The p-Block ElementsNEET Chemistry · Class 12 · NCERT Chapter 13

Solution

Thermal stability order K2CO3>Na2CO3>CaCO3>MgCO3 Therefore MgCO3 releases CO2 most easily. MgCO3 Δ → MgO+CO2

Solution

Strong reducing behaviour of H3PO2 All oxy-acid of phosphorus which contain P – H bond act as reductant. www.vedantu.com 61 Presence of one –OH group and two P – H bonds.

Solution

i. No. of electron in ONF = 24 No. of electron in NO2 −=24 Both are isoelectronic. ii. OF2 is a fluoride of oxygen not oxide of fluorine because EN of fluorine is more than oxygen. OF2=Oxygen difluoride iii. Cl2O7 is an anhydride of perchloric acid. 2HClO4 Δ,−H2O → Cl2O7 iv. O3 molecules is bent shape.

Solution

Stability of +1 oxidation state due to inert pair effect TI<In<Ga<Al.

Solution

Due to strong H-bonding in HF molecule, boiling point is highest for HF. HF>HI>HBr>HCl

Solution

Deuteromycetes - Imperfect fungi which are decomposers of litter and help in mineral cycling. www.vedantu.com 2

Solution

Cu+ 4HNO3 → Cu(NO3)2 + 2NO2 + O2

Solution

Decreasing order of Bond energy, Cl2 > Br2 > F2 > I2 The reason is anomalous behavior due to large electron – electron repulsion among the lone pairs in F2 molecule other than Cl2 and Br2.

Solution

CaC2 + N 2 Calcium Nitrogen Carbide Gas → CaCN 2 + C Calcium Cyanamide CaCN2 is not given in the option so it should be bonus.

Solution

In a solution containing HgCl2, I2 and I–, both HgCl2 and I2 compete for I–. Since formation constant of [HgI 4]2– is 1.9 × 10 30 which is very large as compared with I 3 – (Kf = 700) ∴ I– will preferentially combine with HgCl2. HgCl2 + 2I– → HgI2↓ + 2Cl– Red ppt HgI2 + 2I– → [HgI4]2– soluble

Solution

Inability of ns2 electrons of the valence shell to participate in bonding on moving down the group in heavier p-block elements is called inert pair effect As a result, Pb(II) is more stable than Pb(IV) Sn(IV) is more stable than Sn(II) ∴ Pb(IV) is easily reduced to Pb(II) ∴ Pb(IV) is oxidising agent Sn(II) is easily oxidised to Sn(IV) ∴ Sn(II) is reducing agent

Solution

S S S S OO OO O – O – [S O ]46 2(–) , S O [S O ]23 2(–) O – O – S

Solution

XX′ → Linear XX 3′ → Example : CIF3 → T-shape XX5′ → Example : BrF5 → Square pyramidal XX7′ → Example : IF7 → Pentagonal bipyramidal www.vedantu.com 16

Solution

Z = 114 belong to Group 14, carbon family Electronic configuration = [Rn]5f 146d107s27p2

Solution

Aerosols can cause various problems to agriculture through its direct or indirect effects on plants. However continually increasing air pollution may represent a persistent and largely irreversible threat to agriculture in the future.

Solution

Boron (B) cannot form ${\text{MF}}_6^{3-}$ ions because it lacks enough electrons to achieve an octet in the anion, unlike Al, Ga, and In, which can form such complexes. NCERT XII chapter The p-Block Elements explains that boron's electron configuration prevents it from forming stable hexafluoride ions, so option (a) is correct.

Solution

According to the Ellingham diagram, magnesium (Mg) has a lower standard Gibbs free energy of formation for its oxide compared to aluminum oxide. This makes Mg a suitable reducing agent for alumina, so option (a) is correct.

Solution

Chlorine does not have the highest electron-gain enthalpy among the halogens; fluorine does. NCERT XI chapter The p-Block Elements states that fluorine has the highest electron-gain enthalpy due to its small size and high electronegativity, so option (d) is incorrect.

Solution

In ${\text{ClF}}_3$, the central chlorine atom has two lone pairs of electrons. This is determined by the VSEPR theory, which predicts the molecular geometry based on the number of bond pairs and lone pairs around the central atom, so option (b) is correct.

Solution

Fact

Solution

Factual

Solution

t99%= 1 kln( a0 a0 ×100)= 2 kln10 2×2.303 𝑘 = 4.606 𝑘

Solution

Cl – Cl AlCl3 → Cl⊕[Electrophile]+𝐴𝑙𝐶𝑙4 −

Solution

Fact

Solution

Fact

Solution

No. of atoms in hexagonal lattice (A) = 6 No. of octahedral (c) = 6 × 75 100 = 9 2 ∴ Formula is 𝐶9 2 A6 𝐶9A12 𝐶3A4

Solution

- CO is a neutral oxide.
- BaO is a basic oxide.
- Al${}_2$O${}_3$ is an amphoteric oxide.
- Cl${}_2$O${}_7$ is an acidic oxide.
The correct matching is (b) (ii) (i) (iv) (iii), so option (b) is correct.

Solution

Peroxodisulphuric acid (H${}_2$S${}_2$O${}_8$) contains the peroxide linkage (-O-O-), which is not present in the other oxoacids listed. NCERT XII chapter The p-Block Elements describes the structure of peroxodisulphuric acid, confirming that option (c) is correct.

Solution

Noble gases have low melting and boiling points due to weak dispersion forces, making option (b) incorrect. They are indeed sparingly soluble in water, have weak dispersion forces, and large positive electron gain enthalpies, as described in NCERT XII chapter The p-Block Elements.

Solution

Beryllium chloride is covalent and soluble in organic solvents due to the small size and high polarizing power of the beryllium ion. NCERT XII chapter The p-Block Elements explains that beryllium halides are covalent and exhibit solubility in organic solvents, so option (d) is correct.

Solution

The maximum temperature that can be achieved in a blast furnace is up to 2200 K. This temperature is sufficient for the reduction of iron oxides to iron in the presence of coke and limestone, as described in NCERT XII chapter The p-Block Elements, so option (b) is correct.

Solution

Both statements are true. The acid strength of hydrogen halides increases from HF to HI due to the decreasing bond strength as the size of the halogen increases down the group, as explained in NCERT XII chapter The p-Block Elements. Therefore, option (a) is correct.

Solution

Actinoid contraction is not greater than Lanthanoid contraction element to element. Lanthanoid contraction is more pronounced due to the poor shielding of 4f electrons, making option (a) incorrect. NCERT XII chapter The p-Block Elements discusses these contractions and their effects on properties.

Solution

Iso-electronic species have the same number of electrons. ${\text{Fe}}^{2+}$ has 24 electrons, while ${\text{Mn}}^{2+}$ has 23 electrons, so they are not iso-electronic. Option (d) is correct.

Solution

The correct sequence for increasing acidic character should be ${\text{NH}}_3<{\text{PH}}_3<{\text{AsH}}_3<{\text{SbH}}_3$. However, ${\text{PH}}_3$ is less acidic than ${\text{NH}}_3$, making option (c) incorrect. NCERT XII chapter The p-Block Elements discusses the acidic character of Group 15 hydrides, confirming this trend.

Solution

List – I List – II (Hydrides) (Nature) (a) MgH2 → Ionic (b) GeH4 → Electron precise (c) B2H6 → Electron deficient (d) HF → Electron rich (a) – (iv), (b) – (i), (c) – (ii), (d) – (iii)

Solution

SF4 → sp3d + 1 lone pair XeF2 → sp3d + 3 lone pair ClF3 → sp3d + 2 lone pair IF5 → sp3d2 + 1 lone pair XeF2 having maximum lone pairs, so, it has maximum ‘lone pair-lone pair’ electron repulsions.

Solution

Each boron atoms in diborane uses sp3 hybrid orbitals for bonding.

Solution

Compound Boiling point (K) H2O 373 H2S 213 H2Se 232 H2Te 269 • The boiling points of these hybrids not exactly increases with increase in molar mass. • H2O has maximum boiling point due to intermolecular hydrogen bonding. - 26 - NEET (UG)-2022 (Code-Q1)

Solution

Diamond : • sp3 hybridised carbon atom • Covalent solid • 3-D structure • Cannot be used as dry lubricant Graphite : • sp2 hybridised carbon atom • Covalent solid • 3-D structure • Used as dry lubricant

Solution

In neutral or faintly alkaline solution. ( ) ( ) 7 –1 5 4– – – – 4 2 2 3 iodide iodate 2MnO H O I 2MnO 2OH IO ++ + + + → + + Manganese (Mn) oxidation state change from +7 to +4.

Solution

Oxygen shows –2, –1, +1 and +2 oxidation states Selenium shows –2, +2, +4 and +6 oxidation states Tellurium shows –2, +2, +4 and +6 oxidation states Polonium shows +2 and +4 oxidation states

Solution

All the elements of group 15 form hydrides of $E{H}_3$ type. Nitrogen forms ammonia ($N{H}_3$) while Arsenic forms Arsine ($As{H}_3$)

All the elements of group 15 form two types of oxides : $E_2{O}_3$ and $E_2{O}_5$
Antimony forms antimony pentoxide $S{b}_2{O}_5$

Hence, statement I is correct and statement II is incorrect