MCQ Questions for Class 11 Medical Chemistry The S-Block Elements Quiz 4

The electron affinity of Be is similar to ___________.

0%
He
0%
B
0%
Li
0%
Na

Explanation

$He$ being noble gas has complete 1 st shell It has completed its duplet. While

$Be$ Has electronic configuration

$1s^2$ ,

$2s^2$ so be also has completely filled

$2s$ subshell so we can say that, The electron affinity of

$Be$ is similar to

$Be$ among the given options.

Be and Al resemble in the following but not in :

0%
both form electron decient hydrides
0%
both are rendered passive by $HNO_3$
0%
both form amphoteric oxides
0%
both have sp-hybridisation in their compounds

Explanation

Both form electron deficient hydrides due to incomplete octet.

Both form amphoteric oxides as their oxides react with both acid and base.

Be have sp-hybridization and Al have sp

$^2$ - hybridization in their compounds.

Lithium shows similarities with magnesium in its chemical behaviour because :

0%
similar size, greater electronegativity and lower polarising power
0%
similar size, same electronegativity, and lower polarising power
0%
similar size, same electronegativity and similar high polarising power
0%
none of the above

Which of the following pairs shows diagonal relationship?

0%
$Na$ and $Mg$
0%
$Li$ and $Mg$
0%
$Li$ and $Na$
0%
$Al$ and $C$

Explanation

A diagonal relationship is said to exist between certain pairs of diagonally adjacent elements in the second and third periods of the periodic table. Thus,

$Li$ and

$Mg$ are diagonally related.

Which one is not the property of Lithium by which it differs from the rest of the alkali metals?

0%
Extremely small size of Lithium atom
0%
Greater polarizing power of Lithium ion
0%
Highest electropositive character and highest ionization energy
0%
Non availability of vacant d-orbitals in the valence shell

Which among the given elements form the strongest base?

0%
A
0%
B
0%
C
0%
D
0%
E
0%
F
0%
G

Explanation

In a period from left to right, the basic nature of elements decreases due to decrease in size and decrease in the ability to lose an electron and in a group from top to bottom the basic nature of elements increases due to increase in size and increase in the ability to lose an electron. Generally strong bases donate electrons easily. So the strong basic element is G in the given representation of the periodic table.

Hence, the correct option is

$\text{G}$

What is the order of mobilities of the alkali metal ions in aqueous solution?

0%
$Li^{+}\,>\,Na^{+}\,>\,K^{+}\,>\,Cs^{+}\,>\,Rb^{+}$
0%
$Na^{+}\,<\,Li^{+}\,<\,K^{+}\,<\,Rb^{+}\,<\,Cs^{+}$
0%
$Li^{+}\,>\,K^{+}\,>\,Na^{+}\,>\,Rb^{+}\,>\,Cs^{+}$
0%
$Li^{+}\,<\,Na^{+}\,<\,K^{+}\,<\,Rb^{+}\,<\,Cs^{+}$

Explanation

As smaller the size of the ion, larger the charge of the ion hence greater the hydration and smaller ionic conductance and mobility of the ion. So the increasing order of mobility the alkali metal ions in aqueous solution is

$Li^{+}\,<\,Na^{+}\,<\,K^{+}\,<\,Rb^{+}\,<\,Cs^{+}$ . Hence option D is correct.

Which among the following elements shown in the table forms amphoteric oxide?

0%
$A$
0%
$B$
0%
$C$
0%
$D$
0%
$E$
0%
$F$
0%
$G$

Shine at freshly cut sodium is because of :

0%
due to oscillations of free electrons.
0%
due to weak metallic bonding.
0%
due to absorption of light in crystal lattice.
0%
due to the presence offree valency at the surface.

Among the alkali metals

$Cs$ is the most reactive because :

0%
its incomplete shell is nearest to the nucleus
0%
it has a single electron in the valence shell
0%
it is the heaviest alkali metal
0%
its outermost electron is more loosely bound than the outermost electron ofthe other alkali metals

Explanation

Among the alkali metals,

$Cs$ is the most reactive because its outermost electron is more loosely bound than the outermost electron of the other alkali metals. This is due to the increase in distance between the outermost electron shell and the nucleus.

The characteristic not related to alkali metals is :

0%
their ions are isoelectronic with noble gases.
0%
metallic character.
0%
low electronegativity.
0%
high ionization energy.

Which of the following is/are correct statement(s) ?
(I)

$Ca^{2+}$ ions are important in blood clotting.
(II)

$Ca_3(PO_4)_2$ is a part of bones.
(III)

$3Ca_3(PO_4).CaF_2$ is a part of enamel of teeth.

0%
(I) and (II)
0%
(II) and (III)
0%
(I) and (III)
0%
(I), (II) and (III)

Explanation

All are correct statements.

(I)

$Ca^{2+}$ ions are important in blood clotting and muscle contraction. Low level of

$Ca^{2+}$ ions increases blood clotting time.

(II) Calcium phosphate

$Ca_3(PO_4)_2$ is a part of bones. Calcium deficiency causes osteoporosis where the bones becomes soft due to reduced bone mineral density leading to easy fractures.

(III)

$3Ca_3(PO_4)\cdot CaF_2$ is a part of enamel of teeth

The sequence of ionic mobility in aqueous solution is :

0%
$Rb^+> K^+>Cs^+> Na^+$
0%
$Na^+> K^+> Rb^+ > Cs^+$
0%
$K^+> Na^+> Rb^+> Cs^+$
0%
$Cs^+ > Rb^+> K^+ > Na^+$

Explanation

Hint: In an aqueous solution,

$Ionic\ Radius\ \propto \ mobility$

Step 1: Reason for the decrease in mobility

Consider mobility of ions in an aqueous solution.

The water molecule $H_{2}O$ is a dipole, hence its negative pole is attracted towards the positive ion present in the solution.
Due to this attraction, the positive ions for example $Na^{+}$ ions get surrounded by $H_{2}O$ molecules.
This increases the hydration radius and makes it difficult for the ion to move. Thus it decreases the ionic mobility.

Step 2: Trend in mobility along the group:

Charge per unit volume decreases down the group due to an increase in ionic radius.
This reduces the attractive forces between the dipole and the positive ion.
Thus, mobility increases down the group.
For group $1$ elements, the order for mobility is $Cs^{+}> Rb^{+}> K^{+}> Na^{+}$

Hence, the correct answer is option $D.$

Smallest among these species is :

0%
hydrogen.
0%
helium.
0%
lithium.
0%
lithium ion.

Explanation

The electronic configuration of these elements is as follows:
Hydrogen :

$1s^1$
Helium:

$1s^2$
Lithium:

$1s^2 2s^1$
Lithium ion:

$1s^2$
Three of the elements have one orbital and since hydrogen has only one electron, it is smallest among all the species.

The position in the periodic table to which the following metal belong :

Mg - an alkaline earth metal belongs to _________ .

0%
1-Period
0%
2-Period
0%
3-Period
0%
4-Period

Explanation

Alkaline earth metals are found in the second group of the periodic table. Group II elements include; Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba) and Radium (Ra).

Similar to alkaline metals, these elements also do not occur freely in nature and they are also very reactive.

$Mg$ belongs to the third period.

The aqueous solutions of lithium salts are poor conductors of electricity as compared to that of other alkali metals because of:

0%
high ionization energy
0%
high electronegativity
0%
lower ability of $Li^+$ ions to polarize water molecules
0%
higher degree of hydration of $Li^+$ ions

Explanation

The aqueous solutions of lithium salts are poor conductors of electricity as compared to that of other alkali metals because of higher degree of hydration of

$Li^+$ ions. It decreases the mobility and leads to low conductivity.
Hence, option D is correct.

As the alkaline earth metals (except Be) tend to lose their valence electrons readily. They act as :

0%
weak oxidizing agents.
0%
weak reducing agents.
0%
strong oxidizing agents.
0%
strong reducing agents.

Potash alum is represented by the formula :

0%
$K_2SO_4\cdot Al_2(SO_4)_3\cdot 24H_2O$
0%
$K\cdot Al(SO_4)_2\cdot 12 H_2O$
0%
Both (a) and (b)
0%
None of these

Explanation

Potassium alum or potassium aluminium sulphate is a chemical compound with a chemical formula

$KAl(SO_4)_2$ and it is commonly found in its dodecahydrate form as

$KAl(SO_4)_2.12H_2O$ .

Alum is the common name for this chemical compound.

Which of the following has maximum ionization energy?

0%
$Ba\rightarrow Ba^++e^-$
0%
$Be\rightarrow Be^++e^-$
0%
$Ca\rightarrow Ca^{2+}+2e^-$
0%
$Mg\rightarrow Mg^{2+}+2e^-$

Explanation

$I.E. \propto \cfrac{charge}{size}$

$Be$ and

$Ba$ have lost one electron each, they should have less ionization energy than

$Mg$ and

$Ca$ . But among the

$Mg$ and

$Ca, Mg$ should have more ionization energy even though both lost two electrons because

$Mg$ is smaller in size.

Option D is correct.

Compare

$Li$ and

$Be$ according to the polarisation of cations :

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

Explanation

As the charge on the ion increases and size decreases polarization incresases,so the correct order of polarization

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

Among of alkali metals, the metal with highest ionization potential is :

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

The first ionization energies of alkaline earth metals are higher than those of the alkali metals. This is because:

0%
there is increase in the nuclear charge of the alkaline earth metal
0%
there is decrease in the nuclear charge of the alkaline earth metal
0%
there is no change in the nuclear charge
0%
none ofthe above

$KO_2$ (potassium superoxide) is used in oxygen cylinders in space and submarines because, it :

0%
decomposes to give oxygen
0%
eliminates moisture
0%
absorbs $CO_2$
0%
produces ozone

Explanation

$KO_2$ on decomposition give oxygen and pottassium carbonate.The oxygen is used in oxygen cylinders in space and submarines

$2CO_2+4KO_2\rightarrow 3O_2+2K_2CO_3$ .
Hence option A is correct.

In the reaction :

$K_4[Fe(CN)_6] + K_2CO_3\rightarrow$ the product formed is :

0%
$FeCO_3$
0%
$KCN$
0%
$KFeCO_3$
0%
$K_3 [Fe(CN)_6]$

Explanation

$K_4Fe(CN)_6+K_2CO_3\rightarrow 5KCN+KCNO+Fe+CO_2$ .

In the above reaction only KCN is formed.

Hence, option B is correct.

(X), (Y), (Z) are elements in the third period. Oxide of (X) is ionic, that of (Y) is amphoteric, and of (Z) a giant molecule.

(X), (Y) and (Z) will have atomic numbers in the order :

0%
(X) < (Y) < (Z)
0%
(Z) < (Y) < (X)
0%
(X) < (Z) < (Y)
0%
(Y) < (X) < (Z)

Explanation

For third period, "s" block will form ionic oxide (

$Na_2O,\ MgO$ ), 13th group forms amphoteric oxide (

$Al_2O_3$ ) and 14th group element will form giant oxide

$(SiO_2)$ .

So the correct order of atomic numbers is (X) < (Y) < (Z)

Hence option A is correct.

Among

$LiCl, RbCl, BeCl_2$ , and

$MgCl_2$ the compounds with greatest and least ionic character, respectively are:

0%
LiCl, RbCl
0%
$RbCl, BeCl_2$
0%
$RbCl, MgCl_2$
0%
$MgCl_2, BeCl_2$

Explanation

as the size of the cation increases ionic character increases,so among the given chlorides the Rb is having larger sine and the element having least size is Be. So, the compound having more ionic character is

$RbCl$ .
The compound having more ionic character is

$BeCl_2$ . Hence option B is correct.

Photoelectric effect is maximum in:

0%
$Cs$
0%
$Na$
0%
$K$
0%
$Li$

Explanation

Hint: $\uparrow Ease\ of\ emission\ of\ e^{-} = \uparrow$ Photoelectric effect

Correct Answer: Option A

Explanation

Photoelectric effects involve the emission of electrons from the metal surface due to the incidence of light radiations.

Electrons are easily knocked out of atoms when ionization energy $(I.E.)$ is low.

$I.E.$ is low when there is high shielding and low effective nuclear charge. Therefore, $I.E.$ decreases down the group.

Thus, the order of ease of emission of an $e^-$ is $Cs> K> Na> Li$ .

Statement-1: The ionic mobility of

$Na^+_{(aq)}$ is lower than that of

$K^+_{(aq)}$
Statement-2 :

$r_{Na^+(aq)} > r_{k^+(aq)}$

0%
Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1
0%
Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1
0%
Statement-1 is true, statement-2 is false
0%
Statement-1 is false, statement-2 is true

Explanation

If hydrated size is more, then ionic mobility is less:

$\therefore Na^+_{aq} < K^+ _{aq}$ (order of ionic mobility)
and

$r_{Na^+ (aq.)} > r_{K^+ (aq.)}$ is true a statement
Both statement I and statement II are correct II is correct explanation for I
Hence option A is correct.

Which of the following element(s) is/are not react directly with nitrogen to form nitride ?

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

Explanation

Among all alkaline earth metals only lithium directly react with nitrogen to form nitride due to its small size and high electronegativity(anamalous behaviour of Li).

$6 Li + N_2\rightarrow 2Li_3N$ .

Hence options A,C & D are correct.

Alkaline earth metal salt are :

0%
paramagnetic
0%
diamagnetic
0%
ferromagnetic
0%
all of these

Which among the following statements is incorrect for magnesium metal?

0%
It burns in oxygen with a dazzling white flame
0%
It reacts with cold water to form magnesium oxide and evolves hydrogen gas
0%
It reacts with hot water to form magnesium hydroxide and evolves hydrogen gas
0%
It reacts with steam to form magnesium hydroxide and evolves hydrogen gas

Ripening of fruits can be carried out in the presence of :

0%
$Na_2SO_4$
0%
$NaCl$
0%
$CaCl_2$
0%
$CaC_2$

Be and Al have following resemblance due to diagonal relationship :

0%
have nearly equal electronegativity.
0%
form amphoteric oxides.
0%
have same charge/radius ratio.
0%
both form dimeric halides.

Select the correct statement(s) about barium :

0%
It shows photoelectric effect.
0%
It is silvery white metal.
0%
It forms $Ba(N_3)_2$ which is used in preparation of green fire.
0%
Its ionization energy is less than radium.

Explanation

A) It shows photoelectric effect.False,becuase it has completely filled s orbital so removal of electron is difficult.

B) It is silvery white metal.True.

C) It forms

$Ba(N_3)_2$ which is used in preparation of green fire.True.

D) Its ionization energy is less than radium True.

Hence options B,C & D are correct.

In certain matters, lithium differs from other alkali metals; the main reason for this is :

0%
small size of lithium atom and $Li^+$ .
0%
extremely high electropositivity of $Li$ .
0%
greater hardness of $Li$ .
0%
hydration of $Li^+$ ion.

Explanation

Hint: $Li^{+}$ has high polarizing power.

Correct Answer: Option A

Explanation

Lithium shows anomalous behaviour.
Its compounds are mostly covalent, due to their high polarizing power.
High polarizing power is due to the strong force of attraction by the nucleus on the electron cloud of the anion.
This strong force of attraction is due to the small size of lithium.

State True or False.

In the electrolysis of fused calcium hydride, hydrogen is liberated at cathode.

0%
True
0%
False

Explanation

Electrons come off the cathode and travel to the anode. So the calcium would be attracted to the cathode and deposited there. The

$H^{-}$ would move toward the anode and the extra electron would be released causing

$H_2$ to bubble
At cathode:

$Ca^{2+} + 2e^{-}\rightarrow Ca$
At anode:

$2H^{-}\rightarrow H_2 + 2e^{-}$ .
So the given statement is false.
Hence option B is correct.

${K}_{2}C{S}_{3}$ can be called :

0%
Potassium Sulphocyanide
0%
Potassium Thiocarbide
0%
Potassium Thiocarbonate
0%
Potassium Thiocyanate

Explanation

${K}_{2}C{S}_{3}$ can be called Potassium Thiocarbonate.

Hence, Option "C" is the correct answer.

Which of the following does not give an oxide on heating?

0%
$MgC{O}_{3}$
0%
${Li}_{2}C{O}_{3}$
0%
$ZnC{O}_{3}$
0%
None of the above

Explanation

On thermal decomposition:

$MgC{O}_{3}\rightarrow MgO+CO_2$

${Li}_{2}C{O}_{3}\rightarrow Li_2O+CO_2$

$ZnC{O}_{3}\rightarrow ZnO+CO_2$
All of the above compounds gives oxides.
Hence option D is correct.

State True or False.

$Be{ H }_{ 2 }$ is an ionic hydride.

0%
True
0%
False

Explanation

Beryllium hydride is an inorganic compound with the chemical formula

$BeH_2$

It is a colourless solid that is insoluble in solvents that do not decompose it. Unlike the ionic hydrides of the heavier group 2 elements, beryllium hydride is covalently bonded(due to less electronegativity difference between hydrogen and beryllium).

So the given statement is false.

Hence option B is correct.

As per observation (iii) LiH is more stable than NaH, because :

0%
Due to small size of ${ Li }^{ + }$ , the lattice energy of LiH is greater
0%
Due to greater size of ${ H }^{ - }$ , the lattice energy of LiH is greater
0%
LiH is more covalent than NaH
0%
Due to greater size of ${ Na }^{ + }$ , the lattice energy of NaH is greater.

Explanation

As per observation (iii), LiH is more stable than NaH, because, Due to small size of

${ Li }^{ + }$ , the lattice energy of LiH is greater

As the size of the ion is small it has more force of attraction and, so it requires a lot of energy break down the bond, so it is stable.

Hence, the correct option is

$\text{A}$

Which compound is used for removal of moisture from glass apparatus?

0%
$CaCl_2$
0%
Anhydrous $CaCl_2$
0%
$HCl$
0%
All of these

Explanation

Option (B) is correct.

Anhydrous

$CaCl_2$ is used for removal of moisture from glass apparatus.This can be done by placing anhydrous

$CaCl_2$ in the apparatus which absorbs the moisture.

Note: Solid

$CaCl_2$ cannot be used because it forms a crystalline addition compound.

The metallic lustre exhibited by sodium is explained by:

0%
diffusion of sodium ions
0%
oscillation of loose electrons
0%
excitation of free protons
0%
existence of body centered cubic lattice

The hydration energy of

${Mg^{2+}}$ is greater than that of :

0%
${Al^{3+}}$
0%
${Na^+}$
0%
${Be^{2+}}$
0%
${Mg^{3+}}$

Explanation

Hydration energy

${\propto \frac{1}{size of ions}}$ .
Of all the ions given

$Na^{+}$ is having less size,so

$Na^{+}$ is having hydration energy greater than

$Mg^{+}$ .
Hence option B is correct.

Molten sodium is used in nuclear reactors to:

0%
absorb neutrons in order to control the chain reaction
0%
slow down the fast neutrons
0%
absorb the heat generated by nuclear fission
0%
extract radio -isotopes produced in the reactor

The correct sequence of increasing covalent character is represented by :

0%
${BeCl_2}$ < ${NaCl}$ < ${LiCl}$
0%
${NaCl}$ < ${LiCl}$ < ${BeCl_2}$
0%
${BeCl_2}$ < ${LiCl}$ < ${NaCl}$
0%
${LiCl}$ < ${NaC}$ l< ${BeCl_2}$

Explanation

In a bond between two atoms,

$M$ and

$X$ the amount of covalent character in the M-X bond will depend on the difference in electronegativity.

The correct sequence of increasing covalent character is represented by

${NaCl}$ <

${LiCl}$ <

${BeCl_2}$ .

The covalent character is favoured by small cation with high charge.

Hence the correct option is

$B$

Bone ash contains :

0%
$CaF_2$
0%
${CaSO_4}$
0%
${Ca_3(PO_4)_2}$
0%
${Ca(H_2PO_4)_2}$

Explanation

The approximate formula of bone ash is

$4Ca_3(PO_4)_2.CaO$ .Although raw bone is chemically

$Ca_3(OH)(PO_4)_3$ hydroxyapatite.
Hence option C is correct.

When

$Na$ and

$Li$ placed in dry air we get :

0%
$NaOH, {Na}_{2}O, {Li}_{2}O$
0%
${Na}_{2}C{O}_{3}, {Na}_{2}{O}_{2}, {Li}_{2}O$
0%
${Na}_{2}O, {Li}_{3}N, N{H}_{3}$
0%
${Na}_{2}O, {Li}_{2}O, {Li}_{3}N$

EDTA is used in the estimation of :

0%
${ Mg }^{ 2+ }$ ions
0%
${ Ca }^{ 2+ }$ ions
0%
Both ${ Ca }^{ 2+ }$ and ${ Mg }^{ 2+ }$ ions
0%
${ Mg }^{ 2+ }$ ions but not ${ Ca }^{ 2+ }$ ions

Which of the following groups of elements have chemical properties that are most similar :

0%
$Na, K, Ca$
0%
$Mg, Sr, Ba$
0%
$Be, Al, Ca$
0%
$Be, Ra, Cs$

Explanation

In option B, Mg, Sr, Ba belongs to 2nd group so they are have same outer electronic configuration of

$ns^{2}$ .
In option C, Be and Ca belongs to 2nd group so they are having same properties how ever, Al shows similar properties with Be and Ca due to diagonal relationship of

$Be$ with

$Al$ .
In options A and D, both 1st and 2nd group elements are present together so their properties are different.
Hence, options B and C are correct.

Which of the following imparts violet colouration to the Bunsen burner non-luminous flame?

0%
$NaCl$
0%
${BaCl_2}$
0%
${CaCl_2}$
0%
$KCl$

Explanation

Violet colouration to the Bunsen burner will be produced by potasium, so

$KCl$ imparts violet colouration to the Bunsen burner non-luminous flame.
Hence option D is correct.

CBSE Questions for Class 11 Medical Chemistry The S-Block Elements Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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