JEE Advanced (Single Correct MCQs): The p-Block Elements - JEE MCQ

Temporary hardness of water is due to presence of bicarbonates of Ca and Mg and it is removed by adding Ca(OH)₂ to hard water and precipitating these soluble bicarbonates in the form of insoluble salts.

Due to highest bond dissociation energy.

It is redox reaction. (i) Oxidation n umber in cr eases dur in g oxidation reaction and decreases during reduction reaction.
(ii) In a neutralisation reaction acids and bases  reacts together to form salt and water.

∵ Cl₂ is more reactive than bromine.

All are colourless gases.

None; it reacts with all given compounds. It forms addition compounds with them.
It can be dried over any metal oxide.

HI and HBr (in that order) are the strongest reducing hydracids and hence they reduce H₂SO₄. HCl is quite stable and hence is oxidised by strong oxidising agent like KMnO₄. HF is not a reducing agent. In the smallest F– ion, the electron which is to be removed during oxidation is closest to the nucleus and therefore most difficult to be removed. Therefore, HF is a poor reducing agent.

AlCl₃ exists as a dimer (Al₂Cl₆). It is a strong Lewis acid as it has an incomplete octet and has a tendency to gain electrons. AlCl₃ undergoes hydrolysis easily and forms an acidic solution.
AlCl₃ + 3H₂O → Al(OH)₃ + 3HCl

Option (c) is true that AlCl₃ sublimes at 100ºC under vacuum.
AlCl₃ is a Lewis acid.

Graphite shows moderate electrical conductivity due to the presence of unpaired or free fourth valence electron on each carbon atom.

Bleaching action of chlorine is only in presence of moisture where nascent oxygen is displaced from H2O.
Cl₂ + H₂O → HCl + HClO (unstable) HClO → HCl + | O |

Only nitrates of heavy metals and lithium decompose on heating to produce NO₂.

SO₂ is highly soluble in water and therefore cannot be collected over water.

The structure clearly shows the presence of covalent and co-ordinate bonds.

NO₂ is reddish brown coloured gas. Rest of the oxides are colourless.

Least basic trihalogen of nitrogen is NF₃ because of the highest electronegativity of fluorine.

Chlorine is stronger oxidising agent than bromine therefore, chlorine water will liberate bromine from KBr solution.

2 KBr + Cl₂ → 2 KCl + Br₂

In  there is an S–O–S bond unlike in other.
So S–S bond is absent as can be seen from the structure drawn.

In P₄O₁₀, each P atom is linked to 4 oxygen atoms as can be confirmed by its structure. It is linked to three oxygen atoms by single bond and with one oxygen atom by double bond. [For structure refer to Q.11 of Section A]

HBr is not prepared by heating NaBr with conc. H₂SO₄ because HBr is a strong reducing agent and reduce H₂SO₄ to SO₂ and is itself oxidised to bromine.
NaBr + H₂SO₄ —→ NaHSO₄ + HBr H₂SO₄ + 2HBr —→ SO₂ + Br₂ + 2H₂O

CsBr₃ may be represented as Cs⁺Br₃^–

Potassium fluoride when reacts with hydrogen flouride and produces KF. 
K+(HF₂​)⁻ in fused form. This compound conducts electricity in fused form.

Due to inert pair effect.

CO is an example of neutral oxide.

The species called as pseudohalide ions are these are monovalent and made by electronegative atoms. They possess properties similar to halide ion. The corresponding dimers of these pseudohalide ions are called pseudohalogens. RCOO^– is not is pseudohalide.

Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃ ; i.e 2 moles of phosphine are produced.