401. The ground-state electronic configuration of sulfur, \(Z=16\), is
ⓐ. \(1s^2\,2s^2\,2p^6\,3s^2\,3p^4\)
ⓑ. \(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\)
ⓒ. \(1s^2\,2s^2\,2p^6\,3s^1\,3p^5\)
ⓓ. \(1s^2\,2s^2\,2p^6\,4s^2\,3d^4\)
Correct Answer: \(1s^2\,2s^2\,2p^6\,3s^2\,3p^4\)
Explanation: A neutral sulfur atom contains \(16\) electrons because \(Z=16\). The orbitals fill in the order \(1s\), \(2s\), \(2p\), \(3s\), and \(3p\) for the first \(16\) electrons. The first \(10\) electrons give the neon core, \(1s^2\,2s^2\,2p^6\). The next \(2\) electrons enter \(3s\), and the remaining \(4\) electrons enter \(3p\). The configuration ending in \(3p^6\) would contain \(18\) electrons and belongs to argon, not sulfur.
402. A neutral atom has the configuration \([\mathrm{Ne}]\,3s^2\,3p^3\). The number of valence electrons in this atom is
ⓐ. \(3\)
ⓑ. \(10\)
ⓒ. \(15\)
ⓓ. \(5\)
Correct Answer: \(5\)
Explanation: The noble-gas core \([\mathrm{Ne}]\) represents the inner \(10\) electrons. The valence shell is the shell with the highest principal quantum number \(n\). In \([\mathrm{Ne}]\,3s^2\,3p^3\), the highest shell is \(n=3\). The valence electrons are therefore \(3s^2+3p^3\), giving \(2+3=5\). The core electrons are not counted as valence electrons in this configuration.
403. The electronic configuration \(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1\) represents
ⓐ. an atom with \(18\) electrons
ⓑ. an atom with \(19\) electrons
ⓒ. an atom with \(20\) electrons
ⓓ. an atom with \(21\) electrons
Correct Answer: an atom with \(19\) electrons
Explanation: \( \textbf{Configuration given:} \) \(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1\).
\( \textbf{Add superscripts:} \)
\[
2+2+6+2+6+1=19
\]
\( \textbf{Meaning for a neutral atom:} \) The number of electrons equals the atomic number.
\( \textbf{Core check:} \) \(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\) contains \(18\) electrons, the argon core.
\( \textbf{Remaining electron:} \) The extra electron is in \(4s^1\).
\( \textbf{Final answer:} \) The configuration represents a neutral atom with \(19\) electrons. The last subshell does not alone give the total electron count.
404. A configuration for a neutral atom is written as \([\mathrm{Ar}]\,3d^2\,4s^2\). The total number of electrons in the atom is
ⓐ. \(20\)
ⓑ. \(22\)
ⓒ. \(21\)
ⓓ. \(24\)
Correct Answer: \(22\)
Explanation: \( \textbf{Core electrons:} \) \([\mathrm{Ar}]\) represents \(18\) electrons.
\( \textbf{Electrons outside the core:} \) \(3d^2\) contributes \(2\) electrons.
\( \textbf{Electrons outside the core:} \) \(4s^2\) contributes \(2\) electrons.
\( \textbf{Total electron count:} \)
\[
18+2+2=22
\]
\( \textbf{Neutral atom link:} \) For a neutral atom, this also gives \(Z=22\).
\( \textbf{Final answer:} \) The atom has \(22\) electrons. Noble-gas shorthand must be expanded mentally by adding the core electrons to the written outer electrons.
405. The electronic configuration of \(\mathrm{Na^+}\), given that neutral sodium has \(Z=11\), is
ⓐ. \(1s^2\,2s^2\,2p^6\)
ⓑ. \(1s^2\,2s^2\,2p^6\,3s^1\)
ⓒ. \(1s^2\,2s^2\,2p^5\,3s^2\)
ⓓ. \(1s^2\,2s^2\,2p^6\,3s^2\)
Correct Answer: \(1s^2\,2s^2\,2p^6\)
Explanation: Neutral sodium has \(11\) electrons and the configuration \(1s^2\,2s^2\,2p^6\,3s^1\). The ion \(\mathrm{Na^+}\) is formed by loss of one electron. The electron lost is the outermost \(3s\) electron. After this loss, \(10\) electrons remain. The resulting configuration is \(1s^2\,2s^2\,2p^6\), which has a neon-like arrangement.
406. The configuration of \(\mathrm{Cl^-}\), given that chlorine has \(Z=17\), is
ⓐ. \([\mathrm{Ne}]\,3s^2\,3p^5\)
ⓑ. \([\mathrm{Ar}]\,4s^1\)
ⓒ. \([\mathrm{Ne}]\,3s^1\,3p^6\)
ⓓ. \([\mathrm{Ne}]\,3s^2\,3p^6\)
Correct Answer: \([\mathrm{Ne}]\,3s^2\,3p^6\)
Explanation: A neutral chlorine atom has \(17\) electrons and the configuration \([\mathrm{Ne}]\,3s^2\,3p^5\). The ion \(\mathrm{Cl^-}\) has gained one electron, so it contains \(18\) electrons. The added electron enters the \(3p\) subshell, completing it as \(3p^6\). Thus the configuration becomes \([\mathrm{Ne}]\,3s^2\,3p^6\). The negative charge means electron gain, not electron loss.
407. A neutral magnesium atom has configuration \([\mathrm{Ne}]\,3s^2\). The configuration of \(\mathrm{Mg^{2+}}\) is
ⓐ. \([\mathrm{Ne}]\,3s^1\)
ⓑ. \([\mathrm{Ne}]\)
ⓒ. \([\mathrm{Ne}]\,3s^2\,3p^2\)
ⓓ. \([\mathrm{Ne}]\,2p^4\)
Correct Answer: \([\mathrm{Ne}]\)
Explanation: Neutral magnesium has \(12\) electrons and the outer configuration \(3s^2\). The ion \(\mathrm{Mg^{2+}}\) is formed by losing \(2\) electrons. Both electrons are removed from the outermost \(3s\) subshell. After losing them, magnesium has \(10\) electrons. This gives the noble-gas configuration \([\mathrm{Ne}]\), not a partially filled \(3s\) subshell.
408. A table gives configurations for some species.
| Species | Claimed configuration |
|---|
| P. \(\mathrm{O^{2-}}\) | \(1s^2\,2s^2\,2p^6\) |
| Q. \(\mathrm{F^-}\) | \(1s^2\,2s^2\,2p^6\) |
| R. \(\mathrm{Na^+}\) | \(1s^2\,2s^2\,2p^6\) |
| S. \(\mathrm{Mg^{2+}}\) | \(1s^2\,2s^2\,2p^4\) |
The wrong configuration claim is
ⓐ. S
ⓑ. P
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: \(\mathrm{O^{2-}}\) has \(8+2=10\) electrons, so it has the configuration \(1s^2\,2s^2\,2p^6\). \(\mathrm{F^-}\) has \(9+1=10\) electrons, so it also has this configuration. \(\mathrm{Na^+}\) has \(11-1=10\) electrons, again giving the same neon-like configuration. \(\mathrm{Mg^{2+}}\) has \(12-2=10\) electrons, so it should also be \(1s^2\,2s^2\,2p^6\), not \(1s^2\,2s^2\,2p^4\). Row \(S\) undercounts the electrons after ion formation.
409. The species \(\mathrm{O^{2-}}\), \(\mathrm{F^-}\), \(\mathrm{Na^+}\), and \(\mathrm{Mg^{2+}}\) are called isoelectronic because they have the same
ⓐ. number of protons
ⓑ. number of neutrons
ⓒ. mass number values
ⓓ. number of electrons
Correct Answer: number of electrons
Explanation: Isoelectronic species have the same number of electrons. \(\mathrm{O^{2-}}\) has \(10\) electrons, \(\mathrm{F^-}\) has \(10\) electrons, \(\mathrm{Na^+}\) has \(10\) electrons, and \(\mathrm{Mg^{2+}}\) has \(10\) electrons. They do not have the same number of protons because they are different elements. Their neutron numbers and mass numbers can also differ. The shared feature is the electronic configuration, not the nuclear composition.
410. Among the isoelectronic species \(\mathrm{O^{2-}}\), \(\mathrm{F^-}\), \(\mathrm{Na^+}\), and \(\mathrm{Mg^{2+}}\), the species with the highest nuclear charge is
ⓐ. \(\mathrm{O^{2-}}\)
ⓑ. \(\mathrm{F^-}\) in the notation
ⓒ. \(\mathrm{Mg^{2+}}\)
ⓓ. \(\mathrm{Na^+}\) in the notation
Correct Answer: \(\mathrm{Mg^{2+}}\)
Explanation: Isoelectronic species have the same number of electrons, but their nuclear charges depend on atomic number. Oxygen has \(Z=8\), fluorine has \(Z=9\), sodium has \(Z=11\), and magnesium has \(Z=12\). The species with \(Z=12\) has the greatest nuclear charge. Therefore, \(\mathrm{Mg^{2+}}\) has the highest nuclear charge among the given species. Same electron count does not mean same proton count.
411. A neutral atom has \(15\) electrons. The distribution of its electrons among shells is best written as
ⓐ. \(2,10,3\)
ⓑ. \(8,2,5\)
ⓒ. \(2,8,5\)
ⓓ. \(2,6,7\)
Correct Answer: \(2,8,5\)
Explanation: The first shell can hold a maximum of \(2\) electrons. The second shell can hold up to \(8\) electrons for the early shell filling pattern. After placing \(2+8=10\) electrons, \(5\) electrons remain for the third shell. Therefore, the shell-wise distribution is \(2,8,5\). The arrangement must fill lower shells before placing electrons in higher shells.
412. For a neutral atom with configuration \(1s^2\,2s^2\,2p^6\,3s^2\,3p^2\), the number of unpaired electrons in the \(3p\) subshell is
ⓐ. \(0\)
ⓑ. \(1\)
ⓒ. \(4\)
ⓓ. \(2\)
Correct Answer: \(2\)
Explanation: The configuration ends in \(3p^2\). A \(p\) subshell has three degenerate orbitals. By Hund's rule, the two \(3p\) electrons occupy two different \(p\) orbitals singly before pairing begins. These two electrons have parallel spins in the ground state. Therefore, the number of unpaired electrons is \(2\). Pairing would begin only after each of the three \(p\) orbitals has one electron.
413. The orbital diagram for \(2p^4\) should have
ⓐ. one paired orbital and two singly occupied orbitals
ⓑ. two paired orbitals and one empty orbital
ⓒ. all four electrons in one orbital in the Bohr model
ⓓ. four singly occupied \(2p\) orbitals in the Bohr model
Correct Answer: one paired orbital and two singly occupied orbitals
Explanation: A \(2p\) subshell has three orbitals. According to Hund's rule, the first three electrons enter the three orbitals singly with parallel spins. The fourth electron then pairs with one of the already singly occupied orbitals. This gives one paired orbital and two singly occupied orbitals. There cannot be four singly occupied \(2p\) orbitals because a \(p\) subshell has only three orbitals.
414. A configuration is written as \(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1\). This special stability is associated with
ⓐ. a completely empty \(3d\) subshell
ⓑ. a filled \(4p\) subshell
ⓒ. a half-filled \(3d\) subshell
ⓓ. a paired \(4s\) subshell only
Correct Answer: a half-filled \(3d\) subshell
Explanation: A \(d\) subshell contains five orbitals and can hold a maximum of \(10\) electrons. The arrangement \(3d^5\) has one electron in each \(d\) orbital, making it half-filled. Half-filled subshells have extra stability due to symmetrical distribution and exchange energy. This is why chromium is commonly written as \([\mathrm{Ar}]\,3d^5\,4s^1\) rather than \([\mathrm{Ar}]\,3d^4\,4s^2\). The stability is mainly connected with the \(3d^5\) arrangement.
415. The ground-state configuration of chromium is better represented as \([\mathrm{Ar}]\,3d^5\,4s^1\) rather than \([\mathrm{Ar}]\,3d^4\,4s^2\) because
ⓐ. \(4s\) cannot contain any electron in the electron arrangement
ⓑ. a half-filled \(3d^5\) subshell is especially stable
ⓒ. \(3d\) can hold only \(4\) electrons in the electron arrangement
ⓓ. chromium has only \(18\) electrons
Correct Answer: a half-filled \(3d^5\) subshell is especially stable
Explanation: The expected Aufbau filling for chromium might suggest \([\mathrm{Ar}]\,3d^4\,4s^2\). However, one \(4s\) electron is promoted to the \(3d\) subshell. This gives \([\mathrm{Ar}]\,3d^5\,4s^1\), where the \(3d\) subshell is half-filled. A half-filled \(d\) subshell has special stability because the five \(d\) orbitals can each contain one electron with parallel spin. The change is not because \(4s\) cannot hold electrons, but because the resulting distribution is more stable.
416. Copper has the ground-state configuration \([\mathrm{Ar}]\,3d^{10}\,4s^1\) instead of \([\mathrm{Ar}]\,3d^9\,4s^2\). The main reason is the stability of
ⓐ. half-filled \(4s\) subshell only
ⓑ. completely filled \(3d\) subshell
ⓒ. empty \(3p\) subshell
ⓓ. completely filled \(4p\) subshell
Correct Answer: completely filled \(3d\) subshell
Explanation: A \(d\) subshell can hold a maximum of \(10\) electrons. In copper, the configuration \([\mathrm{Ar}]\,3d^{10}\,4s^1\) gives a completely filled \(3d\) subshell. Filled subshells are especially stable because of symmetrical electron distribution and exchange energy. The alternative \([\mathrm{Ar}]\,3d^9\,4s^2\) leaves the \(3d\) subshell one electron short of being filled. Copper therefore shows an apparent exception to the simple Aufbau expectation.
417. The row that correctly identifies the special configuration is
| Row | Configuration | Stability feature |
|---|
| P | \([\mathrm{Ar}]\,3d^5\,4s^1\) | half-filled \(3d\) |
| Q | \([\mathrm{Ar}]\,3d^{10}\,4s^1\) | filled \(3d\) |
| R | \([\mathrm{Ar}]\,3d^4\,4s^2\) | filled \(3d\) |
| S | \([\mathrm{Ar}]\,3d^9\,4s^2\) | almost filled \(3d\) |
The rows correctly showing extra-stable \(d\)-subshell arrangements are
ⓐ. R and S only
ⓑ. P and R only
ⓒ. P and Q only
ⓓ. Q and S only
Correct Answer: P and Q only
Explanation: A \(d^5\) arrangement is half-filled because the five \(d\) orbitals can each contain one electron. A \(d^{10}\) arrangement is completely filled because all five \(d\) orbitals are paired. These arrangements are associated with extra stability. Row \(R\) is wrong because \(d^4\) is neither half-filled nor filled. Row \(S\) is also not the extra-stable filled arrangement; copper becomes more stable by changing \(3d^9\,4s^2\) into \(3d^{10}\,4s^1\).
418. When forming \(\mathrm{Fe^{2+}}\) from neutral iron, electrons are removed first from
ⓐ. \(4s\) subshell
ⓑ. \(3d\) subshell
ⓒ. \(3p\) subshell
ⓓ. \(2p\) subshell
Correct Answer: \(4s\) subshell
Explanation: Neutral iron is commonly written as \([\mathrm{Ar}]\,3d^6\,4s^2\). Although \(4s\) fills before \(3d\), electrons are removed first from \(4s\) when transition-metal ions form. Thus \(\mathrm{Fe^{2+}}\) is formed by losing the two \(4s\) electrons. The resulting configuration is \([\mathrm{Ar}]\,3d^6\). Filling order and removal order are not always identical for these subshells.
419. The configuration of \(\mathrm{Fe^{3+}}\), given neutral iron as \([\mathrm{Ar}]\,3d^6\,4s^2\), is
ⓐ. \([\mathrm{Ar}]\,3d^6\,4s^1\)
ⓑ. \([\mathrm{Ar}]\,3d^3\,4s^2\)
ⓒ. \([\mathrm{Ar}]\,3d^8\)
ⓓ. \([\mathrm{Ar}]\,3d^5\)
Correct Answer: \([\mathrm{Ar}]\,3d^5\)
Explanation: \( \textbf{Neutral iron configuration:} \) \([\mathrm{Ar}]\,3d^6\,4s^2\).
\( \textbf{Ion formed:} \) \(\mathrm{Fe^{3+}}\) means loss of \(3\) electrons.
\( \textbf{First removal:} \) The two \(4s\) electrons are removed first.
\[
[\mathrm{Ar}]\,3d^6\,4s^2 \rightarrow [\mathrm{Ar}]\,3d^6
\]
\( \textbf{Third electron removal:} \) One electron is then removed from \(3d\).
\[
[\mathrm{Ar}]\,3d^6 \rightarrow [\mathrm{Ar}]\,3d^5
\]
\( \textbf{Final answer:} \) \(\mathrm{Fe^{3+}}\) has configuration \([\mathrm{Ar}]\,3d^5\). The result is also stabilized by the half-filled \(3d\) subshell.
420. A student writes the configuration of \(\mathrm{Cu^+}\) as \([\mathrm{Ar}]\,3d^9\,4s^1\), starting from copper as \([\mathrm{Ar}]\,3d^{10}\,4s^1\). The correction is that \(\mathrm{Cu^+}\) should be
ⓐ. \([\mathrm{Ar}]\,3d^9\,4s^2\)
ⓑ. \([\mathrm{Ar}]\,3d^8\,4s^1\)
ⓒ. \([\mathrm{Ar}]\,3d^{10}\)
ⓓ. \([\mathrm{Ar}]\,4s^2\)
Correct Answer: \([\mathrm{Ar}]\,3d^{10}\)
Explanation: Neutral copper has the configuration \([\mathrm{Ar}]\,3d^{10}\,4s^1\). Formation of \(\mathrm{Cu^+}\) means loss of one electron. The electron is removed from the outer \(4s\) subshell first, not from the filled \(3d\) subshell. Removing the \(4s\) electron gives \([\mathrm{Ar}]\,3d^{10}\). This keeps the filled \(3d\) arrangement intact, which is more stable than making \(3d^9\,4s^1\).
