301. Beryllium and boron have the following first and second ionization enthalpies:
| Element | \(IE_1\) (\(\mathrm{kJ\,mol^{-1}}\)) | \(IE_2\) (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| Be | 900 | 1757 |
| B | 801 | 2427 |
Which explanation correctly accounts for the reversal in their ordering between the first and second ionization enthalpies?
ⓐ. Boron's first electron occupies \(2p\) and is easier to remove than beryllium's \(2s\) electron; after first ionization, \(\mathrm{B^+}\) and \(\mathrm{Be^+}\) are isoelectronic, so boron's second ionization is greater.
ⓑ. Beryllium has a greater first ionization enthalpy because its nuclear charge is greater; boron's second is greater because it contains an additional principal shell.
ⓒ. Boron's first electron occupies \(2p\) and is easier to remove; after first ionization, \(\mathrm{B^+}\) has the relatively stable \(2s^2\) configuration, while \(\mathrm{Be^+}\) has \(2s^1\), contributing to boron's greater second ionization enthalpy.
ⓓ. Beryllium's first ionization is greater because it has a filled \(2p\) subshell, whereas boron's second ionization removes an electron from the \(1s\) core.
Correct Answer: Boron's first electron occupies \(2p\) and is easier to remove; after first ionization, \(\mathrm{B^+}\) has the relatively stable \(2s^2\) configuration, while \(\mathrm{Be^+}\) has \(2s^1\), contributing to boron's greater second ionization enthalpy.
Explanation: \(\textbf{Comparing the neutral atoms:}\)
Beryllium has the configuration \(1s^2\,2s^2\).
Boron has the configuration \(1s^2\,2s^2\,2p^1\).
The first electron removed from beryllium occupies a \(2s\) orbital.
The first electron removed from boron occupies the higher-energy \(2p\) subshell.
The greater penetration of the \(2s\) orbital makes beryllium's outer electron relatively difficult to remove.
Consequently:
\[
IE_1(\mathrm{Be})\gt IE_1(\mathrm{B})
\]
\(\textbf{Examining the ions after first electron removal:}\)
Beryllium forms:
\[
\mathrm{Be^+}:1s^2\,2s^1
\]
Boron forms:
\[
\mathrm{B^+}:1s^2\,2s^2
\]
These ions are not isoelectronic.
Beryllium ion contains three electrons, whereas boron ion contains four.
\(\textbf{Explaining the second-ionization ordering:}\)
The next electron removed from \(\mathrm{Be^+}\) occupies its singly occupied \(2s\) orbital.
Removal produces \(\mathrm{Be^{2+}}\) with the stable helium configuration.
The next electron removed from \(\mathrm{B^+}\) must be taken from its completed \(2s^2\) subshell.
This disrupts a filled subshell, and boron's greater nuclear charge also contributes to the stronger attraction experienced by its electrons.
The supplied values consequently show:
\[
IE_2(\mathrm{B})\gt IE_2(\mathrm{Be})
\]
\(\textbf{Final interpretation:}\)
The first comparison concerns electron removal from different subshells.
The second concerns electron removal from different cations with different \(2s\) occupancies.
The starting electronic species therefore change between the two comparisons.
The third option correctly explains the reversal without falsely treating the monovalent ions as isoelectronic.
302. A researcher reports two measurements involving sodium:
Measurement I: The nearest-neighbour sodium–sodium distance in metallic sodium is \(372\ \mathrm{pm}\).
Measurement II: The sodium–chloride separation in a hypothetical ionic crystal is \(283\ \mathrm{pm}\). Under the specified structural convention, the effective chloride radius is \(181\ \mathrm{pm}\).
Using the conventional metallic-radius relation for Measurement I and the additive ionic-radius approximation for Measurement II, which conclusion is correct?
ⓐ. Metallic sodium radius \(186\ \mathrm{pm}\); sodium-ion radius \(102\ \mathrm{pm}\); the difference is consistent with contraction when sodium loses its \(3s\) electron.
ⓑ. Metallic sodium radius \(186\ \mathrm{pm}\); sodium-ion radius \(102\ \mathrm{pm}\); the difference proves that metallic radius and ionic radius are exactly the same physical quantity.
ⓒ. Metallic sodium radius \(372\ \mathrm{pm}\); sodium-ion radius \(102\ \mathrm{pm}\); both measurements directly give the distance from the nucleus to a sharp electronic boundary.
ⓓ. Metallic sodium radius \(186\ \mathrm{pm}\); sodium-ion radius \(464\ \mathrm{pm}\); electron removal expands the occupied principal shell.
Correct Answer: Metallic sodium radius \(186\ \mathrm{pm}\); sodium-ion radius \(102\ \mathrm{pm}\); the difference is consistent with contraction when sodium loses its \(3s\) electron.
Explanation: \(\textbf{Determining the metallic radius:}\)
In the stated metallic-radius convention, the nearest-neighbour distance between equivalent metal atoms is twice the metallic radius.
Therefore:
\[
r_{\mathrm{metal}}=\frac{d_{\mathrm{Na-Na}}}{2}
\]
Substituting:
\[
r_{\mathrm{metal}}=\frac{372}{2}
\]
\[
r_{\mathrm{metal}}=186\ \mathrm{pm}
\]
\(\textbf{Determining the effective ionic radius:}\)
The additive ionic-radius approximation is:
\[
d_{\mathrm{Na^+-Cl^-}}\approx r_{\mathrm{Na^+}}+r_{\mathrm{Cl^-}}
\]
Rearranging:
\[
r_{\mathrm{Na^+}}\approx d_{\mathrm{Na^+-Cl^-}}-r_{\mathrm{Cl^-}}
\]
Substituting:
\[
r_{\mathrm{Na^+}}\approx283-181
\]
\[
r_{\mathrm{Na^+}}\approx102\ \mathrm{pm}
\]
\(\textbf{Explaining the electronic change:}\)
Neutral sodium has the configuration \([\mathrm{Ne}]\,3s^1\).
The sodium ion has the configuration \([\mathrm{Ne}]\).
Formation of \(\mathrm{Na^+}\) removes the only electron occupying the third principal shell.
The remaining electrons are confined mainly to the first two occupied shells.
This produces substantial contraction of the electronic distribution.
\(\textbf{Interpreting the measurement conventions:}\)
Metallic radius and effective ionic radius are derived from different structural environments.
Neither represents a perfectly sharp physical boundary around an atom or ion.
The numerical difference should therefore be interpreted using both the electronic change and the measurement conventions.
The two computed values are consistent with the expected contraction but are not interchangeable definitions.
\(\textbf{Final answer:}\)
\[
\boxed{r_{\mathrm{metal}}=186\ \mathrm{pm},\qquad r_{\mathrm{Na^+}}=102\ \mathrm{pm}}
\]
303. A fourth-period element \(M\) forms a gaseous trivalent cation with the ground-state configuration \([\mathrm{Ar}]\,3d^4\). Assume that ordinary cation formation removes the \(4s\) electrons before \(3d\) electrons.
Which statement correctly identifies the neutral element, its group and the electron-removal sequence?
ⓐ. \(M=\mathrm{Cr}\), group 6; formation of \(M^{3+}\) removes one \(4s\) electron and two \(3d\) electrons.
ⓑ. \(M=\mathrm{Mn}\), group 7; formation of \(M^{3+}\) removes two \(4s\) electrons and one \(3d\) electron.
ⓒ. \(M=\mathrm{Fe}\), group 8; formation of \(M^{3+}\) removes two \(4s\) electrons and one \(3d\) electron.
ⓓ. \(M=\mathrm{Mn}\), group 7; formation of \(M^{3+}\) removes three \(3d\) electrons while retaining \(4s^2\).
Correct Answer: \(M=\mathrm{Mn}\), group 7; formation of \(M^{3+}\) removes two \(4s\) electrons and one \(3d\) electron.
Explanation: \(\textbf{Counting the electrons in the final cation:}\)
The argon core contains eighteen electrons.
The specified \(3d^4\) subshell contains four additional electrons.
Therefore, the trivalent cation has:
\[
N_e(M^{3+})=18+4
\]
\[
N_e(M^{3+})=22
\]
\(\textbf{Recovering the neutral atomic number:}\)
The \(3+\) charge means that three electrons have been removed from the neutral atom.
Thus:
\[
Z=N_e(M^{3+})+3
\]
\[
Z=22+3
\]
\[
Z=25
\]
Atomic number 25 identifies manganese.
\(\textbf{Establishing the neutral configuration:}\)
Neutral manganese has the ground-state configuration:
\[
\mathrm{Mn}:[\mathrm{Ar}]\,3d^5\,4s^2
\]
Its \(3d^5\,4s^2\) electron population is consistent with group 7 and period 4.
\(\textbf{Checking the ionization sequence:}\)
The first two electrons are removed from the outer \(4s\) subshell:
\[
\mathrm{Mn^{2+}}:[\mathrm{Ar}]\,3d^5
\]
Removal of one additional \(3d\) electron produces:
\[
\mathrm{Mn^{3+}}:[\mathrm{Ar}]\,3d^4
\]
This matches the configuration specified in the question.
\(\textbf{Excluding the alternatives:}\)
Chromium has atomic number 24, not 25.
Iron has atomic number 26.
Retaining manganese's \(4s^2\) electrons while removing three \(3d\) electrons would contradict the stipulated ordinary cation-formation order.
\(\textbf{Final interpretation:}\)
The problem requires working backwards from an ionic configuration to a neutral atomic number.
The parent element's group must then be determined from its neutral configuration rather than from the \(3d^4\) count of the final ion alone.
The second option satisfies the electron count, periodic classification and removal sequence.
\(\textbf{Final answer:}\)
\[
Z=25
\]
304. Two neighbouring elements have the following properties:
| Property | Element P | Element Q |
|---|
| Atomic number | 35 | 36 |
| Outermost configuration | \(4s^2\,4p^5\) | \(4s^2\,4p^6\) |
| Relative atomic mass | 79.9 | 83.8 |
A process changes the isotopic abundances of P without changing its atomic number or the number of electrons in any neutral atom. Its average atomic mass consequently increases.
Which statement remains necessarily correct regardless of the new average atomic mass?
ⓐ. P moves closer to Q in the periodic table if its average atomic mass approaches Q's value.
ⓑ. P becomes a noble gas when its average atomic mass reaches the atomic mass of Q.
ⓒ. P remains a group-17 element immediately preceding the group-18 element Q; its ordinary valence-electron configuration is unchanged.
ⓓ. The electron-gain enthalpy of P must increase by exactly the same percentage as its average atomic mass.
Correct Answer: P remains a group-17 element immediately preceding the group-18 element Q; its ordinary valence-electron configuration is unchanged.
Explanation: Element P has atomic number 35 and is bromine.
Element Q has atomic number 36 and is krypton.
Bromine possesses the outer configuration \(4s^2\,4p^5\) and belongs to group 17.
Krypton possesses the completed outer configuration \(4s^2\,4p^6\) and belongs to group 18.
The elements occupy consecutive positions at the end of period 4.
Changing the relative proportions of isotopes can alter the measured average atomic mass of an element.
However, isotopes of the same element contain identical numbers of protons.
The stipulated process therefore leaves the atomic number of bromine at 35.
Its ordinary neutral electronic configuration also remains unchanged.
Consequently, bromine does not move to a different group or become a noble gas.
The modern periodic arrangement depends on atomic number rather than average isotopic mass.
An isotopic-abundance change also does not justify a numerical prediction for electron-gain enthalpy based simply on the percentage change in mass.
The important invariant is the nuclear charge and the resulting electronic classification.
The third option correctly preserves the periodic identities of both elements.
305. The first ionization enthalpies of sodium and magnesium are \(496\) and \(738\ \mathrm{kJ\,mol^{-1}}\), respectively. Sodium's second ionization enthalpy is \(4562\ \mathrm{kJ\,mol^{-1}}\).
An isolated gaseous ground-state atom is supplied with \(9.0\times10^{-19}\ \mathrm{J}\) of energy. Assume all the supplied energy is available for ionization and that no other energy source is present.
Take \(N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}\).
Which statement is correct?
ⓐ. The supplied energy is sufficient to remove the first electron from either sodium or magnesium, but not the second electron from sodium.
ⓑ. The supplied energy is sufficient to remove the first electron from sodium but not magnesium, and it is insufficient for the second ionization of sodium.
ⓒ. The supplied energy cannot ionize either atom because its numerical value in joules is smaller than the tabulated values in \(\mathrm{kJ\,mol^{-1}}\).
ⓓ. The supplied energy can remove two electrons from sodium because sodium's first ionization enthalpy is lower than magnesium's.
Correct Answer: The supplied energy is sufficient to remove the first electron from sodium but not magnesium, and it is insufficient for the second ionization of sodium.
Explanation: \(\textbf{Identifying the unit mismatch:}\)
The supplied energy refers to one isolated atom.
The ionization enthalpies are expressed per mole of gaseous atoms or ions.
Their numerical values must therefore be converted to a common basis before comparison.
\(\textbf{Converting the supplied energy to molar units:}\)
The equivalent molar energy is:
\[
E_{\mathrm{molar}}=E_{\mathrm{atom}}N_A
\]
Substituting:
\[
E_{\mathrm{molar}}=(9.0\times10^{-19})(6.02\times10^{23})
\]
\[
E_{\mathrm{molar}}=5.418\times10^5\ \mathrm{J\,mol^{-1}}
\]
Converting joules to kilojoules:
\[
E_{\mathrm{molar}}=\frac{5.418\times10^5}{1000}
\]
\[
E_{\mathrm{molar}}=541.8\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Checking first ionization of sodium:}\)
\[
IE_1(\mathrm{Na})=496\ \mathrm{kJ\,mol^{-1}}
\]
Since:
\[
541.8\gt 496
\]
The available energy exceeds sodium's first-ionization requirement.
\(\textbf{Checking first ionization of magnesium:}\)
\[
IE_1(\mathrm{Mg})=738\ \mathrm{kJ\,mol^{-1}}
\]
Since:
\[
541.8\lt 738
\]
The energy is insufficient to ionize magnesium under the stated model.
\(\textbf{Checking further sodium ionization:}\)
Sodium's second ionization begins with \(\mathrm{Na^+}\), whose electrons have the neon configuration.
Its additional ionization requirement is:
\[
IE_2(\mathrm{Na})=4562\ \mathrm{kJ\,mol^{-1}}
\]
The available energy is far below this value.
It also cannot supply the sum of the first and second ionization enthalpies.
\(\textbf{Physical interpretation:}\)
Ionization thresholds must be compared using the correct starting electronic species and consistent energy units.
A lower first ionization enthalpy does not imply that repeated electron removal is energetically easy.
\(\textbf{Final answer:}\)
\[
\boxed{E_{\mathrm{molar}}=541.8\ \mathrm{kJ\,mol^{-1}};\quad \mathrm{Na\ first\ ionization\ only}}
\]
306. Consider four representative elements whose ordinary outer-shell configurations are:
| Element | Outer configuration |
|---|
| P | \(3s^2\,3p^2\) |
| Q | \(3s^2\,3p^3\) |
| R | \(3s^2\,3p^4\) |
| S | \(3s^2\,3p^5\) |
The four elements form representative hydrides, and their highest characteristic oxygen-containing compounds show formal central-element oxidation states of \(+4,+5,+6\) and \(+7\), respectively.
Which conclusion correctly compares their hydride bonding patterns with their highest positive oxidation states?
ⓐ. Their representative hydrides contain four, three, two and one hydrogen atoms per central atom, respectively, while the maximum listed positive oxidation state increases across the sequence.
ⓑ. Their representative hydrides contain four, three, two and one hydrogen atoms per central atom, respectively, so their maximum positive oxidation states must decrease in the same order.
ⓒ. Their representative hydrides contain two, three, four and five hydrogen atoms per central atom, respectively, because valence-electron count directly determines hydrogen-bond count.
ⓓ. All four have identical characteristic hydride valency because they occupy the same principal shell.
Correct Answer: Their representative hydrides contain four, three, two and one hydrogen atoms per central atom, respectively, while the maximum listed positive oxidation state increases across the sequence.
Explanation: Element P has four valence electrons and belongs to group 14.
Its specified configuration corresponds to silicon.
The representative covalent hydride is \(\mathrm{SiH_4}\), containing four Si–H bonds.
Element Q has five valence electrons and belongs to group 15.
It corresponds to phosphorus, whose representative hydride is \(\mathrm{PH_3}\).
Element R has six valence electrons and belongs to group 16.
It corresponds to sulfur, whose representative hydride is \(\mathrm{H_2S}\).
Element S has seven valence electrons and belongs to group 17.
It corresponds to chlorine, whose representative hydride is \(\mathrm{HCl}\).
Across this sequence, the number of hydrogen atoms in the stated representative hydrides decreases from four to one.
However, the listed higher positive oxidation states increase from \(+4\) to \(+7\).
These two patterns describe different aspects of periodic chemical behaviour.
The representative hydride bond count reflects a common valency associated with completion of a stable electron arrangement.
The highest positive oxidation state reflects formal electron assignment in suitable compounds with strongly electronegative partners.
Consequently, the two quantities need not vary in the same direction.
The first option correctly distinguishes the two periodic patterns without treating valency and oxidation state as interchangeable.
307. The approximate first ionization enthalpies and first electron-gain enthalpies of phosphorus and sulfur are:
| Element | \(IE_1\) (\(\mathrm{kJ\,mol^{-1}}\)) | First electron-gain enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| P | 1012 | \(-72\) |
| S | 1000 | \(-200\) |
Which statement correctly reconciles phosphorus's greater first ionization enthalpy with sulfur's more exothermic first electron gain?
ⓐ. Phosphorus has a completed \(3p^6\) subshell, making both electron removal and electron addition impossible.
ⓑ. Sulfur has a half-filled \(3p^3\) subshell before electron gain, while phosphorus has a paired \(3p^4\) arrangement.
ⓒ. Phosphorus has stable \(3p^3\), while removing an electron from sulfur's \(3p^4\) gives \(3p^3\); incoming electrons face different pairing and attraction effects.
ⓓ. The two numerical trends contradict one another because an element with a higher ionization enthalpy must always have a more negative electron-gain enthalpy.
Correct Answer: Phosphorus has stable \(3p^3\), while removing an electron from sulfur's \(3p^4\) gives \(3p^3\); incoming electrons face different pairing and attraction effects.
Explanation: \(\textbf{Establishing the initial configurations:}\)
Phosphorus has the configuration \([\mathrm{Ne}]\,3s^2\,3p^3\).
Its three p-electrons occupy separate orbitals, producing a relatively stable half-filled arrangement.
Sulfur has the configuration \([\mathrm{Ne}]\,3s^2\,3p^4\).
One of its p-orbitals therefore contains a pair of electrons.
\(\textbf{Comparing first ionization:}\)
Removal of an electron from phosphorus disrupts its half-filled p-subshell.
Removing an electron from sulfur produces the half-filled \(3p^3\) arrangement.
Electron-pair repulsion in sulfur also contributes to the relative ease of removing its first electron.
Consequently, phosphorus has the slightly greater first ionization enthalpy despite having the smaller nuclear charge.
\(\textbf{Comparing first electron gain:}\)
Adding an electron to phosphorus produces \(3p^4\).
The added electron must pair with an electron already occupying a p-orbital.
This reduces the energetic advantage associated with the initial half-filled arrangement.
Adding an electron to sulfur produces \(3p^5\).
Although pairing effects are also present, the greater nuclear charge of sulfur and the actual energy balance make its first electron addition more exothermic under the given conditions.
\(\textbf{Interpreting the combined trend:}\)
The first ionization process and the first electron-gain process lead to different final configurations.
Their energetic orderings should therefore not be predicted through one simplistic rule.
A comparatively high first ionization enthalpy does not require a particularly negative electron-gain enthalpy.
The third option correctly explains both observations using the actual initial and final p-subshell occupancies.
308. The following ions are considered under comparable effective-radius conventions:
| Ion | Electron count | Nuclear charge |
|---|
| \(\mathrm{P^{3-}}\) | 18 | 15 |
| \(\mathrm{S^{2-}}\) | 18 | 16 |
| \(\mathrm{Cl^-}\) | 18 | 17 |
An unknown monovalent cation \(M^+\) contains the same number of electrons as these anions and is larger than the corresponding divalent cation \(N^{2+}\), which is also isoelectronic.
Which assignment and decreasing radius order are consistent with the stated information?
ⓐ. \(M=\mathrm{Na}\), \(N=\mathrm{Mg}\); \(\mathrm{P^{3-}\gt S^{2-}\gt Cl^-\gt Na^+\gt Mg^{2+}}\)
ⓑ. \(M=\mathrm{K}\), \(N=\mathrm{Ca}\); \(\mathrm{P^{3-}\gt S^{2-}\gt Cl^-\gt K^+\gt Ca^{2+}}\)
ⓒ. \(M=\mathrm{K}\), \(N=\mathrm{Ca}\); \(\mathrm{Ca^{2+}\gt K^+\gt Cl^-\gt S^{2-}\gt P^{3-}}\)
ⓓ. \(M=\mathrm{Ca}\), \(N=\mathrm{Sc}\); \(\mathrm{P^{3-}\gt S^{2-}\gt Cl^-\gt Ca^+\gt Sc^{2+}}\)
Correct Answer: \(M=\mathrm{K}\), \(N=\mathrm{Ca}\); \(\mathrm{P^{3-}\gt S^{2-}\gt Cl^-\gt K^+\gt Ca^{2+}}\)
Explanation: \(\textbf{Identifying the shared electronic configuration:}\)
Each listed anion contains eighteen electrons.
An eighteen-electron species has the argon electronic configuration.
The unknown cations must also contain eighteen electrons.
\(\textbf{Determining the monovalent cation:}\)
For a monovalent cation:
\[
N_e(M^+)=Z_M-1
\]
Since:
\[
Z_M-1=18
\]
It follows that:
\[
Z_M=19
\]
The element is potassium.
\(\textbf{Determining the divalent cation:}\)
For the divalent cation:
\[
N_e(N^{2+})=Z_N-2
\]
Therefore:
\[
Z_N-2=18
\]
\[
Z_N=20
\]
The element is calcium.
\(\textbf{Applying the isoelectronic radius rule:}\)
All five ions contain the same number of electrons.
Their nuclei contain progressively more protons in the order phosphorus, sulfur, chlorine, potassium and calcium.
Greater nuclear charge contracts the common electron population.
Therefore, the radii decrease as nuclear charge increases.
\(\textbf{Checking the complete order:}\)
\[
\mathrm{P^{3-}\gt S^{2-}\gt Cl^-\gt K^+\gt Ca^{2+}}
\]
The positive ionic charges do not independently determine the order.
They must first be combined with the electron counts to establish the isoelectronic relationship.
The second option correctly identifies both unknown parent elements and the resulting size sequence.
309. Element \(X\) belongs to group 16 and period 3. Element \(Y\) belongs to group 17 and period 3. Their neutral atoms form the molecular compounds \(\mathrm{XY_2}\) and \(\mathrm{XY_4}\) under suitable conditions, with \(Y\) assigned oxidation state \(-1\).
Which statement correctly compares \(X\)'s oxidation states, the numbers of directly bonded \(Y\) atoms and the periodic identities?
ⓐ. \(X=\mathrm{S},\ Y=\mathrm{Cl}\); \(X\) has oxidation states \(+2,+4\) and coordination numbers \(2,4\), respectively.
ⓑ. \(X=\mathrm{S},\ Y=\mathrm{Cl}\); \(X\) has oxidation states \(-2,-4\) and coordination numbers \(2,4\), respectively.
ⓒ. \(X=\mathrm{O},\ Y=\mathrm{F}\); \(X\) has oxidation states \(+2,+4\) and both atoms belong to the third period.
ⓓ. \(X=\mathrm{S},\ Y=\mathrm{Cl}\); \(X\) has oxidation state \(+6\) in both compounds, despite coordination numbers \(2,4\).
Correct Answer: \(X=\mathrm{S},\ Y=\mathrm{Cl}\); \(X\) has oxidation states \(+2,+4\) and coordination numbers \(2,4\), respectively.
Explanation: Element \(X\) belongs to group 16 and period 3, identifying sulfur.
Element \(Y\) belongs to group 17 and period 3, identifying chlorine.
The specified molecular formulas consequently correspond to sulfur chlorides with compositions \(\mathrm{SCl_2}\) and \(\mathrm{SCl_4}\).
For formal oxidation-state accounting, chlorine is assigned \(-1\) in both compounds.
In \(\mathrm{SCl_2}\), two chlorine atoms contribute total oxidation state \(-2\).
Sulfur therefore has oxidation state \(+2\).
In \(\mathrm{SCl_4}\), the four chlorine atoms contribute \(-4\).
Sulfur consequently has oxidation state \(+4\).
Under the stipulated structures, sulfur is directly connected to two chlorine atoms in the first compound and four in the second.
The number of directly bonded atoms and the oxidation state happen to have equal numerical magnitudes in these two examples.
That coincidence does not establish a universal equality between coordination number and oxidation state.
Both compounds contain the same central element, whose periodic position remains unchanged.
The different formulas reflect sulfur's ability to participate in different bonding and oxidation-state arrangements.
The first option correctly identifies the elements and distinguishes the two molecular situations.
310. A third-period non-metal forms two oxygen-containing compounds, \(A\) and \(B\). The central element has oxidation state \(+3\) in \(A\) and \(+5\) in \(B\). Both compounds are acidic oxides under ordinary classification.
A student concludes that an oxide must change from acidic to basic whenever the oxidation state of its central element increases.
Which observation provides a direct counterexample consistent with third-period chemistry?
ⓐ. \(\mathrm{Na_2O}\) and \(\mathrm{MgO}\) are both basic, but sodium and magnesium belong to different groups.
ⓑ. \(\mathrm{Al_2O_3}\) is amphoteric, whereas \(\mathrm{SiO_2}\) is acidic.
ⓒ. Phosphorus(III) and phosphorus(V) oxides are both acidic despite the different phosphorus oxidation states.
ⓓ. \(\mathrm{CO}\) is neutral and \(\mathrm{CO_2}\) is acidic, proving that every higher oxidation state produces a basic oxide.
Correct Answer: Phosphorus(III) and phosphorus(V) oxides are both acidic despite the different phosphorus oxidation states.
Explanation: Phosphorus belongs to group 15 and period 3.
It possesses five valence electrons and can exhibit positive oxidation states including \(+3\) and \(+5\).
The conventional molecular formulas of its corresponding oxides include \(\mathrm{P_4O_6}\) and \(\mathrm{P_4O_{10}}\).
Oxygen has oxidation state \(-2\) in both compounds.
In \(\mathrm{P_4O_6}\), the total oxygen contribution is \(-12\).
The four phosphorus atoms consequently contribute \(+12\), giving oxidation state \(+3\) per phosphorus atom.
In \(\mathrm{P_4O_{10}}\), the ten oxygen atoms contribute \(-20\).
The phosphorus atoms consequently have average oxidation state \(+5\).
Both oxides exhibit acidic behaviour under the usual classification.
Their oxidation states differ without requiring one oxide to become basic.
A tendency for increasing positive oxidation state to increase oxide acidity in some element families is not an exception-free mathematical law.
Oxide character also depends on the chemical identity and bonding of the central element.
The student has therefore converted a useful qualitative relationship into an unsupported universal statement.
The phosphorus oxides provide the direct counterexample requested.
The third option correctly distinguishes the oxidation-state difference from the persistence of acidic oxide character.
311. The internuclear distances of three gaseous halogen molecules are given below. All bond lengths refer to the specified molecular ground states.
| Molecule | Internuclear distance (pm) |
|---|
| \(\mathrm{Cl_2}\) | 198 |
| \(\mathrm{Br_2}\) | 228 |
| \(\mathrm{BrCl}\) | 216 |
A student applies the approximate additivity of single-bond covalent radii to predict the Br–Cl bond length. Which conclusion correctly compares the predicted and observed distances and explains the limitation of the approximation?
ⓐ. The predicted distance is \(213\ \mathrm{pm}\), which is \(3\ \mathrm{pm}\) shorter than the observed distance; covalent-radius additivity is an approximation rather than an exact bond-length law.
ⓑ. The predicted distance is \(213\ \mathrm{pm}\), which is \(3\ \mathrm{pm}\) longer than the observed distance; the discrepancy proves that bromine is smaller than chlorine.
ⓒ. The predicted distance is \(426\ \mathrm{pm}\), which is \(210\ \mathrm{pm}\) longer than the observed distance; heteronuclear bonds require twice the sum of the atomic radii.
ⓓ. The predicted distance is \(216\ \mathrm{pm}\) exactly because the covalent radius of an element changes to enforce additivity in every compound.
Correct Answer: The predicted distance is \(213\ \mathrm{pm}\), which is \(3\ \mathrm{pm}\) shorter than the observed distance; covalent-radius additivity is an approximation rather than an exact bond-length law.
Explanation: \(\textbf{Extracting the single-bond covalent radii:}\)
For a suitable homonuclear single bond, the covalent radius is half the internuclear distance.
Therefore:
\[
r_{\mathrm{Cl}}=\frac{d_{\mathrm{Cl-Cl}}}{2}
\]
\[
r_{\mathrm{Cl}}=\frac{198}{2}
\]
\[
r_{\mathrm{Cl}}=99\ \mathrm{pm}
\]
For bromine:
\[
r_{\mathrm{Br}}=\frac{d_{\mathrm{Br-Br}}}{2}
\]
\[
r_{\mathrm{Br}}=\frac{228}{2}
\]
\[
r_{\mathrm{Br}}=114\ \mathrm{pm}
\]
\(\textbf{Predicting the heteronuclear distance:}\)
The approximate additive relation is:
\[
d_{\mathrm{Br-Cl,predicted}}\approx r_{\mathrm{Br}}+r_{\mathrm{Cl}}
\]
Substituting:
\[
d_{\mathrm{Br-Cl,predicted}}\approx114+99
\]
\[
d_{\mathrm{Br-Cl,predicted}}\approx213\ \mathrm{pm}
\]
\(\textbf{Comparing with the observation:}\)
The measured distance is \(216\ \mathrm{pm}\).
Therefore:
\[
\Delta d=d_{\mathrm{observed}}-d_{\mathrm{predicted}}
\]
\[
\Delta d=216-213
\]
\[
\Delta d=+3\ \mathrm{pm}
\]
The positive difference means that the observed bond is slightly longer than the additive prediction.
\(\textbf{Interpreting the chemical significance:}\)
Bromine has the greater covalent radius, consistent with its position below chlorine in group 17.
However, covalent radii are operational estimates derived from bonding environments.
Actual bond lengths can also depend on the bonding partners and their electronic interactions.
Consequently, radius additivity is useful for estimation but need not reproduce every measured heteronuclear bond length exactly.
The discrepancy does not justify reversing the established chlorine–bromine radius order.
\(\textbf{Final answer:}\)
\[
\boxed{d_{\mathrm{predicted}}=213\ \mathrm{pm},\qquad d_{\mathrm{observed}}-d_{\mathrm{predicted}}=+3\ \mathrm{pm}}
\]
312. Two different nitrogen–nitrogen bond lengths are measured:
| Species | Specified N–N bond | Bond length (pm) |
|---|
| \(\mathrm{N_2}\) | N≡N triple bond | 110 |
| \(\mathrm{N_2H_4}\) | N–N single bond | 146 |
A student calculates half of each bond length and claims that nitrogen must have two different nuclear sizes in the two molecules.
Which statement correctly evaluates the calculation?
ⓐ. The covalent radius of nitrogen is \(55\ \mathrm{pm}\) under the ordinary single-bond definition, while \(73\ \mathrm{pm}\) is its nuclear radius.
ⓑ. The single-bond covalent radius of nitrogen is about \(73\ \mathrm{pm}\); \(55\ \mathrm{pm}\) is half its triple-bond length.
ⓒ. The two measurements prove that nitrogen changes from period 2 to period 3 when it forms hydrazine.
ⓓ. The conventional covalent radius must be the arithmetic mean of \(55\) and \(73\ \mathrm{pm}\), because bond order does not affect internuclear distance.
Correct Answer: The single-bond covalent radius of nitrogen is about \(73\ \mathrm{pm}\); \(55\ \mathrm{pm}\) is half its triple-bond length.
Explanation: \(\textbf{Identifying the relevant radius definition:}\)
The conventional single-bond covalent radius is based on the internuclear distance between identical atoms connected by a suitable single covalent bond.
The N–N bond in hydrazine satisfies the stated bond-order condition.
The N≡N bond in molecular nitrogen has a different bond order.
Therefore, the two distances do not represent identical bonding situations.
\(\textbf{Calculating the single-bond radius:}\)
\[
r_{\mathrm{N,single}}=\frac{d_{\mathrm{N-N}}}{2}
\]
\[
r_{\mathrm{N,single}}=\frac{146}{2}
\]
\[
r_{\mathrm{N,single}}=73\ \mathrm{pm}
\]
\(\textbf{Calculating half the triple-bond length:}\)
\[
\frac{d_{\mathrm{N\equiv N}}}{2}=\frac{110}{2}
\]
\[
\frac{d_{\mathrm{N\equiv N}}}{2}=55\ \mathrm{pm}
\]
This second value is not the conventional single-bond covalent radius.
\(\textbf{Explaining the bond-length difference:}\)
Molecular nitrogen contains three shared electron pairs between its nitrogen atoms.
Hydrazine contains an N–N single bond.
The greater bond order and the corresponding bonding interactions in molecular nitrogen produce a shorter internuclear distance.
This does not mean that the nitrogen nucleus changes size or that its atomic number changes.
\(\textbf{Connecting the result with periodic classification:}\)
Nitrogen retains atomic number 7 in both molecules.
It consequently remains a group-15, period-2 element.
Its electronic configuration as an isolated neutral atom is unaffected by which compound is being discussed.
The measurements illustrate that atomic radius is an operational bonding-dependent quantity rather than a sharply defined universal atomic boundary.
\(\textbf{Final answer:}\)
\[
\boxed{r_{\mathrm{N,single}}=73\ \mathrm{pm},\qquad \frac{d_{\mathrm{N\equiv N}}}{2}=55\ \mathrm{pm}}
\]
313. A graph presents atomic-radius data for selected third-period non-metals and the noble gas argon.
The horizontal axis lists P, S, Cl and Ar in increasing atomic-number order. The vertical axis gives radius in picometres. The plotted values for P, S and Cl are 110, 104 and 99 pm, respectively, each labelled as a covalent radius. The plotted value for Ar is 188 pm and is explicitly labelled as a van der Waals radius. The graph joins all four points with one continuous line, producing a sharp upward segment from Cl to Ar.
Consider the following statements.
Statement I: The covalent-radius data from phosphorus to chlorine are consistent with increasing effective nuclear attraction across the period.
Statement II: The upward segment from chlorine to argon establishes that argon's covalent radius is greater than chlorine's.
Statement III: The plotted argon value represents a different operational radius definition from the three preceding values.
Statement IV: A scientifically justified trend comparison should distinguish changes in electronic structure from changes in the measurement convention.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, III and IV only
ⓓ. Statements I, II and IV only
Correct Answer: Statements I, III and IV only
Explanation: Statement I is correct because phosphorus, sulfur and chlorine belong to the same period.
Their outer electrons occupy the third principal shell.
Nuclear charge increases from phosphorus to sulfur to chlorine.
Shielding does not completely offset the additional nuclear attraction.
The supplied covalent radii consequently decrease from 110 to 104 to 99 pm.
Statement II is incorrect because argon's plotted radius is not a conventional covalent radius.
The preceding three data points describe covalently bonded atoms.
Argon's value is a van der Waals radius obtained from a different type of interatomic separation.
Therefore, the numerical increase from chlorine to argon cannot be interpreted as a measured reversal in the same covalent-radius quantity.
Statement III is correct because the graph itself specifies different operational definitions.
Noble gases ordinarily exist as monatomic species and do not provide the same conventional homonuclear single-bond measurements used for many other non-metals.
Their tabulated radii are commonly based on non-bonded contact distances.
Statement IV is correct because a graph can be numerically accurate while encouraging an invalid comparison if it combines unlike measurements.
The measurement method must be examined before interpreting an apparent discontinuity.
The third-period contraction trend and the noble-gas radius qualification are separate considerations.
Only Statements I, III and IV are correct.
314. Lithium and magnesium lie diagonally from one another in the periodic table.
The following approximate data refer to their neutral atoms under the stated comparable radius convention:
| Property | Li | Mg |
|---|
| Atomic radius (pm) | 152 | 160 |
| First ionization enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) | 520 | 738 |
| Outer electronic configuration | \(2s^1\) | \(3s^2\) |
Which interpretation correctly explains why magnesium can have both a slightly greater atomic radius and a substantially greater first ionization enthalpy than lithium?
ⓐ. Magnesium's larger radius proves that its first electron must be easier to remove, so the supplied ionization data contradict periodicity.
ⓑ. Lithium and magnesium belong to the same period; magnesium's larger radius results from a reduction in nuclear charge.
ⓒ. The additional principal shell in magnesium favours a larger radius, while its greater effective nuclear attraction and filled \(3s^2\) subshell contribute to stronger binding of its first removed electron.
ⓓ. Magnesium's first ionization removes a \(2s\) electron, whereas lithium's removes a \(3s\) electron; the energy difference therefore follows from the opposite shell order.
Correct Answer: The additional principal shell in magnesium favours a larger radius, while its greater effective nuclear attraction and filled \(3s^2\) subshell contribute to stronger binding of its first removed electron.
Explanation: Lithium belongs to group 1 and period 2.
Magnesium belongs to group 2 and period 3.
Moving from lithium to magnesium changes both the group and the period.
Consequently, neither the ordinary across-period trend nor the ordinary down-group trend can be applied in isolation.
Magnesium has an additional occupied principal shell compared with lithium.
This contributes to its slightly greater atomic radius in the supplied data.
However, atomic radius alone does not determine first ionization enthalpy.
Lithium loses its first electron from a singly occupied \(2s\) subshell.
Magnesium loses its first electron from a filled \(3s^2\) subshell.
Magnesium also has substantially greater nuclear charge, with the effective attraction of its outer electrons reflecting the balance of nuclear charge and shielding.
The combined electronic effects produce the supplied greater first ionization enthalpy for magnesium.
Therefore, the element with the larger atomic radius need not invariably possess the lower first ionization enthalpy when comparing different groups and periods.
The comparison does not invalidate the separate general trends across periods or down groups.
It shows that those trends describe patterns governed by multiple electronic factors rather than a universal inverse mathematical relation between radius and ionization enthalpy.
The third option correctly reconciles the two measured properties.
315. Aluminium and gallium belong to group 13. Comparable measurements give the following approximate radii:
| Element or ion | Radius (pm) | Measurement convention |
|---|
| Al | 143 | Metallic radius |
| Ga | 135 | Metallic radius |
| \(\mathrm{Al^{3+}}\) | 53 | Effective ionic radius, coordination number 6 |
| \(\mathrm{Ga^{3+}}\) | 62 | Effective ionic radius, coordination number 6 |
Which statement correctly explains why the neutral-atom size order reverses when the corresponding trivalent ions are compared?
ⓐ. Weak \(3d\)-electron shielding contracts neutral Ga; \(\mathrm{Ga^{3+}}\) nevertheless retains a more extensive electron configuration than \(\mathrm{Al^{3+}}\).
ⓑ. Gallium's nucleus contains fewer protons than aluminium's, but its trivalent ion acquires additional protons during ionization.
ⓒ. Both trivalent ions are isoelectronic, so gallium's larger ionic radius must result from a smaller nuclear charge.
ⓓ. The reversal proves that a metallic radius and an ionic radius represent exactly the same experimental quantity.
Correct Answer: Weak \(3d\)-electron shielding contracts neutral Ga; \(\mathrm{Ga^{3+}}\) nevertheless retains a more extensive electron configuration than \(\mathrm{Al^{3+}}\).
Explanation: \(\textbf{Examining the neutral configurations:}\)
Aluminium has the configuration \([\mathrm{Ne}]\,3s^2\,3p^1\).
Gallium has the configuration \([\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^1\).
Both possess the group-13 outer configuration \(ns^2np^1\).
Gallium's outer electrons occupy the fourth principal shell, which would ordinarily favour a larger atomic radius.
However, its filled inner \(3d^{10}\) subshell shields nuclear charge relatively ineffectively.
The resulting increased effective attraction helps account for the unexpectedly small metallic radius of gallium.
\(\textbf{Constructing the trivalent ions:}\)
Formation of aluminium(III) removes the three outer-shell electrons:
\[
\mathrm{Al^{3+}}:[\mathrm{Ne}]
\]
Formation of gallium(III) removes its outer \(4p^1\) and \(4s^2\) electrons:
\[
\mathrm{Ga^{3+}}:[\mathrm{Ar}]\,3d^{10}
\]
The two ions are not isoelectronic.
Aluminium(III) has ten electrons, whereas gallium(III) has twenty-eight.
\(\textbf{Interpreting the ionic measurements:}\)
The supplied ionic radii use the same coordination-number convention.
Gallium(III) retains occupied third-shell subshells, including \(3d^{10}\).
Aluminium(III) has the compact neon configuration with no occupied third shell.
The gallium ion is consequently larger under the stated comparable convention.
\(\textbf{Resolving the different orders:}\)
The neutral metallic radii follow:
\[
r(\mathrm{Al})\gt r(\mathrm{Ga})
\]
The corresponding trivalent ionic radii follow:
\[
r(\mathrm{Ga^{3+}})\gt r(\mathrm{Al^{3+}})
\]
The reversal reflects different electronic populations and measurement environments.
It does not imply that nuclear charge changes during ion formation.
The first option correctly connects the unusual neutral-atom trend with the different ionic electron configurations.
316. The following data are provided for three gaseous hydrogen halides:
| Molecule | H–X bond length (pm) | Bond dissociation enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| HCl | 127 | 431 |
| HBr | 141 | 366 |
| HI | 161 | 299 |
Which statement correctly combines the halogen atomic-size trend with the supplied bond-length and bond-strength data?
ⓐ. Halogen atomic size decreases from chlorine to iodine, so H–X bond length must increase despite stronger overlap.
ⓑ. The H–X bond becomes shorter down the halogen group, but its bond dissociation enthalpy decreases because nuclear charge increases.
ⓒ. Larger halogen atoms form longer H–X bonds, and weaker orbital overlap at greater distances contributes to lower bond enthalpies.
ⓓ. The decreasing bond dissociation enthalpy proves that chlorine, bromine and iodine possess progressively fewer valence electrons.
Correct Answer: Larger halogen atoms form longer H–X bonds, and weaker orbital overlap at greater distances contributes to lower bond enthalpies.
Explanation: Chlorine, bromine and iodine belong to group 17.
They possess the characteristic valence configuration \(ns^2np^5\).
Moving from chlorine to iodine introduces higher occupied principal shells.
The atomic size therefore generally increases down the group.
The hydrogen atom is the common bonding partner in all three molecules.
The measured H–X bond length increases from HCl to HBr to HI.
This is consistent with the increasing size of the halogen atom.
The supplied bond dissociation enthalpies decrease in the same sequence.
A shorter bond can permit more effective orbital overlap and stronger bonding interactions under comparable circumstances.
As the halogen valence orbitals become larger and more diffuse, their overlap with hydrogen's compact \(1s\) orbital generally becomes less effective.
This contributes to the decreasing strengths of the stated hydrogen-halogen bonds.
The data establish a relationship for the particular H–X series, not an exception-free rule connecting every chemical bond length with bond strength.
All three halogens retain seven valence electrons despite their different bond lengths and dissociation enthalpies.
The third option correctly connects the radius trend, structural observations and bonding-energy data.
317. Beryllium oxide and magnesium oxide are both oxides of group-2 elements. Under ordinary classification, \(\mathrm{BeO}\) is amphoteric, whereas \(\mathrm{MgO}\) is predominantly basic.
Consider the following Assertion and Reason.
Assertion (A): The chemical behaviour of \(\mathrm{BeO}\) cannot be predicted accurately by assuming that every group-2 oxide is equally basic.
Reason (R): Beryllium forms a much smaller divalent ion than magnesium; its greater charge density and polarising influence contribute to increased covalent character and different oxide behaviour.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion.
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
ⓒ. Assertion is true, but Reason is false.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Explanation: The Assertion is true because group-2 elements share a characteristic valence configuration but do not form oxides with identical chemical properties.
Beryllium belongs to period 2, while magnesium belongs to period 3.
Both commonly exhibit oxidation state \(+2\).
However, the beryllium ion is considerably smaller than the magnesium ion.
At the same positive charge, smaller ionic size produces greater charge density.
This contributes to stronger polarisation of neighbouring electron clouds.
Beryllium compounds consequently show relatively greater covalent character than would be predicted using ionic charge alone.
Beryllium oxide can react with suitable acids and strong bases.
It is therefore classified as amphoteric.
Magnesium oxide exhibits predominantly basic character and reacts with acids to form magnesium salts and water.
The Reason is true because the different polarising influences of the divalent cations help explain the chemical distinction.
This comparison illustrates why periodic group membership predicts broad similarity without requiring identical bonding or acid-base behaviour.
The Reason correctly accounts for the Assertion.
Thus, both statements are true, and the Reason provides the appropriate explanation.
318. A graph compares the metallic radii and first ionization enthalpies of lithium, sodium and potassium.
| Element | Metallic radius (pm) | \(IE_1\) (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| Li | 152 | 520 |
| Na | 186 | 496 |
| K | 227 | 419 |
A student proposes that first ionization enthalpy decreases by a fixed amount per picometre of metallic-radius increase throughout this group.
Calculate the magnitude of the ionization-enthalpy decrease per picometre over each interval. Which conclusion is correct?
ⓐ. Li to Na: \(0.71\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}\); Na to K: \(1.88\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}\). The difference proves that atomic radius decreases from sodium to potassium.
ⓑ. Li to Na: \(1.88\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}\); Na to K: \(0.71\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}\). The fixed-rate relation is verified.
ⓒ. Both intervals: \(1.00\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}\). The group trend therefore follows an exact linear law.
ⓓ. Li to Na: \(0.71\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}\); Na to K: \(1.88\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}\). The unequal values disprove the proposed fixed-rate relation for these intervals.
Correct Answer: Li to Na: \(0.71\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}\); Na to K: \(1.88\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}\). The unequal values disprove the proposed fixed-rate relation for these intervals.
Explanation: \(\textbf{Identifying the proposed relation:}\)
The student assumes that a given increase in metallic radius always produces the same decrease in first ionization enthalpy.
This would require equal magnitudes of the ratio:
\[
R=\frac{|\Delta IE_1|}{\Delta r}
\]
for successive intervals.
\(\textbf{Interval I — Lithium to sodium:}\)
The radius increase is:
\[
\Delta r_1=186-152
\]
\[
\Delta r_1=34\ \mathrm{pm}
\]
The magnitude of the ionization-enthalpy decrease is:
\[
|\Delta IE_1|_1=520-496
\]
\[
|\Delta IE_1|_1=24\ \mathrm{kJ\,mol^{-1}}
\]
Therefore:
\[
R_1=\frac{24}{34}
\]
\[
R_1\approx0.71\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}
\]
\(\textbf{Interval II — Sodium to potassium:}\)
The radius increase is:
\[
\Delta r_2=227-186
\]
\[
\Delta r_2=41\ \mathrm{pm}
\]
The ionization-enthalpy decrease is:
\[
|\Delta IE_1|_2=496-419
\]
\[
|\Delta IE_1|_2=77\ \mathrm{kJ\,mol^{-1}}
\]
Thus:
\[
R_2=\frac{77}{41}
\]
\[
R_2\approx1.88\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}
\]
\(\textbf{Testing the proposed invariant:}\)
The two calculated rates are not equal.
Therefore, the supplied observations do not obey the proposed fixed-rate relationship.
\(\textbf{Interpreting the Chemistry:}\)
Metallic radius generally increases down group 1 because additional occupied principal shells are introduced.
First ionization enthalpy generally decreases because the outer electron experiences greater shielding and lies farther from the nucleus.
However, the two properties are not connected by an exact universal linear relation.
Effective nuclear attraction, orbital penetration and electronic structure also influence the ionization energy.
The data support the qualitative periodic trends without supporting the student's additional mathematical assumption.
\(\textbf{Final answer:}\)
\[
\boxed{R_{\mathrm{Li\rightarrow Na}}\approx0.71,\quad R_{\mathrm{Na\rightarrow K}}\approx1.88\ \mathrm{kJ\,mol^{-1}\,pm^{-1}}}
\]
319. The effective ionic radius of \(\mathrm{Mg^{2+}}\) depends on its coordination environment. A simplified structural model uses the following values:
| Coordination number of Mg²⁺ | Effective Mg²⁺ radius (pm) |
|---|
| 4 | 57 |
| 6 | 72 |
For this model, assume that the effective radius of the neighbouring oxide ion is \(140\ \mathrm{pm}\) in both environments and that the Mg–O separation equals the sum of the supplied ionic radii.
What Mg–O separations are predicted, and what conclusion follows concerning the ionic radius of magnesium?
ⓐ. \(197\ \mathrm{pm}\) and \(212\ \mathrm{pm}\); the two environments prove that magnesium belongs to different periodic groups in different crystals.
ⓑ. \(197\ \mathrm{pm}\) and \(212\ \mathrm{pm}\); magnesium must gain two electrons when its coordination number changes from four to six.
ⓒ. \(83\ \mathrm{pm}\) and \(68\ \mathrm{pm}\); increasing coordination number necessarily shortens the Mg–O separation in the stated model.
ⓓ. \(197\ \mathrm{pm}\) and \(212\ \mathrm{pm}\); the effective magnesium radius changes with coordination environment although the ion retains the same charge and electron count.
Correct Answer: \(197\ \mathrm{pm}\) and \(212\ \mathrm{pm}\); the effective magnesium radius changes with coordination environment although the ion retains the same charge and electron count.
Explanation: \(\textbf{Identifying the model relation:}\)
The question explicitly adopts an additive effective-radius approximation.
For either structural environment:
\[
d_{\mathrm{Mg-O}}=r_{\mathrm{Mg^{2+}}}+r_{\mathrm{O^{2-}}}
\]
The oxide-ion radius is held constant by the stated model.
Only the effective magnesium-ion radius changes.
\(\textbf{Four-coordinate environment:}\)
The supplied magnesium radius is:
\[
r_{\mathrm{Mg^{2+},CN4}}=57\ \mathrm{pm}
\]
Therefore:
\[
d_4=57+140
\]
\[
d_4=197\ \mathrm{pm}
\]
\(\textbf{Six-coordinate environment:}\)
The supplied magnesium radius is:
\[
r_{\mathrm{Mg^{2+},CN6}}=72\ \mathrm{pm}
\]
Hence:
\[
d_6=72+140
\]
\[
d_6=212\ \mathrm{pm}
\]
\(\textbf{Comparing the predicted separations:}\)
\[
\Delta d=d_6-d_4
\]
\[
\Delta d=212-197
\]
\[
\Delta d=15\ \mathrm{pm}
\]
The six-coordinate environment has the greater predicted separation under this model.
\(\textbf{Interpreting the ionic-radius convention:}\)
Magnesium retains atomic number 12 in both environments.
Its divalent ion retains the electronic configuration \([\mathrm{Ne}]\).
Changing coordination number does not require changing the ion's nuclear charge or electron count.
An effective ionic radius is obtained from structural measurements using an adopted radius convention.
Its tabulated value can therefore depend on coordination environment.
\(\textbf{Final interpretation:}\)
The numerical difference should not be interpreted as a change in the elemental identity of magnesium.
Nor does it mean that the ion possesses two different sharp electronic boundaries.
The result illustrates why coordination-number information matters when applying ionic-radius tables.
\(\textbf{Final answer:}\)
\[
\boxed{d_4=197\ \mathrm{pm},\qquad d_6=212\ \mathrm{pm},\qquad \Delta d=15\ \mathrm{pm}}
\]
320. Fluorine and chlorine occupy consecutive periods of group 17. The following approximate data are supplied:
| Property | F | Cl |
|---|
| Covalent atomic radius (pm) | 64 | 99 |
| Effective monovalent anion radius (pm) | 133 | 181 |
| First electron-gain enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) | \(-328\) | \(-349\) |
Which conclusion correctly reconciles the radius relationships with the first electron-gain enthalpies?
ⓐ. Fluorine has the smaller atom and anion, so it must have the more negative first electron-gain enthalpy.
ⓑ. Chlorine has the more exothermic first electron gain despite fluorine's smaller size because the compact \(2p\) region offers greater electron-pairing repulsion.
ⓒ. Fluorine and chlorine form equally sized anions because both ions possess completed p-subshells.
ⓓ. Chlorine has a more negative first electron-gain enthalpy because its nuclear charge is smaller than fluorine's.
Correct Answer: Chlorine has the more exothermic first electron gain despite fluorine's smaller size because the compact \(2p\) region offers greater electron-pairing repulsion.
Explanation: \(\textbf{Comparing the neutral atoms:}\)
Fluorine belongs to period 2, whereas chlorine belongs to period 3.
Chlorine possesses an additional occupied principal shell.
Its greater covalent radius is therefore consistent with the ordinary down-group size trend.
\(\textbf{Comparing the anions:}\)
Fluoride has the completed configuration \(1s^2\,2s^2\,2p^6\).
Chloride has the completed configuration \([\mathrm{Ne}]\,3s^2\,3p^6\).
The chloride ion retains an additional occupied principal shell.
It is therefore larger under the supplied comparable effective ionic-radius convention.
\(\textbf{Interpreting electron-gain enthalpy:}\)
The first electron-gain enthalpies describe addition of an electron to neutral gaseous atoms.
Both values are negative, so both processes are exothermic.
The chlorine value is more negative, indicating greater energy release.
This observation is not inconsistent with chlorine's larger size.
\(\textbf{Explaining the second-period exception:}\)
An added electron enters fluorine's compact \(2p\) region.
The resulting electron-electron crowding is substantial.
In chlorine, the added electron enters a more extended \(3p\) region.
The comparatively reduced crowding contributes to chlorine's more exothermic first electron addition.
\(\textbf{Integrating the results:}\)
Atomic radius and ionic radius describe the spatial extent of electronic distributions under specified conventions.
Electron-gain enthalpy measures an energy change between two specified electronic states.
A smaller radius does not guarantee a more negative electron-gain enthalpy.
The correct interpretation requires accounting for both nuclear attraction and electron-electron repulsion.
The second option correctly reconciles all three supplied properties.
