101. Consider isolated gaseous \(\mathrm{Mn^{2+}}\), \(\mathrm{Mn^{3+}}\), \(\mathrm{Ni^{2+}}\) and \(\mathrm{Ni^{3+}}\) ions in their ground states. Manganese and nickel have neutral configurations \([\mathrm{Ar}]\,3d^5\,4s^2\) and \([\mathrm{Ar}]\,3d^8\,4s^2\), respectively. Which statement correctly compares the change in the number of unpaired electrons when each divalent ion loses one more electron?
ⓐ. Both ions gain one unpaired electron because the electron count decreases by one.
ⓑ. Both ions lose one unpaired electron because positive charge increases.
ⓒ. Manganese gains one unpaired electron, whereas nickel loses one unpaired electron.
ⓓ. Manganese loses one unpaired electron, whereas nickel gains one unpaired electron.
Correct Answer: Manganese loses one unpaired electron, whereas nickel gains one unpaired electron.
Explanation: \(\textbf{Constructing the manganese configurations:}\)
Neutral manganese has configuration \([\mathrm{Ar}]\,3d^5\,4s^2\).
Its divalent ion forms by loss of the two \(4s\) electrons.
Therefore:
\[
\mathrm{Mn^{2+}}:[\mathrm{Ar}]\,3d^5
\]
In the isolated-ion ground state, the five d-electrons occupy separate orbitals.
Thus, \(\mathrm{Mn^{2+}}\) has five unpaired electrons.
Further removal of one electron produces:
\[
\mathrm{Mn^{3+}}:[\mathrm{Ar}]\,3d^4
\]
The four remaining d-electrons occupy four separate orbitals.
Consequently, \(\mathrm{Mn^{3+}}\) has four unpaired electrons.
\(\textbf{Examining nickel:}\)
Neutral nickel has configuration \([\mathrm{Ar}]\,3d^8\,4s^2\).
Its divalent ion therefore has configuration:
\[
\mathrm{Ni^{2+}}:[\mathrm{Ar}]\,3d^8
\]
In a \(d^8\) free-ion arrangement, five electrons initially occupy separate orbitals.
The remaining three electrons pair in three of those orbitals.
Two electrons therefore remain unpaired.
Removal of one more electron gives:
\[
\mathrm{Ni^{3+}}:[\mathrm{Ar}]\,3d^7
\]
A \(d^7\) free-ion configuration contains three unpaired electrons.
\(\textbf{Comparing the changes:}\)
For manganese:
\[
5\longrightarrow4
\]
The unpaired-electron count decreases by one.
For nickel:
\[
2\longrightarrow3
\]
The unpaired-electron count increases by one.
\(\textbf{Final inference:}\)
Electron removal does not have one universal effect on the number of unpaired electrons.
Its effect depends on whether the removed electron was previously paired or unpaired.
The specified isolated gaseous ions avoid complications associated with ligand-induced orbital splitting in complexes.
Manganese loses one unpaired electron, whereas nickel gains one when the oxidation state changes from \(+2\) to \(+3\).
102. Chromium, molybdenum and tungsten occupy group 6 in periods 4, 5 and 6. Their atomic numbers are 24, 42 and 74, respectively. Which explanation correctly accounts for the different atomic-number intervals between successive members?
ⓐ. The interval from Cr to Mo includes an f-block region, whereas Mo to W contains only the ten-element d-block region.
ⓑ. Both intervals are eighteen because all three elements occupy the same vertical group.
ⓒ. The intervals are eighteen and thirty-two, respectively; the larger second interval reflects the additional fourteen-element f-filling contribution.
ⓓ. The intervals are eighteen and thirty-two, respectively; the larger second interval proves that tungsten belongs to the f-block.
Correct Answer: The intervals are eighteen and thirty-two, respectively; the larger second interval reflects the additional fourteen-element f-filling contribution.
Explanation: \(\textbf{Determining the atomic-number intervals:}\)
Chromium has atomic number 24, while molybdenum has atomic number 42.
Their separation is:
\[
\Delta Z_1=42-24
\]
\[
\Delta Z_1=18
\]
Molybdenum has atomic number 42, while tungsten has atomic number 74.
Their separation is:
\[
\Delta Z_2=74-42
\]
\[
\Delta Z_2=32
\]
Therefore, the second interval exceeds the first by:
\[
\Delta Z_2-\Delta Z_1=32-18
\]
\[
\Delta Z_2-\Delta Z_1=14
\]
\(\textbf{Explaining the electronic origin:}\)
Chromium belongs to the fourth-period \(3d\) series.
Molybdenum belongs to the fifth-period \(4d\) series.
The fourth and fifth periods each contain eighteen elements.
Accordingly, corresponding group positions in these consecutive periods are separated by eighteen atomic-number units.
Tungsten belongs to the sixth-period \(5d\) series.
The sixth period contains an additional fourteen-position region associated with inner \(4f\) filling.
Its total length is thirty-two elements.
\(\textbf{Interpreting common group membership:}\)
All three elements occupy group 6 despite the unequal atomic-number intervals.
Their common group placement reflects related electronic structures and chemical behaviour.
The additional f-filling region preceding tungsten does not classify tungsten itself as an f-block element.
Tungsten remains a d-block element of group 6.
\(\textbf{Final inference:}\)
The unequal intervals demonstrate that periodic recurrence is not equivalent to a constant difference between successive atomic numbers.
The length of the intervening period determines the separation between corresponding group positions.
The additional fourteen positions in the sixth-period sequence explain the increase from eighteen to thirty-two.
103. Ytterbium and hafnium have atomic numbers 70 and 72, respectively. Their neutral ground-state configurations are:
\[
\mathrm{Yb}:[\mathrm{Xe}]\,4f^{14}\,6s^2
\]
\[
\mathrm{Hf}:[\mathrm{Xe}]\,4f^{14}\,5d^2\,6s^2
\]
Which statement correctly compares the configurations of \(\mathrm{Yb^{2+}}\) and \(\mathrm{Hf^{4+}}\) and their parent elements' block classifications?
ⓐ. The ions are isoelectronic; ytterbium belongs to the f-block, whereas hafnium belongs to the d-block.
ⓑ. The ions are isoelectronic; both parent elements belong to the f-block because their ions contain \(4f^{14}\).
ⓒ. The ions have different electron counts; ytterbium belongs to the f-block, whereas hafnium belongs to the d-block.
ⓓ. The ions have different electron counts; both parent elements belong to the d-block because their outermost shells contain \(6s^2\).
Correct Answer: The ions are isoelectronic; ytterbium belongs to the f-block, whereas hafnium belongs to the d-block.
Explanation: \(\textbf{Determining the ytterbium ion:}\)
Neutral ytterbium has the configuration \([\mathrm{Xe}]\,4f^{14}\,6s^2\).
Removal of its two \(6s\) electrons forms \(\mathrm{Yb^{2+}}\).
The resulting configuration is:
\[
\mathrm{Yb^{2+}}:[\mathrm{Xe}]\,4f^{14}
\]
Its total electron count is:
\[
N_e=70-2=68
\]
\(\textbf{Determining the hafnium ion:}\)
Neutral hafnium contains two \(6s\) electrons and two \(5d\) electrons beyond the filled \(4f\) subshell.
Formation of \(\mathrm{Hf^{4+}}\) removes these four electrons.
The resulting configuration is:
\[
\mathrm{Hf^{4+}}:[\mathrm{Xe}]\,4f^{14}
\]
Its total electron count is:
\[
N_e=72-4=68
\]
Thus, both ions have the same electron count and electronic configuration.
\(\textbf{Distinguishing the parent elements:}\)
Ytterbium belongs to the lanthanoid region associated with \(4f\) occupation.
It is classified as an f-block element.
Hafnium follows the lanthanoid filling region and belongs to the sixth-period \(5d\) series.
It is therefore classified as a d-block element.
\(\textbf{Chemical interpretation:}\)
Both resulting ions contain a completely filled \(4f^{14}\) subshell.
However, block classification is assigned to the parent element according to its periodic position and characteristic differentiating electronic structure.
A filled inner f-subshell can occur in ions originating from different blocks.
Isoelectronicity also does not imply equal nuclear charge or identical chemical behaviour.
The ions are therefore isoelectronic even though their parent elements belong to different blocks.
104. Carbon, silicon, germanium, tin and lead belong to group 14. Their conventional broad classifications and characteristic oxide behaviour change considerably down the group. Which sequence correctly represents their elemental classifications in the order given?
ⓐ. Non-metal → Metalloid → Metal → Metalloid → Metal
ⓑ. Non-metal → Non-metal → Metalloid → Metal → Metal
ⓒ. Metalloid → Metalloid → Metalloid → Metal → Metal
ⓓ. Non-metal → Metalloid → Metalloid → Metal → Metal
Correct Answer: Non-metal → Metalloid → Metalloid → Metal → Metal
Explanation: \(\textbf{Establishing the common group pattern:}\)
All five elements belong to group 14.
Their characteristic valence-shell configuration is \(ns^2np^2\).
They therefore possess four electrons in their outermost shell.
However, their principal quantum numbers increase as one moves down the group.
\(\textbf{Classifying carbon and silicon:}\)
Carbon belongs to the second period and exhibits predominantly non-metallic behaviour.
Its relatively small size and strong tendency to form covalent bonds are characteristic of a non-metal.
Silicon belongs to the third period.
It displays intermediate properties and is conventionally classified as a metalloid.
\(\textbf{Examining germanium:}\)
Germanium belongs to the fourth period.
Like silicon, it exhibits properties intermediate between those of typical metals and non-metals.
It is therefore also classified as a metalloid.
\(\textbf{Examining tin and lead:}\)
Tin and lead belong to periods 5 and 6, respectively.
Both are conventionally classified as metals.
Their metallic character is consistent with the general increase in electropositive behaviour down a main-group family.
\(\textbf{Connecting the classifications with chemical properties:}\)
Carbon forms acidic oxides such as \(\mathrm{CO_2}\).
Silicon dioxide is also acidic, whereas oxides of heavier group-14 elements can exhibit amphoteric character.
The variation in oxide behaviour reflects the changing chemical character within the group.
Nevertheless, all five elements retain the characteristic group-14 valence-electron pattern.
\(\textbf{Final inference:}\)
Common group placement does not imply identical metallic character.
The progression from non-metal through metalloid to metal is an important manifestation of the systematic change in properties down the p-block.
The correct sequence is non-metal, metalloid, metalloid, metal and metal.
105. In the sixth period, the two initial s-block positions are followed by an inner-transition region, a d-block region and six terminal p-block positions. Which statement correctly distinguishes the number of representative-element positions from the total number of elements in this period?
ⓐ. There are eight representative-element positions and thirty-two total positions; the remaining twenty-four arise from f- and d-block regions.
ⓑ. There are eighteen representative-element positions and thirty-two total positions; the remaining fourteen arise entirely from the f-block.
ⓒ. There are eight representative-element positions and eighteen total positions; the remaining ten arise from the d-block.
ⓓ. There are eighteen representative-element positions and thirty-two total positions; the remaining fourteen arise entirely from the d-block.
Correct Answer: There are eight representative-element positions and thirty-two total positions; the remaining twenty-four arise from f- and d-block regions.
Explanation: \(\textbf{Identifying representative elements:}\)
Representative or main-group elements occupy groups 1, 2 and 13–18.
These positions correspond to the characteristic s- and p-block regions of the periodic table.
The sixth period begins with two s-block elements.
It ends with six p-block elements.
Therefore, the number of representative-element positions is:
\[
N_{\mathrm{main}}=2+6
\]
\[
N_{\mathrm{main}}=8
\]
\(\textbf{Accounting for the intervening elements:}\)
The sixth period contains a fourteen-position contribution associated with inner \(4f\) filling.
It also contains ten positions associated with the \(5d\) series.
Together, these contributions account for:
\[
N_{\mathrm{inner}}=14+10
\]
\[
N_{\mathrm{inner}}=24
\]
\(\textbf{Determining the total period length:}\)
The complete period therefore contains:
\[
N_{\mathrm{total}}=N_{\mathrm{main}}+N_{\mathrm{inner}}
\]
\[
N_{\mathrm{total}}=8+24
\]
\[
N_{\mathrm{total}}=32
\]
\(\textbf{Interpreting the group arrangement:}\)
The modern periodic table retains eighteen numbered groups.
The additional fourteen f-filling positions do not create fourteen new conventional group numbers.
They belong within the sixth-period sequence but are displayed separately in the compact table.
The ten d-block positions correspond to the central group region.
\(\textbf{Final inference:}\)
The distinction between representative-element count and total period length is essential.
A period containing thirty-two elements does not require thirty-two numbered groups.
For period 6, eight positions belong to the representative regions, while twenty-four correspond to the d- and f-block contributions.
106. Beryllium and aluminium belong to different groups and periods, yet their oxides \(\mathrm{BeO}\) and \(\mathrm{Al_2O_3}\) are amphoteric. Magnesium belongs to the same group as beryllium, but \(\mathrm{MgO}\) is basic. Which statement correctly explains why these observations do not contradict periodic classification?
ⓐ. Elements in the same group must have identical oxide behaviour, so magnesium is incorrectly placed in group 2.
ⓑ. Similar oxide character can occur across different groups, while small size and polarising effects can produce distinctive behaviour in the first member of a group.
ⓒ. Beryllium and aluminium must belong to the same group because amphoteric oxides necessarily indicate an identical valence-electron count.
ⓓ. The Modern Periodic Law applies only to electronic configurations and excludes chemical properties such as oxide acidity and basicity.
Correct Answer: Similar oxide character can occur across different groups, while small size and polarising effects can produce distinctive behaviour in the first member of a group.
Explanation: \(\textbf{Establishing the periodic positions:}\)
Beryllium belongs to group 2 and period 2.
Magnesium also belongs to group 2 but occupies period 3.
Aluminium belongs to group 13 and period 3.
Thus, beryllium and magnesium have similar characteristic valence configurations, whereas aluminium belongs to a different group.
\(\textbf{Examining beryllium's behaviour:}\)
Beryllium is unusually small compared with heavier group-2 elements.
Its divalent cation has strong polarising influence.
This contributes to distinctive bonding and chemical behaviour.
Beryllium oxide is amphoteric and can react with both acids and suitable strong bases.
\(\textbf{Comparing magnesium:}\)
Magnesium has a larger atomic and cationic size than beryllium.
Its oxide exhibits predominantly basic character.
This is consistent with the more typical metallic-oxide behaviour of magnesium.
\(\textbf{Examining aluminium:}\)
Aluminium belongs to a different periodic group and possesses three valence electrons.
Its oxide \(\mathrm{Al_2O_3}\) is also amphoteric.
Thus, two elements from different groups can form oxides exhibiting the same broad acid-base category.
\(\textbf{Resolving the classification issue:}\)
Periodic classification is based on atomic number and electronic configuration.
It predicts recurring patterns and broad trends, not strict equality of every chemical property.
Chemical behaviour also depends on atomic size, charge density and the characteristics of the resulting compounds.
The amphoteric behaviour of beryllium and aluminium does not imply identical group membership.
Likewise, the difference between beryllium oxide and magnesium oxide does not invalidate their common group placement.
The observations are consistent with periodic trends and first-member qualifications.
107. Chromium and manganese exhibit maximum common oxidation states of \(+6\) and \(+7\), respectively, in suitable oxygen-containing compounds. Their neutral configurations are \([\mathrm{Ar}]\,3d^5\,4s^1\) and \([\mathrm{Ar}]\,3d^5\,4s^2\). Which statement correctly connects their group positions with their formal d-electron populations in these high oxidation states?
ⓐ. Chromium belongs to group 5 and manganese to group 6; both retain \(d^1\) populations in their highest common oxidation states.
ⓑ. Chromium belongs to group 6 and manganese to group 7; both retain \(d^1\) populations in their highest common oxidation states.
ⓒ. Chromium belongs to group 6 and manganese to group 7; their respective \(+6\) and \(+7\) states correspond formally to \(d^0\).
ⓓ. Chromium belongs to group 7 and manganese to group 6; their respective \(+6\) and \(+7\) states correspond formally to \(d^0\).
Correct Answer: Chromium belongs to group 6 and manganese to group 7; their respective \(+6\) and \(+7\) states correspond formally to \(d^0\).
Explanation: \(\textbf{Identifying the neutral electronic populations:}\)
Chromium has five \(3d\) electrons and one \(4s\) electron.
The relevant electron total is:
\[
5+1=6
\]
Therefore, chromium belongs to group 6.
Manganese has five \(3d\) electrons and two \(4s\) electrons.
Its corresponding total is:
\[
5+2=7
\]
Therefore, manganese belongs to group 7.
\(\textbf{Interpreting chromium(VI):}\)
Chromium has six electrons beyond the argon core.
In the formal \(+6\) oxidation state, these six electrons are accounted for as removed in the ionic electron-count model.
The corresponding d-electron count is zero.
Chromium(VI) therefore has the formal \(d^0\) configuration.
\(\textbf{Interpreting manganese(VII):}\)
Manganese has seven electrons beyond the argon core.
In the formal \(+7\) oxidation state, all seven are accounted for in oxidation-state electron bookkeeping.
The corresponding d-electron count is also zero.
Manganese(VII) therefore has the formal \(d^0\) configuration.
\(\textbf{Important chemical distinction:}\)
A formal \(d^0\) assignment does not mean that these oxygen-containing compounds contain free gaseous \(\mathrm{Cr^{6+}}\) or \(\mathrm{Mn^{7+}}\) ions.
The metal–oxygen bonds possess substantial chemical bonding character that cannot be represented solely by separated ions.
Nevertheless, formal oxidation states provide a useful method for comparing electron counts.
\(\textbf{Final inference:}\)
Two neighbouring transition elements can attain different high oxidation states while reaching the same formal d-electron count.
Their group numbers reflect the respective numbers of relevant valence electrons in the neutral atoms.
Thus, chromium(VI) and manganese(VII) are formally \(d^0\) despite belonging to different groups.
108. Two elements \(P\) and \(Q\) are chosen from the third-period p-block. The first has a characteristic outer configuration \(3s^2\,3p^3\), while the second has \(3s^2\,3p^5\). Under suitable chemical conditions, both form covalent chlorides. Which statement correctly compares their group memberships and the electron-pair arrangements of their common trichlorides, \(\mathrm{PCl_3}\) and \(\mathrm{QCl_3}\), where \(P\) and \(Q\) denote the central elements?
ⓐ. Both belong to group 15 and have one lone pair on the central atom in their trichlorides.
ⓑ. Both belong to group 17 and have two lone pairs on the central atom in their trichlorides.
ⓒ. \(P\) belongs to group 15 and \(Q\) to group 17; both central atoms have one lone pair in their trichlorides.
ⓓ. \(P\) belongs to group 15 and \(Q\) to group 17; their central atoms have one and two lone pairs, respectively.
Correct Answer: \(P\) belongs to group 15 and \(Q\) to group 17; their central atoms have one and two lone pairs, respectively.
Explanation: \(\textbf{Identifying the first element:}\)
The configuration \(3s^2\,3p^3\) corresponds to phosphorus.
Phosphorus possesses five valence electrons and belongs to group 15.
In phosphorus trichloride, \(\mathrm{PCl_3}\), the phosphorus atom forms three P–Cl single bonds.
Each bond accounts for one electron contributed by phosphorus in the conventional Lewis representation.
The two remaining valence electrons form one lone pair on phosphorus.
\(\textbf{Identifying the second element:}\)
The configuration \(3s^2\,3p^5\) corresponds to chlorine.
Chlorine possesses seven valence electrons and belongs to group 17.
The corresponding compound \(\mathrm{ClCl_3}\) is not a chemically distinct stable trichloride of chlorine.
A neutral molecule containing one chlorine atom bonded to three other chlorine atoms cannot be treated as a conventional stable molecular chlorine trichloride with the assumed Lewis arrangement.
\(\textbf{Consequent chemical limitation:}\)
Although the hypothetical Lewis electron bookkeeping would leave two lone pairs on a seven-valence-electron central atom forming three single bonds, that calculation does not establish the existence of the proposed molecular species.
The original chemical premise therefore fails for \(Q=\mathrm{Cl}\).
A valid comparison must distinguish formal electron-pair counting from the actual chemical existence and stability of the compound.
Accordingly, the stated trichloride comparison cannot be used as a defensible single-correct NEET question without changing the second compound to a chemically established species.
This proposed item is not accepted into the bank.
109. Boron belongs to group 13 and forms boron trifluoride, \(\mathrm{BF_3}\). The molecule accepts a fluoride ion to form \(\mathrm{BF_4^-}\).
Which statement correctly compares the electron-pair arrangements around boron before and after this change?
ⓐ. Boron has an incomplete octet in both species because it possesses only three valence electrons in its neutral atom.
ⓑ. Boron has a complete octet in \(\mathrm{BF_3}\), but loses one electron pair when \(\mathrm{BF_4^-}\) forms.
ⓒ. Boron has six electrons around it in \(\mathrm{BF_3}\) and eight in \(\mathrm{BF_4^-}\), without requiring participation of valence d-orbitals.
ⓓ. Boron has six electrons around it in \(\mathrm{BF_3}\) and ten in \(\mathrm{BF_4^-}\), because fluoride donation expands its valence shell.
Correct Answer: Boron has six electrons around it in \(\mathrm{BF_3}\) and eight in \(\mathrm{BF_4^-}\), without requiring participation of valence d-orbitals.
Explanation: \(\textbf{Establishing boron's periodic position:}\)
Boron has atomic number 5 and the electronic configuration \(1s^2\,2s^2\,2p^1\).
It belongs to group 13 and possesses three valence electrons.
Because it is a second-period element, its available valence orbitals are \(2s\) and \(2p\).
\(\textbf{Examining the original molecule:}\)
In \(\mathrm{BF_3}\), boron forms three ordinary B–F sigma bonds.
The three shared electron pairs account for six electrons around the boron centre.
Boron therefore has an incomplete octet in the conventional Lewis representation of this molecule.
This electron deficiency allows \(\mathrm{BF_3}\) to behave as a Lewis acid.
\(\textbf{Considering fluoride donation:}\)
A fluoride ion possesses lone pairs that can be donated to an electron-deficient centre.
The reaction is:
\[
\mathrm{BF_3+F^-\rightarrow BF_4^-}
\]
The donated pair establishes an additional B–F bond.
The resulting tetrafluoroborate ion has four shared electron pairs around boron.
Thus, boron is surrounded by eight electrons in the product.
\(\textbf{Checking the valence-shell limitation:}\)
Formation of \(\mathrm{BF_4^-}\) does not require boron to accommodate more than eight valence-shell electrons.
Nor does it require a valence d-orbital, which is unavailable in the second principal shell.
The conversion illustrates how an electron-deficient p-block compound can accept an electron pair while remaining consistent with the octet limitation of a second-period element.
The correct comparison is six electrons in \(\mathrm{BF_3}\) and eight in \(\mathrm{BF_4^-}\).
110. Under the usual conditions used to prepare their characteristic products by direct reaction with excess dry oxygen, lithium, sodium and potassium predominantly form \(\mathrm{Li_2O}\), \(\mathrm{Na_2O_2}\) and \(\mathrm{KO_2}\), respectively.
Which statement correctly identifies the oxygen-containing ions and the formal oxidation states of oxygen in these compounds?
ⓐ. \(\mathrm{Li_2O}\): oxide, \(-2\); \(\mathrm{Na_2O_2}\): peroxide, \(-1\); \(\mathrm{KO_2}\): superoxide, \(-\frac12\)
ⓑ. \(\mathrm{Li_2O}\): peroxide, \(-1\); \(\mathrm{Na_2O_2}\): oxide, \(-2\); \(\mathrm{KO_2}\): superoxide, \(-\frac12\)
ⓒ. \(\mathrm{Li_2O}\): oxide, \(-2\); \(\mathrm{Na_2O_2}\): peroxide, \(-1\); \(\mathrm{KO_2}\): peroxide, \(-1\)
ⓓ. \(\mathrm{Li_2O}\): oxide, \(-2\); \(\mathrm{Na_2O_2}\): superoxide, \(-\frac12\); \(\mathrm{KO_2}\): peroxide, \(-1\)
Correct Answer: \(\mathrm{Li_2O}\): oxide, \(-2\); \(\mathrm{Na_2O_2}\): peroxide, \(-1\); \(\mathrm{KO_2}\): superoxide, \(-\frac12\)
Explanation: \(\textbf{Establishing the common metal oxidation state:}\)
Lithium, sodium and potassium belong to group 1.
Their characteristic outer configuration is \(ns^1\), and they commonly form monovalent positive ions.
Thus, the oxidation state of each metal in the specified compounds is \(+1\).
\(\textbf{Examining lithium oxide:}\)
The formula \(\mathrm{Li_2O}\) contains two lithium atoms for one oxygen atom.
Let the oxidation state of oxygen be \(x\).
Charge balance gives:
\[
2(+1)+x=0
\]
\[
x=-2
\]
The oxygen species is therefore the oxide ion, \(\mathrm{O^{2-}}\).
\(\textbf{Examining sodium peroxide:}\)
In \(\mathrm{Na_2O_2}\), the two sodium ions contribute a total charge of \(+2\).
The oxygen-containing unit must therefore carry charge \(2-\).
It is the peroxide ion, \(\mathrm{O_2^{2-}}\).
If the oxidation state of each oxygen atom is \(y\):
\[
2(+1)+2y=0
\]
\[
y=-1
\]
\(\textbf{Examining potassium superoxide:}\)
In \(\mathrm{KO_2}\), one potassium ion contributes charge \(+1\).
The oxygen-containing unit must consequently carry charge \(1-\).
It is the superoxide ion, \(\mathrm{O_2^-}\).
For the average oxidation state of oxygen:
\[
(+1)+2z=0
\]
\[
z=-\frac12
\]
\(\textbf{Periodic interpretation:}\)
All three metals retain their characteristic \(+1\) oxidation state.
However, they need not form oxygen compounds containing the same oxygen anion.
The product identity depends on the chemical conditions and the relative stability of oxide, peroxide and superoxide species.
Therefore, the simple rule that every group-1 metal forms only \(M_2O\) is incomplete.
The specified products correctly correspond to oxide, peroxide and superoxide, respectively.
111. Consider two pairs of compounds, \(\mathrm{LiF}\) and \(\mathrm{LiI}\), and \(\mathrm{MgF_2}\) and \(\mathrm{MgI_2}\).
Within each pair, the metal cation is unchanged, but the halide anion changes from fluoride to iodide.
Which comparison is consistent with the relative polarizabilities of the halide ions?
ⓐ. \(\mathrm{LiF}\) is more covalent than \(\mathrm{LiI}\), while \(\mathrm{MgI_2}\) is more covalent than \(\mathrm{MgF_2}\).
ⓑ. \(\mathrm{LiI}\) is more covalent than \(\mathrm{LiF}\), while \(\mathrm{MgF_2}\) is more covalent than \(\mathrm{MgI_2}\).
ⓒ. \(\mathrm{LiI}\) is more covalent than \(\mathrm{LiF}\), and \(\mathrm{MgI_2}\) is more covalent than \(\mathrm{MgF_2}\).
ⓓ. \(\mathrm{LiF}\) is more covalent than \(\mathrm{LiI}\), and \(\mathrm{MgF_2}\) is more covalent than \(\mathrm{MgI_2}\).
Correct Answer: \(\mathrm{LiI}\) is more covalent than \(\mathrm{LiF}\), and \(\mathrm{MgI_2}\) is more covalent than \(\mathrm{MgF_2}\).
Explanation: \(\textbf{Identifying the controlled variable:}\)
Within the lithium pair, the metal cation is \(\mathrm{Li^+}\) in both compounds.
Within the magnesium pair, the metal cation is \(\mathrm{Mg^{2+}}\) in both compounds.
Consequently, the decisive comparison within each pair concerns the nature of the halide anion.
\(\textbf{Comparing the halide ions:}\)
Fluoride is a small anion whose electron cloud is relatively compact.
Iodide is considerably larger and contains more occupied electron shells.
Its electron cloud is more readily distorted by the electric field of a neighbouring cation.
Iodide is therefore more polarizable than fluoride.
\(\textbf{Applying polarisation reasoning:}\)
Polarisation of an anion increases the covalent contribution to bonding in a nominally ionic compound.
For the same lithium cation, iodide is distorted more readily than fluoride.
Therefore, \(\mathrm{LiI}\) has greater covalent character than \(\mathrm{LiF}\).
The same reasoning applies to the magnesium compounds.
For the same magnesium cation, iodide is more readily polarized than fluoride.
Consequently, \(\mathrm{MgI_2}\) has greater covalent character than \(\mathrm{MgF_2}\).
\(\textbf{Distinguishing anion and cation effects:}\)
The polarising power of the cation and the polarizability of the anion are related but distinct factors.
Earlier comparisons between smaller and larger metal cations concern changes in cationic polarising power.
Here, the cation is fixed within each comparison, so anion polarizability is the changing factor.
The periodic increase in halide-ion size down group 17 therefore explains both comparisons.
A more covalent compound may still possess a substantial ionic contribution to its bonding.
The correct option identifies greater covalent character for the iodides in both pairs.
112. Three fourth-period elements have the following ground-state configurations and characteristic ions:
| Element | Neutral configuration beyond [Ar] | Ion |
|---|
| X | \(3d^5\,4s^1\) | \(X^{3+}\) |
| Y | \(3d^5\,4s^2\) | \(Y^{2+}\) |
| Z | \(3d^{10}\,4s^2\) | \(Z^{2+}\) |
A student classifies an element as a transition element if its atom or at least one common ion has an incompletely filled d-subshell. Which conclusion is correct?
ⓐ. X and Z qualify, because both possess d-electrons in their neutral atoms.
ⓑ. All three qualify, because each occupies the d-block of the periodic table.
ⓒ. Y and Z qualify, because both neutral atoms contain two \(4s\) electrons.
ⓓ. X and Y qualify, not Z; their ion configurations are \(3d^3\), \(3d^5\) and \(3d^{10}\).
Correct Answer: X and Y qualify, not Z; their ion configurations are \(3d^3\), \(3d^5\) and \(3d^{10}\).
Explanation: \(\textbf{Identify the elements and their positions:}\)
The configuration \(3d^5\,4s^1\) beyond argon identifies chromium, which belongs to group 6.
The configuration \(3d^5\,4s^2\) identifies manganese in group 7.
The configuration \(3d^{10}\,4s^2\) identifies zinc in group 12.
All three are located in the fourth-period d-block, but location alone does not settle the transition-element criterion.
\(\textbf{Form the indicated cations:}\)
In forming cations of these elements, electrons are removed from \(4s\) before the occupied \(3d\) subshell.
Chromium(III) therefore has configuration \([\mathrm{Ar}]\,3d^3\).
Manganese(II) has configuration \([\mathrm{Ar}]\,3d^5\).
Zinc(II) has configuration \([\mathrm{Ar}]\,3d^{10}\).
\(\textbf{Apply the defining distinction:}\)
Chromium and manganese have incompletely filled d-subshells in the listed ions.
Zinc has a completed \(3d^{10}\) subshell in both its neutral atom and its characteristic divalent ion.
Thus zinc is a d-block element but does not meet the usual transition-element definition.
Counting outer \(4s\) electrons or merely locating an element in groups 3–12 would miss this distinction.
113. Three fourth-period atoms have their electron distributions beyond argon given as follows:
Atom P: \(3d^2\,4s^2\)
Atom Q: \(3d^7\,4s^2\)
Atom R: \(3d^{10}\,4s^1\).
All are neutral atoms in their ground states. For a d-block element, use the total of the \((n-1)d\) and \(ns\) electrons to identify its group in the first transition series.
Which statement correctly identifies the groups and the effect of forming the specified ions \(P^{2+}\), \(Q^{2+}\) and \(R^+\)?
ⓐ. Groups 4, 9 and 11; all three ions retain d-electrons, but only \(R^+\) has a completely filled \(3d\) subshell.
ⓑ. Groups 2, 2 and 1; the ions become \(3d^0\), \(3d^0\) and \(3d^{10}\), respectively.
ⓒ. Groups 4, 9 and 11; every ion loses all its \(3d\) electrons before losing its \(4s\) electrons.
ⓓ. Groups 3, 8 and 10; \(P^{2+}\) and \(Q^{2+}\) become isoelectronic with argon.
Correct Answer: Groups 4, 9 and 11; all three ions retain d-electrons, but only \(R^+\) has a completely filled \(3d\) subshell.
Explanation: \(\textbf{Locate the neutral atoms:}\)
The first configuration contains two \(3d\) and two \(4s\) electrons beyond argon.
It corresponds to titanium, which is in group 4.
The second contains seven \(3d\) and two \(4s\) electrons, corresponding to cobalt in group 9.
The third contains ten \(3d\) and one \(4s\) electron, corresponding to copper in group 11.
Their \(4s\) occupation cannot by itself establish the group number.
\(\textbf{Determine the configurations of the ions:}\)
For these elements, the \(4s\) electrons are removed before \(3d\) electrons during ordinary cation formation.
Consequently, \(P^{2+}\) has configuration \([\mathrm{Ar}]\,3d^2\).
The ion \(Q^{2+}\) has configuration \([\mathrm{Ar}]\,3d^7\).
The ion \(R^+\) has configuration \([\mathrm{Ar}]\,3d^{10}\).
All three still contain electrons beyond the argon core.
Only the copper(I)-type ion has a completed d-subshell.
Thus the correct answer combines periodic placement with the proper order of electron removal.
114. The following illustrative ionic-radius data are given for trivalent lanthanoid ions measured using the same coordination-number convention.
| Ion | Atomic number of parent element | Ionic radius (pm) |
|---|
| \(\mathrm{La^{3+}}\) | 57 | 103 |
| \(\mathrm{Gd^{3+}}\) | 64 | 94 |
| \(\mathrm{Lu^{3+}}\) | 71 | 86 |
Which conclusion best reconciles the decreasing radii with the considerable chemical similarities of the trivalent lanthanoid ions?
ⓐ. The ions become smaller because their nuclear charges decrease, while their identical masses produce chemical similarities.
ⓑ. The ions become smaller because f-electrons shield nuclear charge completely, while their different oxidation states produce chemical similarities.
ⓒ. The ions become smaller because their valence shells lose a principal quantum number at each successive element, while their identical electron counts produce chemical similarities.
ⓓ. The ions become smaller because increasing nuclear charge is incompletely shielded by added f-electrons, while their common trivalent state contributes to related chemical behaviour.
Correct Answer: The ions become smaller because increasing nuclear charge is incompletely shielded by added f-electrons, while their common trivalent state contributes to related chemical behaviour.
Explanation: \(\textbf{Identifying the common ionic charge:}\)
All three species are trivalent lanthanoid ions.
The formal positive charge is the same, although their nuclear charges and inner electronic configurations differ.
The lanthanoid region is associated primarily with progressive occupation of the inner \(4f\) subshell.
\(\textbf{Interpreting the radius data:}\)
The ionic radius decreases from 103 pm for \(\mathrm{La^{3+}}\) to 86 pm for \(\mathrm{Lu^{3+}}\).
The total decrease across the supplied series is:
\[
\Delta r=103-86
\]
\[
\Delta r=17\ \mathrm{pm}
\]
This contraction occurs while the atomic number increases from 57 to 71.
\(\textbf{Explaining the electronic origin:}\)
Increasing atomic number means increasing nuclear charge.
The additional electrons in this region occupy predominantly inner f-subshell states.
These f-electrons provide relatively poor shielding of the nuclear charge experienced by outer electrons.
Consequently, effective nuclear attraction increases across the series.
The electron cloud contracts, producing progressively smaller ions.
This systematic decrease is known as lanthanoid contraction.
\(\textbf{Explaining chemical similarity:}\)
The lanthanoids commonly form trivalent ions with related outer electronic arrangements.
They therefore exhibit broadly similar chemical behaviour in many compounds.
However, their ionic radii are not identical, and their chemical properties are not absolutely indistinguishable.
\(\textbf{Final interpretation:}\)
The shared \(+3\) oxidation state contributes to chemical similarity, while increasing effective nuclear attraction produces the observed size trend.
The small but systematic differences in ionic radius can also contribute to differences in chemical behaviour and separation characteristics.
The decreasing radii and related chemistry are therefore compatible consequences of their electronic structures.
115. Europium and gadolinium have atomic numbers 63 and 64, respectively. Their isolated gaseous ions \(\mathrm{Eu^{2+}}\) and \(\mathrm{Gd^{3+}}\) each possess the configuration \([\mathrm{Xe}]\,4f^7\).
Which statement correctly compares these ions and explains the significance of this configuration?
ⓐ. They contain different numbers of electrons, but both possess a completely filled f-subshell.
ⓑ. They are isoelectronic and diamagnetic because their seven f-electrons are paired.
ⓒ. They are isoelectronic and each possesses seven unpaired f-electrons; the half-filled arrangement contributes to the stability of these ionic configurations.
ⓓ. They are isoelectronic and each possesses three unpaired f-electrons because the f-subshell contains three orbitals.
Correct Answer: They are isoelectronic and each possesses seven unpaired f-electrons; the half-filled arrangement contributes to the stability of these ionic configurations.
Explanation: \(\textbf{Checking the electron counts:}\)
Europium has atomic number 63.
Its divalent ion contains:
\[
N_e(\mathrm{Eu^{2+}})=63-2
\]
\[
N_e(\mathrm{Eu^{2+}})=61
\]
Gadolinium has atomic number 64.
Its trivalent ion contains:
\[
N_e(\mathrm{Gd^{3+}})=64-3
\]
\[
N_e(\mathrm{Gd^{3+}})=61
\]
Therefore, the two ions are isoelectronic.
\(\textbf{Interpreting the f-electron configuration:}\)
The xenon core contains 54 electrons.
The remaining seven electrons occupy the \(4f\) subshell.
An f-subshell contains seven orbitals.
According to Hund's rule, the seven electrons occupy these orbitals singly with parallel spins before pairing begins.
Consequently, each specified isolated ion contains seven unpaired f-electrons.
\(\textbf{Examining the stability implication:}\)
The \(4f^7\) configuration is a half-filled f-subshell arrangement.
Half-filled subshells can receive additional stability from their electron distribution.
This contributes to the special stability associated with europium(II).
Gadolinium(III) also possesses the half-filled \(4f^7\) configuration.
\(\textbf{Distinguishing ionic stability from periodic identity:}\)
The two parent elements are consecutive members of the lanthanoid region.
Their ions have different charges but identical electron counts and f-orbital occupancies.
Isoelectronicity does not mean that the ions have equal nuclear charge or identical chemical properties.
The difference in ionic charge affects their interactions with surrounding ions and molecules.
Nevertheless, their shared half-filled configuration explains the specified unpaired-electron counts and provides a useful electronic stability connection.
The correct conclusion is that both ions possess seven unpaired f-electrons in a half-filled \(4f^7\) arrangement.
116. Carbon and silicon belong to group 14. Consider the formation of long chains containing only E–E single bonds, where E represents carbon or silicon.
For comparison, take the approximate bond enthalpies \(D(\mathrm{C-C})=348\ \mathrm{kJ\,mol^{-1}}\) and \(D(\mathrm{Si-Si})=226\ \mathrm{kJ\,mol^{-1}}\).
Which statement correctly interprets their different tendencies toward catenation?
ⓐ. Carbon exhibits greater catenation because its C–C single bonds are substantially stronger, despite carbon and silicon having the same number of valence electrons.
ⓑ. Silicon exhibits greater catenation because its larger atomic size makes Si–Si single bonds stronger than C–C bonds.
ⓒ. Carbon exhibits greater catenation because it belongs to a group containing more valence electrons than silicon.
ⓓ. Silicon exhibits greater catenation because its third-period position requires stronger element–element bonding than carbon's second-period position.
Correct Answer: Carbon exhibits greater catenation because its C–C single bonds are substantially stronger, despite carbon and silicon having the same number of valence electrons.
Explanation: \(\textbf{Establishing the common group pattern:}\)
Carbon has the valence configuration \(2s^2\,2p^2\).
Silicon has the valence configuration \(3s^2\,3p^2\).
Both belong to group 14 and possess four valence electrons.
Their common group placement explains important similarities in bonding capacity.
\(\textbf{Defining the relevant property:}\)
Catenation is the ability of an element to form bonds with atoms of the same element, producing chains or related structures.
The formation of long stable chains depends strongly on the strength of the element–element bonds.
When only ordinary E–E single bonds are considered, the supplied bond enthalpies provide a direct basis for comparison.
\(\textbf{Comparing the bond strengths:}\)
The carbon–carbon single bond has an approximate bond enthalpy of \(348\ \mathrm{kJ\,mol^{-1}}\).
The silicon–silicon single bond has the lower value of \(226\ \mathrm{kJ\,mol^{-1}}\).
The stronger C–C bond is less readily broken under comparable conditions.
This favours the persistence of extended carbon skeletons.
\(\textbf{Connecting the difference with atomic size:}\)
Carbon is smaller than silicon, allowing relatively effective orbital overlap in C–C bonding.
Silicon possesses larger valence orbitals, and Si–Si bonds are generally weaker.
Consequently, silicon shows a less extensive tendency toward stable self-linked chains than carbon.
\(\textbf{Final interpretation:}\)
The difference is not caused by different numbers of valence electrons.
It arises primarily from the different strengths of the bonds formed between like atoms.
This comparison is distinct from the separate question of pi-bond formation, although orbital size influences both phenomena.
Carbon's unusually extensive catenation is therefore an important example of the distinctive behaviour of the first member of a p-block group.
117. Magnesium and zinc both have two electrons in their outermost s-subshells:
\[
\mathrm{Mg}:[\mathrm{Ne}]\,3s^2
\]
\[
\mathrm{Zn}:[\mathrm{Ar}]\,3d^{10}\,4s^2
\]
Their normal oxides are \(\mathrm{MgO}\) and \(\mathrm{ZnO}\), respectively.
Which conclusion correctly combines their group assignments with the acid-base characters of these oxides?
ⓐ. Both belong to group 2; \(\mathrm{MgO}\) is basic and \(\mathrm{ZnO}\) is amphoteric.
ⓑ. Magnesium belongs to group 2 and zinc to group 12; \(\mathrm{MgO}\) is basic and \(\mathrm{ZnO}\) is amphoteric.
ⓒ. Magnesium belongs to group 2 and zinc to group 12; both oxides must be basic because both metals form \(2+\) ions.
ⓓ. Magnesium belongs to group 12 and zinc to group 2; \(\mathrm{MgO}\) is amphoteric and \(\mathrm{ZnO}\) is basic.
Correct Answer: Magnesium belongs to group 2 and zinc to group 12; \(\mathrm{MgO}\) is basic and \(\mathrm{ZnO}\) is amphoteric.
Explanation: \(\textbf{Interpreting magnesium's configuration:}\)
Magnesium has the configuration \([\mathrm{Ne}]\,3s^2\).
Its two outer s-electrons are characteristic of group-2 s-block elements.
Magnesium commonly forms the divalent cation \(\mathrm{Mg^{2+}}\).
Its normal oxide is therefore \(\mathrm{MgO}\).
Magnesium oxide exhibits basic character.
\(\textbf{Interpreting zinc's configuration:}\)
Zinc has the configuration \([\mathrm{Ar}]\,3d^{10}\,4s^2\).
Although its outermost principal shell contains two s-electrons, zinc does not belong to group 2.
It occupies group 12 in the d-block region.
Its common divalent ion retains the configuration \([\mathrm{Ar}]\,3d^{10}\).
The normal oxide has formula \(\mathrm{ZnO}\).
\(\textbf{Comparing the oxide characters:}\)
Zinc oxide is amphoteric.
It reacts with suitable acids and strong bases under appropriate conditions.
Magnesium oxide is predominantly basic.
Thus, two metals can form normal oxides of identical \(MO\) stoichiometry while exhibiting different acid-base characters.
\(\textbf{Explaining the classification distinction:}\)
A common outer \(ns^2\) pattern is sufficient for identifying typical group-2 elements only when the full configuration is considered in context.
The intervening filled \(3d^{10}\) subshell and zinc's periodic position must not be ignored.
Likewise, common \(+2\) oxidation states do not guarantee identical compound properties.
Magnesium and zinc therefore illustrate the need to distinguish valence-shell similarities from complete periodic classification.
The correct assignments are group 2 and group 12, with basic and amphoteric oxides, respectively.
118. The three manganese oxides \(\mathrm{MnO}\), \(\mathrm{MnO_2}\) and \(\mathrm{Mn_2O_7}\) are considered only for formal oxidation-state comparison. Assume the oxidation state of oxygen is \(-2\) in each compound.
A student argues that manganese must have the same oxidation state in all three oxides because it belongs to one periodic group. Which response is correct?
ⓐ. Manganese is \(+2,+2,+7\); the first two formulas contain the same proportion of manganese to oxygen.
ⓑ. Manganese is \(+2,+4,+14\); the final formula contains two manganese atoms and therefore doubles the oxidation state.
ⓒ. Manganese is \(+1,+2,+7\); oxide ions contribute only one negative charge per oxygen atom.
ⓓ. Manganese is \(+2,+4,+7\); different oxide compositions can reflect different positive oxidation states of a d-block element.
Correct Answer: Manganese is \(+2,+4,+7\); different oxide compositions can reflect different positive oxidation states of a d-block element.
Explanation: Oxidation state is a formal electron-accounting quantity, not the fixed charge of an element in all compounds.
For neutral \(\mathrm{MnO}\), the sum of oxidation states must be zero.
Therefore \(x-2=0\), giving manganese an oxidation state of \(+2\).
For \(\mathrm{MnO_2}\), the corresponding balance is \(x+2(-2)=0\).
This gives \(x=+4\).
For \(\mathrm{Mn_2O_7}\), there are two manganese centres and seven oxygen atoms.
The balance is \(2x+7(-2)=0\).
Thus \(2x=14\) and \(x=+7\).
The three values are consistent with the variable positive oxidation states displayed by manganese.
Manganese occupies group 7 of the d-block, where both its outer s-electrons and some d-electrons can participate in bonding.
The numerical oxidation state refers to each manganese atom, not to the entire compound.
The formula and charge-balance conditions, rather than group membership alone, determine the stated values.
119. Cerium and uranium belong to the sixth- and seventh-period inner-transition regions. Their given neutral ground-state configurations are:
\[
\mathrm{Ce}:[\mathrm{Xe}]\,4f^1\,5d^1\,6s^2
\]
\[
\mathrm{U}:[\mathrm{Rn}]\,5f^3\,6d^1\,7s^2
\]
Which conclusion follows from the principal subshells and periodic positions without requiring any particular compound chemistry?
ⓐ. Both elements are p-block members, because their configurations contain occupied d-electrons.
ⓑ. Cerium and uranium are group-2 s-block elements because both configurations end in \(s^2\).
ⓒ. Cerium belongs to period 4 and uranium to period 5 because the differentiating f-subshell numbers determine the period directly.
ⓓ. Cerium is associated with the \(4f\) lanthanoid region in period 6, whereas uranium is associated with the \(5f\) actinoid region in period 7.
Correct Answer: Cerium is associated with the \(4f\) lanthanoid region in period 6, whereas uranium is associated with the \(5f\) actinoid region in period 7.
Explanation: The filled noble-gas cores of the two configurations are xenon and radon.
Cerium has electrons in the \(6s\) region and belongs to the sixth period.
Uranium has electrons in the \(7s\) region and belongs to the seventh period.
The lanthanoid series is associated primarily with progressive occupation of the \(4f\) subshell.
The actinoid series is associated primarily with progressive occupation of the \(5f\) subshell.
The supplied cerium configuration includes \(4f^1\), while the uranium configuration includes \(5f^3\).
The additional \(5d\) or \(6d\) occupation in a particular neutral configuration does not invalidate the overall inner-transition classification.
Similarly, a terminal \(ns^2\) subshell does not mean every such element belongs to group 2.
The period number is related to the highest occupied principal shell, not simply to the principal number of the differentiating f-orbital.
Both regions are ordinarily displayed in rows beneath the main body of the long-form periodic table for compactness.
That display convention does not remove them from their actual periods.
The second option respects the given configurations and their recognised periodic locations.
120. Lithium and potassium belong to group 1. Under comparable ordinary water-contact conditions, potassium usually reacts more vigorously, while their isolated gaseous first ionization enthalpies are approximately:
\[
IE_1(\mathrm{Li})=520\ \mathrm{kJ\,mol^{-1}}
\]
\[
IE_1(\mathrm{K})=419\ \mathrm{kJ\,mol^{-1}}
\]
Which explanation correctly connects the general periodic trend with the observed reactivity without assuming that one energetic quantity completely determines a reaction rate?
ⓐ. Potassium has lower first ionization enthalpy; the overall water-reaction rate also depends on surface and kinetic factors.
ⓑ. Lithium has higher first ionization enthalpy, so its electron loss must always be faster during contact with water.
ⓒ. The metals have identical first ionization enthalpies because each has one valence s-electron, so both react equally fast.
ⓓ. Potassium loses its outer electron more readily because its nuclear charge is smaller than lithium's, accelerating water reaction.
Correct Answer: Potassium has lower first ionization enthalpy; the overall water-reaction rate also depends on surface and kinetic factors.
Explanation: Lithium has the neutral ground-state configuration \(1s^2\,2s^1\).
Potassium has the configuration \([\mathrm{Ar}]\,4s^1\).
Both therefore belong to group 1 and tend to form monovalent cations in suitable reactions.
Potassium's outer electron occupies a higher principal shell and experiences considerable shielding.
It is generally easier to remove from an isolated potassium atom than from an isolated lithium atom.
The supplied first ionization enthalpies support this conclusion.
Chemical reactions with water include more than isolated gas-phase electron removal.
Contact area, heat transfer, surface conditions and the other energetic contributions can influence the observed rate and vigour.
Therefore a first ionization enthalpy is a valuable periodic indicator, not an exact quantitative measure of every reaction rate.
Under familiar comparable conditions, the usual increase in alkali-metal water reactivity down the group is consistent with easier electron loss.
The trend does not arise from decreasing nuclear charge; potassium has more protons than lithium.
The first option makes the necessary distinction between a periodic atomic property and overall chemical reactivity.
