \textbf{Add contributions:} \)
\[
12.00+5.25+3.30=20.55
\]
\( \textbf{Final answer:} \) The average atomic mass is \(20.55\). The value is closer to \(20\) because the lightest isotope is the most abundant.
511. A student compares \({}^{40}_{18}X\), \({}^{40}_{20}Y\), and \({}^{42}_{20}Y\). The correct statement is
ⓐ. \({}^{40}_{18}X\) and \({}^{40}_{20}Y\) are isotopes, while \({}^{40}_{20}Y\) and \({}^{42}_{20}Y\) are isobars
ⓑ. all three are isotopes of one element
ⓒ. \({}^{40}_{18}X\) and \({}^{40}_{20}Y\) are isobars, while \({}^{40}_{20}Y\) and \({}^{42}_{20}Y\) are isotopes
ⓓ. all three have the same neutron number
Correct Answer: \({}^{40}_{18}X\) and \({}^{40}_{20}Y\) are isobars, while \({}^{40}_{20}Y\) and \({}^{42}_{20}Y\) are isotopes
Explanation: Isobars have the same mass number but different atomic numbers. \({}^{40}_{18}X\) and \({}^{40}_{20}Y\) both have \(A=40\), but their \(Z\) values are different, so they are isobars. Isotopes have the same atomic number but different mass numbers. \({}^{40}_{20}Y\) and \({}^{42}_{20}Y\) both have \(Z=20\), but their mass numbers are different, so they are isotopes. The lower number decides element identity, while the upper number helps compare masses.
512. A metal has threshold frequency \(5.0\times10^{14}\,\text{s}^{-1}\). Light of frequency \(8.0\times10^{14}\,\text{s}^{-1}\) falls on it. Taking \(h=6.6\times10^{-34}\,\text{J s}\), the maximum kinetic energy of photoelectrons is closest to
ⓐ. \(3.30\times10^{-19}\,\text{J}\)
ⓑ. \(5.28\times10^{-19}\,\text{J}\)
ⓒ. \(8.58\times10^{-19}\,\text{J}\)
ⓓ. \(1.98\times10^{-19}\,\text{J}\)
Correct Answer: \(1.98\times10^{-19}\,\text{J}\)
Explanation: \( \textbf{Threshold frequency:} \) \(\nu_0=5.0\times10^{14}\,\text{s}^{-1}\).
\( \textbf{Incident frequency:} \) \(\nu=8.0\times10^{14}\,\text{s}^{-1}\).
\( \textbf{Planck constant:} \) \(h=6.6\times10^{-34}\,\text{J s}\).
\( \textbf{Photoelectric relation:} \)
\[
K_{\max}=h(\nu-\nu_0)
\]
\( \textbf{Frequency excess:} \)
\[
\nu-\nu_0=(8.0-5.0)\times10^{14}=3.0\times10^{14}\,\text{s}^{-1}
\]
\( \textbf{Substitution:} \)
\[
K_{\max}=(6.6\times10^{-34})(3.0\times10^{14})\,\text{J}
\]
\( \textbf{Number part:} \)
\[
6.6\times3.0=19.8
\]
\( \textbf{Power part:} \)
\[
10^{-34}\times10^{14}=10^{-20}
\]
\( \textbf{Final answer:} \) \(K_{\max}=1.98\times10^{-19}\,\text{J}\). The kinetic energy depends on the excess frequency above threshold, not on the full incident frequency alone.
513. A table compares shell, subshell, orbital, and electron capacity.
| Row | Item | Claim |
|---|
| P | Shell \(n=2\) | maximum \(8\) electrons |
| Q | \(p\) subshell | \(3\) orbitals |
| R | one orbital | maximum \(2\) electrons |
| S | \(d\) subshell | maximum \(6\) electrons |
The mismatched row is
ⓐ. P
ⓑ. S
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: The shell \(n=2\) can hold \(2n^2=2(2)^2=8\) electrons, so row \(P\) is correct. A \(p\) subshell has \(l=1\), giving \(2l+1=3\) orbitals, so row \(Q\) is correct. One orbital can hold a maximum of two electrons with opposite spins, so row \(R\) is correct. A \(d\) subshell has \(5\) orbitals and can hold \(10\) electrons, not \(6\). Row \(S\) confuses the \(p\)-subshell capacity with the \(d\)-subshell capacity.
514. A hydrogen atom is in \(n=2\). The minimum energy needed to ionize it from this level is
ⓐ. \(1.51\,\text{eV}\)
ⓑ. \(10.2\,\text{eV}\)
ⓒ. \(13.6\,\text{eV}\)
ⓓ. \(3.40\,\text{eV}\)
Correct Answer: \(3.40\,\text{eV}\)
Explanation: \( \textbf{Hydrogen energy relation:} \)
\[
E_n=-\frac{13.6}{n^2}\,\text{eV}
\]
\( \textbf{For }n=2\textbf{:} \)
\[
E_2=-\frac{13.6}{4}=-3.40\,\text{eV}
\]
\( \textbf{Ionized state:} \) At \(n=\infty\), the energy is \(0\,\text{eV}\).
\( \textbf{Energy required:} \)
\[
0-(-3.40)=3.40\,\text{eV}
\]
\( \textbf{Final answer:} \) The minimum ionization energy from \(n=2\) is \(3.40\,\text{eV}\). The electron is less tightly bound in \(n=2\) than in the ground state.
515. A radiation has wave number \(20000\,\text{cm}^{-1}\). Taking \(c=3.0\times10^{10}\,\text{cm s}^{-1}\), the photon frequency and photon energy using \(h=6.626\times10^{-34}\,\text{J s}\) are closest to
ⓐ. \(6.7\times10^{-7}\,\text{s}^{-1}\) and \(4.4\times10^{-40}\,\text{J}\)
ⓑ. \(6.0\times10^{14}\,\text{s}^{-1}\) and \(3.98\times10^{-19}\,\text{J}\)
ⓒ. \(2.0\times10^4\,\text{s}^{-1}\) and \(1.33\times10^{-29}\,\text{J}\)
ⓓ. \(1.5\times10^6\,\text{s}^{-1}\) and \(9.94\times10^{-28}\,\text{J}\)
Correct Answer: \(6.0\times10^{14}\,\text{s}^{-1}\) and \(3.98\times10^{-19}\,\text{J}\)
Explanation: \( \textbf{Given wave number:} \) \(\bar{\nu}=20000\,\text{cm}^{-1}=2.0\times10^4\,\text{cm}^{-1}\).
\( \textbf{Speed of light:} \) \(c=3.0\times10^{10}\,\text{cm s}^{-1}\).
\( \textbf{Frequency-wave number relation:} \)
\[
\nu=c\bar{\nu}
\]
\( \textbf{Frequency calculation:} \)
\[
\nu=(3.0\times10^{10})(2.0\times10^4)=6.0\times10^{14}\,\text{s}^{-1}
\]
\( \textbf{Photon energy relation:} \)
\[
E=h\nu
\]
\( \textbf{Substitution:} \)
\[
E=(6.626\times10^{-34})(6.0\times10^{14})\,\text{J}
\]
\( \textbf{Calculation:} \)
\[
E=39.756\times10^{-20}=3.98\times10^{-19}\,\text{J}
\]
\( \textbf{Final answer:} \) The frequency is \(6.0\times10^{14}\,\text{s}^{-1}\), and the photon energy is about \(3.98\times10^{-19}\,\text{J}\). The \(\text{cm}\) unit cancels in \(c\bar{\nu}\), leaving \(\text{s}^{-1}\).
516. Assertion: A \(4f\) subshell can contain \(14\) electrons.
Reason: For \(f\), \(l=3\), so the number of orbitals is \(2l+1=7\), and each orbital can hold \(2\) electrons.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: An \(f\) subshell corresponds to \(l=3\). The number of orbitals in a subshell is \(2l+1\), so an \(f\) subshell has \(2(3)+1=7\) orbitals. Each orbital can hold a maximum of two electrons with opposite spins. Therefore, the maximum electron capacity is \(7\times2=14\). The Reason gives the orbital-count and electron-capacity logic behind the Assertion.
517. A one-electron species has \(Z=2\). Its \(n=2\) orbit radius is compared with the \(n=4\) orbit radius of hydrogen. The ratio \(\frac{r_{\text{ion},\,n=2}}{r_{\text{H},\,n=4}}\) is
ⓐ. \(\frac{1}{8}\)
ⓑ. \(\frac{1}{4}\)
ⓒ. \(\frac{1}{2}\)
ⓓ. \(2\)
Correct Answer: \(\frac{1}{8}\)
Explanation: \( \textbf{Hydrogen-like radius relation:} \)
\[
r_n=\frac{a_0n^2}{Z}
\]
\( \textbf{Ion data:} \) \(Z=2\), \(n=2\).
\[
r_{\text{ion},\,n=2}=\frac{a_0(2)^2}{2}=2a_0
\]
\( \textbf{Hydrogen data:} \) \(Z=1\), \(n=4\).
\[
r_{\text{H},\,n=4}=\frac{a_0(4)^2}{1}=16a_0
\]
\( \textbf{Ratio:} \)
\[
\frac{r_{\text{ion},\,n=2}}{r_{\text{H},\,n=4}}=\frac{2a_0}{16a_0}=\frac{1}{8}
\]
\( \textbf{Final answer:} \) The ratio is \(\frac{1}{8}\). The square dependence on \(n\) and the inverse dependence on \(Z\) must both be included.
518. A student writes \(\mathrm{Fe^{2+}}\) as \([\mathrm{Ar}]\,3d^4\,4s^2\) from neutral \(\mathrm{Fe}\), \([\mathrm{Ar}]\,3d^6\,4s^2\). The most precise correction is
ⓐ. \(\mathrm{Fe^{2+}}\) is \([\mathrm{Ar}]\,3d^8\), because electrons are added to make a cation
ⓑ. \(\mathrm{Fe^{2+}}\) is \([\mathrm{Ar}]\,3d^6\), because \(4s\) electrons are removed first
ⓒ. \(\mathrm{Fe^{2+}}\) is \([\mathrm{Ar}]\,4s^2\), because all \(3d\) electrons are removed first
ⓓ. \(\mathrm{Fe^{2+}}\) is \([\mathrm{Ar}]\,3d^5\,4s^1\), because every ion must be half-filled
Correct Answer: \(\mathrm{Fe^{2+}}\) is \([\mathrm{Ar}]\,3d^6\), because \(4s\) electrons are removed first
Explanation: Neutral iron is written as \([\mathrm{Ar}]\,3d^6\,4s^2\). Formation of \(\mathrm{Fe^{2+}}\) means loss of two electrons. For transition-metal cations, electrons are removed from \(4s\) before \(3d\). Removing both \(4s\) electrons gives \([\mathrm{Ar}]\,3d^6\). The student's answer removed two \(3d\) electrons while leaving \(4s^2\), which uses the wrong removal order. Cation formation involves electron loss, not electron addition.
519. A node analysis for three orbitals is given below.
| Orbital | Radial nodes | Angular nodes |
|---|
| P. \(2s\) | \(1\) | \(0\) |
| Q. \(3p\) | \(1\) | \(1\) |
| R. \(4d\) | \(1\) | \(2\) |
| S. \(4p\) | \(1\) | \(2\) |
The row with the wrong node analysis is
ⓐ. P
ⓑ. Q
ⓒ. S
ⓓ. R
Correct Answer: S
Explanation: For \(2s\), \(n=2\) and \(l=0\), so radial nodes \(=2-0-1=1\) and angular nodes \(=0\). For \(3p\), \(n=3\) and \(l=1\), so radial nodes \(=3-1-1=1\) and angular nodes \(=1\). For \(4d\), \(n=4\) and \(l=2\), so radial nodes \(=4-2-1=1\) and angular nodes \(=2\). For \(4p\), \(l=1\), so angular nodes should be \(1\), and radial nodes should be \(4-1-1=2\). Row \(S\) reverses the radial and angular node counts.
520. A compact comparison says:
Species P: \(Z=16\), charge \(2-\)
Species Q: \(Z=18\), neutral
Species R: \(Z=20\), charge \(2+\)
The best conclusion about these three species is that they
ⓐ. are isoelectronic with \(18\) electrons each
ⓑ. are isotopes of the same element in the notation
ⓒ. have identical nuclear charges in the notation
ⓓ. have the same number of neutrons
Correct Answer: are isoelectronic with \(18\) electrons each
Explanation: Species \(P\) has \(Z=16\) and charge \(2-\), so it has \(16+2=18\) electrons. Species \(Q\) is neutral with \(Z=18\), so it has \(18\) electrons. Species \(R\) has \(Z=20\) and charge \(2+\), so it has \(20-2=18\) electrons. Since all three have the same electron count, they are isoelectronic. They are not isotopes because their atomic numbers are different, and their nuclear charges are not identical.
