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New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (iv)

BaCO3 will be the most thermal stable. The size of Ba2+ is very large due to which it has low Polarising power and cannot polarise the oxygen atom. As the positive ions gets larger on moving down the group, the effect on the carbonate ion near them decreases. Thus, the carbonate ions of large cations are stable.

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (iii)

Li due to its small size it has a high value of hydration enthalpy. Due to the high hydration enthalpy of Li atom, it is the strongest reducing agent in the aqueous medium.

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10 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (i)

The reactivity of alkali metals increases on moving down the group. So, Li is least reactive. It will react with water least vigorously.

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Payal Gupta

Contributor-Level 10

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (iv)

In alkali metals, on going down the group, as the metallic strength decreases, melting point  decreases. Thus, Cs has the lowest melting point and melts at 30°C.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

In the solid state BeCl2 has a polymeric chain like structure. Be atom is surrounded by four Cl atoms among which two of them are bonded through covalent bond and other two are through co-ordinate bond. Structure of BeCl2 is given by structure A.

In gaseous state beryllium chloride exists as dimer (Be2Cl4) which dissociates to the monomer at about 1200 K temperature which is linear in structure.

Structure of BeCl2 in gaseous state is given by the structure B.

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Assertion Type Questions as classified in NCERT Exemplar

65. Option (i)

i.e.,   Both A and R are correct, and R is the correct explanation of A is the answer since in paper chromatography, a special quality of paper known as chromatography paper is used. Chromatography paper contains water molecules trapped in it which acts as the stationary phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents which acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively

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10 months ago

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

The flame is due to the excitation of the electron from its higher energy state to the lower energy states. Due to the small atomic and ionic size of beryllium and magnesium, electrons are tightly bound to the atom. The electrons of Be and Mg does not gain excitation from the energy provided by the flame. Hence they do not show any flame in the flame test.

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10 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Assertion Type Questions as classified in NCERT Exemplar

64. Option (iii) i.e., Both A and R are not correct is the answer since sulphur is estimated by Carius method in the form of white precipitate of BaSO4 on heating with fuming and BaCl2. If light yellow solid is obtained means impurities are present. It is filtered, washed and then dried to get pure BaSO4

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

The lewis structure of O2 ion is,

Oxygen atom having no charge has 6 electrons, so its oxidation number is zero. Oxygen atoms containing −1 charge have 7 electrons, so its oxidation number is −1. The average oxidation state of oxygen in this ion is, =1/2

New answer posted

10 months ago

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alok kumar singh

Contributor-Level 10

This is a Assertion Type Questions as classified in NCERT Exemplar

63. Option (iv) i.e., A is not correct, but R is correct is the answer since hybridization of C can be found out by counting σ bonds and π bonds present on C atom. This can be shown as-

If C has 3σ bonds then it is sp2 hybridized. If C has 2σ bonds then it is sp hybridized.

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