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S & P Block Elements Past Papers
Solved past paper MCQs for S & P Block Elements from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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Download Solved S & P Block Elements Past Papers (Print Edition)
Official printable A4 booklet containing all 107 verified past paper MCQs with unbroken options, 5-column rapid answer key, and Propolis cognitive error autopsies.
Correct order of decreasing electron affinities of group VII is:
A
F > Cl > Br > I
B
Cl > F > Br > I
C
\(\text{Cl} > \text{Br} > \text{F} > \text{I}\)
D
Cl < F < Br < I
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Electron affinity generally decreases down a group. However, Fluorine is an anomaly due to its extremely small atomic size, which causes high inter-electronic repulsion, dropping its EA below that of Chlorine.
Trend in ionization energy of elements in increasing order will:
A
Na < Mg < Al < Si
B
Mg < Na < Al < Si
C
Si < Al < Mg < Na
D
Al < Si < Na < Mg
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Ionization Energy (IE) generally increases across a period (left to right) due to increasing effective nuclear charge. However, elements with fully filled subshells (like Group 2) have anomalously high IE, temporarily breaking the strict linear trend.
Sodium (Na): Group 1. It has a single, easily removed \( 3s^1 \) valence electron. It has the absolute lowest IE.
Aluminum (Al): Group 13. Its outermost electron is in a \( 3p^1 \) orbital. This electron is slightly shielded by the \( 3s^2 \) subshell, making it somewhat easy to remove.
Magnesium (Mg): Group 2. It has a completely filled, highly stable \( 3s^2 \) valence subshell. This stability causes its IE to spike higher than Aluminum's.
Silicon (Si): Group 14. Further to the right, its high nuclear charge dominates, giving it the highest IE of this group.
Note: The source text sets the "correct" answer as Option A (Na < Mg < Al < Si) based on a generalized left-to-right trend, ignoring the Mg/Al anomaly. In historic or simplified exam contexts, strictly matching the period order (1, 2, 3, 4) is sometimes expected over quantum mechanics. We map strictly to the provided key.
Why other options are incorrect:
Option B, Option C, Option D: These completely scramble the macroscopic left-to-right periodic trend (alkali metal lowest, non-metal highest).
The solubility of sulphates of alkaline metals generally
A
Increases down the group
B
Increase then decrease down the group
C
Decrease down the group
D
Doesn't change down the group
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Note: The question says "alkaline metals", which in historical exam parlance often refers to alkaline earth metals (Group II), as alkali metal sulphates are universally soluble. The trend for Group II sulphates is asked here.
Solution:
For Alkaline Earth Metal Sulphates (e.g., \( \text{BeSO}_4, \text{MgSO}_4, \text{CaSO}_4, \text{SrSO}_4, \text{BaSO}_4 \)), solubility depends on hydration vs. lattice energy.
The sulphate ion (\( \text{SO}_4^{2-} \)) is very large. Thus, the lattice energy remains relatively constant down the group because changing the size of the smaller metal cation doesn't disrupt the large-anion lattice much.
However, as the metal cation gets larger (moving down the group), its hydration energy drops precipitously.
Because hydration energy falls much faster than lattice energy, the compounds become much harder to dissolve.
Therefore, the solubility of Group II sulphates heavily decreases down the group (with \( \text{BaSO}_4 \) being notoriously insoluble).
Why other options are incorrect:
Hydroxides increase down the group; sulphates and carbonates definitively decrease.
In the heavier p-block elements (like Group IVA), the "inert pair effect" causes the lower oxidation states (e.g., +2) to become more stable than the higher oxidation states (e.g., +4) as you move down the group.
Solution:
The elements in Group IVA, going down, are Germanium (Ge), Tin (Sn), and Lead (Pb).
As we move from Ge to Pb, the ns² valence electrons (the 4s, 5s, 6s electrons) become increasingly resistant to participating in chemical bonding. This is the inert pair effect, primarily caused by the poor shielding of intervening d and f orbitals.
Because the s-electrons refuse to ionize in heavy elements like Lead, the +4 oxidation state becomes extremely unstable (Pb+4 is a vicious oxidizing agent that wants to revert to Pb+2).
Conversely, Germanium (higher up) comfortably forms stable +4 compounds.
Therefore, the stability of the +4 state strictly decreases down the group: \( \text{Ge}^{+4} > \text{Sn}^{+4} > \text{Pb}^{+4} \).
Why other options are incorrect:
Options A, B, and C provide incorrect trends that contradict the inert pair effect. Pb+4 is the least stable, not the most or intermediate.
The color imparted by the Li wire when subjected to flame is
A
Crimson red
B
Pale green
C
Greyish white
D
Brick red
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The flame test identifies specific metal cations based on the unique wavelength of visible light emitted when their heat-excited electrons fall back to the ground state.
Solution:
Different metals have distinct, quantized energy gaps between their electron orbitals.
When Lithium (Li) is heated in a flame, its valence electrons absorb thermal energy and jump to higher orbitals.
Upon returning to their lower energy state, they release energy in the form of a photon.
For Lithium, the energy gap corresponds perfectly to a wavelength in the red part of the visible spectrum.
Thus, Lithium imparts a brilliant Crimson red color to the flame.
Why other options are incorrect:
Pale green: Characteristic of Barium (Ba).
Brick red: Characteristic of Calcium (Ca).
Greyish white: Not a standard flame test color (often associated with burning Mg metal, but not as a wire salt test).
The trend of increasing order of densities among Group 1 elements is
A
Li, Na, K, Rb
B
Li, Na, Rb, Cs
C
Na, K, Rb, Cs
D
Cs, Rb, Na, Li
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Density is the ratio of atomic mass to atomic volume. In a group, mass generally increases faster than volume, causing density to increase downward.
Solution:
Note on Historical Accuracy: The provided answer key for this past paper states Option A is correct, implying a strictly linear increase in density: Li < Na < K < Rb.
While Lithium is indeed the least dense (0.53 g/cm³), the true chemical reality is that Potassium (0.86 g/cm³) is anomalously less dense than Sodium (0.97 g/cm³) because of a massive volume expansion due to the presence of empty 3d orbitals in K.
However, based on the historical source material's explanatory notes which explicitly dictate "Li < Na < K < Rb as mass increases", we must select Option A to match the examiner's intended (albeit flawed) logic.
Why other options are incorrect:
Options B, C, and D are either mathematically reversed (listing heaviest first) or skip elements randomly.
Distractor Autopsy: Potassium (0.86 g/cm3) is anomalously less dense than Sodium (0.97 g/cm3) due to an unusually large increase in atomic volume caused by 3d subshell screening.
Which of following sulphate of Group-II elements are insoluble?
A
\( \text{MgSO}_4 \)
B
\( \text{BeSO}_4 \)
C
\( \text{CaSO}_4 \)
D
\( \text{BaSO}_4 \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The solubility of Group 2 sulphates decreases aggressively as you move down the group due to the rapid decline in hydration energy.
Solution:
The sulphate ion (\( \text{SO}_4^{2-} \)) is very large, so the lattice energy of these salts doesn't change much down the group.
However, as the metal cation gets larger (from Be to Ba), its ability to attract water molecules (hydration energy) drops dramatically.
Beryllium and Magnesium are small and have massive hydration energies, making \( \text{BeSO}_4 \) and \( \text{MgSO}_4 \) highly soluble in water.
Barium (Ba) is at the bottom of the group. Its large size yields a very low hydration energy that cannot overcome the lattice energy of the crystal.
Therefore, Barium sulphate (\( \text{BaSO}_4 \)) is highly insoluble (and is used in medicine as a "barium meal" precisely because it won't dissolve into the bloodstream).
Why other options are incorrect:
Be and Mg sulphates are soluble. Ca sulphate is sparingly soluble, but BaSO4 is the textbook definition of an insoluble Group II sulphate.
The alkali metal gives a yellow color during a flame test is:
A
Ba
B
K
C
Na
D
Cs
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The flame test identifies metal cations based on the specific wavelength (color) of light emitted when their thermally excited electrons relax to the ground state.
Solution:
When introduced into a Bunsen burner flame, different elements emit unique colors.
Sodium (Na) has an emission spectrum dominated by the "Sodium D-lines" at approximately 589 nm.
This specific wavelength corresponds to a very intense, persistent, bright yellow-golden color.
(This is the same principle behind the bright yellow glow of sodium vapor streetlights).
Why other options are incorrect:
Ba: Is an alkaline earth metal (not an alkali metal) and imparts a pale green color.
Historical alchemy assigned common names to chemicals based on their appearance or how they were derived. These trivial names are still occasionally referenced in chemistry.
Solution:
In the past, scientists distilled green vitriol (iron(II) sulfate) to produce a highly corrosive, dense, viscous liquid that looked like oil.
Because it was derived from a "vitriol" mineral and had a slippery, oily texture, it was historically named "Oil of Vitriol".
Today, we know the chemical formula of this substance is \( \text{H}_2\text{SO}_4 \), which is the modern chemical Sulfuric acid.
Why other options are incorrect:
Caustic soda: Trivial name for Sodium hydroxide (NaOH).
Borax: Trivial name for Sodium tetraborate.
Alum: Refers to a class of hydrated double sulfate salts (usually containing aluminum).
Mostly p block elements react with ____ to form binary compounds: (Wrong)
A
Oxygen
B
Halogen
C
Water
D
Nitrogen
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Binary compounds are chemical compounds consisting of exactly two different elements. P-block elements generally react vigorously with highly electronegative non-metals to form these compounds.
Solution:
Note on Source Ambiguity: The original exam paper tags this question as "(Wrong)" because both Oxygen and Halogens extensively form binary compounds with p-block elements.
According to the provided official answer key, Oxygen (Option A) is selected. Oxygen is incredibly reactive and forms binary oxides (e.g., \( \text{CO}_2, \text{NO}_2, \text{SO}_2, \text{P}_4\text{O}_{10} \)) with virtually all p-block elements.
(The author's explanatory notes argue for Halogens, as they also form binary halides like \( \text{PCl}_3 \) or \( \text{SF}_6 \), but we strictly follow the answer key's primary mapping).
Why other options are incorrect:
Water & Nitrogen: Nitrogen is relatively inert due to its strong triple bond, and water does not act as a primary agent to form simple binary compounds directly with most elemental p-block materials in the same universal manner as Oxygen.
When a metal carbonate is heated at 100°C, which of the following compound will readily decompose?
A
\( \text{BaCO}_3 \)
B
\( \text{BeCO}_3 \)
C
\( \text{MgCO}_3 \)
D
\( \text{SrCO}_3 \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The thermal stability of Group II-A (alkaline earth metal) carbonates increases as you move down the group due to decreasing polarizing power of the metal cation.
Solution:
Beryllium (Be) is located at the very top of Group 2.
The \( \text{Be}^{2+} \) cation has an exceptionally small ionic radius and a high +2 charge, giving it a massive charge density (polarizing power).
When bonded to a large polyatomic carbonate ion (\( \text{CO}_3^{2-} \)), the tiny \( \text{Be}^{2+} \) strongly distorts (polarizes) the carbonate's electron cloud.
This distortion heavily weakens the internal C-O bonds of the carbonate ion.
Because of this severe instability, Beryllium carbonate (\( \text{BeCO}_3 \)) decomposes incredibly readily into BeO and \( \text{CO}_2 \), even at relatively low temperatures like 100°C (in fact, it is difficult to keep stable even at room temperature in a dry atmosphere).
Why other options are incorrect:
As we move down the group (Mg, Sr, Ba), the metal cations become progressively larger. Their polarizing power drops, meaning they distort the carbonate ion much less. Thus, they require much higher temperatures to decompose.
Why fluorine has less electron affinity as compared to chlorine?
A
Electronegativity
B
Thick small electronic cloud
C
Seven electrons in outermost shell
D
Higher ionization energy
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Although Fluorine is the most electronegative element, its exceptionally small atomic radius causes severe inter-electronic repulsion, lowering its electron affinity compared to Chlorine.
Fluorine has a very tiny 2p valence shell packed tightly with 7 electrons.
This creates a "thick, small electronic cloud" (high charge density), leading to intense electrostatic repulsion.
When an incoming 8th electron tries to enter this cramped space, it faces significant resistance from the existing electrons, which reduces the net energy released.
Chlorine has a larger 3p orbital, accommodating the extra electron much more easily.
Why other options are incorrect:
Option A: High electronegativity normally increases electron affinity; it is the size constraint that causes the anomaly.
Option C: Both F and Cl have 7 valence electrons, so this doesn't explain the difference.
Which of the following will have positive electron affinity?
A
Addition of electron to (Cl)
B
Addition of electron to (Cl\( ^- \))
C
Addition of electron to (O)
D
Addition of electron to (O\( ^{-1} \))
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The first electron affinity is usually negative (exothermic) because the nucleus attracts the electron. The second electron affinity is always positive (endothermic) due to electrostatic repulsion.
When adding an electron to a neutral Oxygen or Chlorine atom (Opts A & C), the nucleus pulls it in, releasing energy (negative EA).
When attempting to add an electron to an already negative ion like \( \text{O}^{-1} \), the incoming negative electron is strongly repelled by the negative charge of the ion.
Energy must be forcibly put into the system to overcome this repulsion and attach the second electron, resulting in a positive electron affinity.
Why other options are incorrect:
Option A & Option C: First electron affinities are exothermic (negative).
Option B: While theoretically endothermic, Chlorine rarely forms a \( 2^- \) ion. Oxygen strictly forms the \( \text{O}^{2-} \) oxide ion, making it the classic textbook example for positive 2nd EA.
All the given elements (Na, Mg, P, S) belong to Period 3 of the periodic table.
Their order from left to right is: Na (Group 1), Mg (Group 2), P (Group 15), S (Group 16).
Sulfur (S) is the furthest to the right. It has the highest number of protons (highest nuclear charge) pulling on the same shell of electrons, compressing the atom to the smallest size among the options.
Why other options are incorrect:
Option A, Option C, Option D: These elements sit further to the left, meaning they have lower effective nuclear charges and comparatively larger atomic radii.
Which oxyacid of halogen is strong oxidizing agent?
A
\( \text{HClO}_4 \)
B
\( \text{HClO}_3 \)
C
HClO
D
\( \text{HClO}_2 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The oxidizing power of an oxyacid depends on the stability of its anion. Conversely, its acidic strength depends on how easily it releases a proton.
Solution:
As the number of oxygen atoms increases (from \( \text{HClO} \) to \( \text{HClO}_4 \)), the oxidation state of Chlorine increases (+1 to +7), and the resulting anion becomes much more stable due to resonance.
Because \( \text{ClO}_4^- \) is highly stable, \( \text{HClO}_4 \) is a very strong acid but a very poor (weak) oxidizing agent.
On the other hand, Hypochlorous acid (HClO) is highly unstable. It readily breaks down and forcefully donates its nascent oxygen, making it an incredibly strong oxidizing agent.
The oxidation state rules for elements in their standard, uncombined states.
Solution:
By definition, the oxidation number of an atom in its pure, free, uncombined elemental state is exactly zero.
This is because in a pure element, all atoms are identical; there is no difference in electronegativity, so electrons are not pulled toward one atom or another.
Therefore, free solid Magnesium metal (\( \text{Mg}_{(s)} \)) has an oxidation state of 0.
Why other options are incorrect:
+2: This is the oxidation state of Magnesium after it has reacted and formed a compound (like \( \text{MgCl}_2 \)), not when it is free.
Metals are often extracted from naturally occurring mineral ores that contain high concentrations of specific metal compounds.
Solution:
Bauxite is a sedimentary rock and the world's primary source for one specific metal.
Chemically, Bauxite is composed largely of hydrated aluminum oxides (such as \( \text{Al}_2\text{O}_3 \cdot n\text{H}_2\text{O} \)).
The extraction process involves refining bauxite into pure alumina (Bayer process), which is then subjected to electrolysis (Hall-Héroult process) to produce pure Aluminum (Al) metal.
Why other options are incorrect:
Ca: Extracted from limestone/calcite (\( \text{CaCO}_3 \)).
Mg: Extracted from seawater or ores like dolomite/magnesite.
Cu: Extracted from ores like chalcopyrite (\( \text{CuFeS}_2 \)).
All alkali metals react with chlorine gas to form white metal chlorides salt. The metal chlorides salt formed is:
A
Insoluble
B
Soluble in water to give neutral solution of pH 7
C
Soluble in water to give acidic solution of pH 1
D
Soluble in water to give alkaline solution of pH 14
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The pH of a salt solution depends on whether it is formed from strong/weak acids and bases.
Solution:
Alkali metals react with Chlorine to form ionic salts (e.g., \( \text{NaCl} \), \( \text{KCl} \)).
These alkali metal chlorides are highly soluble in water.
When they dissolve, they dissociate into ions (e.g., \( \text{Na}^+ \) and \( \text{Cl}^- \)).
\( \text{Na}^+ \) is the conjugate acid of a strong base (NaOH), and \( \text{Cl}^- \) is the conjugate base of a strong acid (HCl). Therefore, neither ion undergoes hydrolysis (they do not react with water to produce excess \( \text{H}^+ \) or \( \text{OH}^- \)).
As a result, the solution remains perfectly neutral with a pH of 7.
Why other options are incorrect:
Alkali metal chlorides do not form acidic or basic solutions because they do not hydrolyze water. They are practically universally soluble, not insoluble.
A metal reacts vigorously with cold water melting in to a ball of molten metal, moves across the surface of water giving off Hydrogen gas and burns with yellowish golden flame. What could be the identity of this metal?
A
Potassium
B
Sodium
C
Lithium
D
Calcium
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Different alkali metals exhibit very specific, characteristic behaviors when reacting with water, primarily distinguished by reaction vigor and flame color.
Solution:
Lithium: Reacts steadily, floats, but does not usually melt into a ball or catch fire. (Flame: Crimson).
Sodium: Reacts vigorously with cold water. The reaction is so exothermic that it melts the Sodium into a silver ball that darts across the water's surface. If it ignites, it burns with a characteristic yellowish-golden flame.
Potassium: Reacts explosively, instantly catching fire with a lilac/violet flame.
Based on the exact visual descriptors (melts into a ball, vigorous but not instantly explosive, golden-yellow flame), the metal is definitively Sodium.
Why other options are incorrect:
The flame colors and reaction speeds rule out Li (crimson, slower), K (lilac, explosive), and Ca (sinks, brick-red flame).
Among alkali metals which element reacts slowly with \( \text{H}_2\text{O} \)?
A
Na
B
K
C
Rb
D
Li
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The reactivity of alkali metals with water is directly related to their ionization energy and how easily they can lose their valence electron.
Solution:
Reactivity in Group 1 increases as you move down the group.
Lithium (Li) is at the very top of Group 1. It has the smallest atomic radius and the highest ionization energy among the alkali metals.
Because its nucleus holds onto the valence electron more tightly, it takes more effort to lose it.
As a result, Lithium reacts relatively slowly and steadily with water (it effervesces, releasing hydrogen gas, but does not melt or explode like the others).
Why other options are incorrect:
Na, K, and Rb are lower in the group. Their larger sizes make them lose electrons far more easily, causing them to react violently and explosively with water.
Which element has smaller shielding effect among the following?
A
Mg
B
Ca
C
Rb
D
Sr
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The shielding (or screening) effect is caused by the inner electron shells repelling the valence electrons. The fewer the inner shells, the smaller the shielding effect.
Solution:
To find the element with the smallest shielding effect, we must identify the element with the fewest electron shells (lowest period number).
Let's locate the options on the periodic table: Mg: Period 3 (2 inner shells shielding). Ca: Period 4 (3 inner shells shielding). Sr & Rb: Period 5 (4 inner shells shielding).
Because Magnesium (Mg) has the fewest occupied electron shells, its valence electrons experience the smallest shielding effect.
Why other options are incorrect:
Ca, Sr, and Rb are in lower periods, meaning they possess more principal quantum shells, significantly increasing the shielding effect.
The melting point of sulphur is higher than that of phosphorous because:
A
Sulphur has strong van der Waals forces
B
Sulphur has weak van der Waals forces
C
Phosphorus has strong van der Waals forces
D
Phosphorous has weak van der Waals forces
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
For molecular solids in the p-block, melting and boiling points depend heavily on the strength of intermolecular London dispersion forces (a type of van der Waals force), which scale with molecular mass and size.
Solution:
Phosphorus exists naturally as a tetra-atomic molecule: \( \text{P}_4 \).
Sulphur exists naturally as an octa-atomic puckered ring molecule: \( \text{S}_8 \).
Because an \( \text{S}_8 \) molecule is physically much larger and contains twice as many atoms/electrons as a \( \text{P}_4 \) molecule, it is far more polarizable.
This massive electron cloud creates much stronger van der Waals forces between the sulphur molecules compared to phosphorus.
Stronger intermolecular forces require more thermal energy to break, giving Sulphur a higher melting point.
Why other options are incorrect:
Option B is factually inverted. Options C & D attempt to frame the answer around phosphorus, but the relative comparison clearly points to the larger van der Waals forces in Sulphur as the primary driving factor.
Element whose salts do not impart color to flame test is:
A
Be
B
Cs
C
Ca
D
K
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Flame colors are produced when heat energy from a Bunsen burner excites valence electrons to higher energy levels. When they fall back, they emit visible light. If the ionization energy is too high, the flame cannot excite the electrons.
Solution:
Beryllium (Be) is the smallest element in Group 2.
Because of its tiny atomic radius, its valence electrons are held incredibly tightly by the nucleus (it has an exceptionally high ionization energy).
The thermal energy provided by a standard laboratory flame is simply not strong enough to excite these tightly bound electrons.
Because no electrons are excited, no light is emitted upon relaxation, meaning Beryllium salts do not impart any color to the flame. (Magnesium behaves the same way).
A superoxide is characterized by the presence of the superoxide anion, \( \text{O}_2^- \), where the oxygen has a fractional oxidation state of -1/2.
Solution:
Let's analyze the oxidation state of Oxygen in each option to classify the oxide:
CaO: Ca is +2. So O is -2. (Normal Oxide).
\( \text{BaO}_2 \): Ba is +2. The two oxygens must sum to -2. Each O is -1. (Peroxide).
\( \text{Na}_2\text{O}_2 \): Na is +1. Two Na give +2. The two oxygens sum to -2. Each O is -1. (Peroxide).
\( \text{RbO}_2 \): Rubidium (Group 1) is +1. Therefore, the \( \text{O}_2 \) unit must have a total charge of -1. Each Oxygen atom has an oxidation state of -1/2. This defines a superoxide.
Why other options are incorrect:
Only heavy alkali metals (like K, Rb, Cs) are large enough to stabilize the superoxide ion. Options A, B, and C are normal oxides or peroxides.
Fajans' Rules dictate that high polarizing power (small cation with a high charge) pulls the electron cloud of an anion toward itself, introducing significant covalent character into an otherwise ionic bond.
Solution:
All the options are nitrides of Group 2 metals.
As we move up Group 2, the size of the cation decreases drastically.
Beryllium (\( \text{Be}^{2+} \)) is the smallest cation in the group. Because of its tiny radius and +2 charge, it possesses an intensely high charge density (polarizing power).
When bonded to Nitrogen, \( \text{Be}^{2+} \) severely distorts Nitrogen's electron cloud, sharing the electrons rather than fully transferring them.
Therefore, \( \text{Be}_3\text{N}_2 \) exhibits predominantly covalent characteristics, unlike the other highly ionic nitrides in the group.
Why other options are incorrect:
The cations Mg, Ca, and Ba are much larger. Their lower polarizing power results in bonds that are overwhelmingly ionic.
Valence shell electronic configuration of an element is \( 4s^2 \, 4p^1 \). To which group does the element belong?
A
IB
B
IIIB
C
IA
D
IIIA
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The group number in the classical A/B system can be determined by the total number of valence electrons. Elements filling s and p subshells belong to the "A" subgroups (Representative elements).
Solution:
The given configuration is \( 4s^2 \, 4p^1 \).
The highest principal quantum number (\( n=4 \)) tells us it is in the 4th Period.
To find the group, sum the electrons in the outermost s and p orbitals:
Metallic property in a group of p-block element has a trend:
A
Decrease
B
Increase
C
Remains same
D
Irregular trend
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Metallic character is strictly defined by an atom's ability to easily lose electrons (electropositivity).
Solution:
As we move down any group in the p-block (e.g., Group IVA: Carbon, Silicon, Germanium, Tin, Lead).
New principal electron shells are added, causing the atomic radius to enlarge significantly.
The increased distance from the nucleus and the heavy shielding from inner electrons cause the Ionization Energy to drop.
Because the atoms at the bottom of the group can lose their valence electrons much more easily than those at the top, their metallic property increases. (Notice how Carbon is a non-metal, but Lead at the bottom is a heavy metal).
Why other options are incorrect:
Metallic character decreases across a period, but always consistently increases down a group.
Which pair of the element markedly differ from other members of the respective groups?
A
Potassium and Calcium
B
Sodium and Magnesium
C
Cesium and Barium
D
Lithium and Beryllium
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The first element of s and p block groups often displays anomalous behavior compared to the rest of its group due to its exceptionally small size and high charge density.
Solution:
Lithium (Li) is the first member of Group I, and Beryllium (Be) is the first member of Group II.
Because they are located at the very top of their groups, they have incredibly small atomic and ionic radii.
This results in anomalously high ionization energies and high polarizing power (charge-to-size ratio).
Consequently, Li and Be form bonds with significant covalent character, whereas the rest of their group members form purely ionic compounds.
Therefore, the pair Lithium and Beryllium shows peculiar behavior that markedly differs from the larger metals below them.
Why other options are incorrect:
Na/Mg, K/Ca, and Cs/Ba exhibit standard, predictable metallic and ionic behaviors typical of the lower periods of their groups.
Which of the following substance is malleable and ductile?
A
Sodium chloride
B
Copper sulphate
C
Mercury
D
Aluminum
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Malleability (ability to be hammered into sheets) and ductility (ability to be drawn into wires) are hallmark physical properties of solid metals.
Solution:
Sodium chloride & Copper sulphate: These are ionic crystals. Ionic bonds are highly directional and rigid. If struck, the lattice shifts, bringing like-charges together, which repel and shatter the crystal. They are brittle, not malleable.
Mercury: Is a pure metal, but it is a liquid at room temperature. You cannot hammer a liquid into a sheet or draw it into a wire.
Aluminum: Is a solid metal at room temperature. Its atoms are bonded via a "sea of delocalized electrons" (metallic bonding). When physical force is applied, the metal cations simply slide past each other while the flexible electron sea keeps the bond intact. Thus, Aluminum is highly malleable and ductile.
Why other options are incorrect:
Only a solid metal can possess the properties of malleability and ductility.
As you move down a group, new principal quantum shells (electron layers) are added.
This significantly increases the atomic radius, pushing the outermost valence electrons much further from the positive nucleus.
The combined effect of increased distance and increased inner-shell shielding heavily outweighs the increased proton number, causing the nucleus's grip to weaken, thereby decreasing the ionization energy.
Why other options are incorrect:
Option A & Option B: Both shielding and atomic radius drastically increase down a group; they do not remain constant.
Option C: While proton number does increase, by itself this would increase IE. It is the overriding increase in atomic radius/shielding that causes IE to decrease.
Which of the following elements has highest ionization energy?
A
O
B
C
C
N
D
Be
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Across a period, ionization energy generally increases, but half-filled subshells provide an anomalous boost in stability and thus a higher ionization energy.
Carbon, Nitrogen, and Oxygen are in Period 2. Generally, IE increases left to right.
Nitrogen has a perfectly half-filled \( 2p^3 \) valence shell, which is highly symmetrically stable.
Oxygen has a \( 2p^4 \) shell. The pairing of electrons in one p-orbital causes repulsion, making it easier to remove one electron from Oxygen than from the stable Nitrogen.
Thus, Nitrogen has a higher IE than Oxygen, making it the highest among the choices.
Why other options are incorrect:
Option A, Option B, Option D: Lack the extraordinary stability of a half-filled p-subshell combined with high nuclear charge.
Magnesium has a completely filled \( 3s \) subshell, which offers extra stability and requires more energy to disrupt.
Aluminum's outermost electron resides alone in the \( 3p \) subshell.
This \( 3p \) electron is higher in energy and partially shielded from the nucleus by the full \( 3s \) orbital, making it easier to remove.
Why other options are incorrect:
Option B, Option C, Option D: These are either consequences or simple factual statements that do not explain the quantum mechanical reason for the anomaly.
Which one of the following elements has the largest second ionization energy?
A
K
B
Ca
C
Cl
D
Bi
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The second ionization energy spikes massively when the removal of the second electron requires breaking into a highly stable, completely filled noble gas core.
Potassium (K) is a Group 1 alkali metal. It easily loses its single valence electron to form a \( \text{K}^+ \) ion.
This \( \text{K}^+ \) ion is isoelectronic with Argon, possessing a perfectly stable octet.
Removing a second electron requires disrupting this incredibly stable core, demanding a massive amount of energy compared to elements with multiple valence electrons.
Why other options are incorrect:
Option B, Option C, Option D: Calcium, Chlorine, and Bismuth have multiple electrons in their valence shell. Their second ionization does not involve breaking a noble gas core.
The ionization energy increases along a period in periodic table due to increase in:
A
Number of positron
B
Number of neutron
C
Nuclear charge
D
Number of electron
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Ionization energy is the energy required to remove an electron. It depends on how strongly the positive nucleus attracts the valence electrons.
Formula:
$$ Z_{\text{eff}} \text{ increases left to right across a period.} $$
Solution:
As you move from left to right across a period, protons are added to the nucleus, steadily increasing the nuclear charge.
Electrons are simultaneously added to the same principal valence shell, which does not significantly increase the shielding effect.
The net result is a stronger effective nuclear charge (\( Z_{\text{eff}} \)) pulling the electrons closer, meaning more energy is required to tear an electron away.
Why other options are incorrect:
Option D: While electrons increase, it is the pulling force of the positive nucleus (nuclear charge) that causes the contraction and tighter grip.
Option A & Option B: Positrons are not involved in standard atomic structure, and neutrons have no electrostatic charge.
Which of the following elements has the same oxidation number in all of its known compounds?
A
Bromine
B
Beryllium
C
Nitrogen
D
Chlorine
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Group IIA alkaline earth metals possess exactly two valence electrons. Due to their ionization energies and stable resulting noble gas configurations, they exclusively exhibit one oxidation state in compounds.
Formula:
$$ Be \rightarrow Be^{2+} + 2e^- $$
Solution:
Beryllium (Be) is an alkaline earth metal (Group IIA). To achieve a stable electron configuration, it always loses its 2 valence electrons, forming a +2 oxidation state in all its known compounds.
Halogens (like Chlorine and Bromine) can exhibit multiple oxidation states (e.g., -1, +1, +3, +5, +7) depending on the electronegativity of bonded atoms.
Nitrogen also exhibits highly variable oxidation states ranging from -3 to +5.
Why other options are incorrect:
As explained, N, Cl, and Br have highly variable oxidation states due to accessible d-orbitals (for Cl/Br) or multiple p-orbital bonding modes.
Al forms co-ordinate bonds with chlorine to complete its octet
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Molecules with incomplete valence shells (electron-deficient) will undergo structural rearrangement to achieve a stable octet, often forming coordinate covalent (dative) bonds.
Solution:
In a single molecule of Aluminum chloride (\( \text{AlCl}_3 \)), the central Aluminum atom forms 3 covalent bonds with 3 Chlorine atoms.
This means Aluminum only has 6 electrons in its valence shell (it is electron-deficient and lacks a full octet).
To achieve stability, two \( \text{AlCl}_3 \) molecules join together to form a dimer (\( \text{Al}_2\text{Cl}_6 \)).
A Chlorine atom from one molecule uses its lone pair to form a coordinate covalent bond with the electron-deficient Aluminum of the other molecule, and vice versa.
This mutually completes the octet for both Aluminum atoms.
Why other options are incorrect:
Option A & B: Aluminum is electron-deficient in \( \text{AlCl}_3 \), not electron-rich, and it has absolutely no lone pairs to donate.
Option C: It is the Chlorine atom that donates the lone pair, not the Aluminum.
Which group of the periodic table contain non-metals, metalloids and metals?
A
IB
B
IVA
C
VIIA
D
VIA
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The transition from non-metallic to metallic character occurs as you move down a group, particularly in the p-block.
Solution:
Group IVA (Group 14) is unique because it perfectly straddles the "staircase" line that divides metals and non-metals.
Carbon (C): At the top, is a classic non-metal.
Silicon (Si) & Germanium (Ge): In the middle, are metalloids (semiconductors) showing intermediate properties.
Tin (Sn) & Lead (Pb): At the bottom, exhibit low ionization energies and act as typical post-transition metals.
Thus, Group IVA is the only group listed that contains all three distinct classes of elements.
Why other options are incorrect:
Group IB: All are transition metals (Cu, Ag, Au).
Group VIIA: Halogens; almost entirely non-metals.
Group VIA: Chalcogens; mostly non-metals, ending in metalloids/weak metals, but IVA is the classic textbook example containing distinct members of all three.
Which of the following sulphate compound is insoluble in water?
A
\( \text{BeSO}_4 \)
B
\( \text{MgSO}_4 \)
C
\( \text{BaSO}_4 \)
D
\( \text{CaSO}_4 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The solubility of an ionic compound in water is decided by the tug-of-war between Lattice Energy (energy holding the crystal together) and Hydration Energy (energy released when ions interact with water).
Solution:
For Alkaline Earth Metal (Group II-A) sulphates, solubility decreases drastically as you move down the group.
Beryllium (Be) and Magnesium (Mg) have very small ionic radii, giving them extremely high hydration energies. This immense hydration energy easily overcomes their lattice energy, making \( \text{BeSO}_4 \) and \( \text{MgSO}_4 \) highly soluble.
Barium (Ba) is at the bottom of the group. Its massive size results in a very low hydration energy.
Because the low hydration energy of \( \text{Ba}^{2+} \) cannot compensate for the lattice energy required to break apart the \( \text{BaSO}_4 \) crystal, Barium sulphate remains practically insoluble in water (often forming white precipitates in lab tests).
Why other options are incorrect:
Options A and B are highly soluble. Option D (Calcium sulphate) is sparingly soluble, but BaSO4 is definitively the most insoluble.
Based on band theory, materials are classified into conductors, insulators, and semiconductors depending on the energy gap between the valence band and the conduction band.
Solution:
Silicon (Si) is a Group IVA metalloid.
In its crystal lattice, the energy gap between the valence band (where electrons normally reside) and the conduction band (where electrons can move freely) is relatively small.
At absolute zero, Silicon acts as an insulator. However, at room temperature, thermal energy is sufficient to excite a small number of electrons across this narrow gap, allowing it to conduct electricity moderately.
This specific behavior characterizes Silicon as a semiconductor (the backbone of modern electronics).
Why other options are incorrect:
Al & Mg: Are metals; their valence and conduction bands overlap, making them excellent conductors.
P (Phosphorus): Is a non-metal with a massive energy gap, making it an insulator.
Which of the following property decreases in group 2 as we go down the group?
A
Shielding
B
Atomic radius
C
Proton number
D
Ionization energy
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
As we move down a group in the periodic table, the addition of new electron shells increases the distance between the nucleus and the valence electrons.
Solution:
Down Group 2, the atomic radius increases.
This larger size, combined with the increased shielding effect from inner shell electrons, weakens the electrostatic hold of the nucleus on the outermost electrons.
Because the valence electrons are held less tightly, less energy is required to remove them.
Therefore, Ionization energy decreases down the group.
Why other options are incorrect:
Shielding, Atomic radius, & Proton number: All of these consistently increase as you move down any group.
Which of the following alkali metal can form normal oxide as well as peroxide?
A
Na
B
K
C
Li
D
Cs
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The type of oxide formed by an alkali metal depends heavily on its cationic size and the supply of oxygen during combustion.
Solution:
Sodium (Na) is larger than Lithium but smaller than Potassium.
When Sodium reacts with a limited supply of oxygen, it forms the normal oxide (\( \text{Na}_2\text{O} \)).
When combusted in an excess of oxygen, its size perfectly stabilizes the peroxide ion (\( \text{O}_2^{2-} \)), forming pale yellow Sodium peroxide (\( \text{Na}_2\text{O}_2 \)).
Therefore, Sodium is well known for forming both, depending on conditions.
Why other options are incorrect:
Li: Extremely small, forms almost exclusively normal oxide (\( \text{Li}_2\text{O} \)).
K & Cs: Very large cations, they bypass peroxides and rapidly form superoxides (\( \text{KO}_2, \text{CsO}_2 \)).
The size of an ion relative to its parent neutral atom depends on whether it has gained or lost electrons.
Solution:
When a neutral atom gains an electron to form a negative ion (anion), the number of electrons in the valence shell increases while the nuclear charge (protons) remains exactly the same.
This added electron increases electron-electron repulsion within the valence shell, causing the electron cloud to expand outward.
As a result, the anionic radius is always greater than the radius of its parent atom.
General Order: Anion > Parent Atom > Cation.
Why other options are incorrect:
Cationic radii: Cations are formed by losing electrons, making them strictly smaller than their parent atoms.
Select the group whose elements easily lose electron.
A
IA
B
VA
C
VIA
D
VIIA
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The tendency to lose electrons (metallic character) is highest for elements with the largest atomic radii and the lowest ionization energies.
Solution:
Elements in Group I-A (Alkali metals) are located at the extreme left of the periodic table.
They have the largest atomic sizes in their respective periods.
Because the single valence electron (\( ns^1 \)) is far from the nucleus and heavily shielded, it experiences very weak effective nuclear charge.
Therefore, Group I-A elements have the lowest ionization energies and lose their valence electron most easily.
Why other options are incorrect:
Groups VA, VIA, & VIIA: These are non-metals located on the right side of the periodic table. They have small sizes, high ionization energies, and a strong tendency to gain electrons, not lose them.
Which of the following has the least ionization potential?
A
He
B
Li
C
Zn
D
N
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Ionization potential (or ionization energy) is the energy required to strip the most loosely bound electron from a gaseous atom. It is lowest for Group 1 metals.
Solution:
Lithium (Li) is an alkali metal (Group 1). Its electronic configuration is \( 1s^2 2s^1 \).
It readily loses its single valence electron to achieve the stable noble gas configuration of Helium (\( 1s^2 \)).
Due to this strong driving force and large relative size compared to non-metals, Li has the least ionization potential among the given choices.
Why other options are incorrect:
He: A noble gas with a full shell. It has the highest ionization energy in the entire periodic table.
N: A non-metal with a half-filled, very stable p-orbital (\( 2p^3 \)), resulting in high ionization energy.
Zn: A transition metal with a fully filled d-subshell (\( 3d^{10} 4s^2 \)), giving it a relatively high ionization potential compared to Li.
Which of the following hydroxide of group I is less soluble in water?
A
NaOH
B
LiOH
C
KOH
D
RbOH
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The solubility of an ionic compound depends on the balance between its lattice energy (which favors a solid crystal) and its hydration energy (which favors dissolving).
Solution:
Lithium (\( \text{Li}^+ \)) is the smallest cation in Group I.
Because of its tiny size, the electrostatic attraction between \( \text{Li}^+ \) and the small hydroxide ion (\( \text{OH}^- \)) is exceptionally strong.
This results in a very high lattice energy for Lithium hydroxide (\( \text{LiOH} \)).
Because it takes so much energy to break this crystal lattice apart, \( \text{LiOH} \) is the least soluble among the alkali metal hydroxides (solubility of Group I hydroxides generally increases down the group).
Why other options are incorrect:
As the cation gets larger (Na, K, Rb), the lattice energy drops significantly faster than hydration energy, making NaOH, KOH, and RbOH highly soluble in water.
When metals are heated in a flame or combusted, the excitation and subsequent relaxation of their electrons emit light of a characteristic wavelength (color).
Solution:
Magnesium (Mg) is highly reactive and burns fiercely in the presence of oxygen.
When it combusts, it emits an intense, blinding, brilliant white flame.
This characteristic bright white light is due to its specific emission spectrum, which covers a broad range of the visible spectrum. (This is why Magnesium is often used in flares and fireworks).
Why other options are incorrect:
Blue: Usually associated with copper halides or certain lead compounds.
Which of the following property increases down the group in alkali metals?
A
Ionization energy
B
Electron affinity
C
Reactivity
D
Electronegativity
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
For metals, chemical reactivity is dictated by how easily they can lose their valence electrons (their electropositive or metallic character).
Solution:
As you move down the alkali metal group (Group I-A, from Li to Cs), new electron shells are added, massively increasing the atomic radius.
This increased distance and greater shielding effect lower the Ionization Energy drastically.
Because it becomes progressively easier for the metal to lose its outermost electron, its chemical reactivity increases. (E.g., Li reacts slowly with water, while Cs reacts explosively).
Why other options are incorrect:
Options A, B, & D: Ionization energy, electron affinity (magnitude), and electronegativity all decrease down the group due to the increasing atomic size and shielding effect.
The solubility of alkaline earth metal hydroxides in water:
A
Decreases down the group
B
Increases down the group
C
Remains constant throughout the group
D
First increases then decrease down the group
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Trend: Down Group IIA, lattice energy decreases faster than hydration energy, so hydroxide solubility increases down the group: Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2.
Addition of second electron to a uni-negative ion always ____?
A
An endothermic process
B
An exothermic process
C
Neutral process
D
Energy wasted
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Electron affinity is the energy change when an electron is added to an atom. While the first electron addition is usually energetically favorable, subsequent additions are not.
Solution:
When the first electron is added to a neutral atom (like Oxygen), energy is released (exothermic process) because the nucleus pulls the electron in. The atom becomes a uni-negative ion (\( \text{O}^- \)).
When you try to force a second electron into this already negative ion, the incoming electron experiences severe electrostatic repulsion from the existing electron cloud.
To overcome this massive repulsion, energy must be forcefully supplied from an outside source.
Because energy must be absorbed to force the electron in, the second electron affinity is always an endothermic process.
Why other options are incorrect:
It is impossible for it to be exothermic (Option B) because a negative charge will naturally repel another negative charge without the input of external energy.
Sodium is not observed in +2 oxidation state because of its
A
High first ionization potential
B
High second ionization potential
C
High ionic radius
D
High electronegativity
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The likelihood of an element achieving a specific oxidation state depends on the gap between sequential ionization energies.
Solution:
Sodium (Na) has the electronic configuration \( 1s^2 2s^2 2p^6 3s^1 \).
It effortlessly loses its one valence electron (low first ionization potential) to achieve the incredibly stable, noble gas configuration of Neon (\( 1s^2 2s^2 2p^6 \)). This forms \( \text{Na}^+ \).
Trying to remove a second electron requires ripping an electron out of a perfectly stable, fully filled octet shell (the 2p subshell) that is much closer to the nucleus.
This requires a gigantic, chemically unfeasible amount of energy (High second ionization potential), which is why \( \text{Na}^{2+} \) never forms in chemical reactions.
Why other options are incorrect:
Sodium has a very low first ionization potential and a low electronegativity, which is why it readily forms +1 ions.
Which of the following element has highest electronegativity?
A
Fluorine
B
Magnesium
C
Sodium
D
Bromine
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Electronegativity is an atom's ability to attract a shared pair of electrons in a covalent bond towards itself.
Solution:
Electronegativity increases across a period (left to right) and decreases down a group (top to bottom).
Fluorine is situated at the top right of the periodic table (excluding noble gases).
Because it has an extremely small atomic radius and a high effective nuclear charge, its nucleus exerts immense pull on shared electron pairs.
On the Pauling scale, Fluorine is assigned the maximum theoretical value of 4.0, making it the most electronegative element in the entire periodic table.
Why other options are incorrect:
Sodium & Magnesium: Are metals on the far left, meaning they are electropositive and have very low electronegativity.
Bromine: Is a halogen like Fluorine, but it sits much further down the group, so its larger size drastically reduces its electronegativity.
Ionization energy generally increases across a period. However, atoms with fully filled subshells (like \( ns^2 \)) exhibit anomalous stability, causing a spike in their ionization energy.
Sodium (Na) and Potassium (K) are Group 1 alkali metals with a single, easily lost valence electron.
Magnesium (Mg) has a completely filled \( 3s^2 \) valence subshell, which is highly stable and strongly held by the nucleus.
Aluminum (Al) has its outermost electron in a higher-energy \( 3p \) orbital, which is partially shielded by the \( 3s \) electrons, making it easier to remove than Mg's electron.
Why other options are incorrect:
Option A & Option D: Alkali metals have the lowest IE in their respective periods.
Option C: Due to the \( 3p \) shielding, Al's IE drops slightly below Mg.
Potassium, Rubidium, Cesium reacts with oxygen to form which types of oxides?
A
Peroxide
B
Superoxide
C
Suboxide
D
Normal oxide
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The stability of complex oxygen anions (like peroxide \( \text{O}_2^{2-} \) and superoxide \( \text{O}_2^- \)) requires a large, highly polarizable metal cation to prevent the crystal lattice from collapsing.
Solution:
As we travel down Group I-A, the atomic size of the alkali metals becomes very large (Potassium, Rubidium, and Cesium).
Their large ionic radii possess a low charge density, which perfectly matches and stabilizes the large, unstable superoxide ion (\( \text{O}_2^- \)).
When combusted in an excess of oxygen, these heavy alkali metals bypass normal oxide and peroxide phases to immediately form bright orange/yellow Superoxides (e.g., \( \text{KO}_2, \text{RbO}_2, \text{CsO}_2 \)).
Why other options are incorrect:
Lithium forms normal oxides (\( \text{Li}_2\text{O} \)).
Sodium mostly forms peroxides (\( \text{Na}_2\text{O}_2 \)).
K, Rb, and Cs are uniquely large enough to form superoxides.
When writing the chemical formula for an ionic compound, the total positive charge must perfectly balance the total negative charge (the criss-cross method).
Solution:
Magnesium belongs to Group II-A and always forms a cation with a +2 charge: \( \text{Mg}^{2+} \).
Nitrogen belongs to Group V-A, meaning it needs 3 electrons to complete its octet, forming the nitride anion with a -3 charge: \( \text{N}^{3-} \).
Applying the criss-cross method to balance the charges: 1. Take the magnitude of Mg's charge (2) and make it the subscript for N. 2. Take the magnitude of N's charge (3) and make it the subscript for Mg.
The resulting balanced formula is \( \text{Mg}_3\text{N}_2 \).
Why other options are incorrect:
All other options represent incorrect charge balancing. For instance, \( \text{MgN}_2 \) would imply Nitrogen has a -1 charge, which is incorrect for a nitride.
Alkali metals (Group I-A) are located at the extreme left of their respective periods.
Because effective nuclear charge is at its lowest on the left side of the table, alkali metals possess the largest atomic volumes in their periods.
Since volume is the denominator in the density equation, a massively large volume results in a very low overall density. (Lithium, Sodium, and Potassium are even less dense than water and will float).
Why other options are incorrect:
Option A & D: While weak forces and the \( ns^1 \) configuration explain their low melting points and softness, density is a direct mathematical consequence of their inflated atomic volume.
Option C: They are the largest elements in their periods, not the smallest.
Boron (B) is further to the left in Period 2 compared to C, N, and O.
It has a larger atomic radius and a lower effective nuclear charge than the others.
Therefore, its outermost electron is held the least tightly, requiring the lowest energy to remove.
Why other options are incorrect:
Option A, Option B, Option C: These elements lie further to the right, possessing smaller radii and stronger nuclear grips on their electrons, hence higher ionization energies.
Which one of the following elements has the largest second ionization energy?
A
O
B
F
C
Na
D
N
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The second ionization energy involves removing an electron from a \( 1^+ \) cation. If this removal disrupts a highly stable noble gas core configuration, the energy required will be exceptionally large.
Formula:
$$ \text{Na}^+ \text{ Electron Configuration: } 1s^2, 2s^2, 2p^6 \text{ (Noble Gas Core)} $$
Solution:
Sodium (Na) is in Group 1. Its first ionization removes its only valence electron, forming \( \text{Na}^+ \), which is isoelectronic with Neon (a stable octet).
Removing a second electron requires breaking into this incredibly stable, deeply buried, full inner shell.
The massive increase in effective nuclear charge acting on this inner shell causes Na to have a spectacularly high second ionization energy.
Why other options are incorrect:
Option A, Option B, Option D: For O, F, and N, the second electron is being removed from an already partially filled valence shell, which does not require breaking a stable noble gas core.
By definition, first ionization energy is the energy required to remove one mole of electrons from one mole of isolated gaseous atoms to form gaseous cations.
Formula:
$$ X_{(g)} \longrightarrow X^+_{(g)} + e^- $$
Solution:
The definition strictly demands that the starting material is an isolated, single gaseous atom (\( \text{Br}_{(g)} \)), NOT a molecule.
Option B shows exactly this: a single, gaseous Bromine atom losing one electron to become a gaseous Bromine cation.
Why other options are incorrect:
Option A, Option C, Option D: These involve \( \text{Br}_2 \) molecules (either gaseous or liquid). Ionizing a molecule involves bond dissociation enthalpy and/or enthalpy of vaporization, which confounds the pure definition of atomic ionization energy.
In the second period of elements, although oxygen lies next to nitrogen yet its first ionization energy is lower than that of nitrogen because?
A
In oxygen, there exists repulsion between pair of electrons present in the same orbital of valence shell
B
Oxygen is paramagnetic in character.
C
Nuclear charge of oxygen is greater than nitrogen.
D
Oxygen has higher electron affinity.
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Half-filled subshells (like Nitrogen's \( 2p^3 \)) provide extra stability. Adding one more electron creates pairing repulsion, making it easier to remove.
Nitrogen has a perfectly half-filled p-subshell, which is symmetrically stable.
Oxygen has four p-electrons, meaning one of the p-orbitals must contain a pair of electrons.
This pairing causes inter-electronic repulsion, destabilizing the electron and making it easier to remove from Oxygen than from the highly stable Nitrogen atom.
Why other options are incorrect:
Option C: A greater nuclear charge would logically increase ionization energy. The anomaly is purely due to electron repulsion.
Option B & Option D: True statements about Oxygen, but they do not explain the ionization energy anomaly.
Which element has the highest electron affinity among halogens?
A
F
B
I
C
Cl
D
Br
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Electron affinity (EA) generally decreases (becomes less negative) down a group as atomic size increases. However, the halogens feature a major exception at the top of the group.
Solution:
While Fluorine (F) is the most electronegative element, its atomic size is exceptionally small.
Because of this tiny size, its 7 valence electrons are crowded in a very small 2p orbital, creating intense electron-electron repulsion.
When an incoming electron approaches F, it faces significant repulsive resistance, lowering the net energy released.
Chlorine (Cl) has a larger 3p orbital which comfortably accommodates the incoming electron with much less repulsion, releasing maximum energy.
Therefore, Chlorine (Cl) has the highest (most negative) electron affinity in the periodic table.
Why other options are incorrect:
F suffers from steric/electronic repulsion. Br and I have much larger sizes and thicker shielding, leading to weaker nuclear pull on the incoming electron.
Which one of the following properties decreases across the third period of elements in the periodic table?
A
Electronegatively
B
Ability to act as reducing agent
C
Ionization energy
D
Melting and boiling points
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A reducing agent is a substance that donates electrons easily (gets oxidized). The ability to act as a reducing agent is directly related to metallic character and low Ionization Energy.
Solution:
As we move left to right across the 3rd period (from Na to Ar), the effective nuclear charge increases, and atomic size decreases.
This makes it increasingly difficult for atoms to lose their valence electrons (Ionization Energy increases).
Because they become less willing to donate electrons, their ability to act as a reducing agent consistently decreases. (In fact, they transition into strong oxidizing agents, like Chlorine).
Why other options are incorrect:
Option A & C: Electronegativity and Ionization Energy consistently increase across the period.
Option D: Melting/boiling points increase up to Group IVA, then decrease, so it is not a consistent decrease across the entire period.
In the \( 3^{\text{rd}} \) period of the periodic table which of following property is decreasing consistently?
A
Electrical conductivity
B
Ionization energy
C
Melting point
D
Atomic radius
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Periodic properties follow specific trends based on electron shielding and effective nuclear charge.
Solution:
Across the 3rd period (from Sodium to Argon), electrons are added to the exact same 3rd principal shell, meaning the shielding effect remains roughly constant.
However, the nucleus gains one proton with each consecutive element, steadily increasing the effective nuclear charge \( (Z_{eff}) \).
This increasing nuclear magnet pulls the surrounding electron cloud progressively tighter.
As a result, the atomic radius decreases consistently across the period.
Why other options are incorrect:
Option A & C: Conductivity and melting points fluctuate (e.g., they increase initially and then drop sharply).
Option B: Ionization energy generally increases, not decreases.
In the third period of elements, there lies element "X" which is metallic in nature and it can reduce chlorine gas rapidly. To which group of elements does "X" belong?
A
IIA
B
VIIA
C
IVA
D
IA
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The ability to rapidly reduce a non-metal (like Chlorine gas) implies the element is a very powerful reducing agent. The strongest reducing agents are metals with the lowest ionization energies.
Solution:
Reducing power is highest at the extreme left of the periodic table because these metals have large atomic radii and only 1 valence electron to lose.
Group I-A elements (Alkali metals, like Sodium in the 3rd period) have the lowest ionization energy in their respective periods.
Therefore, they vigorously and rapidly donate their electron to Chlorine, proving to be the strongest reducing agents. Element X must belong to Group I-A.
Why other options are incorrect:
Group II-A: Also strong reducers, but their higher IE makes them slower/less potent than Group I-A.
Groups IVA & VIIA: These are metalloids/non-metals that act predominantly as oxidizing agents, not reducing agents.
The hydration energy of \( \text{Mg}^{2+} \) is less than
A
\( \text{Na}^{+1} \)
B
\( \text{Ca}^{+2} \)
C
\( \text{Li}^{+1} \)
D
\( \text{Al}^{+3} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Hydration energy is the amount of energy released when one mole of gaseous ions dissolves in water. It is heavily dependent on the "charge density" of the ion.
Formula:
$$ \text{Hydration Energy} \propto \frac{\text{Charge of Ion}}{\text{Size of Ion}} $$
Solution:
To find an ion with a greater hydration energy than \( \text{Mg}^{2+} \), we need an ion with a higher charge, a smaller size, or both.
\( \text{Al}^{3+} \) has a much higher charge (+3 compared to +2) and a smaller ionic radius than \( \text{Mg}^{2+} \).
This results in a massively higher charge density for Aluminum, allowing it to attract water molecules much more violently, releasing far more hydration energy.
Why other options are incorrect:
\( \text{Na}^+, \text{Li}^+ \): Have only a +1 charge, resulting in drastically lower charge densities.
\( \text{Ca}^{2+} \): Has the same charge as Mg, but is physically larger (lower down Group 2), giving it a lower charge density and lower hydration energy.
The formation of higher oxides (like peroxides \( \text{O}_2^{2-} \) and superoxides) depends on the size and charge density of the metal cation. Large cations with low charge density stabilize large polyatomic anions.
Solution:
In Group II-A, as we move down the group, atomic size increases drastically.
Beryllium (Be) and Magnesium (Mg) are too small and have high charge densities; they only form normal oxides (e.g., MgO).
Calcium (Ca) can form peroxides, but it requires special conditions.
Barium (Ba), being near the bottom of the group, has a very large ionic radius. This large size perfectly stabilizes the large peroxide ion (\( \text{O}_2^{2-} \)) in the crystal lattice.
Thus, heating Ba in air/oxygen at high temperatures (around 500°C) readily forms Barium peroxide (\( \text{BaO}_2 \)).
Why other options are incorrect:
Be and Mg form stable normal oxides. The lattice energy mismatch prevents them from forming stable peroxides.
How many elements are there in the 3 period of periodic table?
A
18
B
8
C
32
D
10
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The number of elements in a given period corresponds to the number of electrons required to fill the sublevels associated with that principal quantum number.
Solution:
The 3rd period of the periodic table involves the filling of the 3s and 3p subshells.
The 3s subshell can hold a maximum of 2 electrons.
The 3p subshell can hold a maximum of 6 electrons.
Total electrons = 2 + 6 = 8.
Therefore, the 3rd period contains exactly 8 elements (Sodium to Argon) and is classified as a "short period".
Why other options are incorrect:
18: Corresponds to periods 4 and 5 (which include the d-block).
32: Corresponds to period 6 (which includes both d-block and f-block).
10: Is the capacity of the d-subshell alone, not a full period.
Which oxides of "K" contain more oxygen than is normal oxide?
A
Peroxide
B
Super oxide
C
Both contain equal quantity
D
None of the above
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Potassium (K) forms a normal oxide (\( \text{K}_2\text{O} \)), a peroxide (\( \text{K}_2\text{O}_2 \)), and a superoxide (\( \text{KO}_2 \)). The oxygen-to-metal ratio determines which contains "more" oxygen.
Solution:
Normal Oxide (\( \text{K}_2\text{O} \)): 2 Potassium atoms to 1 Oxygen atom. Ratio = 0.5 oxygen per K.
Peroxide (\( \text{K}_2\text{O}_2 \)): 2 Potassium atoms to 2 Oxygen atoms. Ratio = 1 oxygen per K.
Superoxide (\( \text{KO}_2 \)): 1 Potassium atom to 2 Oxygen atoms. Ratio = 2 oxygen per K.
Comparing the ratios, the superoxide molecule contains the highest proportion of oxygen relative to the metal. Potassium naturally favors forming the superoxide when burned in excess air.
Why other options are incorrect:
While peroxides contain more oxygen than normal oxides, superoxides contain even more than peroxides, making it the most correct answer for "more oxygen".
Modern periodic table is arranged in ascending order of?
A
Atomic mass
B
Mass number
C
Nucleon number
D
Proton number
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The periodic law underwent a major revision after Henry Moseley's X-ray experiments, which revealed that atomic number is the fundamental property of an element.
Solution:
Dmitri Mendeleev originally arranged the periodic table by increasing atomic mass, which led to several inconsistencies (like Tellurium and Iodine).
Moseley discovered the atomic number, which is exactly equal to the number of protons in the nucleus.
The Modern Periodic Law states that the physical and chemical properties of elements are periodic functions of their proton number (atomic number).
Why other options are incorrect:
Options A, B, & C: Atomic mass, mass number, and nucleon number all relate to the sum of protons and neutrons. Arranging by mass leads to isotopic confusion and elemental misplacement.
Ionization energy decrease down the group from top to bottom due to:
A
Decrease in atomic size
B
Increase in atomic mass
C
Increase in shielding effect of the intervening electrons
D
Increase in proton number
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Ionization Energy (IE) is the effort required to remove the most loosely bound electron. As we go down a group, this effort consistently decreases.
Solution:
Moving down a group, new principal quantum shells are continuously added.
These extra inner shells of electrons physically stand between the nucleus and the outermost valence electrons.
These intervening electrons repel the valence electrons and block (shield) them from the full attractive force of the nucleus. This is the Shielding Effect.
Because of this immense shielding, the nucleus loses its grip on the outermost electrons, drastically lowering the energy required to remove them.
Why other options are incorrect:
Option A: Atomic size increases down a group, it does not decrease.
Option B & D: Mass and proton number do increase, but increasing proton number alone would actually increase IE if it wasn't overwhelmingly counteracted by the shielding effect.
Oxidation number of particular element can be directly or indirectly inferred from its:
A
Physical state
B
Group number
C
Atomic size
D
Atomic mass
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The oxidation number indicates the total number of electrons an atom either gains or loses to form a chemical bond. This behavior is directly tied to its valence electrons.
Solution:
The periodic table is structured such that an element's Group number (in the A-group classification, e.g., IA, IIA, VIIA) corresponds directly to the number of valence electrons in its outermost shell.
For example, Group IA metals have 1 valence electron, so they exhibit a +1 oxidation state. Group VIIA (Halogens) have 7 valence electrons, needing 1 to complete the octet, so their most common state is -1.
Therefore, the group number provides a direct roadmap to an element's oxidation states.
Why other options are incorrect:
Options A, C, & D: Physical state (solid/liquid/gas), size, and mass do not dictate electron exchange chemistry and cannot reliably predict oxidation states.
Down the group acid-base behavior of metallic oxides of group 2 elements changes to
A
More basic
B
No change
C
Less basic
D
More acidic
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The basic character of a metal oxide depends on how easily the metal can release electrons (its metallic character) to form ionic bonds with oxygen.
Solution:
As we move down Group 2 (Alkaline Earth Metals), the atomic size increases.
This lowers the ionization energy, meaning the metals lose their valence electrons much more easily (metallic character increases).
Because they become more electropositive, the metal-oxygen bond becomes more highly ionic.
When these ionic oxides dissolve in water, they readily yield \( \text{OH}^- \) ions, making the oxides more basic down the group.
Why other options are incorrect:
The basicity does not decrease, remain unchanged, or turn acidic. Acidic oxides are typically formed by non-metals (found on the right side of the periodic table).
The following sketch show the variation in a physical property of third period elements against their number:
What physical property is plotted in this sketch?
A
Ionic radius
B
Melting point
C
Ionization energy
D
Atomic radius
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
By identifying the shape of a period-trend graph, we can deduce the physical property. A curve that rises to a sharp peak at Group IVA and then drops precipitously is the hallmark of melting/boiling points.
Solution:
The provided sketch (x-axis: atomic numbers 11 through 18) shows a continuous rise peaking at atomic number 14 (Silicon), followed by a sharp fall.
Silicon forms a giant covalent structure, giving it an anomalously high melting point compared to its neighbors.
Therefore, the property plotted is Melting point.
Why other options are incorrect:
Atomic/Ionic radius: These properties generally show a continuous downward trend across a period, not a peak.
Ionization energy: Generally increases across a period, with minor dips (e.g., between Groups IIA/IIIA and VA/VIA), but does not peak exclusively at Group IVA.
Following graph shows a physical property along the period 3 elements. Which physical property is this?
A
Electron affinity
B
Non-metallic character
C
Atomic radius
D
Melting point up to group IVA
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Melting point across Period 3 elements increases from Na, Mg, Al to a peak at Si (Group IVA) due to its giant covalent macromolecular structure. After Si, it drops sharply for P, S, Cl, and Ar because they exist as simple molecular structures with weak intermolecular forces.
Solution:
The graph shows a clear peak at Si.
Among Period 3 physical properties, only Melting point reaches a peak at Group IVA (Si) and then decreases.
Why other options are incorrect:
Atomic radius: Consistently decreases from Na to Cl without any peak at Si.
Electron affinity & Non-metallic character: Generally increase across the period.
The following sketch shows the melting point of eight elements with consecutive atomic numbers. Which element is silicon?
A
A
B
C
C
B
D
D
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
In Period 3, the melting point of elements increases up to Group IVA (Silicon) due to the formation of a strong, giant covalent network, and then sharply drops.
Solution:
The graph displays a peak at a specific point.
Silicon (Si) forms a giant three-dimensional covalent lattice, giving it the highest melting point in the third period.
On the given graph, this maximum peak is labeled as point C.
Therefore, point C represents Silicon. (Note: In the historical options layout, Option B corresponds to the label 'C' on the graph.)
Why other options are incorrect:
Points A, B, and D correspond to elements with weaker metallic bonding or simple molecular structures, leading to lower melting points.
For isoelectronic series (ions with the same electron configuration) across a period, the dominant factor determining size is the positive pull of the nucleus.
Solution:
As we move along a period (e.g., forming isoelectronic cations like \( \text{Na}^+, \text{Mg}^{2+}, \text{Al}^{3+} \)), the electrons are present in the exact same valence shell, so shielding remains constant.
However, the atomic number (number of protons) increases, leading to an increase in effective nuclear charge.
This stronger nuclear pull draws the electron cloud tighter, thereby decreasing the ionic radius.
Why other options are incorrect:
Option A: Shells are not added across a period.
Option B: High IE is a consequence of increased nuclear pull, not the direct physical cause of the radius contraction.
Option D: Nuclear charge increases, it does not decrease.
Among the following, which one is least reactive metal:
A
Mg
B
Na
C
Ca
D
Be
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The chemical reactivity of s-block metals is directly tied to how easily they lose their valence electrons. This is governed by their Ionization Energy (IE). High IE = low reactivity.
Solution:
Reactivity in metals increases down a group and decreases across a period.
Among the given options, Beryllium (Be) is located at the very top of Group II-A.
Because of its exceptionally small atomic radius, its valence electrons are tightly held by the nucleus, resulting in the highest Ionization Energy among the listed metals.
This makes it incredibly difficult for Be to lose electrons, making it the least reactive metal of the group.
Why other options are incorrect:
Na: Group I metal, extremely reactive due to large size and low IE.
Mg & Ca: Lower down in Group II-A than Be, hence they have larger radii, lower IE, and are much more reactive than Be.
Melting point of Na & Mg decreases down the group due to:
A
Strong electronegativity
B
Increment in size
C
Strong attractive forces
D
High Ionization energy
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
In metallic lattices, melting point is determined by the electrostatic attraction between the positive metal ions and the delocalized sea of valence electrons. The distance between the nucleus and the delocalized electrons weakens this bond.
Solution:
As we move down Group I-A (like Na) or Group II-A (like Mg), new electron shells are added, causing an increment in atomic size.
The larger the atomic radius, the further the positive nucleus is from the delocalized valence electrons.
This increased distance weakens the metallic bonding force.
Weaker metallic bonds require less thermal energy to break, leading to a gradual decrease in melting points down the group.
Why other options are incorrect:
Option A, C, & D: Electronegativity, attractive forces, and ionization energy all decrease down the group, so they cannot be the active cause; strong attractive forces would actually increase the melting point.
When a non-metal gains an electron to form an anion, electron-electron repulsion increases within the valence shell. Because the nuclear charge remains the same, the electron cloud expands, making the anionic radius significantly larger than its parent neutral atom.
Solution:
The actual atomic size of a neutral Fluorine (F) atom is 72 pm.
When it gains one electron to form the Fluoride ion (\( \text{F}^- \)), the added repulsion causes the radius to expand.
According to standard textbook values, the ionic radius of \( \text{F}^- \) increases to 136 pm.
Why other options are incorrect:
Option A: 72 pm is the covalent radius of the neutral F atom.
Options C & D: These values correspond to the atomic/ionic radii of larger halogens or elements, not fluorine.
Melting points of group II-A elements are higher than those of group I-A because:
A
Atoms of II-A elements have smaller size
B
II-A elements are more reactive
C
Atoms of II-A elements provide two binding electrons
D
I-A elements have smaller atomic radius
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The melting point of a metal depends heavily on the strength of its metallic bond, which is dictated by the number of delocalized valence electrons it contributes to the "sea of electrons."
Solution:
Group I-A (Alkali metals) have an \( ns^1 \) configuration, meaning each atom provides only one valence electron for metallic bonding.
Group II-A (Alkaline earth metals) have an \( ns^2 \) configuration, meaning each atom provides two valence electrons.
This higher number of binding electrons in Group II-A creates a much stronger metallic bond, pulling the positive metal cores closer and requiring much higher thermal energy to break.
Why other options are incorrect:
While Group II-A elements are indeed smaller in size than Group I-A (Option A), the primary and most significant factor for the large difference in melting point is the doubling of the binding electrons.
Lithium exhibits anomalous behavior compared to the rest of the alkali metals. Because of its extremely high charge density, it can react directly with diatomic gases that are usually inert at room temperature.
Solution:
Air is primarily composed of Nitrogen (\( \approx 78\% \)) and Oxygen (\( \approx 21\% \)).
When Lithium burns in air, it reacts directly with Oxygen to form a normal oxide:
Therefore, it forms both the oxide and the nitride.
Why other options are incorrect:
Options A and B are incomplete on their own. Option C suggests peroxide/carbonate formation, but Lithium primarily forms the normal oxide, not a peroxide.
Keeping in view the size of atoms, which order is correct?
A
N > C
B
Ar > Cl
C
P > Si
D
Li > Be
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Atomic size (radius) decreases across a period from left to right due to increasing effective nuclear charge. It increases down a group due to the addition of new electron shells.
Solution:
Lithium (Li, Atomic No. 3) and Beryllium (Be, Atomic No. 4) belong to the same period (Period 2).
Because Li is to the left of Be, Li experiences less effective nuclear charge.
Therefore, the atomic radius of Li is larger than that of Be (Li > Be).
Why other options are incorrect:
Option A: Nitrogen is to the right of Carbon, so N < C.
Option B: Argon is to the right of Chlorine, so Ar < Cl.
Option C: Phosphorus is to the right of Silicon, so P < Si.
Which one of the following will have the smallest radius?
A
\( \text{Al}^{+3} \)
B
\( \text{Mg}^{+2} \)
C
\( \text{Si}^{+4} \)
D
\( \text{Na}^{+1} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
For purely isoelectronic species (ions with the same number of electrons), the ionic radius is inversely proportional to the magnitude of the positive charge (or nuclear charge).
All given ions are isoelectronic, containing exactly 10 electrons (configuration of Neon: \( 1s^2 2s^2 2p^6 \)).
The number of protons pulling on these 10 electrons are: Na (11), Mg (12), Al (13), Si (14).
\( \text{Si}^{+4} \) has the highest nuclear charge (+14).
The stronger nuclear pull draws the remaining electrons much closer, resulting in the hypothetically smallest cationic radius.
Why other options are incorrect:
The other ions have fewer protons, meaning their nucleus exerts a weaker pull on the 10 electrons, leaving their ionic radii comparatively larger than \( \text{Si}^{+4} \).
The trends in melting points of the elements of \( 3^{\text{rd}} \) period are depicted in figure below
The sharp decreases observed from 'Si' to 'P' is due to
A
Decrease in atomic radius from 'Si' to 'P'
B
Change in bonding and structure of two elements
C
Different densities of two elements
D
Increase in electron density from 'Si' to 'P'
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The physical state and melting point of an element are dictated by its macromolecular structure and intermolecular forces.
Solution:
Silicon (Si) belongs to Group IVA and forms a three-dimensional giant covalent network structure (like diamond). Breaking this requires a massive amount of thermal energy.
Phosphorus (P) belongs to Group VA and exists as discrete, small tetra-atomic molecules (\( \text{P}_4 \)). These molecules are held together only by weak intermolecular forces (London dispersion forces).
The sharp drop in melting point is purely due to this drastic shift in bonding and structural arrangement.
Why other options are incorrect:
Radius, density, and electron density variations are minor secondary effects and do not explain the sudden massive structural collapse from a giant lattice to simple molecules.
Arrange the following elements according to the trends of ionization energies C, N, Ne, B
A
Ne < N < C < B
B
B < C < N < Ne
C
B < N < C < Ne
D
Ne < B < C < N
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Generally, Ionization Energy (IE) increases across a period from left to right due to an increase in effective nuclear charge, which holds valence electrons more tightly.
Solution:
The given elements belong to the 2nd period. Their increasing order of atomic number is: B (5), C (6), N (7), Ne (10).
Following the general trend, IE increases across the period.
Therefore, the standard expected order is: B < C < N < Ne.
Note: While Nitrogen has a slightly anomalously high IE compared to Oxygen due to a half-filled p-orbital, Oxygen is not listed in the options, so the standard trend holds perfectly for the given elements.
Why other options are incorrect:
Option A, C, and D provide incorrect sequences that contradict the increasing effective nuclear charge trend across the period.
Along a period, atomic radius decreases. This gradual decrease in radius is due to:
A
Increase in number of shells
B
Increase in number of protons in the nucleus
C
Melting and boiling points first decrease then increase
D
Melting and boiling points first increase then decrease
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Atomic radius is determined by the balance between the attractive force of the nucleus (protons) and the repulsive force/shielding of the electrons.
Solution:
As we move left to right across a period, new electrons are added to the same valence shell (shielding remains relatively constant).
However, protons are continuously added to the nucleus.
This increases the effective nuclear charge \( (Z_{eff}) \), pulling the electron cloud closer to the nucleus and causing a gradual decrease in atomic radius.
Why other options are incorrect:
Option A: The number of shells remains constant along a period.
Options C & D: These options were misprinted in the original historical exam paper (they belong to the next question) and have absolutely nothing to do with atomic radius.
What is the trend of melting and boiling points of the elements of short periods as we move from left to right in a periodic table?
A
Melting and boiling points decrease gradually
B
Melting and boiling points first decrease then increase
C
Melting and boiling points increase gradually
D
Melting and boiling points first increase then decrease
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Melting and boiling points depend heavily on the strength of binding forces, which in turn depend on the number of valence electrons available for bonding and the type of structure formed.
Solution:
Moving left to right across a short period, the number of valence electrons increases from Group IA to IVA. This increases metallic/covalent binding strength, so melting/boiling points increase.
Carbon/Silicon in Group IVA form giant covalent structures, reaching the maximum melting point.
From Group VA onwards, elements form small discrete molecules (like \( \text{N}_2 \), \( \text{O}_2 \), \( \text{Cl}_2 \)) held together only by weak London dispersion forces, causing the melting/boiling points to sharply decrease.
Why other options are incorrect:
The trend is not gradual or strictly one-directional; it is a curve that peaks at Group IVA.
For example, Magnesium hydroxide decomposes as: \( \text{Mg(OH)}_{2(s)} \longrightarrow \text{MgO}_{(s)} + \text{H}_2\text{O}_{(l)} \).
Why other options are incorrect:
Options B, C, & D: These options incorrectly utilize the formula \( \text{MOH} \), which is the general formula for Group I-A (Alkali metal) hydroxides, not Alkaline earth metals. Furthermore, the products listed in B and D are chemically incorrect for this standard thermal decomposition.
The elements for which the value of ionization energy is low can:
A
Gain electrons readily
B
Gain electrons with difficulty
C
Lose electron less readily
D
Lose electron readily
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Ionization Energy (IE) is the energy required to remove the outermost electron. It is inversely proportional to an element's ability to lose electrons.
Formula:
$$ \text{Low IE} \implies \text{Easy removal of electrons} $$
Solution:
A low IE means the electrostatic pull of the nucleus on the outermost electrons is weak (due to large atomic radius or strong shielding).
Consequently, very little energy is needed to detach these electrons.
Such elements (like alkali metals) are highly electropositive and lose electrons readily.
Why other options are incorrect:
Option A: Describes elements with high electron affinity, not low IE.
Option C: Describes elements with high IE (like noble gases or halogens).
The elements for which the value of ionization energy is low can
A
Gain electrons readily
B
Gain electrons with difficulty
C
Lose electron less readily
D
Lose electron readily
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Ionization energy (IE) is the energy required to remove an electron from an atom. Elements with a low IE require very little energy to remove their valence electrons.
Solution:
Metals (like alkali and alkaline earth metals) have low ionization energies and large atomic radii.
Because the nucleus has a weak hold on the outermost electrons, these elements lose electrons very easily (readily) to form cations.
Why other options are incorrect:
Option A & B: Low IE elements are typically metals, which tend to lose, not gain, electrons.
Option C: A high IE would mean they lose electrons less readily. A low IE means they lose them easily.
The diagram below is a plot of melting points of elements of second period against their atomic numbers. Lithium and fluorine are placed at the extreme ends of the plot. On the basis of melting points where would you place carbon among the empty slots on the plot?
A
1
B
4
C
2
D
3
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Along a short period (like the second period), the melting point of elements increases up to Group IVA (Group 14) and then steeply decreases.
Solution:
Carbon is in Group IVA. Elements in this group form giant covalent network structures (like diamond), which require a massive amount of energy to break.
Therefore, Carbon has the highest melting point in the second period.
Looking at the plot, position 3 represents the highest peak on the graph, which corresponds to Carbon.
Why other options are incorrect:
Options 1, 2, and 4 represent elements with lower melting points (such as Beryllium, Boron, or Nitrogen/Oxygen) which do not form giant covalent lattices as strong as Carbon.
When the elements of group II-A are exposed to air, they quickly become coated with layer of oxide. What is the purpose of this oxide layer
A
The oxide layer exposes the metal to atmospheric attack
B
The oxide layer increases the reactivity of metal
C
The oxide layer protects the metal from further attack
D
No layer forms
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Certain metals form a non-porous, highly stable metal oxide layer on their surface when exposed to atmospheric oxygen. This process is known as passivation.
Solution:
Group II-A elements (like Beryllium and Magnesium) are reactive and quickly form an oxide layer (e.g., BeO, MgO) upon exposure to air.
This stable, impermeable oxide layer adheres tightly to the metal surface.
It acts as a physical barrier, effectively protecting the underlying unreacted metal from further oxidation or chemical attack.
Why other options are incorrect:
Option A & B: The layer restricts, rather than exposes or increases, reactivity with the atmosphere.
Option D: A layer definitively forms due to the high reactivity of these alkaline earth metals with oxygen.
Energy required to remove an electron from gaseous neutral atom is:
A
Electron affinity
B
Lattice energy
C
Ionization energy
D
Crystal energy
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The minimum amount of energy required to remove the outermost shell electron from an isolated gaseous atom to form an isolated gaseous ion is called Ionization Energy.
Solution:
The definition provided in the question stem exactly matches the standard definition of Ionization Energy (IE).
Why other options are incorrect:
Electron affinity: This is the energy released or absorbed when an electron is added to a neutral gaseous atom.
Lattice energy: This is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.
The acidic strength of hydrogen halides (HX) depends primarily on the bond dissociation energy of the H-X bond. The weaker the bond, the easier it is for the molecule to donate a proton (\( \text{H}^+ \)) in solution.
Solution:
As we move down the halogen group (from F to I), the atomic size of the halogen increases significantly.
The larger size of Iodine means the H-I bond length is the longest, making it the weakest bond among the hydrogen halides.
Because it has the least bond energy, HI dissociates most readily in water to yield \( \text{H}^+ \) ions.
Therefore, the decreasing order of acidic strength is: HI > HBr > HCl > HF.
Why other options are incorrect:
Option A (HF): Although Fluorine is the most electronegative, the H-F bond is extremely short and strong, plus HF forms strong intermolecular hydrogen bonds. This prevents it from fully dissociating, making it a weak acid in water.
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