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Science (English Medium) Class 11 - CBSE Question Bank Solutions for Chemistry

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Chemistry
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The molecular formula of the compound formed from \[\ce{B}\] and \[\ce{C}\] will be ______.

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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The electronic configuration of the outer most shell of the most electronegative element is ______.

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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Amongst the following elements whose electronic configurations are given below, the one having the highest ionisation enthalpy is ______.

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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Dimagnetic species are those which contain no unpaired electrons. Which among the following are dimagnetic?

(i) \[\ce{N2}\]

(ii) \[\ce{N^{2-}2}\]

(iii) \[\ce{O2}\] 

(iv) \[\ce{O^{2-}2}\]

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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Explain the shape of \[\ce{BrF5}\].

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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Explain why \[\ce{PCl5}\] is trigonal bipyramidal whereas \[\ce{IF5}\] is square pyramidal.

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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Which of the following statements is correct?

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these? Give reasons.

[8] Organic Chemistry - Some Basic Principles and Techniques
Chapter: [8] Organic Chemistry - Some Basic Principles and Techniques
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Match Column I with Column II.

Column I Column II
(i) Dumas method (a) AgNO3
(ii) Kjeldahl’s method (b) Silica gel
(iii) Carius method (c) Nitrogen gas
(iv) Chromatography (d) Free radicals
(v) Homolysis (e) Ammonium sulphate
[8] Organic Chemistry - Some Basic Principles and Techniques
Chapter: [8] Organic Chemistry - Some Basic Principles and Techniques
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Which of the following reactions of methane is incomplete combustion:

[9] Hydrocarbons
Chapter: [9] Hydrocarbons
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The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol–1, 363.7 kJ mol–1 and 296.8 kJ mol–1 respectively. What will be the order of reactivity of these halogen acids?

[9] Hydrocarbons
Chapter: [9] Hydrocarbons
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Predict the major product (s) of the following reactions and explain their formation.

\[\ce{H3C - CH = CH2 ->[(Ph.CO.O)2][HBr]}\]

\[\ce{H3C - CH = CH2 ->[HBr]}\]

[9] Hydrocarbons
Chapter: [9] Hydrocarbons
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Nucleophiles and electrophiles are reaction intermediates having electron-rich and electron-deficient centres respectively. Hence, they tend to attack electron-deficient and electron-rich centres respectively. Classify the following species as electrophiles and nucleophiles.

(i) H3CO 

(ii) \[\begin{array}{cc}
\phantom{.}\ce{O}\\
\phantom{.}||\\
\ce{H3C - C - O-}
\end{array}\]

(iii) \[\ce{\overset{\bullet}{C}l}\]

(iv) Cl2C:

(v) (H3C)3C+ 

(vi) Br– 

(vii) H3COH

(viii) R – NH – R

[9] Hydrocarbons
Chapter: [9] Hydrocarbons
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Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.

[9] Hydrocarbons
Chapter: [9] Hydrocarbons
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0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

[8] Organic Chemistry - Some Basic Principles and Techniques
Chapter: [8] Organic Chemistry - Some Basic Principles and Techniques
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In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

[8] Organic Chemistry - Some Basic Principles and Techniques
Chapter: [8] Organic Chemistry - Some Basic Principles and Techniques
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Discuss the principle of estimation of halogens present in an organic compound.

[8] Organic Chemistry - Some Basic Principles and Techniques
Chapter: [8] Organic Chemistry - Some Basic Principles and Techniques
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\[\ce{CO}\] is isoelectronic with

(i) \[\ce{NO+}\]

(ii) \[\ce{N2}\]

(iii) \[\ce{Sncl2}\]

(iv) \[\ce{NO^{-}2}\]

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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Give reasons for the following:

Ethyne molecule is linear.

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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Assertion (A): Among the two \[\ce{O - H}\] bonds in \[\ce{H2O}\] molecule, the energy required to break the first \[\ce{O - H}\] bond and the other \[\ce{O - H}\] bond is the same.

Reason (R): This is because the electronic environment around oxygen is the same even after breakage of one \[\ce{O - H}\] bond.

[4] Chemical Bonding and Molecular Structure
Chapter: [4] Chemical Bonding and Molecular Structure
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