Periodic Table & Periodicity

JEE Chemistry · 270 questions · Page 19 of 27 · Click an option or "Show Solution" to reveal answer

Q181

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Beryllium has less negative value of reduction potential compared to the other alkaline earth metals. Reason R : Beryllium has large hydration energy due to small size of Be

^{2+}

but relatively large value of atomization enthalpy In the light of the above statements, choose the most appropriate answer from the options given below :

A Both A and R are correct but R is NOT the correct explanation of A

B Both A and R are correct and R is the correct explanation of A

C A is correct but R is not correct

D A is not correct but R is correct

Correct Answer

Option B

Solution

The most appropriate answer is Both A and R are correct and R is the correct explanation of A.

Assertion A states that beryllium has a less negative value of reduction potential compared to the other alkaline earth metals.

This assertion is correct, as beryllium has higher ionization energy and a smaller atomic radius than the other alkaline earth metals, which makes it more difficult to reduce Be2+ to Be metal.

Reason R provides an explanation for Assertion A.

Beryllium has a relatively large value of atomization enthalpy due to the strong Be-Be bonds in the solid state, but it also has a large hydration energy due to the small size of Be2+ ions.

The strong Be-Be bonds make it difficult to reduce Be2+ to Be metal, and the small size of Be2+ ions leads to a large hydration energy.

This makes it energetically unfavorable to remove the Be2+ ion from the aqueous phase and reduce it to Be metal, which results in a less negative value of reduction potential for beryllium.

Therefore, both Assertion A and Reason R are correct, and Reason R provides the correct explanation for Assertion A.

Q182

Match the following compounds (Column-I) with their uses (Column-II) : .tg .tg S.No. Column-I S.No. Column-II (I) Ca(OH)2 (A) casts of statues (II) NaCl (B) white wash (III) CaSO4,

{1 \over 2}

H2O (C) antacid (IV) CaCO3 (D) washing soda preparation

A (I)-(B), (II)-(C), (III)-(D), (IV)-(A)

B (I)-(B), (II)-(D), (III)-(A), (IV)-(C)

C (I)-(C), (II)-(D), (III)-(B), (IV)-(A)

D (I)-(D), (II)-(A), (III)-(C), (IV)-(B)

Correct Answer

Option B

Solution

(I) Ca(OH)2 is used in white wash due to its disinfectant nature.

(II) NaCl is used in preparation of washing soda(Na2CO3).

(1) 2NH3 + H2O + CO2 $\to$ (NH4 )2CO3 (2) (NH4 )2CO3 + H2O + CO2 $\to$ 2NH4HCO3 (3) NH4HCO3 + NaCl $\to$ NH4Cl + NaHCO3 (4) 2NaHCO3 $\to$ Na2CO3 + CO2 + H2O (III) CaSO4.

{1 \over 2}

H2O (Plaster of Paris) Used for making casts of statues. (IV) CaCO3 is used as an Antacid.

Q183

Identify the correct statements about alkali metals. A. The order of standard reduction potential (M

^+

| M) for alkali metal ions is Na > Rb > Li. B. CsI is highly soluble in water. C. Lithium carbonate is highly stable to heat. D. Potassium dissolved in concentrated liquid ammonia is blue in colour and paramagnetic. E. All the alkali metal hydrides are ionic solids. Choose the correct answer from the options given below :

A C and E only

B A and E only

C A, B, D only

D A, B and E only

Correct Answer

Option B

Solution

The order of standard reduction potential (M+ | M) for alkali metal ions is Na > Rb > Li.

This statement is correct.

The standard reduction potential generally increases as we move down the group from Li to Cs so order should be Cs > Rb > K > Na > Li but here Na and K are exception, Na has highest standard reduction potential in the entire group and then K.

So right order is Na > K > Cs > Rb > Li.

That is why right order is Na > Rb > Li.

Standard reduction potential (M+ | M) for alkali metal ions are - (1).

Li = -3.04 V (2).

Na = -2.714 V (3).

K = -2.925 V (4).

Rb = -2.930 V (5).

Cs = -2.927 V B.

CsI is highly soluble in water.

This statement is not correct.

CsI is the least soluble alkali metal halide in water, and its solubility decreases as the size of the anion increases.

Lithium carbonate is highly stable to heat.

This statement is not correct.

Lithium carbonates decomposes easily on heating to form the lithium oxide and CO2.

Potassium dissolved in concentrated liquid ammonia is blue in colour and paramagnetic.

This statement is not correct.

In concentrated solution, the blue colour changes to bronze colour and the solution becomes dimagnetic.

All the alkali metal hydrides are ionic solids.

This statement is correct.

All alkali metal hydrides are ionic solids, and their properties are determined by the ionic nature of the compounds.

Q184

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion (A) :

\mathrm{BeCl}_{2}

and

\mathrm{MgCl}_{2}

produce characteristic flame Reason (R) : The excitation energy is high in

\mathrm{BeCl}_{2}

and

\mathrm{MgCl}_{2}

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

\mathrm{(A)}

is true but

(\mathrm{R})

is false

B Both (A) and

(\mathrm{R})

are true and

(\mathrm{R})

is the correct explanation of

(A)

C (A) is false but (R) is true

D Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

Correct Answer

Option C

Solution

BeCl2 and MgCl2 do not produce characteristic flame because excitation energy is high in BeCl2 and MgCl2.

Hence, the correct answer is option (C).

Q185

Better method for preparation of

\mathrm{BeF}_{2}

, among the following is :

\mathrm{BeH}_{2}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}

\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}

\mathrm{Be}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}

\mathrm{BeO}+\mathrm{C}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}

Correct Answer

Option B

Solution

According to the NCERT (s block) textbook, the better method for the preparation of BeF₂ is heating (NH₄)₂BeF₄.

Option B :

\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2} + 2\mathrm{NH}_{4}\mathrm{F}

This method involves the thermal decomposition of ammonium beryllium fluoride ((NH₄)₂BeF₄) to form beryllium fluoride (BeF₂) and ammonium fluoride (NH₄F).

The reaction is more efficient compared to other options and results in a purer product.

Q186

Identify the correct order of standard enthalpy of formation of sodium halides :

\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaCl}<\mathrm{NaF}

\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaF}<\mathrm{NaCl}

\mathrm{NaF}<\mathrm{NaCl}<\mathrm{NaBr}<\mathrm{NaI}

\mathrm{NaCl}<\mathrm{NaF}<\mathrm{NaBr}<\mathrm{NaI}

Correct Answer

Option A

Solution

Lattice energy is directly proportional to the charges of the ions and inversely proportional to the sum of their radii: Lattice energy ∝ (Q₁Q₂) / (r₁ + r₂) As we move down the group, the size of the halide ions increases, resulting in a decrease in lattice energy.

This leads to a decrease in the exothermicity of the lattice energy release during the formation of sodium halides.

Thus, the correct order of standard enthalpy of formation of sodium halides, considering both lattice energy and electron affinity, is:

\mathrm{NaF}>\mathrm{NaCl}>\mathrm{NaBr}>\mathrm{NaI}

Q187

\mathrm{Be}(\mathrm{OH})_{2}

reacts with

\mathrm{Sr}(\mathrm{OH})_{2}

to yield an ionic salt. Choose the incorrect option related to this reaction from the following :

A Be is tetrahedrally coordinated in the ionic salt.

B The element

\mathrm{Be}

is present in the cationic part of the ionic salt.

C Both

\mathrm{Sr}

and

\mathrm{Be}

elements are present in the ionic salt.

D The reaction is an example of acid - base neutralization reaction.

Correct Answer

Option B

Solution

$\mathrm{Be}(\mathrm{OH})_2$ is amphoteric in nature. $\mathrm{Sr}(\mathrm{OH})_2$ is basic in nature.

These two undergo acid - base reaction to form a salt.

\mathrm{Be}(\mathrm{OH})_2+\mathrm{Sr}(\mathrm{OH})_2 \rightarrow \underset{(\text{salt})}{\mathrm{Sr}\left[\mathrm{Be}(\mathrm{OH})_4\right]}

Q188

The density of alkali metals is in the order :

\mathrm{Na}<\mathrm{K}<\mathrm{Cs}<\mathrm{Rb}

\mathrm{K}<\mathrm{Na}<\mathrm{Rb}<\mathrm{Cs}

\mathrm{K}<\mathrm{Cs}<\mathrm{Na}<\mathrm{Rb}

\mathrm{Na}<\mathrm{Rb}<\mathrm{K}<\mathrm{Cs}

Correct Answer

Option B

Solution

Alkali metals are part of Group 1 on the periodic table and include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), and cesium (Cs).

As you move down the group, the atomic radius increases due to the addition of energy levels (shells).

While you might think that the increased atomic mass would correspond to an increase in density, the increase in atomic radius generally outpaces the increase in atomic mass, leading to a decrease in density.

The densities (at room temperature) of the mentioned alkali metals are : Sodium (Na) : 0.97 g/cm³ Potassium (K) : 0.86 g/cm³ Rubidium (Rb) : 1.53 g/cm³ Cesium (Cs) : 1.93 g/cm³ So, the order of densities is : K < Na < Rb < Cs

Q189

The correct order of metallic character is :

A K > Be > Ca

B Ca > K > Be

C K > Ca > Be

D Be > Ca > K

Correct Answer

Option C

Solution

As you move from top to bottom in a group (vertical column) on the periodic table, the metallic character increases.

This is because elements at the bottom of a group have more electron shells than those at the top.

These extra shells shield the outermost electrons from the pull of the nucleus, which makes it easier for these electrons to be lost, a characteristic of metals.

On the other hand, as you move from left to right across a period (horizontal row) on the periodic table, the metallic character decreases.

This is because the nuclear charge increases as you move across a period, which pulls the electrons closer to the nucleus and makes them less likely to be lost.

Given this, for the elements K (Potassium), Be (Beryllium), and Ca (Calcium), the order of metallic character should be K > Ca > Be.

Potassium (K) is furthest down (group 1, period 4), calcium (Ca) is above potassium (group 2, period 4), and beryllium (Be) is above calcium (group 2, period 2).

So, the correct order should be Option C: K > Ca > Be.

Q190

Lime reacts exothermally with water to give 'A' which has low solubility in water. Aqueous solution of 'A' is often used for the test of CO

_2

, a test in which insoluble B is formed. If B is further reacted with CO

_2

then soluble compound is formed, 'A' is :

A Quick lime

B White lime

C Lime water

D Slaked lime

Correct Answer

Option D

Solution

Lime (Calcium oxide, CaO) reacts exothermally with water to form slaked lime (Calcium hydroxide, Ca(OH) $_2$ ): CaO (s) + H $_2$ O (l) $\rightarrow$ Ca(OH) $_2$ (s) This slaked lime has low solubility in water and is used for the test of carbon dioxide (CO $_2$ ).

In this test, the carbon dioxide reacts with the calcium hydroxide to form calcium carbonate (CaCO $_3$ ), an insoluble white precipitate: Ca(OH) $_2$ (aq) + CO $_2$ (g) $\rightarrow$ CaCO $_3$ (s) + H $_2$ O (l) Further reaction of calcium carbonate with carbon dioxide forms calcium bicarbonate [Ca(HCO $_3$ ) $_2$ ], a soluble compound: CaCO $_3$ (s) + CO $_2$ (g) + H $_2$ O (l) $\rightarrow$ Ca(HCO $_3$ ) $_2$ (aq) Thus, 'A' in this case is slaked lime.

So, the correct option is: Option D : Slaked lime

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