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Class 12th Chemistry in Everyday Life

Match List I with List II

List I		List II
(Anion)		(gas evolved on reaction with dil H₂SO₄)
A.	CO₃^2-	I.	Colourless gas which turns lead acetate paper black
B.	S²^-	II.	Colourless gas which turns acidified potassium dichromate solution green
C.	SO₃²-	III.	Brown fumes which turns acidified Kl solution containing starch blue.
D.	NO₂^-	IV.	Colourless gas evolved with brisk effervescence, which turns lime water milky

Choose the correct answer from the option given below:

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Payal Gupta

Contributor-Level 10

A. $\underset{S a l t}{N a_{2} C O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a S O_{4} + H_{2} O + C O_{2} ↑$

CO₂ is colourless gas with brisk effervescence and turns lime water milky

$C O_{2} + C a {(O H)}_{2} \to C a C O_{3} ↓ + H_{2} O$

B. $\underset{S a l t}{N a_{2} S} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} S ↑$

H₂S + (CH₃COO)₂Pb → $\underset{B l a c k}{P b S} + 2 C H_{3} C O O H$

C. $\underset{S a l t}{N a_{2} S O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} O + S O_{3} ↑$

SO₃ is colourless gas which turns acidified potassium dichromate solution green as

$S O_{3} + K M n O_{4} + H^{+} \to \underset{G r e e n}{M n S O_{4}} + S O_{4}^{2 -} + H_{2} O$

D. $\underset{S a l t}{2 N a N O_{2}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + 2 H N O_{2}$ &nbs

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A. $\underset{S a l t}{N a_{2} C O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a S O_{4} + H_{2} O + C O_{2} ↑$

CO₂ is colourless gas with brisk effervescence and turns lime water milky

$C O_{2} + C a {(O H)}_{2} \to C a C O_{3} ↓ + H_{2} O$

B. $\underset{S a l t}{N a_{2} S} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} S ↑$

H₂S + (CH₃COO)₂Pb → $\underset{B l a c k}{P b S} + 2 C H_{3} C O O H$

C. $\underset{S a l t}{N a_{2} S O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} O + S O_{3} ↑$

SO₃ is colourless gas which turns acidified potassium dichromate solution green as

$S O_{3} + K M n O_{4} + H^{+} \to \underset{G r e e n}{M n S O_{4}} + S O_{4}^{2 -} + H_{2} O$

D. $\underset{S a l t}{2 N a N O_{2}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + 2 H N O_{2}$

$3 H N O_{2} \to N H O_{3} + 2 N O + H_{2} O$

$2 N O + O_{2} \to \underset{B r o w n g a s}{2 N O_{2}}$

$2 H N O_{2} + 2 K l + 2 C H_{3} C O O H \to 2 C H_{3} C O O K + 2 H_{2} O + 2 N O + I_{2}$

I₂ + starch → Blue colour complex.

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Class 12th Chemistry in Everyday Life

Match List I with List II

List I List II

A. Antipyretic I. Reduces pain

B. Analgesic II. Reduces stress

C. Tranquilizer &nbs

...more

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Payal Gupta

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A. Antipyretic reduces fever.

B. Analgesic reduces pain.

C. Tranquilizer reduces stress.

D. Antacid reduces acidity of stomach.

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Class 12th Biomolecules

Given below are two statements.

Statements I: Maltose has two a-D-glucose units linked at C₁ and C₄ and is a reducing sugar:

Statements II: Maltose has two manosaccharides: a-D-glucose and b-D-glucose linked at C₁ and C₆ and its is a non-reducing sugar.

In the light of the above statements, choose the correct answer from the options given below.

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Payal Gupta

Contributor-Level 10

$C_{12} H_{22} O_{11} + H_{2} O \to α - D - g l u c o s e + α - D - g l u c o s e$

In maltose, two units of a-D-glucose are linked by glycosidic linkage at C₁ and C₄ as shown.

a - D – glucose a - D - glucose

Maltose has hemiacetal link so it is reducing sugar.

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Class 12th Polymers

Which is true about Buna-N

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Payal Gupta

Contributor-Level 10

Buna-N is obtained by copolymerization of 1, 3-butadiene and acrylonitrile,

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Class 12th Amines

Decarboxylation of all six possible forms of diaminobenzoic acid $C_{6} H_{3} {(N H_{2})}_{2} C O O H$ yields three product A, B and C. Three acids give a products 'A' two acids gives a products 'B' and one acids give a products 'C'. The melting point of product 'C' is

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Payal Gupta

Contributor-Level 10

Above are the six possible forms of diaminobenzoic acid.

On decarboxylation we get;

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Class 12th Biomolecules

Match List I with List II

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Payal Gupta

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...more

A. Phenol is converted to salicylaldehyde by Riemer-Tiemann reaction

B. Phenol is converted to benzene using Zn powder

C. Phenol is converted to benzoinone using oxidizing agent which is $N a_{2} C r_{2} O_{7} / H^{+}$

D. Phenol can be converted to p-bromophenol using Br₂/CS₂

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Class 12th Environmental chemistry

On the surface of polar stratospheric clouds, hydrolysis of chlorine nitrate given A and B while its reaction with HCl produces B and C.A, B and C are, respectively.

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Payal Gupta

Contributor-Level 10

Chlorine nitrate as hydrolysis gives hypohalous acid and nitric acid as,

$C l O N O_{2} + H_{2} O \to \underset{A}{H O C l} + \underset{B}{H N O_{3}}$

Chlorine nitrate on reaction with HCl produces Cl₂ and HNO₃ as,

$C l O N O_{2} + H C l \to \underset{C}{C l_{2}} + \underset{B}{H N O_{3}}$

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Class 12th Coordination Compounds

Arrange the following coordination compounds in the increasing order of magnetic moments (Atomic numbers: Mn = 25; Fe = 26)

(A) ${[F e F_{6}]}^{3 -}$ (B) ${[F e {(C N)}_{6}]}^{3 -}$ (C) ${[M n C l_{6}]}^{3 -}$ (high spin) (D) ${[M n {(C N)}_{6}]}^{3 -}$

Choose the correct answer from the options given below

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Payal Gupta

Contributor-Level 10

A. ${[F e F_{6}]}^{3 -}$

Fe³⁺ - 3d⁵4s⁰

F^- is weak field ligand, so no pairing of electrons :

Number of unpaired electrons, n = 5

B. ${[F e {(C N)}_{6}]}^{3 -}$

$F e^{3 +} - 3 d^{5} 4 s^{0}$

CN^- is strong field ligand, so pairing of electrons takes place.

Number of unpaired electrons, n = 1

C. ${[M n C l_{6}]}^{3 -}$

...more

A. ${[F e F_{6}]}^{3 -}$

Fe³⁺ - 3d⁵4s⁰

F^- is weak field ligand, so no pairing of electrons :

Number of unpaired electrons, n = 5

B. ${[F e {(C N)}_{6}]}^{3 -}$

$F e^{3 +} - 3 d^{5} 4 s^{0}$

CN^- is strong field ligand, so pairing of electrons takes place.

Number of unpaired electrons, n = 1

C. ${[M n C l_{6}]}^{3 -}$

$M n^{3 +} - 3 d^{3} 4 s^{0}$

Number of unpaired electrons, n = 3

D. ${[M n {(C N)}_{6}]}^{3 -}$

$M n^{3 +} - 3 d^{3} 4 s^{0}$

Number of unpaired electrons, n = 3

Higher the number of unpaired electrons. Then higher be the magnetic moment as;

Magnetic moment, $μ = \sqrt[]{n (n + 2)} B . M$

Hence, order of magnetic moment is ;

B < D = C < A

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Class 12th The d and f Block Elements

The 'f' orbitals are half and completely filled, respectively in lanthanide ions

[Given: Atomic no. Eu, 63; Sm, 62; Tm, 69; Tb, 65; Yb, 70; Dy, 66]

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Payal Gupta

Contributor-Level 10

Electronic configuration of the given ions is :

$E u^{2 +} - 4 f^{7} 6 s^{0}$

$\begin{array}{l} T m^{2 +} - 4 f^{13} 6 s^{0} \\ S m^{2 +} - 4 f^{6} 6 s^{0} \\ T m^{3 +} - 4 f^{12} 6 s^{0} \end{array}$

$\begin{array}{l} T b^{4 +} - 4 f^{7} 6 s^{0} \\ Y b^{2 +} - 4 f^{14} 6 s^{0} \\ D y^{3 +} - 4 f^{9} 6 s^{0} \\ y b^{3 +} - 4 f^{13} 6 s^{0} \end{array}$

Hence, Eu²⁺ and Tb⁴⁺ have half | filled f-orbitals and Yb²⁺ have completely filled f-orbitals.

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Class 12th The d and f Block Elements

In 3d series, the metal having the highest M²⁺ / M standard electrode potential is

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Payal Gupta

Contributor-Level 10

In 3d-series, all metals except Cu have negative value of $E_{M^{2 +} / M}^{0}$ due to low hydration enthalpy and high I.E and sublimation enthalpy.

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