p-Block Groups 13 to 18

8 MCQs9-step worked example
Source: NCERT p-Block ElementsPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025, 2026Official key: NTA-verifiedLast updated: 27 Sep 2026

p-Block Groups 13 to 18, explained for NEET

The group survey is where a single counting slip costs a mark. Group 13 sits in the periodic table's thirteenth column, but its elements do not carry thirteen valence electrons — they carry three: ns² np¹. Subtract 10 for the filled d block whenever the group number exceeds 12. Group 15 gives five (ns² np³), Group 18 gives eight (ns² np⁶), and helium alone breaks the np⁶ pattern with 1s².

NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, page 315 sets the frame: the p-block spans Groups 13 to 18, and every member's valence configuration is ns² np¹⁻⁶. That one formula generates the whole survey. Read the superscript on p and you have the group; read the group and you have the superscript.

Oxidation states follow, but with a caveat repeaters keep losing marks on. The group oxidation state for a p-block group is (group number − 10) — +3 for Group 13, +4 for Group 14, +5 for Group 15. Heavier members frequently prefer a state two units lower, because the ns² pair becomes reluctant to bond. So Group 13's survey answer is "+3 and +1," not "+3." Thallium's preference for +1 is not an exception to be memorised separately; it is the bottom end of a trend the group description must already carry.

Chapter 11 page 323 records that non-metallic character weakens down each p-block group while metallic character strengthens — Group 14 running carbon (non-metal) through silicon and germanium (metalloids) to tin and lead (metals) inside one column.

Preparing chlorine (Group 17). NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, page 202 gives two laboratory methods. (i) Heat manganese dioxide with concentrated hydrochloric acid: MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O; a mixture of common salt and concentrated H₂SO₄ is used in place of HCl: 4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂. (ii) The action of HCl on potassium permanganate: 2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂. Read the oxidation states off these equations before judging any statement about them: manganese is reduced (+4 in MnO₂, +7 in KMnO₄, to +2 in MnCl₂) while chloride is oxidised to Cl₂. The same page lists Deacon's process (HCl oxidised by air over CuCl₂ at 723 K) and the electrolysis of brine, where chlorine is liberated at the anode, as the manufacturing routes.

Watch-out: the survey question usually asks for common oxidation states, plural. An option listing only the group state is incomplete for any group whose heavier members drop by two.

Can you answer these p-Block Groups 13 to 18 MCQs?

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The general outer electronic configuration of p-block elements is:

Show answer and why every option is right or wrong

Answer: A. A is correct. NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, page 315 defines the p-block as those elements whose last electron enters an np orbital, giving the general valence configuration ns² np¹⁻⁶ across Groups 13 to 18.

Why B is wrong: B is wrong because it allows ns¹, which describes no p-block ground state; the s subshell is filled before any np electron enters.

Why C is wrong: C is wrong because (n−1)d¹⁻¹⁰ ns¹⁻² is the d-block (transition element) configuration, not the p-block.

Why D is wrong: D is wrong because it places electrons in nd orbitals of the same shell; p-block valence electrons occupy only ns and np.

MCQ 2Direct ApplicationPractice

An element has the outer electronic configuration 4s² 4p³. It belongs to:

Show answer and why every option is right or wrong

Answer: C. C is correct. The principal quantum number 4 fixes the period; for p-block elements the group is 10 + (s + p electrons) = 10 + 5 = 15, per the configuration-to-group rule in NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, page 315.

Why A is wrong: A is wrong because Group 13 requires np¹, giving 4s² 4p¹ — this element has three p electrons, not one.

Why B is wrong: B is wrong because it counts only the valence electrons and stops there, omitting the +10 offset that the intervening d block imposes on p-block group numbering.

Why D is wrong: D is wrong because the period is read from n in the valence shell; n = 4 places the element in period 4, not period 3.

MCQ 3Easy RecallPractice

The number of valence electrons in a Group 16 element is:

Show answer and why every option is right or wrong

Answer: B. B is correct. Group 16 elements have the configuration ns² np⁴, giving 2 + 4 = 6 valence electrons; the group number minus 10 recovers the np occupancy (NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, page 315).

Why A is wrong: A is wrong because four is the np occupancy alone; the two filled ns electrons are also valence electrons and must be added.

Why C is wrong: C is wrong because it reads the group number directly as the electron count, which only works for Groups 1 and 2, not for any p-block group.

Why D is wrong: D is wrong because ten is the d-block offset subtracted from the group number, not an electron count for any p-block member.

MCQ 4Direct ApplicationPractice

The group oxidation state of Group 14 elements is:

Show answer and why every option is right or wrong

Answer: A. A is correct. The group oxidation state equals the total number of valence electrons available, which for Group 14 (ns² np²) is 4. NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, page 315 ties the maximum positive oxidation state of a p-block group to its valence-electron count.

Why B is wrong: B is wrong because it transcribes the group number as the oxidation state; no element exhibits +14.

Why C is wrong: C is wrong because +2 is the lower state preferred by the heavier members, arising when the ns² pair does not participate — it is not the group oxidation state itself.

Why D is wrong: D is wrong because it fixes carbon's negative state as the group state; Group 14 elements show +4 as the group oxidation state and the heavier members also show +2.

MCQ 5Easy RecallPractice

Helium is placed in Group 18 although its valence configuration is:

Show answer and why every option is right or wrong

Answer: D. D is correct. Helium has only two electrons, both in 1s, so it cannot show the ns² np⁶ configuration of the other noble gases; it is grouped with them on the strength of a completely filled valence shell (NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, page 315).

Why A is wrong: A is wrong because 1s² 2s² is beryllium's configuration (four electrons), a Group 2 element.

Why B is wrong: B is wrong because 1s¹ is hydrogen's configuration; helium has two electrons, completing the 1s subshell.

Why C is wrong: C is wrong because 2s² 2p⁶ is neon's valence configuration — that is the ns² np⁶ pattern helium precisely does not follow.

MCQ 6Concept TrapPractice

For a Group 13 element, an examiner asks for the common oxidation states. The most complete answer is:

Show answer and why every option is right or wrong

Answer: B. B is correct. Group 13 shows the group state +3, and the heavier members also show +3 − 2 = +1 because the ns² pair becomes increasingly reluctant to participate in bonding down the group; a survey answer for the group must carry both.

Why A is wrong: A is wrong because it reports only the group oxidation state and omits the lower state the heavier congeners show — incomplete for a question asking for common states, plural.

Why C is wrong: C is wrong because it universalises the heavier members' preference; the lighter members (boron, aluminium) use all three valence electrons and show +3.

Why D is wrong: D is wrong because it treats the group number as the oxidation state; the group state is (group number − 10) = +3.

MCQ 7Direct ApplicationPractice

An element has the outer electronic configuration 5s² 5p⁵. Its group oxidation state is

Show answer and why every option is right or wrong

Answer: B. B is correct. The group oxidation state equals the total count of ns and np valence electrons: 2 + 5 = 7, a single application of the rule tying oxidation state to valence-electron count (NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, page 315).

Why A is wrong: A is wrong because it counts only the np electrons (5) and drops the ns² pair, which is also part of the valence shell.

Why C is wrong: C is wrong because it transcribes the group number (10 + 7 = 17) as the oxidation state instead of the valence-electron count itself.

Why D is wrong: D is wrong because it counts only the ns² electrons (2) and drops the five np electrons — the opposite slip from option A, which drops the ns² pair instead.

MCQ 8CalculationPractice

Two elements, P (3s² 3p²) and Q (3s² 3p⁴), lie in the same period. The difference between the group oxidation state of P and the number of valence electrons in Q is:

Show answer and why every option is right or wrong

Answer: D. D is correct. P has 4 valence electrons, so its group oxidation state is +4; Q has 2 + 4 = 6 valence electrons. The difference 4 − 6 = −2, following the configuration-to-group rules of NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, page 315.

Why A is wrong: A is wrong because it subtracts in the reverse order (6 − 4) or discards the sign; the stem specifies P's value minus Q's.

Why B is wrong: B is wrong because it uses the d-block offset of 10 rather than computing either quantity from the configurations given.

Why C is wrong: C is wrong because it treats Q's valence count as equal to P's oxidation state; P has four valence electrons and Q has six, so the two values cannot coincide.

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How do you solve a p-Block Groups 13 to 18 question? A worked example

  1. 1

    Given

    An element E has ground-state electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p¹. Atomic number is to be read off the configuration. Counting integers here (electron counts, the 10 in the group rule) are exact and do not carry significant figures.

  2. 2

    Required

    (i) the atomic number of E; (ii) its block, period and group; (iii) its group oxidation state and the other common oxidation state its group displays.

  3. 3

    Concept

    For a p-block element, the last electron enters an np orbital. The period is the principal quantum number of the valence shell. The group is 10 + (number of ns and np valence electrons), the 10 accounting for the intervening d block. The group oxidation state equals the count of those valence electrons; heavier congeners may show a state two units lower when the ns² pair does not bond (NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, pages 315 and 323).

  4. 4

    Formula

    Group number = 10 + (ns + np electrons), for p-block elements only. Group oxidation state = (group number − 10). Lower common state = (group oxidation state − 2).

  5. 5

    Substitution

    Total electrons = 2 + 2 + 6 + 2 + 6 + 10 + 2 + 1 = 31 (exact count). Valence shell is n = 4; valence electrons = 4s² + 4p¹ = 2 + 1 = 3 (exact count). Group = 10 + 3.

  6. 6

    Calculation

    Atomic number Z = 31. Group = 10 + 3 = 13. Group oxidation state = 13 − 10 = +3. Lower common state = 3 − 2 = +1. Every number in this calculation is an exact counting integer — electron counts and the constant 10 in the group rule — so none of them limits the significant figures of the result; the answers are exact integers, not measured quantities.

  7. 7

    Final answer

    (i) Z = 31; (ii) p-block, period 4, Group 13; (iii) group oxidation state +3, with +1 also shown by the group's heavier members. All values exact.

  8. 8

    Common trap

    Adding the 3d¹⁰ electrons to the valence count. They belong to the penultimate shell and are already absorbed into the +10 offset of the group rule — count them once as the offset or not at all, never both. A student who adds them reaches "13 valence electrons" and then reports Group 23, which does not exist.

  9. 9

    Similar NEET-style question

    An element has the configuration [Ar] 3d¹⁰ 4s² 4p⁴. State its group, its group oxidation state, and the oxidation state its heaviest congener prefers. *(Answer: Group 16; group oxidation state +6; the heaviest congener prefers +4 — two units lower, the ns² pair staying unbonded.)*

What to remember before solving p-Block Groups 13 to 18 questions

In the NEET syllabus; removed from current NCERT.

In p-block elements the last electron enters the outermost p orbital. As we know that the number of p orbitals is three and, therefore, the maximum number of electrons that can be accommodated in a set of p orbitals is six. Consequently there are six groups of p–block elements in the periodic table numbering from 13 to 18. Boron, carbon, nitrogen, oxygen, fluorine and helium head the groups. Their valence shell electronic configuration is ns2np1-6(except for He). The inner core of the electronic configuration may, however, differ. The difference in inner core of elements greatly influences their physical properties (such as atomic and ionic radii, ionisation enthalpy, etc.) as well as chemical properties. Consequently, a lot of variation in properties of elements in a group of p-block is observed. The maximum oxidation state shown by a p-block element is equal to the total number of valence electrons (i.e., the sum of the s- and p-electrons). Clearly, the number of possible oxidation states increases towards the right of the periodic table. In addition to this so called group oxidation state, p-block elements may show other oxidation states which normally, but not necessarily, differ from the total number of valence electrons by unit of two. The important oxidation states exhibited by p-block elements are shown in Table 11.1. In boron, carbon and nitrogen families the group oxidation state is the most stable state for the lighter elements in the group. However, the oxidation state two unit less than the group oxidation state becomes progressively more stable for the heavier elements in each group. The occurrence of oxidation states two unit less than the group oxidation states are sometime attributed to the ‘inert pair effect’.

-- NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, p. 315

In the NEET syllabus; removed from current NCERT.

11.1 GROUP 13 ELEMENTS: THE BORON FAMILY This group elements show a wide variation in properties. Boron is a typical non-metal, aluminium is a metal but shows many chemical similarities to boron, and gallium, indium, thallium and nihonium are almost exclusively metallic in character. … 11.1.1 Electronic Configuration The outer electronic configuration of these elements is ns2np1. A close look at the electronic configuration suggests that while boron and aluminium have noble gas core, gallium and indium have noble gas plus 10 d-electrons, and thallium has noble gas plus 14 f- electrons plus 10 d-electron cores. … 11.1.2 Atomic Radii On moving down the group, for each successive member one extra shell of electrons is added and, therefore, atomic radius is expected to increase. However, a deviation can be seen. Atomic radius of Ga is less than that of Al. This can be understood from the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect (Unit 2) for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic radius of gallium (135 pm) is less than that of aluminium (143 pm). 11.1.3 Ionization Enthalpy The ionisation enthalpy values as expected from the general trends do not decrease smoothly down the group. The decrease from B to Al is associated with increase in size. The observed discontinuity in the ionisation enthalpy values between Al and Ga, and between In and Tl are due to inability of d- and f-electrons ,which have low screening effect, to compensate the increase in nuclear charge. The order of ionisation enthalpies, as expected, is ∆iH1 <∆iH2 <∆iH3. The sum of the first three ionisation enthalpies for each of the elements is very high. Effect of this will be apparent when you study their chemical properties. 11.1.4 Electronegativity Down the group, electronegativity first decreases from B to Al and then increases marginally (Table 11.2). This is because of the discrepancies in atomic size of the elements. Table 11.2 (B, Al, Ga, In, Tl): Ionization enthalpy ∆iH1 (kJ mol–1) 801, 577, 579, 558, 589; Electronegativity (Pauling scale) 2.0, 1.5, 1.6, 1.7, 1.8; Melting point / K 2453, 933, 303, 430, 576.

-- NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, p. 317

In the NEET syllabus; removed from current NCERT.

11.5.1 Electronic Configuration The valence shell electronic configuration of these elements is ns2np2. The inner core of the electronic configuration of elements in this group also differs. 11.5.2 Covalent Radius There is a considerable increase in covalent radius from C to Si, thereafter from Si to Pb a small increase in radius is observed. This is due to the presence of completely filled d and f orbitals in heavier members. 11.5.3 Ionization Enthalpy The first ionization enthalpy of group 14 members is higher than the corresponding members of group 13. The influence of inner core electrons is visible here also. In general the ionisation enthalpy decreases down the group. Small decrease in ∆iH from Si to Ge to Sn and slight increase in ∆iH from Sn to Pb is the consequence of poor shielding effect of intervening d and f orbitals and increase in size of the atom. 11.5.4 Electronegativity Due to small size, the elements of this group are slightly more electronegative than group 13 elements. The electronegativity values for elements from Si to Pb are almost the same. 11.5.5 Physical Properties All members of group14 are solids. Carbon and silicon are non-metals, germanium is a metalloid, whereas tin and lead are soft metals with low melting points. Melting points and boiling points of group 14 elements are much higher than those of corresponding elements of group 13. 11.5.6 Chemical Properties Oxidation states and trends in chemical reactivity The group 14 elements have four electrons in outermost shell. The common oxidation states exhibited by these elements are +4 and +2. Carbon also exhibits negative oxidation states. Since the sum of the first four ionization enthalpies is very high, compounds in +4 oxidation state are generally covalent in nature. In heavier members the tendency to show +2 oxidation state increases in the sequence Ge<Sn<Pb. It is due to the inability of ns2 electrons of valence shell to participate in bonding. The relative stabilities of these two oxidation states vary down the group. Carbon and silicon mostly show +4 oxidation state. Germanium forms stable compounds in +4 state and only few compounds in +2 state. Tin forms compounds in both oxidation states (Sn in +2 state is a reducing agent). Lead compounds in +2 state are stable and in +4 state are strong oxidising agents.

-- NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, p. 323

In the NEET syllabus; removed from current NCERT.

Group 15 includes nitrogen, phosphorus, arsenic, antimony, bismuth and moscovium. As we go down the group, there is a shift from non-metallic to metallic through metalloidic character. Nitrogen and phosphorus are non-metals, arsenic and antimony metalloids, bismuth and moscovium are typical metals. … 7.1.2 Electronic Configuration The valence shell electronic configuration of these elements is ns2np3. The s orbital in these elements is completely filled and p orbitals are half-filled, making their electronic configuration extra stable. 7.1.3 Atomic and Ionic Radii Covalent and ionic (in a particular state) radii increase in size down the group. There is a considerable increase in covalent radius from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due to the presence of completely filled d and/or f orbitals in heavier members. 7.1.4 Ionisation Enthalpy Ionisation enthalpy decreases down the group due to gradual increase in atomic size. Because of the extra stable half-filled p orbitals electronic configuration and smaller size, the ionisation enthalpy of the group 15 elements is much greater than that of group 14 elements in the corresponding periods. The order of successive ionisation enthalpies, as expected is ∆iH1 < ∆iH2 < ∆iH3 (Table 7.1). 7.1.5 Electronegativity The electronegativity value, in general, decreases down the group with increasing atomic size. However, amongst the heavier elements, the difference is not that much pronounced. 7.1.6 Physical Properties All the elements of this group are polyatomic. Dinitrogen is a diatomic gas while all others are solids. Metallic character increases down the group. Nitrogen and phosphorus are non-metals, arsenic and antimony metalloids and bismuth is a metal. This is due to decrease in ionisation enthalpy and increase in atomic size. The boiling points, in general, increase from top to bottom in the group but the melting point increases upto arsenic and then decreases upto bismuth. Except nitrogen, all the elements show allotropy. Table 7.1 (N, P, As, Sb, Bi): Ionisation enthalpy I (kJ mol–1) 1402, 1012, 947, 834, 703; Electronegativity 3.0, 2.1, 2.0, 1.9, 1.9.

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 170

In the NEET syllabus; removed from current NCERT.

Ionisation enthalpy decreases down the group. It is due to increase in size. However, the elements of this group have lower ionisation enthalpy values compared to those of Group15 in the corresponding periods. This is due to the fact that Group 15 elements have extra stable half-filled p orbitals electronic configurations. Because of the compact nature of oxygen atom, it has less negative electron gain enthalpy than sulphur. However, from sulphur onwards the value again becomes less negative upto polonium. Next to fluorine, oxygen has the highest electronegativity value amongst the elements. Within the group, electronegativity decreases with an increase in atomic number. This implies that the metallic character increases from oxygen to polonium. Oxidation states and trends in chemical reactivity The elements of Group 16 exhibit a number of oxidation states (Table 7.6). The stability of -2 oxidation state decreases down the group. Polonium hardly shows –2 oxidation state. Since electronegativity of oxygen is very high, it shows only negative oxidation state as –2 except in the case of OF2 where its oxidation state is + 2. Other elements of the group exhibit + 2, + 4, + 6 oxidation states but + 4 and + 6 are more common. Anomalous behaviour of oxygen The anomalous behaviour of oxygen, like other members of p-block present in second period is due to its small size and high electronegativity. One typical example of effects of small size and high electronegativity is the presence of strong hydrogen bonding in H2O which is not found in H2S.

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 187

In the NEET syllabus; removed from current NCERT.

All these elements have seven electrons in their outermost shell (ns2np5) which is one electron short of the next noble gas. The halogens have the smallest atomic radii in their respective periods due to maximum effective nuclear charge. The atomic radius of fluorine like the other elements of second period is extremely small. Atomic and ionic radii increase from fluorine to iodine due to increasing number of quantum shells. They have little tendency to lose electron. Thus they have very high ionisation enthalpy. Due to increase in atomic size, ionisation enthalpy decreases down the group. Halogens have maximum negative electron gain enthalpy in the corresponding periods. This is due to the fact that the atoms of these elements have only one electron less than stable noble gas configurations. Electron gain enthalpy of the elements of the group becomes less negative down the group. However, the negative electron gain enthalpy of fluorine is less than that of chlorine. It is due to small size of fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2p orbitals of fluorine and thus, the incoming electron does not experience much attraction. Table 7.8 (F, Cl, Br, I): Electronegativity 4, 3.2, 3.0, 2.7; Bond dissociation enthalpy /(kJ mol–1) 158.8, 242.6, 192.8, 151.1. … They have very high electronegativity. The electronegativity decreases down the group. Fluorine is the most electronegative element in the periodic table. … One curious anomaly we notice from Table 7.8 is the smaller enthalpy of dissociation of F2 compared to that of Cl2 whereas X-X bond dissociation enthalpies from chlorine onwards show the expected trend: Cl – Cl > Br – Br > I – I. A reason for this anomaly is the relatively large electron-electron repulsion among the lone pairs in F2 molecule where they are much closer to each other than in case of Cl2. Oxidation states and trends in chemical reactivity All the halogens exhibit –1 oxidation state. However, chlorine, bromine and iodine exhibit + 1, + 3, + 5 and + 7 oxidation states also as explained below: … The ready acceptance of an electron is the reason for the strong oxidising nature of halogens. F2 is the strongest oxidising halogen and it oxidises other halide ions in solution or even in the solid phase. In general, a halogen oxidises halide ions of higher atomic number.

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 198

In the NEET syllabus; removed from current NCERT.

All noble gases have general electronic configuration ns2np6 except helium which has 1s2 (Table 7.12). Many of the properties of noble gases including their inactive nature are ascribed to their closed shell structures. Due to stable electronic configuration these gases exhibit very high ionisation enthalpy. However, it decreases down the group with increase in atomic size. Atomic radii increase down the group with increase in atomic number. Since noble gases have stable electronic configurations, they have no tendency to accept the electron and therefore, have large positive values of electron gain enthalpy. Physical Properties All the noble gases are monoatomic. They are colourless, odourless and tasteless. They are sparingly soluble in water. They have very low melting and boiling points because the only type of interatomic interaction in these elements is weak dispersion forces. Helium has the lowest boiling point (4.2 K) of any known substance. It has an unusual property of diffusing through most commonly used laboratory materials such as rubber, glass or plastics. Chemical Properties In general, noble gases are least reactive. Their inertness to chemical reactivity is attributed to the following reasons: (i) The noble gases except helium (1s2) have completely filled ns2np6 electronic configuration in their valence shell. (ii) They have high ionisation enthalpy and more positive electron gain enthalpy. … After this discovery, a number of xenon compounds mainly with most electronegative elements like fluorine and oxygen, have been synthesised. … The structures of the three xenon fluorides can be deduced from VSEPR and these are shown in Fig. 7.9. XeF2 and XeF 4 have linear and square planar structures respectively. XeF6 has seven electron pairs (6 bonding pairs and one lone pair) and would, thus, have a distorted octahedral structure as found experimentally in the gas phase. Fig. 7.9 The structures of (a) XeF2 (b) XeF4 (c) XeF6 (d) XeOF4 and (e) XeO3: (a) Linear (b) Square planar (c) Distorted octahedral (d) Square pyramidal (e) Pyramidal

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 209

In the NEET syllabus; removed from current NCERT.

The absence of d orbitals in oxygen limits its covalency to four and in practice, rarely exceeds two. On the other hand, in case of other elements of the group, the valence shells can be expanded and covalence exceeds four. (i) Reactivity with hydrogen: All the elements of Group 16 form hydrides of the type H2E (E = O, S, Se, Te, Po). Some properties of hydrides are given in Table 7.7. Their acidic character increases from H2O to H2Te. The increase in acidic character can be explained in terms of decrease in bond enthalpy for the dissociation of H–E bond down the group. Owing to the decrease in enthalpy for the dissociation of H–E bond down the group, the thermal stability of hydrides also decreases from H2O to H2Po. All the hydrides except water possess reducing property and this character increases from H2S to H2Te. Table 7.7: Properties of Hydrides of Group 16 Elements (H2O, H2S, H2Se, H2Te) — m.p/K: 273, 188, 208, 222; b.p/K: 373, 213, 232, 269; H–E distance/pm: 96, 134, 146, 169; HEH angle (°): 104, 92, 91, 90; ΔfH/kJ mol–1: –286, –20, 73, 100; ΔdissH (H–E)/kJ mol–1: 463, 347, 276, 238; Dissociation constant (aqueous solution, 298 K): 1.8×10–16, 1.3×10–7, 1.3×10–4, 2.3×10–3.

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 188

In the NEET syllabus; removed from current NCERT.

Chlorine can be prepared by heating manganese dioxide with concentrated hydrochloric acid, MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O. A mixture of common salt and concentrated H2SO4 is used in place of HCl: 4NaCl + MnO2 + 4H2SO4 → MnCl2 + 4NaHSO4 + 2H2O + Cl2.

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 202

More in p-Block Elements: 12 exam traps and mistakes from its other lessons.

p-Block Groups 13 to 18 questions from past NEET papers

9 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025, 2026. Answers verified against NTA official keys.

NEET 2026

Given below are two statements: Statement-I : Heating NaCl with concentrated H₂SO₄ and MnO₂ results in oxidation of Mn. Statement-II : Heating NaI with concentrated H₂SO₄ and MnO₂ results in reduction of Mn. In light of the above statements, choose the most appropriate answer from the options given below:

1Both Statement-I and Statement-II are correct.
2Both Statement-I and Statement-II are incorrect.
3Statement-I is correct but Statement-II is incorrect.
4Statement-I is incorrect but Statement-II is correct.
NTA Answer: Option 4(final)
NEET 2025

Given below are two statements : Statement I : Like nitrogen that can form ammonia, arsenic can form arsine. Statement II : Antimony cannot form antimony pentoxide. In the light of the above statements, choose the most appropriate answer from the options given below :

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 4(final)
NEET 2023

Match List-I with List-II : List-I (Oxoacids of Sulphur) List-II (Bonds) A. Peroxodisulphuric acid I. Two S–OH, Four S=O, One S–O–S B. Sulphuric acid II. Two S–OH, One S=O C. Pyrosulphuric acid III. Two S–OH, Four S=O, One S–O–O–S D. Sulphurous acid IV. Two S–OH, Two S=O Choose the correct answer from the options given below.

1A–III, B–IV, C–I, D–II
2A–I, B–III, C–IV, D–II
3A–III, B–IV, C–II, D–I
4A–I, B–III, C–II, D–IV
NTA Answer: Option 1(final)

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