Q: The heating of NH _4 NO _2 at 60-70^ C and NH _4 NO _3 at 200-250^ C is associated with the formation of nitrogen containing compounds X and Y , respectively. X and Y , respectively, are

aligned & NH _4 NO _2 [60-70^ C ] N _2+2 H _2 O \\ & NH _4 NO _3 [200-250^ C ] N _2 O +2 H _2 O aligned

Q: The reaction of Xe and O 2 F 2 gives a Xe compound P . The number of moles of HF produced by the complete hydrolysis of 1 mol of P is _____ .

Xe + 2 O 2 F 2 → XeF 4 P + 2 O 2 6 XeF 4 P + 12 H 2 O ⟶ 4 Xe + 2 XeO 3 + 24 HF + 3 O 2 Number of moles of HF formed by complete hydrolysis of 1 mole of P = 4

Q: Dissolving 1 . 24 g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas Q . The amount of CuSO 4 (in g ) required to completely consume the gas Q is____[Given

P 4 1 . 24 g + 3 NaOH + 3 H 2 O → PH 3 + 3 NaH 2 PO 2 Number of moles of P 4 = Given mass molar mass = 1 . 24 124 = 0 . 01 moles As NaOH is present in excess. So, amount of phosphine formed is 0 . 01 mole (as P 4 is limiting reagent) 2 PH 3 0 . 01 mole + 3 CuSO 4 → Cu 3 P 2 + 3 H 2 SO 4 Since 2 moles of phosphine required 3 moles of CuSO 4 . Amount of CuSO 4 required = 3 × 0 . 01 2 mole Mass

Question 1

Correct statement(s) about the compounds X , Y and Z is(are)

Accepted Answer

From the given reactions, we can identify the compounds X , Y , and Z as follows: Reaction 1: MnO _2 + 4 HCl (conc.) MnCl _2 + Cl _2 + 2 H _2 O The greenish-yellow gas X is Cl _2. Reaction 2: NH _3 + 3 Cl _2 (excess) NCl _3 + 3 HCl Compound Y is NCl _3. Reaction 3: Cl _2 + 3 F _2 (excess) 573 K 2 ClF _3 Compound Z is ClF _3. Evaluating the given statements: (A) Cl _2 is widely used as a disinfectant for sterilizing drinking water. This statement is correct. (B) In NCl _3, the nitrogen atom is sp

Question 2

Correct statement(s) about the compounds P , Q and R is(are)

Accepted Answer

From the given reaction conditions: Xe(g) + 2 F _2 (g) [ 1 : 5 ratio ] 873 K, 7 bar XeF _4 Thus, P is XeF _4. XeF _4 + O _2 F _2 143 K XeF _6 + O _2 Thus, Q is XeF _6. XeF _6 + 3 H _2 O complete hydrolysis XeO _3 + 6 HF Thus, R is XeO _3. Evaluating the given options: (A) P ( XeF _4) has 8 valence electrons on the central Xe atom. It forms 4 single bonds with F, leaving 4 electrons which form 2 lone pairs. This statement is correct. (B) Q ( XeF _6) has 6 bond pairs and 1 lone pair on the central

Question 3

The heating of NH _4 NO _2 at 60-70^ C and NH _4 NO _3 at 200-250^ C is associated with the formation of nitrogen containing compounds X and Y , respectively. X and Y , respectively, are

Accepted Answer

aligned & NH _4 NO _2 [60-70^ C ] N _2+2 H _2 O \ & NH _4 NO _3 [200-250^ C ] N _2 O +2 H _2 O aligned

Question 4

During sodium nitroprusside test of sulphide ion in an aqueous solution, one of the ligands coordinated to the metal ion is converted to

Accepted Answer

Na _2 [ Fe ( CN )_5( NO ) ]+ Na _2 ~S Na _4 [ Fe ( CN )_5( NOS ) ] Sodium nitroprusside purple solution

Question 5

The complete hydrolysis of ICl _2 ClF _3 and BrF _5, respectively, gives

Accepted Answer

Answer option index: 0

Question 6

The compound(s) with P - H bond(s) is(are)

Accepted Answer

Answer option index: 3

Question 7

In the following reaction sequences, the major product Q is: L-Glucose $ [ (ii) Cr _2 O _3, 775 ~K , 10-20 atm ] (i) HI , P [ UV ] Cl _2 (excess) Q $

Accepted Answer

Answer option index: 3

Question 8

The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are :

Accepted Answer

PCl _5 when fluorinated in a polar organic solvent, ionic isomers are formed. [ PCl _4^ + ] [ PCl _4 ~F _2 ]^ - (colorless crystals) [ PCl _4^ + ] [ PF _6 ]^ - (white crystals)

Question 9

The compound (s) having peroxide linkage is(are) :

Accepted Answer

Answer option index: 3

Question 10

The reaction of Xe and O 2 F 2  gives a Xe compound P . The number of moles of HF produced by the complete hydrolysis of 1 &#160; mol of P is _____ .

Accepted Answer

Xe + 2 O 2 &#160; F 2 &#8594; XeF 4 P + 2 O 2 6 XeF 4 P + 12 H 2 O &#10230; 4 Xe + 2 XeO 3 + 24 HF + 3 O 2 Number of moles of HF formed by complete hydrolysis of 1 mole of P = 4

Question 11

The compound(s) which react(s) with NH 3  to give boron nitride BN is(are)

Accepted Answer

The compound(s) which react with NH 3  to form Boron nitride BN is/are: (A) Amorphous B + NH 3 &#10230; &#160; Tempreature &#160; BN + H 2 (B) B 2 H 6 + NH 3 excess &#10230; &#160; HighTemp .&#160; BN x (C) B 2 O 3 + 2 NH 3 &#10230; 900 &#176; C 2 BN + 3 H 2 O (D) HBF 4 + NH 3 &#10230; NH 4 &#8853; BF 4 &#8854; Hence, (A,B,C) are correct

Question 12

The reaction of HClO 3  with HCl gives a paramagnetic gas, which upon reaction with O 3  produces

Accepted Answer

HClO 3  reacts with HCl according to the following equation, 2 HClO 3 + 2 HCl &#10230; 2 ClO 2 + Cl 2 + 2 H 2 O ClO 2  molecule is paramagnetic, as it contains odd number of electrons. 2 ClO 2 + 2 O 3 &#10230; Cl 2 O 6 + 2 O 2 ClO 2 is used as a bleaching agent for paper pulp and textiles and in water treatment

Question 13

The reaction of Pb NO 3 2  and NaCl in water produces a precipitate that dissolves upon the addition of HCl of appropriate concentration. The dissolution of the precipitate is due to the formatio

Accepted Answer

Pb 2 + on reaction with Cl - , produces white precipitate of PbCl 2 Pb 2 + + 2 Cl - &#10230; PbCl 2 &#8595; This precipitate is soluble in concentrated hydrochloric acid due to formation of tetrachloroplumbate II  ion PbCl 2 &#8595; + 2 Cl - &#10230; PbCl 4 2 - On addition of potassium chromate (K2CrO4 ) solution to lead chloride a yellow precipitate of lead chromate is obtained.

Question 14

Dissolving 1 . 24 &#160; g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas Q . The amount of CuSO 4 (in g ) required to completely consume the gas Q  is____[Given

Accepted Answer

P 4 1 . 24 &#160; g + 3 NaOH + 3 H 2 O &#8594; PH 3 + 3 NaH 2 PO 2 Number of moles of  P 4 &#160; = &#160; Given &#160; mass molar &#160; mass &#160; = &#160; 1 . 24 124 &#160; = &#160; 0 . 01 &#160; moles As NaOH is present in excess. So, amount of phosphine formed is 0 . 01  mole (as P 4 is limiting reagent) 2 PH 3 0 . 01 mole + 3 CuSO 4 &#8594; Cu 3 P 2 + 3 H 2 SO 4 Since 2 moles of phosphine required 3 moles of  CuSO 4 . Amount of CuSO 4 required = 3 &#215; 0 . 01 2 mole Mass

Question 15

The correct statement(s) related to oxoacids of phosphorous is(are)

Accepted Answer

4 H 3 PO 3 &#8594; &#8710; 3 H 3 PO 4 + PH 3  (Disproportionation) Oxidation state of  ' P '  in  H 3 PO 4  is  + 5  which is maximum oxidation state of  ' P '  so  H 3 PO 4  cannot act as a reducing agent whereas  H 3 PO 3  can act as a reducing agent because the oxidation state is + 3 and it can increase upto + 5 . &#8658;   H 3 PO 3 :  It is dibasic acid. &#8658;  Only  H - atom attached with&

Question 16

Paragraph: The reaction of K _ 3 [ Fe ( CN )_ 6 ] with freshly prepared FeSO _ 4 solution produces a dark blue precipitate called Turnbull's blue. Reaction of K _ 4 [ Fe ( CN )_ 6 ] with the FeSO _ 4

Accepted Answer

3 Fe 2 + + NO 3 - + 4 H + &#8594; NO + 3 Fe 3 + + 2 H 2 O Fe H 2 O 6 2 + + NO &#8594; Fe H 2 O 5 NO 2 + Brown &#160; ring + H 2 O

Question 17

Ozonolysis of ClO 2 produces an oxide of chlorine. The average oxidation state of chlorine in this oxide is ______ .

Accepted Answer

Reaction :  ClO 2 + O 3 &#8594; ClO 3 + O 2 ClO 3 x + &#8722; 2 &#215; 3 = 0 x = + 6 Here  x  is oxidation state of chlorine in  ClO 3 .

Question 18

With respect to hypochlorite, chlorate and perchlorate ions, choose the correct statement(s).

Accepted Answer

( A ) since HOCl is weakest oxyacid of chlorine. Therefore ClO - become strongest conjugate base. ( B ) ( C ) 3 ClO - ⟶ 2 Cl - + ClO 3 - K eq = 10 27 In hot solution rate of disproportionation is fairly rapid. 4 ClO 3 - ⟶ Cl - + 3 ClO 4 - Keq = 10 29 It is thermodynamically possible but kinetically, it is very slow reaction. ( D ) Lower oxyacids are good oxidizing agents. SO 3 2 - + O ¯ Cl ⟶ SO 4 2 - + Cl -

Question 19

With reference to aqua regia , choose the correct option(s).

Accepted Answer

Aqua regia is 3 past HCl and 1 part H N O 3 3 HCl + HNO 2 &#8594; NOCl + Cl 2 + 2 H 2 O Yellow&#8201;colour&#8201;is&#8201;due&#8201;to&#8201;NOCl&#8201;and&#8201; Cl 2 Au + Aqua&#8201;Regia &#8594; HAuCl 4 +3&#8201;O.S&#8201;of&#8201;Au

Question 20

The amount of water produced ( in &#160; g ) in the oxidation of 1 mole &#160; of rhombic sulphur by conc. H N O 3 to a compound with the highest oxidation state of sulphur is (Given data: Molar mass

Accepted Answer

S 8 + 48 HNO 3 &#8594; 8 H 2 SO 4 + 48 NO 2 + 16 H 2 O &#8757; 1 mol of S 8 produces 16 mole of H 2 O ; &#8756; Mass of H 2 O produced &#8658; 16 &#215; 18 &#8658; 288 &#160; g

p Block Elements (Group 15, 16, 17 & 18) — Previous Year Questions & Solutions