NEET Practice Questions (MCQs) with Answers & Solutions

Hydrogen tends to gain one electron to achieve a stable helium configuration (1s^2). In this respect, it resembles the halogens, which also need one electron to complete their outermost shell and attain a stable noble gas configuration.

Among the given halogens, iodine ( ext{I}_2) can be purified by sublimation. This process involves the transition of iodine directly from a solid to a gaseous state without passing through a liquid phase.

ext{CCl}_4 is actually resistant to hydrolysis due to the absence of vacant d-orbitals in the carbon atom, while 	ext{BCl}_3 can be easily hydrolyzed because it has vacant p-orbitals in boron, allowing it to accept lone pairs from water molecules. Thus, the statement that 	ext{CCl}_4 is hydrolyzed whereas 	ext{BCl}_3 is inert is not true.

$BCl_3$ (Boron trichloride) can form a dative bond with ammonia ($NH_3$) because boron has an incomplete octet and can accept a pair of electrons from the nitrogen atom in ammonia.

Concentrated $H_2SO_4$ cannot be used to prepare HBr from NaBr because it oxidizes HBr to bromine gas ($Br_2$). $H_2SO_4$ is a strong oxidizing agent, and in the presence of $H_2SO_4$, $HBr$ gets oxidized.

When fluoride compounds are heated with concentrated $H_2SO_4$ and $MnO_2$, hydrofluoric acid (HF) is produced. The reaction can be represented as: $$MnO_2 + 2NaF + 2H_2SO_4 → MnSO_4 + Na_2SO_4 + 2HF$$. Therefore, the correct answer is HF.

Fluorine does not show higher oxidation states because it lacks d-orbitals. All other halogens can utilize their d-orbitals to achieve higher oxidation states, but fluorine, being in the second period, has no d-orbitals available. Therefore, the correct answer is 'It has no d-orbital'.

The molecule $OF_4$ is theoretically not possible. Oxygen cannot form four bonds with fluorine due to its inability to expand its octet. Oxygen can form a maximum of two bonds in stable compounds. Therefore, the correct answer is $OF_4$.