First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family: (1) 1012 kJ mol-1 (2) 1402 kJ mol-1 (3) 834 kJ mol-1 (4) 947 kJ mol-1

The main element of the apatite mineral family (e.g., fluorapatite Ca5(PO4)3F) is Phosphorus (P). The first four elements of Group 15 are N, P, As, Sb. First ionization enthalpy decreases regularly down the group: Sorting the given enthalpy data values in decreasing order: Assigning these to the elements: * N = 1402 kJ mol-1 * P = 1012 kJ mol-1 * As = 947 kJ mol-1 * Sb = 834 kJ mol-1 Apatite family = Phosphorus reference. Match the elements down a column directly to a monotonic numerical array. Class 12 Chemistry: p-Block Elements

First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family: (1) 1012 kJ mol-1 (2) 1402 kJ mol

The main element of the apatite mineral family (e.g., fluorapatite Ca5(PO4)3F) is Phosphorus (P). The first four elements of Group 15 are N, P, As, Sb. First ionization enthalpy decreases regularly down the group: Sorting the given enthalpy data values in decreasing order: Assigning these to the elements: * N = 1402 kJ mol-1 * P = 1012 kJ mol-1 * As = 947 kJ mol-1 * Sb = 834 kJ mol-1 Apatite family = Phosphorus reference. Match the elements down a column directly to a monotonic numerical array.

First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family: (1) 1012 kJ mol-1 (2) 1402 kJ mol-1 (3) 834 kJ mol-1 (4) 947 kJ mol-1

The main element of the apatite mineral family (e.g., fluorapatite Ca5(PO4)3F) is Phosphorus (P). The first four elements of Group 15 are N, P, As, Sb. First ionization enthalpy decreases regularly down the group: Sorting the given enthalpy data values in decreasing order: Assigning these to the elements: * N = 1402 kJ mol-1 * P = 1012 kJ mol-1 * As = 947 kJ mol-1 * Sb = 834 kJ mol-1 Apatite family = Phosphorus reference. Match the elements down a column directly to a monotonic numerical array. Class 12 Chemistry: p-Block Elements

First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family: (1) 1012 kJ mol-1 (2) 1402 kJ mol

The main element of the apatite mineral family (e.g., fluorapatite Ca5(PO4)3F) is Phosphorus (P). The first four elements of Group 15 are N, P, As, Sb. First ionization enthalpy decreases regularly down the group: Sorting the given enthalpy data values in decreasing order: Assigning these to the elements: * N = 1402 kJ mol-1 * P = 1012 kJ mol-1 * As = 947 kJ mol-1 * Sb = 834 kJ mol-1 Apatite family = Phosphorus reference. Match the elements down a column directly to a monotonic numerical array.

First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the…

Year	2026	2025	2024	Total
Questions	16	13	13	42

Q51 jee_main_2026_21_jan_morning Reactions of Lead Compounds

Consider the following reactions.

PbCl₂ + K₂CrO₄ arrow A + 2KCl

(Hot solution)

A + NaOH leftharpoons B + Na₂CrO₄

PbSO₄ + 4CH₃COONH₄ arrow (NH₄)₂SO₄ + X

In the above reactions, A, B and X are respectively

A. $Na₂[Pb(OH)₂] , PbCrO₄ and (NH₄)₂[Pb(CH₃COO)₄]$
B. $PbCrO₄ , Na₂[Pb(OH)₄] and [Pb(NH₃)₄]SO₄$
C. $Na₂[Pb(OH)₂] , PbCrO₄ and [Pb(NH₃)₄]SO₄$
D. $PbCrO₄ , Na₂[Pb(OH)₄] and (NH₄)₂[Pb(CH₃COO)₄]$

Solution

Core Logic

The precipitation and complex formation reactions of Lead are:

PbCl₂ + K₂CrO₄ arrow PbCrO₄ + 2KCl (Hot solution) so, A is PbCrO₄

PbCrO₄ + 4NaOH (excess) arrow Na₂[Pb(OH)₄] + Na₂CrO₄ so, B is Na₂[Pb(OH)₄]

PbSO₄ + 4CH₃COONH₄ arrow (NH₄)₂ [Pb(CH₃COO)₄] + (NH₄)₂SO₄ so, X is (NH₄)₂[Pb(CH₃COO)₄]

Pattern Recognition

Lead forms a yellow precipitate of lead chromate (A), which is amphoteric and dissolves in excess NaOH to form soluble plumbate(II) complex (B). It also forms a stable soluble complex with ammonium acetate (X).

Chapter Mix

Class 11 Chemistry: p-Block Elements Class 12 Chemistry: d and f Block Elements

Q55 jee_main_2026_21_jan_morning Group 13 and 14 Compounds

Given below are two statements : Statement I : The number of pairs among

[SiO₂, CO₂]

,

[SnO, SnO₂]

,

[PbO, PbO₂]

and

[GeO, GeO₂]

, which contain oxides that are both amphoteric is 2. Statement II :

BF₃

is an electron deficient molecule can act as a lewis acid, forms adduct with

NH₃

and has a trigonal planar geometry. In the light of the above statement, choose the correct answer from the option given below.

A. $Both Statement I and Statement II are true.$
B. $Both Statement I and Statement II are false.$
C. $Statement I is true but Statement II is false.$
D. $Statement I is false Statement II is true.$

Solution

Core Logic

Evaluating Statement I:

$SiO₂$ $SiO_2$, $CO₂$ $CO_2$, $GeO$ $GeO$, $GeO₂$ $GeO_2$ are acidic in nature.
$SnO$ $SnO$, $SnO₂$ $SnO_2$, $PbO$ $PbO$, $PbO₂$ $PbO_2$ are amphoteric in nature.

Therefore, the pairs $[SnO, SnO₂]$ $[SnO, SnO_2]$ and $[PbO, PbO₂]$ $[PbO, PbO_2]$ contain oxides that are both amphoteric. Number of such pairs = 2. Statement I is True.

Evaluating Statement II:

$BF₃$ $BF_3$ has 6 electrons in the outermost shell of the central Boron atom. It is electron-deficient and acts as a Lewis acid.
It accepts a lone pair from Lewis bases like $NH₃$ $NH_3$ to form an adduct.
In $BF₃$ $BF_3$, Boron is $sp²$ $sp^2$ hybridized, resulting in a trigonal planar geometry. Statement II is True.

Step 1: Conclusion

Both statements are factually correct.

Pattern Recognition

Oxides of heavier Group 14 elements (Sn, Pb) are typically amphoteric in both their +2 and +4 oxidation states. $BF₃$ $BF_3$ is the quintessential Lewis acid.

Chapter Mix

Class 11 Chemistry: p-Block Elements Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q70 jee_main_2026_21_jan_evening Chromyl Chloride Test

On heating a mixture of common salt and

K₂Cr₂O₇

in equal amount along with concentrated

H₂SO₄

in a test tube, a gas is evolved. Formula of the gas evolved and oxidation state of the central metal atom in the gas respectively are: (1)

CrO₂Cl₂

and

+5

(2)

CrO₂Cl₂

and

+6

(3)

Cr₂O₂Cl₂

and

+6

(4)

Cr₂O₂Cl₂

and

+3

A. $(1) CrO₂Cl₂ and +5$
B. $(2) CrO₂Cl₂ and +6$
C. $(3) Cr₂O₂Cl₂ and +6$
D. $(4) Cr₂O₂Cl₂ and +3$

Solution

Core Logic

This is the classic Chromyl Chloride test:

4NaCl + K₂Cr₂O₇ + 6H₂SO₄ 2KHSO₄ + 2CrO₂Cl₂ + 4NaHSO₄ + 3H₂O

In chromyl chloride ( $CrO₂Cl₂$ $\text{CrO}_2\text{Cl}_2$), chromium is in the $+6$ $+6$ oxidation state.

Step 1: Final Conclusion

The gas is $CrO₂Cl₂$ $\text{CrO}_2\text{Cl}_2$ and the oxidation state of Cr is $+6$ $+6$, corresponding to option (2).

Pattern Recognition

Sees: qualitative analysis test for chloride ions (chromyl chloride test). Trap: Confusing oxidation state of chromium in dichromate versus chromyl chloride.

Chapter Mix

Class 12 Chemistry: p-Block Elements

Q55 jee_main_2026_22_january_morning Ionization Enthalpy Trends

A 'p'-block element (E) and hydrogen form a binary cation

(EHₓ)⁺

, while

EH₃

on treatment with

K₂HgI₄

in alkaline medium gives a precipitate of basic mercury(II)amido-iodine. Given below are first ionisation enthalpy values (

kJ mol⁻¹

) for first element each from group 13, 14, 15 and 16. Identify the correct first ionisation enthalpy value for element E.

A. $1312$ $\text{1312}$
B. $1086$ $\text{1086}$
C. $1402$ $\text{1402}$
D. $801$ $\text{801}$

Solution

Related Formula

NH₃ + K₂HgI₄ + KOH arrow HgO\cdot Hg(NH₂)I + KI + H₂O

Core Logic

The reagent $K₂HgI₄$ $K_2HgI_4$ in alkaline medium is Nessler's reagent. It gives a brown precipitate (iodide of Millon's base) with ammonia ( $NH₃$ $NH_3$). Therefore, the compound $EH₃$ $EH_3$ is $NH₃$ $NH_3$, and the element (E) is Nitrogen (N). The binary cation is the ammonium ion $(NH₄^+)$ $(NH_4^+)$.

We need to find the first ionization enthalpy of Nitrogen among the first elements of groups 13 (B), 14 (C), 15 (N), and 16 (O). Due to its stable half-filled $2p³$ $2p^3$ configuration, Nitrogen has a remarkably high first ionization energy, higher than Oxygen. The order is: $B < C < O < N$ $B < C < O < N$. Among the given values (801, 1086, 1312, 1402), 1402 $kJ mol⁻¹$ $kJ\text{ mol}^{-1}$ is the highest and corresponds to Nitrogen.

Step 1: Final Conclusion

Element E is Nitrogen, and its correct first ionization enthalpy is 1402 $kJ mol⁻¹$ $kJ\text{ mol}^{-1}$.

Pattern Recognition

Always remember the ionization energy exception across period 2: N > O and Be > B due to stable half-filled and fully-filled orbitals.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: p-Block Elements

Q64 jee_main_2026_22_january_morning Hydrides and Bond Properties

Given below are two statements: Statement I: The halogen that makes longest bond with hydrogen in HX, has the smallest covalent radius in its group. Statement II: A group 15 element's hydride

EH₃

has the lowest boiling point among corresponding hydrides of other group 15 elements. The maximum covalency of that element E is 4. In the light of the above statements, choose the correct answer from the options given below.

A. $Both Statement I and Statement II are true.$
B. $Statement I is false but Statement II is true.$
C. $Both Statement I and Statement II are false.$
D. $Statement I is true but Statement II is false.$

Solution

Core Logic

Evaluate Statement I: The bond length of HX increases down the group ( $HF < HCl < HBr < HI$ $HF < HCl < HBr < HI$) due to the increasing atomic radius of the halogen. Therefore, the halogen making the longest bond is Iodine (I). However, Iodine has the largest covalent radius in the group, not the smallest. Thus, Statement I is false.

Evaluate Statement II: In group 15 hydrides, the boiling point order is $PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃$ $PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$. The hydride with the lowest boiling point is $PH₃$ $PH_3$ (Phosphine). The maximum covalency of Phosphorus is 6 (as seen in $PF₆^-$ $PF_6^-$) because it has empty d-orbitals, not 4. Only Nitrogen has a maximum covalency of 4. Since the element is P, statement II is also false.

Step 1: Final Conclusion

Both Statement I and Statement II are false.

Pattern Recognition

Standard p-block trends: Boiling points of hydrides show anomalies due to hydrogen bonding. $NH₃$ $NH_3$, $H₂O$ $H_2O$, and HF jump out of the trend. P, S, Cl hydrides mark the lowest points.

Chapter Mix

Class 12 Chemistry: p-Block Elements

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Core Logic

Pattern Recognition

Chapter Mix

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