P-Block Elements MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: The configuration \([\mathrm{Ne}]\,3s^2 3p^1\) represents aluminium, which has atomic number \(13\). The highest principal quantum number in the configuration is \(n=3\), so the element belongs to Period 3. For a p-block element, the outermost \(s\) and \(p\) electrons are counted together to determine the number of valence electrons. Here, the \(3s\) subshell contains two electrons and the \(3p\) subshell contains one electron. Thus, the total number of valence electrons is \(2+1=3\). The general valence-shell configuration of Group 13 elements is \(ns^2np^1\). The group number is \(13\), not \(3\), even though the atom has three valence electrons. Counting only the single \(3p\) electron would incorrectly ignore the two outermost \(3s\) electrons. Hence, aluminium belongs to Group 13 and Period 3 and possesses three valence electrons.

Explanation: \(\textbf{Determine the valence-electron count:}\) The general valence-shell configuration of Group 13 elements is \(ns^2np^1\). The outermost \(s\) subshell contains two electrons, while the \(p\) subshell contains one electron. Therefore, each neutral Group 13 atom possesses three valence electrons. The number of valence electrons must not be confused with the number of electrons occupying only the \(np\) subshell. \(\textbf{Examine the possible oxidation states:}\) The +3 oxidation state involves formal participation of all three valence electrons. It is particularly important for lighter Group 13 elements such as boron and aluminium. The +1 oxidation state corresponds formally to participation of the \(np^1\) electron while the \(ns^2\) pair remains occupied. Both oxidation states are possible for certain members, but their relative stabilities change down the group. \(\textbf{Apply the inert-pair effect:}\) Among heavier Group 13 elements, participation of the outermost \(ns^2\) electron pair becomes increasingly unfavourable. This behaviour is associated with the inert-pair effect. Consequently, the +1 oxidation state becomes relatively more stable towards the heavier members. Thallium provides a characteristic example because Tl(I) is generally more stable than Tl(III). The inert-pair effect does not imply that the +3 oxidation state becomes completely impossible. It describes a change in relative energetic stability rather than a change in the number of valence electrons. \(\textbf{Eliminate the incorrect interpretations:}\) Option A incorrectly associates the greatest importance of the +1 state with the lighter members. Option C incorrectly counts only the \(np^1\) electron as a valence electron. Option D reverses the general trend in relative oxidation-state stability. Therefore, Group 13 elements possess three valence electrons, while the +1 state becomes increasingly important towards the heavier members.

Explanation: Boron and aluminium are both Group 13 elements with the general valence-shell configuration \(ns^2np^1\). Consequently, both possess three valence electrons, so a difference in valence-electron count cannot explain their contrasting metallic characters. Boron is considerably smaller and has a stronger attraction for its valence electrons. Its relatively high ionization enthalpy makes the formation of a simple \(\mathrm{B^{3+}}\) ion energetically unfavourable. Boron therefore shows predominantly covalent bonding and non-metallic or metalloid-like behaviour. Aluminium has a larger atomic size and greater electropositive character. Its valence electrons are less strongly attracted than those of boron, contributing to its metallic behaviour. Both elements remain members of the p-block despite this difference. The broad increase in metallic character down Group 13 is therefore compatible with an unchanged general valence-shell configuration. This illustrates why membership of the same group does not imply identical physical or bonding behaviour.

Explanation: \(\textbf{Electronic starting point:}\) The neutral Group 13 atom has the outermost configuration \(ns^2np^1\). The \(np^1\) electron is the first outer-shell electron considered for removal in a simplified electron-counting description. Removal of this electron gives the formal +1 state while leaving the \(ns^2\) pair occupied. Removal of all three outer-shell electrons corresponds formally to the +3 state. \(\textbf{Why the lower state becomes important:}\) In heavier Group 13 elements, the \(ns^2\) electron pair increasingly resists participation in bonding. This is described by the inert-pair effect. Consequently, the +1 oxidation state can become more stable relative to the +3 state as one moves towards the heavier members. The effect does not mean that the \(ns^2\) electrons have disappeared or that they are absolutely incapable of participating in bonding. It means that their participation becomes energetically less favourable in comparison with the lower oxidation state. Thallium illustrates this behaviour because Tl(I) is generally more stable than Tl(III). The decisive distinction is therefore between the number of available valence electrons and the number that participate in producing a particular oxidation state.

Explanation: The assertion is correct because Tl(I) is generally more stable than Tl(III). Thallium belongs to Group 13 and has three outermost electrons, represented by the general arrangement \(ns^2np^1\). The +3 oxidation state formally involves participation of all three valence electrons. The +1 state corresponds to participation of only the \(np^1\) electron while the \(ns^2\) pair remains occupied. The reason is also correct because the inert-pair effect becomes important among heavier p-block elements. In thallium, the reluctance of the \(6s^2\) electron pair to participate in bonding favours the lower oxidation state. This gives a direct chemical explanation for the relatively greater stability of Tl(I). The +3 state is not impossible, but its stability is reduced relative to that of the +1 state. It would therefore be incorrect to assume that the highest accessible oxidation state must always be the most stable. Since the reason identifies the underlying cause of the observation stated in the assertion, both statements are true and R correctly explains A.

Explanation: \(\textbf{Identify P from its chemical character:}\) Boron is the first member of Group 13 and differs significantly from the more metallic members. Its small atomic size and high ionization enthalpy favour covalent bonding rather than ordinary metallic behaviour. Boron commonly exhibits the +3 oxidation state, so P is boron. \(\textbf{Distinguish Q from R:}\) Aluminium is metallic and commonly forms compounds in the +3 oxidation state. Although it has the same number of valence electrons as boron and thallium, the relative stability of its oxidation states is different. For aluminium, the +3 oxidation state is particularly important. Thallium is also metallic, but the inert-pair effect strongly favours its +1 oxidation state relative to +3. Thus, R corresponds to thallium and Q corresponds to aluminium. This assignment simultaneously satisfies the chemical-character evidence and the oxidation-state evidence. Assigning thallium to Q would neglect the important stability difference between Tl(I) and Tl(III). The correct sequence is therefore boron, aluminium and thallium for P, Q and R, respectively.

Explanation: \(\textbf{Determine the oxidation-state change:}\) Thallium changes from oxidation state +3 to oxidation state +1. A decrease in oxidation state indicates reduction rather than oxidation. The change is represented by the following formal electron-transfer relation: \[ \mathrm{Tl^{3+}+2e^-\longrightarrow Tl^+} \] The oxidation state decreases by two units, so two electrons are gained. \(\textbf{Interpret the chemical role:}\) A species that gains electrons undergoes reduction. When it accepts electrons from another species, it acts as an oxidizing agent. Therefore, thallium(III) can behave as an oxidizing agent while being reduced to thallium(I). The greater relative stability of Tl(I) is consistent with the tendency of Tl(III) compounds to undergo reduction under suitable conditions. This behaviour is related to the inert-pair effect, which favours retention of the outermost \(6s^2\) pair. The reverse conversion, Tl(I) to Tl(III), would require oxidation and participation of two additional valence electrons. Hence, electron-transfer direction, oxidizing behaviour and oxidation-state stability all support the same conclusion.

Explanation: \(\textbf{Evaluate the oxide formulas:}\) For an element M in oxidation state +1, two M atoms are required to balance one oxide ion carrying charge \(-2\). Thus, the neutral formula \(\mathrm{M_2O}\) is consistent with oxidation state +1, making Statement I correct. For oxidation state +3, two M atoms contribute a total formal oxidation-state sum of +6. Three oxygen atoms contribute \(-6\), giving the neutral formula \(\mathrm{M_2O_3}\). Therefore, Statement II is also correct. Statement III is incorrect because a chemically possible oxidation state need not be the most stable oxidation state. A compound containing Tl(III) can exist even though Tl(I) is generally more stable. Statement IV is correct because the inert-pair effect favours retention of the \(ns^2\) pair in heavier Group 13 elements. In thallium, this favours the +1 state relative to +3. The formula \(\mathrm{Tl_2O}\) corresponds to Tl(I), whereas a Tl(III) oxide has the formula \(\mathrm{Tl_2O_3}\). The crucial distinction is between satisfying oxidation-state charge balance and predicting relative chemical stability. Thus, Statements I, II and IV are correct.

Explanation: Boron has three valence electrons and commonly exhibits the formal oxidation state +3 in its compounds. However, the oxidation state is an electron-accounting quantity and is not necessarily the actual charge on an atom. Formation of an isolated \(\mathrm{B^{3+}}\) ion would require removal of three electrons from a very small atom. The exceptionally high energy requirement makes such simple ionic behaviour unfavourable under ordinary conditions. Consequently, boron generally forms compounds involving shared electron pairs. For example, boron trichloride contains covalent B–Cl bonds while boron has the formal oxidation state +3. The oxidation state is assigned using electronegativity-based electron counting, rather than by claiming complete physical transfer of electrons. Thus, covalent bonding and a positive formal oxidation state are chemically compatible. This is particularly important when comparing boron with more electropositive Group 13 members. The misconception is to interpret an oxidation-state number as proof that a corresponding free cation must exist. The correct explanation is therefore based on boron's small size, high ionization enthalpy and preference for covalent bonding.

Explanation: Boron has three valence electrons and ordinarily forms three covalent bonds in \(\mathrm{BCl_3}\). Each B–Cl single bond contributes a shared electron pair around the central boron atom. The three bonding pairs therefore account for six electrons around boron. Thus, the central atom does not possess a complete octet in the usual Lewis structure of \(\mathrm{BCl_3}\). Boron has an available orbital that can accept electron density from a suitable donor. A Lewis base possesses an electron pair that can be donated to an electron-pair acceptor. Consequently, \(\mathrm{BCl_3}\) can behave as a Lewis acid. Electron-pair acceptance can increase the coordination number of boron from three to four. This behaviour does not require formation of a free \(\mathrm{B^{3+}}\) ion. It also does not mean that boron initially possesses eight electrons around it. The key reasoning is that three covalent bonds satisfy boron's usual trivalent bonding pattern without necessarily satisfying the octet rule.

Explanation: \(\textbf{Examine the initial molecule:}\) In \(\mathrm{BF_3}\), boron is bonded to three fluorine atoms. Boron has the formal oxidation state +3 because each fluorine is assigned oxidation state \(-1\). Three B–F bonds provide six electrons around the central boron atom in the ordinary Lewis structure. Thus, boron is electron-deficient and can accept a lone pair from a donor. \(\textbf{Interpret adduct formation:}\) Ammonia has a lone pair on nitrogen that can be donated to the electron-deficient boron centre. The resulting coordinate bond increases the number of atoms directly bonded to boron from three to four. Therefore, boron's coordination number increases from three to four, and its immediate electron-counting environment becomes octet-complete. This is a Lewis acid–base interaction rather than a redox reaction. The formal oxidation state of boron remains +3 because electron-pair donation does not produce the electron transfer characteristic of reduction to +1 or to zero. In the adduct, oxidation-state assignments remain consistent with boron at +3 and nitrogen at \(-3\). The essential distinction is that coordination number and oxidation state describe different chemical properties. Formation of an additional coordinate bond does not, by itself, prove that an atom's oxidation state has changed.

Explanation: Boron and aluminium both belong to Group 13 and commonly show the +3 oxidation state. Consequently, their ordinary oxides can have the same general stoichiometric formula, \(\mathrm{M_2O_3}\). However, identical stoichiometry does not imply identical acid–base behaviour. Boron is a small, relatively non-metallic element whose compounds commonly show pronounced covalent character. Its oxide, \(\mathrm{B_2O_3}\), is acidic and can react with suitable bases. Aluminium has greater metallic character, but its oxide is not simply a strongly basic oxide. \(\mathrm{Al_2O_3}\) is amphoteric, meaning that it can react with both acids and bases under appropriate conditions. Amphoterism must not be confused with chemical neutrality. The distinction reflects the different chemical characters of boron and aluminium despite their common group membership. Their oxidation states establish an electron-accounting similarity, whereas their bonding and acid–base properties reveal an important difference. The student's conclusion is therefore incorrect because group membership and oxidation-state equality are insufficient to predict identical oxide behaviour.

Explanation: \(\textbf{Classify the first oxide:}\) Oxide P reacts with a suitable strong base but does not show the characteristic acid–base reaction of a basic or amphoteric oxide with the supplied acid. This behaviour is consistent with an acidic oxide. Among the two specified Group 13 oxides, \(\mathrm{B_2O_3}\) is acidic. Therefore, P is identified as \(\mathrm{B_2O_3}\). \(\textbf{Classify the second oxide:}\) Oxide Q reacts with both a suitable acid and a suitable strong base. Such dual acid–base behaviour is the defining feature of an amphoteric oxide. Aluminium oxide, \(\mathrm{Al_2O_3}\), is amphoteric, so Q is identified as \(\mathrm{Al_2O_3}\). The two observations for Q must be considered together; reaction with an acid alone would not establish amphoterism. \(\textbf{Connect the observations to periodic character:}\) Boron is more non-metallic and its oxide shows acidic character. Aluminium is more metallic, but its oxide retains amphoteric behaviour rather than behaving as an exclusively basic oxide. The identical +3 oxidation state of the two elements does not determine the oxide classification by itself. Thus, the response pattern supports P as boron oxide and Q as aluminium oxide.

Explanation: All three observations are chemically valid, but they describe different consequences of electronic structure. Statement I is correct because boron commonly forms covalent compounds owing to its small size and high ionization enthalpy. In \(\mathrm{BCl_3}\), boron's formal oxidation state is +3 without requiring the compound to contain free \(\mathrm{B^{3+}}\) ions. The oxidation-state number therefore cannot be used as a direct measurement of ionic bonding character. Statement II is correct because aluminium oxide is amphoteric. Aluminium's metallic character does not require every aluminium compound to behave exclusively as a basic substance. Oxide acid–base behaviour depends on more than the broad metal/non-metal classification of the parent element. Statement III is also correct because Tl(I) is relatively more stable than Tl(III). The inert-pair effect increasingly favours non-participation of the outermost \(ns^2\) electrons in heavier Group 13 elements. Thus, three valence electrons do not guarantee that +3 will always be the preferred oxidation state. The three observations are linked by a central principle: electronic configuration provides a starting framework, but atomic size, bonding and relative energetic stability determine actual chemical behaviour. A successful prediction must therefore distinguish oxidation state from bonding character and chemical stability.

Explanation: \(\textbf{Establish the common valence framework:}\) Both X and Y belong to Group 13, so their general valence-shell configuration is \(ns^2np^1\). Each neutral atom therefore has three outermost electrons. The observation that X exhibits a particularly important +3 state is consistent with substantial participation of these three electrons in bonding. For a lighter member such as aluminium, the +3 oxidation state is commonly favoured. \(\textbf{Interpret the heavier member:}\) The +1 oxidation state corresponds formally to participation of the \(np^1\) electron while the \(ns^2\) pair remains occupied. In heavier Group 13 members, the inert-pair effect makes participation of that pair less favourable. Consequently, the lower oxidation state becomes relatively more stable. Thallium provides an important example because Tl(I) is generally more stable than Tl(III). This difference does not arise because thallium has fewer valence electrons than aluminium. It arises from the changing relative tendency of the available electrons to participate in chemical bonding. The correct prediction must therefore preserve the common valence-electron count while allowing oxidation-state stability to change. This is why X can favour +3 compounds while Y exhibits greater relative stability in +1 compounds.

Explanation: \(\textbf{Identify X using two related properties:}\) An electron-deficient trihalide capable of accepting an electron pair is characteristic of boron trihalide behaviour. In a molecule such as \(\mathrm{BCl_3}\), the central boron atom has six electrons around it in the ordinary Lewis structure. It can accept a suitable donor lone pair and thereby act as a Lewis acid. Boron also forms the acidic oxide \(\mathrm{B_2O_3}\). Both the electron-deficiency evidence and the oxide evidence therefore identify X as boron. \(\textbf{Identify Y from amphoterism:}\) The oxide of Y reacts with both acids and strong bases. This establishes amphoteric behaviour rather than exclusively acidic or basic behaviour. Aluminium oxide, \(\mathrm{Al_2O_3}\), is amphoteric. Therefore, Y is aluminium. \(\textbf{Identify Z through oxidation-state stability:}\) The remaining element is thallium, but its assignment must also agree with the independent oxidation-state evidence. Thallium can exhibit +1 and +3 oxidation states, with Tl(I) generally more stable. The inert-pair effect helps explain this preference for the lower oxidation state. Thus, Z is thallium. The three identifications are mutually consistent with the common Group 13 valence configuration \(ns^2np^1\). Their differences arise from changes in atomic size, chemical character, bonding and the relative stability of oxidation states. The correct assignment is therefore X = boron, Y = aluminium and Z = thallium.

Explanation: \(\textbf{Compare the electronic configurations:}\) Aluminium has the configuration \([\mathrm{Ne}]\,3s^23p^1\), whereas gallium has \([\mathrm{Ar}]\,3d^{10}4s^24p^1\). Both elements possess three valence electrons and belong to Group 13. Gallium has an additional principal shell, so a simple shell-number argument would predict a larger atomic radius. However, the supplied radii show that gallium is slightly smaller than aluminium. \(\textbf{Account for the shielding difference:}\) The ten \(3d\) electrons in gallium do not shield the increasing nuclear charge as effectively as an equivalent number of electrons in more penetrating inner orbitals. Consequently, the outer electrons experience a relatively strong effective nuclear attraction. This increased attraction offsets the size increase expected from the additional principal shell. The result is a contraction sufficient to make gallium's covalent radius slightly smaller than aluminium's. The observation cannot be explained by claiming that gallium has fewer valence electrons or belongs to a lower period. It illustrates why increasing principal quantum number does not guarantee a perfectly regular increase in atomic radius down every p-block group. Thus, poor shielding by the filled \(3d\) subshell explains the observed exception.

Explanation: \(\textbf{Determine the physical states:}\) A pure substance is normally solid below its equilibrium melting point and liquid above it, provided no unusual metastable condition is maintained. The stated temperature of \(30^\circ\mathrm{C}\) is far below aluminium's melting point of approximately \(660^\circ\mathrm{C}\). Therefore, aluminium remains solid under the specified conditions. Gallium melts at approximately \(29.8^\circ\mathrm{C}\), which is slightly below the stated temperature. Consequently, gallium is liquid at equilibrium at \(30^\circ\mathrm{C}\). \(\textbf{Explain the physical-property difference:}\) Both elements possess three valence electrons and are metallic, but this does not imply identical cohesive forces in their solids. Melting depends on the bonding and structural arrangement of atoms in the condensed phase. Gallium has unusual solid-state bonding and packing characteristics that contribute to its low melting point. Aluminium has a different metallic crystal structure and substantially greater resistance to melting. Atomic size alone cannot provide a reliable quantitative prediction of these melting points. Likewise, the melting points of all Group 13 elements cannot be assumed to decrease uniformly with increasing atomic number. The correct conclusion follows from interpreting the supplied temperatures and recognising the influence of solid-state structure.

Explanation: \(\textbf{Interpret the common oxide formula:}\) In \(\mathrm{M_2O_3}\), each oxygen atom is assigned oxidation state \(-2\). The three oxygen atoms contribute a total oxidation-state sum of \(-6\). Therefore, the two atoms of M must together contribute +6, giving M an oxidation state of +3. This electron-accounting result applies to both \(\mathrm{B_2O_3}\) and \(\mathrm{In_2O_3}\). However, identical oxidation states do not require identical oxide acid–base behaviour. \(\textbf{Compare the parent elements:}\) Boron is much less metallic than indium and forms predominantly covalent compounds. Its oxide, \(\mathrm{B_2O_3}\), is acidic. Indium is a metallic member of Group 13 and forms an oxide with appreciably greater basic character. \(\mathrm{In_2O_3}\) can also exhibit amphoteric behaviour under suitable conditions, so its increased basic tendency should not be interpreted as proof that it is exclusively basic. The trend is consistent with the broad increase in metallic character down the group. The essential distinction is that oxidation-state equality determines formal electron balance, whereas oxide acid–base behaviour depends strongly on chemical bonding and the character of the parent element. Hence, indium oxide has a greater basic tendency than boron oxide without requiring a different oxidation state.

Explanation: \(\textbf{Identify the elements:}\) The configuration of X contains two electrons in the \(4s\) subshell and two electrons in the \(4p\) subshell. Its complete ground-state configuration corresponds to atomic number \(32\), identifying X as germanium. The highest principal quantum number is four, placing germanium in Period 4. The configuration of Y corresponds to atomic number \(50\), identifying Y as tin. Its outermost electrons occupy the \(5s\) and \(5p\) subshells, placing it in Period 5. \(\textbf{Connect configuration with character:}\) Both configurations end in \(ns^2np^2\), the characteristic pattern of Group 14. Each element therefore has four valence electrons. However, identical general valence configurations do not imply identical metallic character. Germanium is commonly classified as a metalloid, whereas tin is metallic. Moving down Group 14, increasing atomic size and changes in electron attraction generally favour increasing metallic character. Tin consequently shows more pronounced metallic behaviour than germanium. The filled inner \(d\) subshells do not convert either element into a d-block element because the differentiating electron enters a \(p\) subshell. Thus, the correct identification must combine electronic configuration, periodic position and the down-group change in chemical character.