NEET and JEE Questions p-Block Elements

Que 1. Which of the following elements does not form stable diatomic molecules.
(a) Iodine                                                                                (b) Phosphorus
(c) Nitrogen                                                                            (d) Oxygen
Ans 1. (b) Phosphorus
Explanation: Phosphorous forms tetra atomic molecule it does not form stable diatomic molecule whereas the other given elements form stable diatomic molecules(Due to larger atomic size P does not form pi bonds and so it exists as a tetra-atomic molecule in which each P atom is bonded with 3 other P atoms by 3 sigma bond. But, due to smaller atomic size N forms 1 sigma and 2 pi bonds i.e. triple bonds with another N atom, and exists as a diatomic molecule.

Que 2. Among NH₃, PH₃, AsH_3, and SbH₃ which one is a stronger reducing agent?
(a) NH₃                                                                                               (b) PH₃
(c) AsH₃                                                                                              (d) SbH₃
Ans 2. (d)
Explanation: When we go down the group the size of the atom increase also the distance between The metal hydrogen bond increases in hydrides. So the strength of the metal hydrogen bond decreases therefore hydrogen is easily released. A reducing agent is one that supplies hydrogen atoms.

Que 3. The mixed anhydride of nitrogen is
(a) N₂O₂                      (b) N₂O₄                      (c) N₂O₅                      (d) N₂O₃
Ans 3. (b)
Explanation: Mixed anhydride or double acid anhydride is the compound that reacts with water to form a mixture of two acidic anhydrides.
Two oxides of nitrogen are acid anhydrides; that is, they react with water to form two nitrogen-containing oxyacids.
2NO₂​ + H₂​O ⟶ HNO₃ ​+ HNO₂​

Que 4. Which of the following statements is wrong?
(a) The stability of hydries increases from NH₃ to BiH₃ in group 15 of the Periodic Table
(b) Nitrogen can’t from dπ-pπ bond.
(c) Single N—N bond is weaker than the single P—P bond.
(d) N₂O₄ has two resonance structure.
Ans 4. (a)
Explanation: Thermal stability of the hydrides decrease as we go down the group in the periodic table for group 15, due to increase in E—H bond length down the group.

Que 5. Which of the following is the most basic oxide?
(a) Al₂O₃                                                                                 (b) SeO₂
(c) Bi₂O₃                                                                                 (d) Sb₂O₃
Ans 5. (b)
Explanation: More the oxidation state of the central atom (metal) more is its acidity. Hence  SeO₂ (O. S. of Se = +4) is acidic.

NEET and JEE Questions p-Block Elements

Que 6. In addition to conc.H₂SO₄ to a chloride salt, colourless fumes are evolved but in the case of iodise salt, violet fumes come out. This is because
(a) H₂SO₄ reduces Hl to l₂
(b) Hl is of violet colour
(c) Hl gets oxidised to l₂
(d) Hl changes to HlO₃
Ans 6. (c)
Explanation: Hydrogen iodide is a stronger reducing agent than sulphuric acid so it reduces H₂SO₄ to SO₂ and HI to I₂​.
Chloride salts when treated with sulphuric acid to give colourless gas due to formation of HCl gas.
NaCl + H₂SO₄ ⟶ HCl + Na₂​SO₄​
Due to the formation of iodine (I₂) gas, the violet fumes are evolved during the reaction.
2NaI + H₂SO₄ ⟶ Na₂​SO₄ + ​ 2HI
Na₂​SO₄ + 2HI ⟶ 2H₂​O + SO₂ ​+ I₂

Que 7. Arrange the following in order of increasing acidity H₃PO₄, H₂CO₃, HCl, HI
(a) H₃PO₄, HCl , H₂CO₃, HI
(b) H₃PO₄, H₂CO₃, HCl, HI
(c) H₂CO₃, H₃PO₄, HCl, HI
(d) None of these
Ans 7. (c)
Explanation: H₂CO₃ ˂ H₃PO₄ ˂ HCl ˂ HI

Que 8. Producer gas is a mixture of
(a) CO and N₂                                                                                           (b) CO₂ and H₂
(c) CO and H₂                                                                                           (d) CO₂ and N₂
Ans 8. (a)
Explanation: A combustible mixture of nitrogen, carbon monoxide, and hydrogen, generated by passing air with steam over burning coke or coal in a furnace and used as fuel. Also called air gas.

Que 9. Among the following statement which one is true?
(a). NH₃ is less soluble than PH₃ in water
(b) NH₃ is a stronger base and stronger reducing agent than PH₃
(c) NH₃ has a higher boiling point than PH₃ and has a lower melting point than PH₃
(d) PH₃ is a stronger reducing agent than NH₃ and it has a lower critical temperature than NH₃
Ans 9. (d)
Explanation: Due to the larger atomic size of P the bond length of the P−H bond is larger and bond dissociation energy is smaller. Hence, PH₃​ easily gets dissociated to give up H₂​ gas. Therefore, PH₃​ acts as a stronger reducing agent than NH₃​.

Que 10. Which of the following statements regarding N₂O₄ is not correct?
(a) It is a planar molecule
(b) It is used as a non-aqueous solvent
(c) It involves N—N bond which is shorter than the N—N bond in hydrazine
(d) Ammonium nitrate in N₂O₄ acts as a base.
Ans 10. (c)
Explanation: Ammonium nitrate in N₂​O₄​ acts as a base.
It is a planar molecule.
It involves an N-N bond which is longer than the N-N bond in hydrazine.

NEET and JEE Questions p-Block Elements

Que 11. In which of the following arrangements, the sequence is not strictly according to the property written against it?
(a) CO₂ < SiO₂ < SnO₂ < PbO₂ increasing oxidising power
(b) HF < HCl < HBr < HI  increasing acid strength
(c) NH₃ > PH₃ < AsH₃ < SbH₃  increasing basic strength
(d) B < C < O < N  increasing first ionisation enthalpy
Ans 11. (c)
Explanation: NH₃ < PH₃ < AsH₃ < SbH₃  increasing basic strength

Que 12. Which one of the following orders is not in accordance with the property stated against it?
(a) F₂ > Cl₂ > Br₂ > I₂ ; electronegativity
(b) F₂ > Cl₂ > Br₂ > I₂ ; bond dissociation energy
(c) F₂ > Cl₂ > Br₂ > I₂ oxidizing power
(d) HI > HBr > HCl > HF acidic property in water
Ans 12. (b)
Explanation: In the case of diatomic molecules (X₂) of halogens the bond dissociation energy decreases in the order: Cl₂ > Br₂ > F₂ > I₂
The oxidizing power, electronegativity and reactivity decrease in the following order: F₂ > Cl₂ > Br₂ > I₂ Electron gain enthalpy of halogens decrease in the following order: Cl₂ > F₂ > Br₂ > I₂
The low value of electron gain enthalpy of fluorine is due to the small size of the fluorine atom.

Que 13. In the preparation of compounds of Xe, Bartlett had taken O₂ + PtF₆ ^– as a base compound. This is because
(a) both O₂ and Xe have the same size
(b) both O₂ and Xe have the same electron gain enthalpy
(c) both O₂ and Xe have almost the same ionisation enthalpy
(d) both Xe and O₂ are gases.
Ans 13. (c) both O₂ and Xe have almost the same ionisation enthalpy.

Que 14. Each of the following is true for white and red phosphorus except that they
(a) Are both soluble in CS₂
(b) Can be oxidised by heating in air
(c) Consists of the same kind of atoms
(d) Can be converted into one another
Ans 14. (a)
Explanation: White phosphorus, yellow phosphorus, or simply tetra-phosphorus (P₄​) exists as molecules made up of four atoms in a tetrahedral structure. As the phosphorous molecule is non-polar, so it dissolves in a non-polar solvent like carbon disulfide, CS₂​.

Que 15. The formation of PH₄⁺ is difficult compared to NH₄⁺ because:
(a) lone pair of phosphorus is optically inert
(b) lone pair of phosphorus resides in almost pure p-orbital
(c) lone pair of phosphorus resides at sp3 orbital
(d) lone pair of phosphorus resides in almost pure s-orbital
Ans 15. (d) lone pair of phosphorus resides in almost pure s-orbital.

NEET and JEE Questions p-Block Elements

Que 16. Which of the following will be formed when HNO₂ disproportionate in an aqueous medium?
(a) NH₃                                                                                               (b) N₂
(c) NO                                                                                                 (d) HNO₃
Ans 16. (c, d)
Explanation: HNO₂ → HNO₃ + NO + H₂O

Que 17. Which one of the following reactions of xenon compounds is not feasible?
(a) XeO₃ + 6HF → XeF₆ + 3H₂O
(b) 3XeF₄ + 6H₂O → 2Xe + XeO₃ + 12HF + 1/5O₂
(c) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
(d) XeF₆ + RbF → Rb[XeF₇]
Ans 17. (a)
Explanation: XeO₃ + 6HF → XeF₆ + 3H₂O

Que 18. The variation of the boiling point of the hydrogen halides is in the order HF > HI > HBr > HCl. What explains the higher boiling point of hydrogen Fluoride?
(a) There is strong hydrogen bonding between HF molecules
(b) The electronegativity for fluorine is much higher than for other elements in the group
(c) The bond energy of HF molecules is greater than in other hydrogen halides
(d) The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule
Ans 18. (a)
Explanation: Due to strong H-bonding in HF it has very high BP.

Que 19. Which one is not a property of ozone?
(a) It acts as an oxidising agent in a dry state
(b) Oxidation of KI into KIO₂
(c) PbS is oxidised to PbSO₄
(d) Hg is oxidised to Hg₂O
Ans 19. (b)
Explanation: Oxidation of KI by O₃​ gives potassium hydroxide, iodine, and oxygen and not KIO₂​. KI + O₃ ​+ H₂​O ⟶ KOH + O₂ ​+ I₂​

Que 20. Which of the following is not isostructural with SiCl₄?
(a) PO₄^3–                                                                                             (b) NH₄⁺
(c) SCl₄                                                                                               (d) SO₄^2–
Ans 20. (c)
Explanation: SiCl₄​ is tetrahedral as it has four chlorine atoms attached tetrahedrally whereas SCl₄​ has four chlorine atoms and a lone pair on sulfur which makes its structure see-saw and the remaining molecules are having tetrahedral geometry.
So SCl₄​ is not isostructural with SiCl₄.​

ELECTROCHEMISTRY NEET IMPORTANT SOLVED QUESTIONS
https://www.sciencemotive.com/class-12-chemistry/electrochemistry-neet-important-solved-questions/

NEET and JEE Questions p-Block Elements