Quiz about Chemistry in Class 11

Question 1

An unidentified bivalent metal M reacts with unidentified halogen X to form an unknown compound of halogen gas. When 1.12g of it is heated 0.72g of MX is obtained along with 56 ml of halogen gas at 1atm and 273 K. Identify the metal.

Accepted Answer

NA

Question 2

An unbalanced chemical equation is against the law of

Accepted Answer

NA

Question 3

An S_(N)1 reaction at an asymmetric carbon of compound always gives

Accepted Answer

NA

Question 4

An queous solution of a substance gives a white precipitate on treating with dilute hydrochloric acid which dissolves on heating. When hydrogen sulphide is passed through the hot solution a black precipitate is obtained. The substance is

Accepted Answer

NA

Question 5

An oxygen containing organic compound was found to contain 53% carbon and 13% hydrogen. Its vapour density is 23. The compound reacts with sodium metal to liberate hydrogen. Identify the functional isomer of this compound.

Accepted Answer

NA

Question 6

An oxidising agent is a substance which ......... electrons whereas a reducing agent is the substance which ......... Electrons

Accepted Answer

NA

Question 7

An oxide X in the normal form is almost non-reactive due to very high X - O bond enthalpy. It resists the attack by halogens hydrogen and most of acids and metals even at elevated temperature. It is only attacked by HF and NaOH. The oxide X is.

Accepted Answer

NA

Question 8

An oxide which gives H_2O_2on treatment with dilute acid is.....

Accepted Answer

NA

Question 9

An oxide of nitrogen contains 36.8% by weight of nitrogen. The formula of the compound is

Accepted Answer

NA

Question 10

An oxide of nitrogen contains 25.9% of nitrogen. What is its empirical formula

Accepted Answer

NA

Question 11

An oxide of nitrogen contains 30.43% nitrogen. The molecular mass of the compound is 92 amu. Find the molecular formula of the given oxide.

Accepted Answer

NA

Question 12

An oxide of metal M has 40% by mass of oxygen. Metal M has relative atomic mass of 24. The empirical formula of the oxide is

Accepted Answer

NA

Question 13

An oxide of element A was analysed and found to have mass ratio of A to oxygen equal to 73. Then formular of oxide can be [Atomic mass of A = 56]

Accepted Answer

NA

Question 14

An oxide of manganese contains 2.29g of manganese per gram of oxygen. What is the empirical formular of this compound?

Accepted Answer

NA

Question 15

An oxide of an element is a gas and dissolves in water to give an acidic solution. The element belongs to

Accepted Answer

NA

Question 16

An oxide of alkaline earth metals X reacts with C and Cl_(2) to give a compound Y. Y is found in polymeric chain structure and is electron deficient molecule. The compound Y is

Accepted Answer

NA

Question 17

Anoxideof alkalineearthmetal on heatingwithcarbonin thepresence of chlorinegivesBeCl_(2).Theoxideis _________.

Accepted Answer

NA

Question 18

An organometallic compound on analysis gave the following results C = 64.4 % H = 5.5 % Fe = 29.9 %. Determine the empirical formula.

Accepted Answer

NA

Question 19

An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals carbenes nitrenes etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals positively and negatively charged species) on the substrate molecule). The substrate molecule although as a whole electrically neutral has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductiveelectromeric occur through inductive electromeric resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects on the other hand involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions however follows opposite order.  The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures the one which has greater number of covalent bonds has less separation of opposite Charges a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others.  For 1-methoxy-1 3-butadiene which of the following resonating structure is the least stable ?

Accepted Answer

NA

Question 20

An organic substance from its aqueous solution can be separated by

Accepted Answer

NA

Question 21

An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals carbenes nitrenes etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals positively and negatively charged species) on the substrate molecule). The substrate molecule although as a whole electrically neutral has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductiveelectromeric occur through inductive electromeric resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects on the other hand involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions however follows opposite order.  The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures the one which has greater number of covalent bonds has less separation of opposite Charges a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others.  Which of the following is the most stable cation ?

Accepted Answer

NA

Question 22

An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals carbenes nitrenes etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals positively and negatively charged species) on the substrate molecule). The substrate molecule although as a whole electrically neutral has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductiveelectromeric occur through inductive electromeric resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects on the other hand involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions however follows opposite order.  The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures the one which has greater number of covalent bonds has less separation of opposite Charges a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others.  The most stable carbanion among the following is

Accepted Answer

NA

Question 23

An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals carbenes nitrenes etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals positively and negatively charged species) on the substrate molecule). The substrate molecule although as a whole electrically neutral has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductiveelectromeric occur through inductive electromeric resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects on the other hand involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions however follows opposite order.  The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures the one which has greater number of covalent bonds has less separation of opposite Charges a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others.  The C-C bond length in propene s little shorter (1.49 Å) than the C-C bond length (1.54 Å) in ethane. This is due to

Accepted Answer

NA

Question 24

An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals carbenes nitrenes etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals positively and negatively charged species) on the substrate molecule). The substrate molecule although as a whole electrically neutral has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductiveelectromeric occur through inductive electromeric resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects on the other hand involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions however follows opposite order.  The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures the one which has greater number of covalent bonds has less separation of opposite Charges a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others.  Out of the following the one containing only nucleophiles is

Accepted Answer

NA

Question 25

An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals carbenes nitrenes etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals positively and negatively charged species) on the substrate molecule). The substrate molecule although as a whole electrically neutral has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductiveelectromeric occur through inductive electromeric resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects on the other hand involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions however follows opposite order.  The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures the one which has greater number of covalent bonds has less separation of opposite Charges a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others.  What is the decreasing order of stability of the ions ?  I. CH_(3)-overset(+)(C)H-CH_(3)II. CH_(3)overset(+)(C)H-OCH_(3)  III. CH_(3)-overset(+)(C)H-COCH_(3)

Accepted Answer

NA

Question 26

An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals carbenes nitrenes etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals positively and negatively charged species) on the substrate molecule). The substrate molecule although as a whole electrically neutral has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductiveelectromeric occur through inductive electromeric resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects on the other hand involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions however follows opposite order.  The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures the one which has greater number of covalent bonds has less separation of opposite Charges a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others.  The most stable free radical among the following is

Accepted Answer

NA

Question 27

An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals carbenes nitrenes etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals positively and negatively charged species) on the substrate molecule). The substrate molecule although as a whole electrically neutral has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductiveelectromeric occur through inductive electromeric resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects on the other hand involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions however follows opposite order.  The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures the one which has greater number of covalent bonds has less separation of opposite Charges a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others.  Which of the following series contains only electrophiles ?

Accepted Answer

NA

Question 28

An organic ompound contains C = 48 H = 8%. 0.48g of the compound was Kjeldahlised and the liberated ammonia required 19.2mL N2 H_(2)SO_(4). Find the empirical formula of the compound.

Accepted Answer

NA

Question 29

An organic liquid vaporises at a temperature below its boiling point in steam distillation. Assign reason.

Accepted Answer

NA

Question 30

An organic liquid of the composition C_(4)H_(8)O_(2) yields a sodium salt of an acid C_(3)H_(6)O_(2) and methanol on boiling with NaOH solution. The given liqud is

Accepted Answer

NA

Question 31

An organic liquid decomposes below its boiling point. How will you purify ?

Accepted Answer

NA

Question 32

An organic fruit smelling compound on analysis has the following composition by mass C = 54.54% H = 9.09% O = 36.36%. Find out the molecular formula of the compound. The vapour density of the compound was found to be 44

Accepted Answer

NA

Question 33

An organic compund (A) has a molecular formula C_(2)H_(6)O it is one of the primary alcohol. A reacts with acidified potassium dichromate to give B.B on further undergoes to oxidation reaction to give C.C on reacts with SOCI_(2) to give D which is chlorinated product.Identify ABC and D explain the equation.

Accepted Answer

NA

Question 34

An organiccompounds which decomposes below its boiling point can be purified by........

Accepted Answer

NA

Question 35

An organic compounds A of a molecular formula C_6H_6Which is simple aromatic hydrocarbon. A react with Cl_2 in pressure of FeCl_(3) to give B. B reacts with NaOH at 350^(@)C and 300 atm pressure to give C. B again reacts with ammonia at 250^(@)C and 50 atm pressure to give D. Identify ABC and Dexplain the reaction.

Accepted Answer

NA

Question 36

An organic compound X(C_(6)H_(13)Br) is optically active X on treatment with (CH_(3))_(3)COK in (CH_(3))_(3)COH gives Y(C_(6)H_(12)) a major product. Y on treatment with Br_(2) - C Cl_(4) in the presence of FeBr_(3) gives a dibromide which on further treatment with NaNH_(2) givesC_(6)H_(10) which is still optically active. Hence X and Y respectively are

Accepted Answer

NA

Question 37

An organic compound (X) with molecular formula C_(7)H_(8)O is insoluble NaHCO_(3) but dissolves in NaOH. When treated with bromine water (X) rapidly gives (Y) C_(7)H_(5)OBr_(3). The compound (X) and (Y) respectively are

Accepted Answer

NA

Question 38

An organic compound X on treatment with acidified K_(2)Cr_(2)O_(7) gives a compound Y which reacts with I_(2) and sodium carbonate to form Tridomethane. The compound X is

Accepted Answer

NA

Question 39

An organic compound x having molecular mass 60 is found to contain C = 20% H = 6.67% and N = 46.67% while rest is oxygen. On heating it gives ammonia along with a solid residue. The solid residue gave violet colour with alkaline copper sulphate solution. The compound X is

Accepted Answer

NA

Question 40

An organic compound X contains Y and Z impurities. Their solubilities differ in a chosen solvent. They may be separated by

Accepted Answer

NA

Question 41

An organic compound X (C_(4)H_(8)O_(2)) gives positive test with NaOH and phenophtalein. Structure of X will be

Accepted Answer

NA

Question 42

An organic compound with the formula C_6 H_(12) O_6 forms a yellow crystalline solid with phenylhydrazine and gives a mixture of sorbitol and mannitol when reduced with sodium. Which among the following could be the compound?

Accepted Answer

NA

Question 43

An organic compound with the formula C_(10)H_(8) absorbs 5H_2 during hydrogenation. The number ofrings present in it is________

Accepted Answer

NA

Question 44

An organic compound with moleular formula C_(2)H_(6)O_(2) evolves H_(2) when treated with sodium metal and gives two moles of formaldehyde on oxidation with HIO_(4). The compound is

Accepted Answer

NA

Question 45

An organic compound with molecular formula C_(7)H_(8)O dissolves in NaOH and gives characteristic colour with FeCl_(3). On treatment with Br_(2) it gives tribromo product C_(7)H_(5)OBr_(3). The compound is

Accepted Answer

NA

Question 46

An organic compound with molecular formula C_(6)H_(10) is not reduced by H_(2)PdBaSO_(4). From the given options the compound may be

Accepted Answer

NA

Question 47

An organic compound with 68.9% of C and 4.92 % of H is aromatic and gives CO_(2) with NaHCO_(3). The organic compound is

Accepted Answer

NA

Question 48

An organic compound which produces a bluish green coloured flame on heating in presence of copper is

Accepted Answer

NA

Question 49

An organic compound weighing 0.15 g gave on carius estimation 0.12 g of silver bromide. The percentage of bromine in the compound will be close to ............... .

Accepted Answer

NA

Question 50

An organic compound was found to have contained carbon = 40.65% hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour - density was found to be 29.5. What is the molecular formula of the compound?

Accepted Answer

NA

Explore topic-wise InterviewSolutions in Current Affairs.

An unidentified bivalent metal M reacts with unidentified halogen X to form an unknown compound of halogen gas. When 1.12g of it is heated, 0.72g of MX is obtained along with 56 ml of halogen gas at 1atm and 273 K. Identify the metal.

An unbalanced chemical equation is against the law of

An S_(N)1 reaction at an asymmetric carbon of compound always gives

An queous solution of a substance gives a white precipitate on treating with dilute hydrochloric acid which dissolves on heating. When hydrogen sulphide is passed through the hot solution, a black precipitate is obtained. The substance is :

An oxygen containing organic compound was found to contain 53% carbon and 13% hydrogen. Its vapour density is 23. The compound reacts with sodium metal to liberate hydrogen. Identify the functional isomer of this compound.

An oxidising agent is a substance which ......... electrons whereas a reducing agent is the substance which ......... Electrons

An oxide X in the normal form is almost non-reactive due to very high X - O bond enthalpy. It resists the attack by halogens, hydrogen and most of acids and metals even at elevated temperature. It is only attacked by HF and NaOH. The oxide X is.

An oxide which gives H_2O_2on treatment with dilute acid is.....

An oxide of nitrogen contains 36.8% by weight of nitrogen. The formula of the compound is

An oxide of nitrogen contains 25.9% of nitrogen. What is its empirical formula

An oxide of nitrogen contains 30.43% nitrogen. The molecular mass of the compound is 92 amu. Find the molecular formula of the given oxide.

An oxide of metal M has 40% by mass of oxygen. Metal M has relative atomic mass of 24. The empirical formula of the oxide is:

An oxide of element A was analysed and found to have mass ratio of A to oxygen equal to 7:3. Then formular of oxide can be : [Atomic mass of A = 56]

An oxide of manganese contains 2.29g of manganese per gram of oxygen. What is the empirical formular of this compound?

An oxide of an element is a gas and dissolves in water to give an acidic solution. The element belongs to

An oxide of alkaline earth metals X reacts with C and Cl_(2) to give a compound Y. Y is found in polymeric chain structure and is electron deficient molecule. The compound Y is

Anoxideof alkalineearthmetal on heatingwithcarbonin thepresence of chlorinegivesBeCl_(2).Theoxideis _________.

An organometallic compound on analysis gave the following results : C = 64.4 %, H = 5.5 %, Fe = 29.9 %. Determine the empirical formula.

An organic substance from its aqueous solution can be separated by

An organic ompound contains C = 48, H = 8%. 0.48g of the compound was Kjeldahlised and the liberated ammonia required 19.2mL N//2 H_(2)SO_(4). Find the empirical formula of the compound.

An organic liquid vaporises at a temperature below its boiling point in steam distillation. Assign reason.

An organic liquid of the composition C_(4)H_(8)O_(2) yields a sodium salt of an acid C_(3)H_(6)O_(2) and methanol on boiling with NaOH solution. The given liqud is

An organic liquid decomposes below its boiling point. How will you purify ?

An organic fruit smelling compound on analysis has the following composition by mass: C = 54.54%, H = 9.09%, O = 36.36%. Find out the molecular formula of the compound. The vapour density of the compound was found to be 44

An organiccompounds which decomposes below its boiling point can be purified by........

An organic compound (X) with molecular formula C_(7)H_(8)O is insoluble NaHCO_(3) but dissolves in NaOH. When treated with bromine water (X) rapidly gives (Y), C_(7)H_(5)OBr_(3). The compound (X) and (Y) respectively are

An organic compound 'X' on treatment with acidified K_(2)Cr_(2)O_(7) gives a compound 'Y' which reacts with I_(2) and sodium carbonate to form Tridomethane. The compound 'X' is

An organic compound x having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67%, while rest is oxygen. On heating, it gives ammonia along with a solid residue. The solid residue gave violet colour with alkaline copper sulphate solution. The compound X is

An organic compound X contains Y and Z impurities. Their solubilities differ in a chosen solvent. They may be separated by :

An organic compound X (C_(4)H_(8)O_(2)) gives positive test with NaOH and phenophtalein. Structure of X will be :

An organic compound with the formula C_6 H_(12) O_6 forms a yellow crystalline solid with phenylhydrazine and gives a mixture of sorbitol and mannitol when reduced with sodium. Which among the following could be the compound?

An organic compound with the formula C_(10)H_(8) absorbs 5H_2 during hydrogenation. The number ofrings present in it is________

An organic compound with moleular formula, C_(2)H_(6)O_(2) evolves H_(2) when treated with sodium metal and gives two moles of formaldehyde on oxidation with HIO_(4). The compound is

An organic compound with molecular formula C_(7)H_(8)O dissolves in NaOH and gives characteristic colour with FeCl_(3). On treatment with Br_(2) it gives tribromo product C_(7)H_(5)OBr_(3). The compound is

An organic compound with molecular formula C_(6)H_(10) is not reduced by H_(2)//Pd//BaSO_(4). From the given options, the compound may be :

An organic compound with 68.9% of C and 4.92 % of H, is aromatic and gives CO_(2) with NaHCO_(3). The organic compound is

An organic compound which produces a bluish green coloured flame on heating in presence of copper is

An organic compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the compound will be close to ............... .

An organic compound was found to have contained carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour - density was found to be 29.5. What is the molecular formula of the compound?

1.	An unidentified bivalent metal M reacts with unidentified halogen X to form an unknown compound of halogen gas. When 1.12g of it is heated, 0.72g of MX is obtained along with 56 ml of halogen gas at 1atm and 273 K. Identify the metal.
Answer» Zn Cu Fe Pb Answer :B

2.	An unbalanced chemical equation is against the law of
Answer» The LAW of GASEOUS volumes The law of CONSTANT proportions The law of MASS action The law of CONSERVATION of mass Answer :D

3.	An S_(N)1 reaction at an asymmetric carbon of compound always gives
Answer» an enantiomer of the substance a product with opposite OPTICAL rotation a MIXTURE of diastereomers a product with 100% INVERSION Solution :a product with 100% inversion

4.	An queous solution of a substance gives a white precipitate on treating with dilute hydrochloric acid which dissolves on heating. When hydrogen sulphide is passed through the hot solution, a black precipitate is obtained. The substance is :
Answer» `Hg^(2+)` salt `Cr^(3+)` salt `Ag^(+)` salt `Pb^(2+)` salt Solution :`Hg^(2+)` salt GIVES white precipitate with dilute HCl. `HgO + 2HCl rarr UNDERSET(("White ppt"))(HgCl_(2)) + H_(2)O` `HgCl_(2) + H_(2)S rarr underset(("Black ppt"))(HgS) + 2HCl`

5.	An oxygen containing organic compound was found to contain 53% carbon and 13% hydrogen. Its vapour density is 23. The compound reacts with sodium metal to liberate hydrogen. Identify the functional isomer of this compound.
Answer» Solution :% O = 100 - (53 + 13) = 34 `:.` Ratio of atoms, `C : H : O = (53)/(12) : (13)/(1) : (34)/(16) = 4.4 : 13 : 2.125 = 2 : 6 : 1` `:. E.F. = C_(2)H_(6)O` andE.F. wt `= 2 XX 12 + 6 xx 1 + 1 xx 16 = 46` Mol. wt. `= 2 xx V.D. = 2 xx 23 = 46` `:.` n = Mol. wt/E.F. wt. `= (46)/(46) = 1` Thus, M.F. `= 1 xx E.F. = C_(2)H_(6)O` Since the given organic compound reacts with sodium o liberate `H_(2)` gas, therefore, it must be ethanol. `underset("Ethanol")(2CH_(3)CH_(2)OH)+2Na rarr 2CH_(3)CH_(2)ONa+H_(2)` The functional ISOMER of ethanol is methoxymethane, i.e., `CH_(3)OCH_(3)`.

6.	An oxidising agent is a substance which ......... electrons whereas a reducing agent is the substance which ......... Electrons
Answer» SOLUTION :GAINS, LOSES

7.	An oxide X in the normal form is almost non-reactive due to very high X - O bond enthalpy. It resists the attack by halogens, hydrogen and most of acids and metals even at elevated temperature. It is only attacked by HF and NaOH. The oxide X is.
Answer» `SiO_(2)` `CO_(2)` `SnO_(2)` `PbO_(2)` SOLUTION :`SiO_(2)` is attacked by HF and NaOH but not by chemicals under NORMAL conditions

8.	An oxide which gives H_2O_2on treatment with dilute acid is.....
Answer» `PbO_2` `Na_2O_2` `MnO_2` `TiO_2` ANSWER :B

9.	An oxide of nitrogen contains 36.8% by weight of nitrogen. The formula of the compound is
Answer» `N_(2)O` `N_(2)O_(3)` `NO` `NO_(2)` SOLUTION :CALCULATE relative No: of atoms of Nitrogen and oxygen `N: O = (26.8)/(14): (100 - 36.8)/(18)`

10.	An oxide of nitrogen contains 25.9% of nitrogen. What is its empirical formula
Answer» SOLUTION :`N_2O_5`

33.	An organic compund (A) has a molecular formula C_(2)H_(6)O it is one of the primary alcohol. A reacts with acidified potassium dichromate to give B.B on further undergoes to oxidation reaction to give C.C on reacts with SOCI_(2) to give D which is chlorinated product.Identify A,B,C and D, explain the equation.
Answer» Solution :(i) `C_(2)H_(6)O` is `CH_(3)-CH_(2)-OH` which is a primary alcohol (A). (II) `CH_(3)-CH_(2) -OH (A)` reacts with `H^(+)/K_(2)Cr_(2)O_(7)` to give ACETALDEHYDE (B) `underset"(Ethanol (A)" (CH_(3)- CH_(2) - OH) overset (H^(+)//K_(2)Cr_(2)O_(7)) to underset ("Acetaldehyde (B)")(CH_(3)CHO )` (iii) `CH_(3)CHO(B)` which on further oxidation to give acetic ACID (C). `underset("Acetaldehyde (B)")(CH_(3)CHO) overset("O") to underset("Acetic Acid (C)")(CH_(3)COOH)` (iv) Acetic acid reacts with `SOCI(2)` to give ACETYL chloride (D). `underset("Acetic acid (C)") (CH_(3)COOH)overset("O") to underset ("Acetyl chloride (D)") (CH_(3)COCI+ SO_(2) + HCI)` `{: ( "(A)`CH_(3)-CH_(2)-OH`", "Ethanol"),( "(B)`CH_(3)CHO`", "Acetaldehyde") , ("(C)`CH_(3)COOH`" , "Acetic acid"),("(D)`CH_(3)COCI` " ,"Acetyl chloride") :}`

11.	An oxide of nitrogen contains 30.43% nitrogen. The molecular mass of the compound is 92 amu. Find the molecular formula of the given oxide.
Answer» ANSWER :`N_(2)O_(4)`

12.	An oxide of metal M has 40% by mass of oxygen. Metal M has relative atomic mass of 24. The empirical formula of the oxide is:
Answer» `M_(2)O` `M_(2)O_(3)` MO `M_(3)O_(4)` Answer :C

13.	An oxide of element A was analysed and found to have mass ratio of A to oxygen equal to 7:3. Then formular of oxide can be : [Atomic mass of A = 56]
Answer» `A_(2)O_(2)` `A_(2)O_(3)` AO `A_(2)O` ANSWER :B

14.	An oxide of manganese contains 2.29g of manganese per gram of oxygen. What is the empirical formular of this compound?
Answer» MnO `MnO_(2)` `Mn_(2)O_(3)` `MnO_(3)` ANSWER :C

15.	An oxide of an element is a gas and dissolves in water to give an acidic solution. The element belongs to
Answer» II GROUP IV group VIII group Zero group ANSWER :B

16.	An oxide of alkaline earth metals X reacts with C and Cl_(2) to give a compound Y. Y is found in polymeric chain structure and is electron deficient molecule. The compound Y is
Answer» BeO `BeCl_(2)` `Be(OH)_(2)` `BeCO_(3)` Solution :`underset((X))(BeO)+C+Cl_(2) to underset((Y))(BeCl_(2))+CO` `BeCl_(2)` is polymeric and ELECTRON DEFICIENT MOLECULE.

17.	Anoxideof alkalineearthmetal on heatingwithcarbonin thepresence of chlorinegivesBeCl_(2).Theoxideis _________.
Answer» `BEO` `BeO_(2)` `Be_(2)O` `Be_(2)O_(2)` ANSWER :A

18.	An organometallic compound on analysis gave the following results : C = 64.4 %, H = 5.5 %, Fe = 29.9 %. Determine the empirical formula.
Answer» Solution :EMPIRICAL formula may be calculated as : `{:("Element","Percentage","Atomic MASS","Gram ATOMS (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",64.4,12,(64.4)/(12)=5.4,(5.4)/(0.53)=10.2,10),("H",5.5,1,(5.5)/(1)=5.5,(5.5)/(0.53)=10.3,10),("Fe",29.9,56,(29.9)/(56)=0.53,(0.53)/(0.53)=1.0,1):}` Empirical formula of the compound `= C_(10)H_(10)Fe`.

19.	An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals, carbenes, nitrenes, etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals, positively and negatively charged species) on the substrate molecule). The substrate molecule, although as a whole electrically neutral, has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductive,electromeric occur through inductive, electromeric, resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects, on the other hand, involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order: 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions, however, follows opposite order. The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures, the one which has greater number of covalent bonds, has less separation of opposite Charges, a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others. For 1-methoxy-1, 3-butadiene, which of the following resonating structure is the least stable ?
Answer» `H_(2)overset(-)(C)-overset(+)(C)H-CH=CH-O-CH_(3)` `H_(2)overset(-)(C)-CH=CH-CH=overset(+)(O)-CH_(3)` `H_(2)C = CH-overset(+)(C)H-overset(-)(C)H-O-CH_(3)` `H_(2)C=CH-overset(-)(C)H-CH=overset(+)(O)-CH_(3)` Solution :All the atoms in STRUCTURES (b) and (d) have complecyed octets and HENCE are more stable than structure (a) and (c) in which the POSITIVELY charged carbon has only sextet of ELECTRONS. Out of (a) and (c), (a) is stabilized by +R-effect of the oxygen atom but (c) is not. Thus, (c) is th LEAST stable structure.

20.	An organic substance from its aqueous solution can be separated by
Answer» EXTRACTION with solvent Steam distillation Fractional distillation Vacuum distillation Answer :A

21.	An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals, carbenes, nitrenes, etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals, positively and negatively charged species) on the substrate molecule). The substrate molecule, although as a whole electrically neutral, has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductive,electromeric occur through inductive, electromeric, resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects, on the other hand, involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order: 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions, however, follows opposite order. The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures, the one which has greater number of covalent bonds, has less separation of opposite Charges, a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others. Which of the following is the most stable cation ?
Answer» `F_(3)C - CH_(2)^(+)` `(CH_(3))_(2)CH^(+)` `CH_(3)^(+)` `CF_(3)^(-)` Solution :CARBOCATION are stabilized by +I -effect of `CH_(3)` GROUPS and destabilized by +I-effect of F ATOMS. Therefore, `(CH_(3))_(2)overset(+)(C)H` is the most stable carbocation.

22.	An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals, carbenes, nitrenes, etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals, positively and negatively charged species) on the substrate molecule). The substrate molecule, although as a whole electrically neutral, has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductive,electromeric occur through inductive, electromeric, resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects, on the other hand, involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order: 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions, however, follows opposite order. The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures, the one which has greater number of covalent bonds, has less separation of opposite Charges, a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others. The most stable carbanion among the following is
Answer» ANSWER :D

23.	An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals, carbenes, nitrenes, etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals, positively and negatively charged species) on the substrate molecule). The substrate molecule, although as a whole electrically neutral, has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductive,electromeric occur through inductive, electromeric, resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects, on the other hand, involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order: 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions, however, follows opposite order. The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures, the one which has greater number of covalent bonds, has less separation of opposite Charges, a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others. The C-C bond length in propene s little shorter (1.49 Å) than the C-C bond length (1.54 Å) in ethane. This is due to
Answer» `+I` EFFECT of `CH_(3)` GROUP mesomeric effect electromeric effect hyperconjugation effect Answer :D

24.	An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals, carbenes, nitrenes, etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals, positively and negatively charged species) on the substrate molecule). The substrate molecule, although as a whole electrically neutral, has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductive,electromeric occur through inductive, electromeric, resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects, on the other hand, involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order: 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions, however, follows opposite order. The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures, the one which has greater number of covalent bonds, has less separation of opposite Charges, a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others. Out of the following, the one containing only nucleophiles is
Answer» `NH_(3), CN^(-), CH_(3)OH` `AlCl_(3), BF_(3), NH_(3)` `AlCl_(3), NH_(2)^(-),H_(2)O` `RNH_(2), :CX_(2), H^(-)` ANSWER :A

25.	An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals, carbenes, nitrenes, etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals, positively and negatively charged species) on the substrate molecule). The substrate molecule, although as a whole electrically neutral, has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductive,electromeric occur through inductive, electromeric, resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects, on the other hand, involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order: 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions, however, follows opposite order. The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures, the one which has greater number of covalent bonds, has less separation of opposite Charges, a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others. What is the decreasing order of stability of the ions ? I. CH_(3)-overset(+)(C)H-CH_(3)II. CH_(3)overset(+)(C)H-OCH_(3) III. CH_(3)-overset(+)(C)H-COCH_(3)
Answer» `I GT II gt III` `II gt III gt I` `III gt I gt II` `II gt I gt III` ANSWER :D

26.	An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals, carbenes, nitrenes, etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals, positively and negatively charged species) on the substrate molecule). The substrate molecule, although as a whole electrically neutral, has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductive,electromeric occur through inductive, electromeric, resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects, on the other hand, involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order: 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions, however, follows opposite order. The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures, the one which has greater number of covalent bonds, has less separation of opposite Charges, a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others. The most stable free radical among the following is
Answer» `C_(6)H_(5)CH_(2)overset()(C)H_(2)` `C_(6)H_(5)overset()(C)HCH_(3)` `CH_(3)overset()(C)H_(2)` `CH_(3)overset()(C)HCH_(3)` Answer :B

27.	An organic reaction occurs through making and breaking of bonds. The breaking of bonds may occur either homolytically leading to the formation of radicals or heterolytically generating positively and negatively charged species. The neutral species (free radicals, carbenes, nitrenes, etc.) and positively charged species being electron deifcient are collectively called electrophiles while neutral and negatively charged species which are electron rich are called nucleophiles. An organic reaction usually involves the attack of a reagent (radicals, positively and negatively charged species) on the substrate molecule). The substrate molecule, although as a whole electrically neutral, has centres of low and high electron density due to displacement of bonding electrons. These electrons displacements occur through inductive,electromeric occur through inductive, electromeric, resonance and hyperconjugation effects. Whereas inductive effects involve displacement transfer of n-and pi-electrons along a conjugated system. Hypercongation effects, on the other hand, involve sigma-pi-conjugation. Both inductive and hyperconjugation effects can be used to explain the stability of carbocations and free radicals which follow the stability order: 3^(@) gt 2^(@) gt 1^(@). The stability of carbanions, however, follows opposite order. The stability of a molecule can be judged on the basis of contribution of its resonance structures. Resonance structures have same position of nuclei and have same number of unpaired electrons. Among resonance structures, the one which has greater number of covalent bonds, has less separation of opposite Charges, a negative charge on more electronegative and a positive charge on a more electropostive atom are more stable than others. Which of the following series contains only electrophiles ?
Answer» `H_(2)O, SO_(3), H_(3)O^(+)` `NH_(3), H_(2)O, AlCl_(3)` `AlCl_(3), SO_(3), OVERSET(+)(N)O_(2)` `H_(2)O, overset(+)(C)L, NH_(3)` ANSWER :C

28.	An organic ompound contains C = 48, H = 8%. 0.48g of the compound was Kjeldahlised and the liberated ammonia required 19.2mL N//2 H_(2)SO_(4). Find the empirical formula of the compound.
Answer» Solution :Percentage of nitrogen `= (1.4 xx N_(1) xx V_(1))/(W)` `W =` Mass of ORGANIC compound `= 0.48g` `N_(1) = (1)/(2) = 0.5, V_(1) = 19.2 mL` Percentage of `N = (1.4 xx 0.5 xx 19.2)/(0.48) = 28` Percentage of oxygen `= 100 - (48 +8 +28) = 16` Calculation of EMPIRICAL FORMULA. `{:("Element","Percentage","At.mass","Relative number of ATOMS","Simplest ratio of atoms"),("Carbon",48.0,12,(48)/(12)=4,(4)/(1)=4),("Hydrogen",8.0,1,(8)/(1)=8,(8)/(1)=8),("Nitrogen",28.0,14,(28)/(14)=2,(2)/(1)=2),("Oxygen",16.0,16,(16)/(16)=1,(1)/(1)=1):}` Empirical formula `= C_(4)H_(8)N_(2)O`

29.	An organic liquid vaporises at a temperature below its boiling point in steam distillation. Assign reason.
Answer» Solution :Steam distillation is ACTUALLY distillation under REDUCED PRESSURE. The vapour pressure of both water vapours and ORGANIC liquid placed in the distillation flask BECOME equal to the atmospheric pressure. This means that both of them will vaporise at a temperature which is less than their normal boiling point temperatures.

30.	An organic liquid of the composition C_(4)H_(8)O_(2) yields a sodium salt of an acid C_(3)H_(6)O_(2) and methanol on boiling with NaOH solution. The given liqud is
Answer» `CH_(3)CH_(2)COOCH_(3)` `CH_(3)COOC_(2)H_(5)` `HCOOC_(3)H_(7)` `CH_(3)CH_(2)CH_(2)COOH` Solution :`H_(3)C-CH_(2)-OVERSET(O)overset("\|\|")(C)-O-CH_(3) overset(NaOH)(rarr) H_(3)C-CH_(2)-overset(O)overset("\|\|")(C)-ONa+CH_(3)OH`.