posses low ionisation potential

have a high affinity towards non-metals.

MCQ Questions for Class 11 Medical Chemistry The S-Block Elements Quiz 3

The alkali metal which acts as a nutrient for plants is:

0%
$Na$
0%
$K$
0%
$Li$
0%
$Rb$

Alkaline earth metals act as:

0%
reducing agents
0%
oxidising agent
0%
amphoteric
0%
acidic

In the following, the element with the highest electropositivity is:

0%
copper
0%
caesium
0%
barium
0%
chromium

Which of the following statements is/are false?

0%
The halides of beryllium are electron deficient and are polymeric with halogen bridges.
0%
$BeF_2$ is very soluble in water owing to high solvation energy of Be but fluorides of other alkaline earth metals are almost insoluble.
0%
CaO, SrO and BaO react exothermically with water forming soluble hydroxides.
0%
The bicarbonates of alkaline earth metals are only stable in solutions.

Which of the following statements is true?

0%
BeO and $Be(OH)_2$ both are purely basic in nature.
0%
Be can expand its covalency to six.
0%
Be largely forms covalent compounds which get easily hydrolysed.
0%
Be combines with hydrogen upon heating to form $BeH_2$ .

Explanation

(A) BeO and

$Be(OH)_2$ are amphoteric in nature.

(B) Be cannot expand its covalency greater than four as it does not have vacant d-orbitals.

(D) Do not combine directly with

$H_2$ on heating to form a hydride.

Method used is:

$2BeCl_2+LiAlH_4\rightarrow 2BeH_2+LiCl+AlCl_3$ .

Which amongst the following is soluble in water?

0%
$BeSO_4$
0%
BeO
0%
$Be(OH)_2$
0%
$BeCl_2$

Explanation

$BeSO_4$ is soluble in water as its hydration enthalpy because of the larger size of the sulphate ion.

In which case first has lower melting point than second?

0%
Mg, Be
0%
Be, Ca
0%
Cs, Li
0%
K, Li

The ionic radii of alkali metal ions in water (hydrated radii) are in the order :

0%
$Li^+> Na^+ > K^+ > Rb^+ < Cs^+$
0%
$Li^+(aq.)> Na^+(aq.) > K^+ (aq.)> Rb^+ (aq.)>Cs^+(aq.)$
0%
$Li^+< Na^+ > K^+ > Rb^+ > Cs^+$
0%
$Li^+(aq.) > Na^+(aq.) < K^+ (aq.) < Rb^+ (aq.) < Cs^+(aq.)$

Explanation

Ion	$Li^+$	$Na^+$	$K^+$	$Rb^+$	$Cs^+$
Ionic radius (pm)	76	102	138	152	167
Hydrated radius (pm)	340	276	232	228	226
Ionic mobility $(ohm^{-1}cm^2mol^{-1})$	33.5	43.5	64.5	67.5	68.0

Hydration of ions: The alkali metal ions are highly hydrated. The smaller the size of the ion, the greater is the degree of hydration. Thus,

$Li^+$ ion gets much more hydrated than

$Na^+$ ion which in turn is more hydrated than

$K^+$ ion and so on. The extent of hydration decreases down the group.

Which of the following statements is false?

0%
Both $Be(OH)_2$ and $Al(OH)_3$ dissolve in excess of alkali to give beryllate ion $[Be(OH)_4]^{2-}$ and aluminate ion $[Al(OH)_4]^-$ espectively.
0%
Chlorides of Be and Al are soluble in organic solvents and are strong Lewis acid.
0%
Be and Al have strong tendency to from complexes with F-ligands.
0%
The chlorides of Be and Al have $Cl^-$ bridged structure in liquid phase.

Explanation

$BeCl_2$ and

$AlCl_3$ have bridged structure in vapour phase and not in liquid phase.
1158088_114975_ans_8895435e166f405892919cb0c435d477.png

On heating in air, sodium burns violently with golden yellow flame forming :

0%
monoxide and hydroxide
0%
monoxide and carbonate
0%
peroxide and hydroxide
0%
monoxide and peroxide

Explanation

Reaction of process:

$Na+air\rightarrow Na_2O+Na_2O_2$ (pale yellow)

$2Na_2O_2+2CO_2\rightarrow Na_2CO_3+O_2$

In which case second has lower boiling point than first?

0%
Li, K
0%
Be, Mg
0%
Li, Cs
0%
Ca, Mg

Explanation

(A) The boiling point of K is

$1032$ K which is lower than that of Li (

$1615$ K)
(B) The boiling point of Mg is

$1363$ K which is lower than that of Be (

$2745$ K).
(C) The boiling point of Cs is

$944$ K which is lower than that of Li (

$1615$ K).
(D) The boiling point of Mg is

$1363$ K is which is lower than that of Ca (

$1767$ K).
Hence, all are correct.

Which is the major constituent of gun powder?

0%
Nitre
0%
Sulphur
0%
Charocal
0%
Chile salt petre

A metal which does not react with nitrogen is:

0%
$Li$
0%
$K$
0%
$Ca$
0%
$Mg$

Explanation

Only

$K$ doesn't react with nitrogen directly, while others do.
The reactions are:

$Li+N_2\rightarrow Li_3N$

$Mg+N_2\rightarrow Mg_3N_2$

$3Ca(s)+N_2(g)\rightarrow Ca_3N_2(s)$

Lithium is the smallest atom among alkali metals. ... When Lithium reacts with nitrogen gas at room temperature (N2), it forms Lithium Nitride (Li3N) which is stable, because the lattice energy released from the formation of Li3N is high enough to make the overall reaction exothermic.

As K is a group 1 element, it will not react with N

$_2$

Hence, the correct option is

$\text{B}$

The alkali metals:

0%
form salts like hydrides
0%
posses low ionisation potential
0%
have a high affinity towards non-metals.
0%
have low density

Nitrate can be converted into metal oxide on heating in case of :

0%
Li
0%
Na
0%
both (a) and (b)
0%
None of these

Explanation

Reaction on heating:

$2LiNO_{3} \rightarrow Li_{2}O+2NO_{2}+ \frac{1}{2} O_{2}$

$Mg(NO_{3})_{2} \rightarrow MgO+2NO_{2}+ \frac{1}{2} O_{2}$

$2NaNO_3\rightarrow 2NaNO_2+O_2$

Calgon used as a water softener is:

Explanation

Sodium hexa meta phosphate $Na_6(PO_3)_6$ or $Na_6P_6O_{18}$ is called Calgon. It is used for softening hard water. Calgon ionizes to give a complex anion: The addition of Calgon to hard water causes the calcium and magnesium ions of hard water to displace sodium ions from the anion of Calgon. This results in the removal of calcium and magnesium ions from hard water in the form of a complex with Calgon. The water is softened and sodium ions are released into the water.

$Na_2[Na_4(PO_3)_6] + 2Ca^{2+} \rightarrow Na_2[Ca_2(PO_3)₆_6+ 4Na^{+}$ .

Hence option A is correct.

There is loss in weight when mixture of

$Li_{2}CO_{3}$ and

$Na_{2}CO_{3}\cdot10H_{2}O$ is heated strongly. This loss is due to :

0%
$Li_{2}CO_{3}$
0%
$Na_{2}CO_{3}\cdot10H_{2}O$
0%
both (a) and (b)
0%
none of these

Explanation

Loss in weight due to following reactions:

$Li_{2}CO_{3} \rightarrow Li_{2}O+CO_{2} \uparrow$

$Na_{2}CO_{3}\cdot10H_{2}O \rightarrow Na_{2}CO_{3} + 10H_{2}O \uparrow$
As sodium carbonate is thermally stable so it will not dissociates on heating.

The common name, "cream of tartar" refers to :

0%
$KHC_{4}H_{4}O_{6}$
0%
$K(SbO)C_{4}H_{4}O_{6}$
0%
$KNaC_{4}H_{4}O_{6}$
0%
$Mg_{2} (OH)_{2}(C_{4}H_{4}O_{6})H_{2}O$

Explanation

Cream of tartar : Potassium bitartrate(potassium hydrogen tartrate)

$KHC_4H_4O_6$

Which are involved in various physiological functional in animals?

0%
$Na^{+}$
0%
$K^{+}$
0%
Both (a) and (b)
0%
None of these

Explanation

$Na^{+}$ ,

$K^{+}$ ,

$Mg^{+2}$ and

$Ca^{+2}$ ions are concentrated in animals.

Arrange Li, Na, K, Rb, Cs in order of increasing ionization energy:

0%
$Cs\, <\, Rb\, <\, K\, <\, Na\, <\, Li$
0%
$K\, <\, Rb\, <\, Cs\, <\, Na\, <\, Li$
0%
$Cs\, <\, K\, <\, Rb\, <\, Na\, <\, Li$
0%
None of these

Explanation

Concept: As we move from top to bottom in group the ionization energy will be decreased,because the atomic radius will be increased so the removal electron will be easy.
Li, Na, K, Rb, Cs are IA group elements from Li to Rb the ionization energy
will be decreased. So, the correct order of ionization energy is

$Cs\, <\, Rb\, <\, K\, <\, Na\, <\, Li$
Hence, option A is correct.

Arrange Li, Na, K, Rb, Cs in order of increasing electron-affinity:

0%
$Cs\, <\, Rb\, <\, K\, <\, Na\, <\, Li$
0%
$K\, <\, Na\, <\, Cs\, <\, Rb\, <\, Li$
0%
$Cs\, <\, K\, <\, Rb\, <\, Li\, <\, Na$
0%
None of these

Explanation

Electron affinity is defined as the change in energy (in kJ/mole) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion.
As we move from top to bottom in group the electron-affinity will be decreased, because the atomic radius will be increased so distance from outer electrons to the nucleus will be increased,so the electrons will be difficultly added Li, Na, K, Rb, Cs are IA group elements from Li to Rb the electronegativity will be decreased.so the correct order of electronegativity is

$Cs\, <\, Rb\, <\, K\, <\, Na\, <\, Li$ .

Increasing order of atomic radii
Li, Na, K, Rb, Cs :

0%
Li < Na < K < Rb < Cs
0%
Li < Na < K < Cs < Rb
0%
Li < K < Na< Rb < Cs
0%
None of these

Flame test is not given by :

0%
Ca
0%
Ba
0%
Be
0%
Li

Hydration energy of :

0%
$Be^{2+}\, >\, Na^{+}$
0%
$Be^{2+}\, <\, Na^{+}$
0%
$Be^{2+}\, =\, Na^{+}$
0%
None of these

Explanation

The heat energy released when new bonds are made between the ions and water moleculesis known as the hydration enthalpy of the ion. The hydration enthalpy is the enthalpy change when 1 mole of gaseous ions dissolve in sufficient water to give an infinitely dilute solution. Hydration enthalpies are always negative. Larger the atomic size lesser the hydration energy or the energy decreases.
The size of

$Na^{+}$ greater than

$Be^{2+}$ So the hydration energy of

$Be^{2+}$ is greater than

$Na^{+}$ .
So the correct oreder of hydration energy is

$Be^{2+}\, >\, Na^{+}$
Hence option A is correct.

Arrange Li, Na, K, Rb, Cs in order of increasing electronegativity:

0%
$Cs\, <\, Rb\, =\, K\, <\, Na\, <\, Li$
0%
$Rb\, <\, Cs\, =\, Li\, <\, Na\, <\, L$
0%
$Cs\, <\, Rb\, =\, Na\, <\, Li\, <\, K$
0%
None of these

Explanation

As we move from top to bottom in a group the

$electronegativity$ (measure of the tendency of an atom to attract a bonding pair of electrons) will be decreased because the atomic radius will be increased so distance from outer electrons to the nucleus will be increased, so the electrons cannot be attracted easily.

Li, Na, K, Rb, Cs are IA group elements from Li to Rb the electronegativity will be decreased.

So, the correct order of electronegativity is

$Cs\, <\, Rb\, =\, K\, <\, Na\, <\, Li$
Hence, option A is correct.

The correct order of covalent nature is:

0%
$BeCl_2 > LiCl$
0%
$BeCl_2 < LiCl$
0%
$BeCl_2 = LiCl$
0%
None of these

Explanation

As the size of cation (metal atom) less the covalent nature of metal halides increases.

So order of the size of metal atoms is

$Be>Li$

So the correct order of covalent nature of metal halides is

$BeCl_2>LiCl$ .

Hence option A is correct.

Choose the correct options regarding the nature of oxides formed by

$Li$ and

$Be$ :

0%
$Li_2O$ is acidic while $BeO$ is basic.
0%
$BeO$ is amphoteric.
0%
$Li_2O$ is basic.
0%
$Li_2O$ and $BeO$ both are amphoteric.

Explanation

In a period from left to right acidic natutre increases basic nature decreases.
So

$Li_2O$ is basic.

$BeO$ is amphoteric.
So options A & B are correct.

Compare thermal stability of

$Li_2CO_3$ and

$BeCO_3$ :

0%
$Li_2CO_3\, > \,BeCO_3$
0%
$Li_2CO_3\, < \,BeCO_3$
0%
$Li_2CO_3\,= \,BeCO_3$
0%
None of these

Explanation

Thermal stability is the stability of a molecule at high temperatures.

$Li$ and

$Be$ are 2nd-period elements , so as we move from left to right in a period the thermal stability of carbonates, decreased due to a decrease in the size of the ions. Hence the thermal stability of

$Li_2CO_3\, > \,BeCO_3$ . Option A is correct.

Compare

$Li$ and

$Be$ according to their

$(IE)_2$ :

0%
$Li\, >\, Be$
0%
$Li\, <\, Be$
0%
$Li\, =\, Be$
0%
None of these

Explanation

Electronic configuration of

$Li$ is

$1s^{2}2s^{1}$ so

$(IE)_2$ we have to remove the second electron present in

$1s$ shell which is completely filled and stable so it requires more energy to remove an electron from

$1s$ shell of

$Li$
Electronic configuration of

$Be$ is

$1s^{2}2s^{2}$ so

$(IE)_2$ we have to remove the second electron present in

$2s$ shell which is partially filled hence it requires less energy to remove an electron from

$2s$ shell of

$Be$ when compared with

$Li$ so

$(IE)_2$ of

$Li$ is greater than

$Be$ . Hence option A is correct.

Size of hydrated ion of

$Li^+ and Be^{2+}$ is :

0%
$Li^{+}\, <\, Be^{2+}$
0%
$Li^{+}\, >\, Be^{2+}$
0%
$Li^{+}\, = \, Be^{2+}$
0%
None of these

Explanation

The hydration energies of alkali metal ions decrease with increase in ionic size. The size of alkali metal ions in aqueous solution(hydrated ion) is in the following order

$Cs^{+}\, <\, Rb^{+}\, <\, K^{+}\, <\, Na^{+}\, <\, Li^{+}$

so as the size of the ion increases the hydrated ion size decreases.

So the correct order

$Li^{+}\, <\, Be^{2+}$ Hence, option A is correct,

Which has no effect in terms of action under normal heating?

$Na_2CO_3 \ and \ CaCO_3 \\$

0%
$Na_2CO_3 \quad$
0%
$CaCO_3 \quad \\$
0%
$Na_2CO_3 \ and \ CaCO_3 \ both\\$
0%
$None \ of \ these \\$

Explanation

$Na_2CO_3$ has no effect in terms of action under normal heating

Where as

$CaCO_3$ on thermal decomposition forms

$CaO,CO_2$ .

$\quad Na_{ 2 }CO_{ 3 } \ \ \ \ \underrightarrow { \triangle }\ \ \ no \quad effect$

$\quad CaCO_{ 3 }\ \ \ \ \ \ \underrightarrow { \triangle } \ \ \ \ CaO+CO_{ 2 }\\$

Hence option A is correct.

The bicarbonate of alkali metal which does not exist in solid state is:

0%
$LiHCO_3$
0%
$NaHCO_3$
0%
$KHCO_3$
0%
$RbHCO_3$

Explanation

Li differs from the rest of the family members and shows anomalous behaviour. Lithium bicarbonate found in solution form while other bicarbonates of alkali metals exist in solid-state. Its anomalous behaviour is due to the following reasons:-

1.Its exceptionally small size
2.High polarizing power.

So $LiHCO_3$ does not exist in solid-state.

Hence option A is correct.

Statement: The lattice energy of

$LiCl$ is smaller than

$NaCl$ .

State whether the given statement is true or false.

0%
True
0%
False

Explanation

$Li^{ + }$ and

$Na^{ + }$ both have same charge but the

$Na^{ + }$ ion has a large radius.Since it has an extra shell to accomadate the extra electrons.Since the distance between the centre of

$Na^{ + }$ and Cl- is greater than that between

$Li^{ + }$ and Cl-,the attractive force between

$Li^{ + }$ and Cl- is greater than that between

$Na^{ + }$ and Cl-.Therefore ,

$LiCl$ has a greater lattice lattice energy than

$NaCl$ .

Hence, the given statement is

$false$

Compare

$Li^{+}$ and

$Be^{+}$ in ionic conductance.

0%
$Be^{+}$ > $Li^{+}$
0%
$Be^{+}$ < $Li^{+}$
0%
$Be^{+}$ = $Li^{+}$
0%
None of these

Explanation

As smaller the size of the ion, larger will be the charge of the ion. Hence, greater the hydration and smaller ionic conductance of the ion. As

$Be^{+}$ has less size, therefore has less ionic conductance than

$Li^{+}$ . So, the order is

$Be^{+} < Li^{+}$ .
Hence, option B is the correct answer.

Estimation of calcium and magnesium is done by :

0%
EDTA.
0%
oxalate.
0%
phosphate.
0%
None of these.

Select correct statement(s):

0%
$Mg^{2+}$ ions are necessary for the activation of phosphate transfer enzymes.
0%
Mg is present in chlorophyll used in photosynthesis in green plants.
0%
Operation of $Na^{+}\, -\, K^{+}$ pumps is biological.
0%
All the above are correct statements.

Explanation

All phosphate transfer enzymes (kinases), such as creatine kinase, require the essential presence of

$Mg^{2+}$ ions.

Mg is central atom in chlorophyll,the green pigment in plant

$Na^+$ OR

$K^+$ pump found in the membrane of animal cell which uses the hydrolysis of ATP to pump

$3Na^+$ out of the cell and

$2K^+$ in the cell.

Among

$Na^+, Na, Mg$ and

$Mg^{2+}$ the largest atom is :

0%
$Mg^{2+}$
0%
Mg
0%
Na
0%
$Na^+$

Be and Al show diagonal relationship hence, both have :

0%
same degree of electronegativity.
0%
amphoteric nature of oxides.
0%
approximately same charge/radius ratio.
0%
all the properties above.

II A (alkaline earth metals) and II B (zinc family) resemble :

0%
$MgSO_4 .7H_2O$ is isomorphous with $ZnSO_4. 7H_2O$
0%
II A and II B cations are not precipitated by $H_2S$ in acidic medium
0%
both (a) and (b)
0%
None of the above

Explanation

$MgSO_4. 7H_2O$ is is Amorphous with

$ZnSO_4 \quad 7H_2O$ as they both have same lattice.

Zn cations precipitates by

$H_2S$ in presence of

$NH_4Cl+NH_4OH$ .

Which does not exist in solid state?

0%
$NaHCO_3$
0%
$Na_{2}CO_{3}$
0%
$LiHCO_3$
0%
$CaCO_3$

Explanation

$LiHCO_3$ is found in solution(aqueous) form while rest alkali metals hydrogen carbonates are solid due to the anomalous properties of Li and it's diagonal relationship with the Mg.

Following are the points of difference between Li and other alkali metals.

(i) it's exceptionally small size.

(ii)it's high polarizing power(charge/radius ratio).

Following are the ionisation potential values of :

$(I_1) 899 kJ mol^{-1}, (I_2) 1757 kJ mol^{-1}, (I_3) 15000 kJ mol^{-1}$

0%
Na
0%
K
0%
Be
0%
Ne

Explanation

The above values of ionization enthalpies represent the

$IP$ values of

$Be$ as the charge increases the IP values increases and the

$IP_3$ value is suddenly increased a lot due to attainment of duplet configuration of

$Be$ .

So, it requires more energy to remove the electron from an element of duplet configuration.

Hence option C is correct.

Which is used for fixing atmospheric nitrogen?

0%
$CaCN_2$ (nitrolim)
0%
$Li_3N$
0%
$Mg_3N_2$
0%
All of these

Explanation

For fixing atmospheric nitrogen -nitrolim is used and further

$CaCN_2+C$ is used as fertilizer.

Li resembles to element X which is not the member of alkali metals.
The element X is :

0%
Mg
0%
Be
0%
Ca
0%
B

Lightest alkaline earth metal is :

0%
Be
0%
Mg
0%
Ca
0%
Sr

Alkaline earth metals are :

0%
reducing agent
0%
dehydrating agent
0%
amphoteric
0%
oxidising agent

Explanation

Alkaline earth metals configuration:

$ns^2$

So, they have tendency to loose valence electrons so they are reducing agents.

Li resembles to element X which is not the member of alkali metals. The group of X is?

0%
II
0%
III
0%
IV
0%
I

Li has chemical properties most similar to :

0%
Al
0%
Ca
0%
Mg
0%
Ti

Explanation

Li has chemical properties most similar to Mg due to diagonal relationship.
Both exhibits same properties like
(i).

$MgCO_3\, \overset{\Delta}{\rightarrow}\, MgO\, +\, CO_2$

$Li_2CO_3\, \overset{\Delta}{\rightarrow}\, Li_2O\, +\, CO_2$

(ii).

$3Mg\, +\, N_2\, \rightarrow Mg_3N_2$

$6Li+ N_2 \rightarrow 2Li_3N$

Hence option C is correct answer.

Which one of the following has magnesium?

0%
Vitamin $B_{12}$
0%
Chlorophyll
0%
Haemocyanin
0%
Carbonic anhydrase

Explanation

$Mg$ is present in chlorophyll which is a green pigment and is used in photosynthesis in green plants.

Hence option B is correct.

$Mg$ and

$Li$ and similar in their properties due to:

0%
same $\dfrac {e}{m}$ ratio
0%
same electron affinity
0%
same group
0%
same radius

Explanation

$Li$ and

$Mg$ show diagonal relationship due to the similarity in size (radius), electronegativity, ionisation enthalpy and charge/radius ratio.

The activity of alkaline earth metals as reducing agent :

0%
decreases from Be to Ba
0%
increases from Be to Ba
0%
increase from Be to Ca and decreases from Ca to Ba
0%
decreases from Be to Ca and increases from Ca to Ba

Explanation

Alkaline earth metals configuration:

$ns^2$

So, they have tendency to loose electron so they are reducing agents.

As we move down, electronegativity decreases so reducing nature increases.

CBSE Questions for Class 11 Medical Chemistry The S-Block Elements Quiz 3 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry The S-Block Elements Quiz 3 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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