(a) Give reasons for the following: (i) Sulphur in vapour state shows paramagnetic behaviour. (ii) N-N bond is

Question

Q. (a) Give reasons for the following:

(i) Sulphur in vapour state shows paramagnetic behaviour.

(ii) N-N bond is weaker than P-P bond.

(iii) Ozone is thermodynamically less stable than oxygen.

(b) Write the name of gas released when Cu is added to

(i) dilute

HNO_3

and

(ii) conc.

HNO_3

(OR)

(a)

(i) Write the disproportionation reaction of

H_3PO_3

(ii) Draw the structure of

XeF_4

(b) Account for the following:

(i) Although Fluorine has less negative electron gain enthalpy yet

F_2

is strong oxidizing agent.

(ii) Acidic character decreases from

N_2O_3

Bi_2O_3

in group 15.

CBSECBSE Class XII Board 2019Subjective· 5mImportance★★★★★est

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Group 15 Nitrogen Family

The Nitrogen Family – From Air to Metal

Think about the air you breathe. Nearly 78% of it is nitrogen gas — colourless, odourless, and so unreactive that it just sits there. Now imagine a grey, shiny solid that glows in the dark. That's phosphorus. And then there's arsenic, once used as a poison; antimony, used in fireproofing; and bismuth, a heavy metal with a rainbow oxide layer.

These five elements — nitrogen, phosphorus, arsenic, antimony, bismuth — make up Group 15 of the periodic table. They start as a gas, become a non-metal, then a metalloid, and finally a metal. That's the single most important pattern: metallic character increases as you go down the group.

The Common Thread: Five Valence Electrons

Every element in Group 15 has the electron configuration $ns^2 np^3$ in its outermost shell. Five valence electrons means they need three more to complete an octet — hence the common oxidation state of −3 (as in $\ce{NH3}$ , ammonia, or $\ce{PH3}$ , phosphine).

But because the $ns^2$ pair can also be lost or shared, you get +3 (e.g., $\ce{NCl3}$ , $\ce{PCl3}$ ) and +5 (e.g., $\ce{HNO3}$ , $\ce{PCl5}$ ). The +5 state becomes less stable as you go down — bismuth hardly ever shows +5 because its 6s electrons are too tightly held (the inert pair effect).

\text{General outer configuration: } ns^2 np^3

The Downward Trend: From Gas to Metal

Property	N	P	As	Sb	Bi
Nature	Non-metal	Non-metal	Metalloid	Metalloid	Metal
State at room temp	Gas	Solid (white/red)	Solid	Solid	Solid
Electronegativity	3.0	2.1	2.0	1.9	1.8
Ionisation enthalpy	High	↓	↓	↓	Lowest

Why does metallic character increase? As you go down, atomic size increases. The valence electrons are farther from the nucleus and more shielded. They become easier to lose (lower ionisation enthalpy) and harder to gain. So the elements start behaving less like electron-grabbers (non-metals) and more like electron-donors (metals). …

Figure — Structure of XeF4 (square planar). Central Xe with FOUR Xe-F single bonds pointing to the corners of

(a) $S_2$ vapour paramagnetic (2 $\pi^*$ unpaired e $^-$ ); N–N weaker than P–P (small-N lone-pair repulsion); $O_3$ thermodynamically less stable than $O_2$ ; Cu + dil./conc. $HNO_3$ → NO / $NO_2$ .

(b) $4H_3PO_3\to3H_3PO_4+PH_3$ ; $XeF_4$ square planar; $F_2$ strongest oxidiser; oxide acidity falls $N_2O_3\to Bi_2O_3$ ; $SO_2$ decolourises acidified $KMnO_4$ .

(i) Sulphur vapour is paramagnetic

Solid sulphur is $S_8$ (all electrons paired, diamagnetic), but at high temperature it forms $S_2$ molecules, isoelectronic with $O_2$ . Molecular-orbital filling places two unpaired electrons in the degenerate $\pi^*$ antibonding orbitals, so $S_2$ (and hence sulphur vapour) is paramagnetic.

(ii) N–N weaker than P–P

The N atom is very small, so in an N–N single bond the non-bonding lone pairs and bonding electrons on the two adjacent atoms are close together and repel strongly, weakening the bond (~159 kJ mol $^{-1}$ ). P atoms are larger, so this repulsion is much less and the P–P bond is stronger (~213 kJ mol $^{-1}$ ).

(iii) Ozone less stable than oxygen

2O_3(g) \rightarrow 3O_2(g),\quad \Delta H < 0,\ \Delta S > 0 \Rightarrow \Delta G < 0

The decomposition is spontaneous; $O_3$ has a positive enthalpy of formation ( $+142$ kJ mol $^{-1}$ ) and O–O bond order 1.5 versus 2 in $O_2$ , so it is thermodynamically less stable and readily reverts to $O_2$ .

Start your 14-day free trial to see the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set ANNUAL1 mark
$PCl_5$
›Reveal solutionSolution

Forming a pentahalide requires expanding the central atom's octet using d-orbitals; phosphorus (n=3) has vacant 3d orbitals available for this, while nitrogen (n=2) has no d-orbitals at all in its valence shell, so it is structurally incapable of forming $NCl_5$ .

Octet expansion and $d$ -orbital availability

To form $PCl_5$ , the central phosphorus atom must bond to 5 chlorine atoms, requiring 5 hybrid orbitals ( $sp^3d$ , trigonal bipyramidal geometry) — i.e. an expanded octet of 10 electrons around P. This is possible because phosphorus is in period 3 and its valence shell ( $n=3$ ) includes empty $3d$ orbitals that are energetically accessible enough to participate in hybridisation, accommodating the extra electron pairs.

Why nitrogen cannot do the same

…

Start your 14-day free trial to see the full solution →
CBSE 2019Set HE1 mark
$Q. Fill in the blank: White phosphorus possesses ______ composition in vapour state.$
›Reveal solutionSolution

White phosphorus is molecular; in the vapour state it exists as P₄ molecules, with the four phosphorus atoms arranged at the corners of a regular tetrahedron.

White (yellow) phosphorus is the most reactive allotrope of phosphorus and consists of discrete P₄ tetrahedral units, both in the solid and in the vapour state. Each P atom is bonded to three other P atoms by single P–P bonds, with a P–P–P bond angle of about 60° (strained), which is one rea …

Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 markMCQ
$Q. Second period of periodic table contains elements :$
(a) 2
(b) 8
(c) 18
(d) 32
›Reveal solutionSolution

Second period = 8 elements (Li → Ne).

The second period corresponds to filling the n = 2 shell (2s and 2p orbitals), which hold a maximum of 8 electrons. The elements are Li, Be, B, C, N, O, F, Ne — a total o …

Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 markMCQ
$Q. Group I elements of Periodic Table are known as -$
(a) Alkali metals
(b) Alkaline earth metals
(c) Inert gases
(d) None
›Reveal solutionSolution

Group 1 elements are the alkali metals.

The elements of Group 1 (Li, Na, K, Rb, Cs, Fr) are highly reactive metals whose hydroxides are strong bases (alkalis) that dissolve readily in water — hence 'alkali metals'. Group 2 elements a …

Start your 14-day free trial to see the full solution →
CBSE 2018Set ANNUAL1 markMCQ
$Q. The molecule which has zero dipole moment is -$
(a) NF3
(b) BF3
(c) ClO2
(d) CH2Cl2
›Reveal solutionSolution
Only the symmetric trigonal-planar BF₃ has μ = 0.

BF₃: sp², trigonal planar, F at 120°; the three equal B→F dipoles cancel → μ = 0.

NF₃: sp³ pyramidal with a lone pair → μ ≠ 0. …
Start your 14-day free trial to see the full solution →
CBSE 2018Set ANNUAL1 markMCQ
$Q. Which of the following has the most stable +2 oxidation state?$
(a) Sn
(b) Ag
(c) Fe
(d) Pb
›Reveal solutionSolution

Inert-pair effect makes Pb²⁺ the most stable +2 state.

For heavy p-block elements the ns² pair becomes reluctant to bond (inert-pair effect), stabilising the lower (+2) oxidation state. In group 14 this is strongest for lead, so Pb²⁺ is very stable while Pb⁴⁺ is oxidising. …

Start your 14-day free trial to see the full solution →
CBSE 2017Set ANNUAL1 mark
$Q. What is laughing gas?$
›Reveal solutionSolution

N2O is called laughing gas because inhaling it produces a mild euphoric/intoxicating sensation, and it is used medically as a weak anaesthetic (e.g., in dentistry).

Laughing gas is the common name for nitrous oxide, $\text{N}_2\text{O}$ , an oxide of nitrogen in which nitrogen is in the +1 oxidation state.

…

Start your 14-day free trial to see the full solution →

Q.(a) Give reasons for the following:

Concept understanding — Group 15 Nitrogen Family

The Nitrogen Family – From Air to Metal

The Common Thread: Five Valence Electrons

The Downward Trend: From Gas to Metal

(i) Sulphur vapour is paramagnetic

(ii) N–N weaker than P–P

(iii) Ozone less stable than oxygen

Gases with copper and nitric acid

(Alternative) Part (a)

(i) Disproportionation of $H_3PO_3$

(ii) Structure of $XeF_4$

(Alternative) Part (b): Accounting

(i) $F_2$ a strong oxidising agent …

Q.(a) Give reasons for the following:

Concept understanding — Group 15 Nitrogen Family

The Nitrogen Family – From Air to Metal

The Common Thread: Five Valence Electrons

The Downward Trend: From Gas to Metal

(i) Sulphur vapour is paramagnetic

(ii) N–N weaker than P–P

(iii) Ozone less stable than oxygen

Gases with copper and nitric acid

(Alternative) Part (a)

(i) Disproportionation of H3PO3H_3PO_3H3​PO3​

(ii) Structure of XeF4XeF_4XeF4​

(Alternative) Part (b): Accounting

(i) F2F_2F2​ a strong oxidising agent …

🎓Unlock everything free for 14 days

(i) Disproportionation of $H_3PO_3$

(ii) Structure of $XeF_4$

(i) $F_2$ a strong oxidising agent …

Unlock everything free for 14 days