JEE Advanced (Matrix Match & Integer Answer): The p-Block Elements | Chapter-wise Tests for JEE Main & Advanced PDF Download

Match the Following

DIRECTIONS (Q. 1 to 3) : Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :

If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s then the correct darkening of bubbles will look like the given.

Q.1. Match gases under specified conditions listed in Column I with their properties/laws in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.

Column I                                        Column II

(A) Explosive                                      (p) NaN₃

(B) Artificial gem                                (q) Fe₃O₄

(C) Self reduction                               (r) Cu

(D) Magnetic material                        (s) Al₂O₃

                                                            (t) Pb(N3)2

                                                            (u) Fe2O3

                                                            (v) Cu

                                                             (w) SiC

Ans. (A)-t; (B)-s; (C)-v; (D)-u

Solution. (A)-(t),   Pb(N3)2 is an explosive
(B)-(s),  Al₂O₃ is used to prepare artificial gun
(C)-(v),  Extraction of copper involves self-reduction process.
(D)-(u),  Fe₂O₃ is a magnetic material.

Q.2. Match the following :

Ans. (A)-q; (B)-s; (C)-p; (D)-r

Solution. (A)-(q), Bi³⁺ hydrolyses to yield BiO⁺ ion
(B)-(s), AlO₂^– on dilution yields a white ppt. of Al(OH)₃
(C)-(p), When heated (SiO₄)^4– changes to (Si₂O₇)^6–
(D)-(r), When acidified (B₄O₇)^2– gives B(OH)₃ (or H₃BO₃)

Q.3. Match each of the diatomic molecules in Column I with its property/properties in Column II.

Column I                                              Column II

(A) B₂                                      (p) Paramagnetic

(B) N₂                                     (q) Undergoes oxidation

(C) O^–₂                                  (r) Undergoes reduction

(D) O₂                                    (s) Bond order > 2

                                               (t) Mixing of  's' and 'p' orbital

Ans. (A)-p, r, t; (B)-s, t ; (C)-p, q, r ; (D)-p, r, s

Solution. 

DIRECTIONS (for Q. 4) : Following question has matching lists. The codes for the lists have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Q.4. The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are provided in List II. Match List I with List II and select the correct answer using the code given below the lists

Ans. (d)

Solution.

Integer Value Correct Type

Q.1. The coordination number of Al in the crystalline state of AlCl₃ is                  (2009)

Ans. 6

Solution. Coordination number of Al is 6. It exists in ccp lattice with 6 coordinate layer structure.

Q.2. The value of n in the molecular formula BenAl₂Si₆O₁₈ is              (2010)

Ans. 3

Solution. Total cationic charge = Total anionic charge

2n + 6 + 24 = 36 ⇒ n = 3

Q.3. Reaction of Br₂ with Na₂CO₃ in aqueous solution gives sodium bromide and sodium bromate with evolution of CO₂ gas. The number of sodium bromide molecules involved in the balanced chemical equation is           (2011)

Ans. 5

Solution. 3Br₂ + 3Na₂CO ₃ → 5NaBr + NaBrO₃+ 3CO₂

Q.4. Among the following, the number of compounds than can react with PCl₅ to give POCl₃ is  (2011) 

O₂, CO₂, SO₂, H₂O, H₂SO₄, P₄O₁₀

Ans. 4

Solution. PCl₅ + SO₂ → POCl₃ + SOCl₂
PCl₅ + H₂O →POCl₃ + 2HCl
PCl₅ + H₂SO₄ → POCl₃ + SO₂Cl₂ + 2HCl
6PCl₅ + P₄O₁₀ → 10POCl₃

Q.5. The total number of lone pairs of electrons in N₂O₃ is                (JEE Adv. 2015)

Ans. 8

Solution. 

Number of lone pairs = 8

Q.6. Three moles of B₂H₆ are completely reacted with methanol. The number of moles of boron containing product formed is    (JEE Adv. 2015)

Ans. 6

Solution. 3B₂H₆ + 18CH₃OH → 6B(OCH₃)₃ + 18H₂