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

State whether the following statement is true or false.

Reactivity and reducing power decreases on going from group 1 to group 2.

0%
True
0%
False

The hydration energy of

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

0%
$A1^{3+}$
0%
$Na^{+}$
0%
$Be^{3+}$
0%
$Mg^{3+}$

Explanation

$Mg^{2+}$ has the larger hydration energy, because hydration energy increases with the charge on the cation and decreases as the size of the cation increases. Therefore answer is

$Na^+$ .

The alkali metal that reacts with nitrogen directly to form nitride is:

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

Explanation

Alkali metals that react with nitrogen directly to form nitride is

$Li$ . This anamolous behaviour is due to its small size and high polarizing power.

$6 Li + N_2(g)\rightarrow 2 Li_3N$
Hence, option A is correct.

Out of the elements marked A, B, C, D, E, F, G and H
which form super oxide?

0%
E and G
0%
C, E and G
0%
A and D
0%
G and B

Be and Al show diagonal relationship and thus :

0%
their oxides are soluble in alkali solution forming $[Be(OH)_4]^{3-}$
0%
they form complex anion $BeF_4^{2-}$ and $AlF_6^{3-}$
0%
$BeCl_2$ and $AlCl_3$ are Lewis acid
0%
oxides are basic

Explanation

1) $Be^{2+}$ and $Al^{3+}$ have almost same and smaller size and thus favour for covalent bonding.

2) Both these form covalent compounds having low melting point and soluble in organic solvent.

3) Both have same value of electronegativity.

4) Their oxides are soluble in alkali solution forming $[Be(OH)_4]^{3-}$

5) Both form many stable complexes for example $BeH_3^-, \ AlH_4^-$ .

6) They form complex anion $BeF_4^{2-}$ and $AlF_6^{3-}$ .

7) $BeCl_2$ and $AlCl_3$ are Lewis acid.

Hence options A,B & C are correct.

The material used in photoelectric cells contains:

0%
$Cs$
0%
$Si$
0%
$Sn$
0%
$Ti$

Explanation

$Cs$ and

$Ti$ is used in photoelectric cells because of low ionization enthalpy. electrons could be emitted very easily when exposed to light.

When sodium is heated in moist air, the ultimate product obtained is:

0%
$\displaystyle Na_{2}O$
0%
$NaOH$
0%
$\displaystyle Na_{2}CO_{3}$
0%
$\displaystyle Na_{2}O_{2}$

Explanation

When sodium is heated with moist air, the ultimate product obtained is

$\displaystyle Na_2CO_3$ .
In moist air, a layer of sodium oxide, hydroxide and carbonate is formed on its surface due to which, it loses its lustre.

The complete reaction sequence is as follows:

$\displaystyle 4Na + O_2 \rightarrow 2Na_2O \xrightarrow {2H_2O} 4NaOH \xrightarrow {2CO_2} 2 Na_2CO_3 + 2H_2O$

Alkali metals readily react with oxyacids forming corresponding salts like

${ M }_{ 2 }{ CO }_{ 3 }$ ,

${ M }H{ CO }_{ 3 }$ ,

${ M }N{ O }_{ 3 }$ ,

${ M }_{ 2 }{ SO }_{ 4 }$ , etc. with evolution of hydrogen. They also dissolve in liquid

${ NH }_{ 3 }$ but without the evolution of hydrogen. The color of its dilute solution is blue but when it is heated and concentrated then its color becomes bronze.

Which of the following statement about the sulphate of alkali metal is correct :

0%
except ${ Li }_{ 2 }{ SO }_{ 4 }$ all sulphate of other alkali metals are soluble in water.
0%
all sulphates of alkali metals except lithium sulphate forms alum.
0%
the sulphates of alkali metals cannot be hydrolysed.
0%
all of these.

Explanation

Except

${ Li }_{ 2 }{ SO }_{ 4 }$ all sulphate of other alkali metals are soluble in water and forms alum. Thus, they form double salts with the sulphates of trivalent metals like

$Al, Fe, Cr$ etc. The sulphates of alkali metals cannot be hydrolysed as they are salts of a strong acid and strong base.

Select the correct order of ionic conductance of the following cations in an aqueous solution.

0%
${ Li }^{ + }<{ Na }^{ + }<{ K }^{ + }<{ Rb }^{ + }$
0%
${ Li }^{ + }>{ Na }^{ + }>{ K }^{ + }>{ Rb }^{ + }$
0%
${ Li }^{ + }<{ Na }^{ + }>{ K }^{ + }>{ Rb }^{ + }$
0%
${ Li }^{ + }={ Na }^{ + }<{ K }^{ + }<{ Rb }^{ + }$

Explanation

Smaller is the size of cation, greater will be the mobility and hence, the conductivity. The size of the cation increases as we move from

$Li^{+}$ to

$Rb^{+}$ . Hence, the conductivity should decrease. However, in aqueous solution, due to very small size of

$Li^+$ , it is strongly hydrated and hence, its effective hydrated size is greater than that of

$Rb^+$ .
So, the correct order of increasing ionic conductance is as follows:

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

Among the following hydroxides, the one which has the lowest value of

$K_{sp}$ at ordinary temperature is:

0%
$Be(OH)_2$
0%
$Mg(OH)_2$
0%
$Ca(OH)_2$
0%
$Ba(OH)_2$

Explanation

Option

$(A)$ is correct.

$Be(OH)_2$ has the lowest value of

$K_{sp}$ at ordinary temperature because

$Be^{2+}$ ion is smaller than the other metal ions in the group, which results in a tighter bond with the

$0H^-$ ions, thus much lower solubility.

The solubility of a hydroxide of group 2 elements increases down the group because as you go down the group size of metal increases thereby increasing the bond length and decreasing bond energy.

Which of the following does not decompose water?

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

Explanation

Option

$(A)$ is correct.

$Be$ does not decompose water due to its small size and high ionisation energy but it will react with the oxygen of

$H_2O$ .

Which of the following has the maximum coordinate number not more than 4?

0%
Beryllium
0%
Calcium
0%
Barium
0%
Radium

Explanation

Option

$(A)$ is correct.

Beryllium has the maximum coordinate number not more than 4 as it has four orbitals in the valence shell. The other members of this group has coordination number 6 by making use of

$d$ - orbitals.

A metal

$M$ forms water-soluble sulphate

$MSO_4$ , water-insoluble hydroxide

$M(OH)_2$ and oxide

$MO$ which becomes inert on heating. The hydroxide is soluble in

$NaOH$ . Then

$M$ is:

0%
$Be$
0%
$Mg$
0%
Cannot be determine as data is incomplete
0%
$Sr$

Explanation

Option

$(A)$ is correct.

$Be$ forms water soluble

$BeSO_4$ , water insoluble

$Be(OH)_2$ and

$BeO$ .

$Be(OH)_2$ is insoluble in

$NaOH$ giving sodium beryllate,

$Na_2BeO_2$ .

On moving down, water solubility of alkaline earth metals decreases.Oxides and hydroxides of alkaline earth metal are basic except

$Be$ which is amphoteric.Hence, the metal is

$Be$ .

Which of the following alkaline earth metal does not impart characteristic colour to flame?

0%
$Ca$
0%
$Sr$
0%
$Ba$
0%
$Be$

Explanation

Option

$(D)$ is correct.

$Be$ does not impart characteristic colour to flame because its atoms are comparatively smaller and the valance electrons are strongly attached to the nucleus. Therefore its ionisation energies is high and needs large amount of energy for the excitation of their valence electrons to higher energy level which is usually not available in the Bunsen flame. So beryllium do not impart any colour to the flame.

$Ca, Sr$ and

$Ba$ impart brick red, blood red and apple green colours respectively to the flame.

Which of the following alkaline earth metal indirectly form a polymeric hydride with three centre bond?

0%
$Mg$
0%
$Be$
0%
Both $A$ and $B$
0%
none of these

Explanation

Option

$(B)$ is correct.

$Be$ indirectly form a polymeric hydride with three centre bond.

$BeH_2$ is covalent and has polymeric structure involving multicentre bonds (three-centre two-electron bond).Each beryllium atom is bonded to two hydrogen atoms and each hydrogen atom to two beryllium atoms. Now, beryllium has only two valence electrons and hydrogen only one, it is apparent that there are not sufficient electrons to form so many electron pair bonds. The monomeric

$BeH_2$ , if formed with normal bonds, would have only four electrons in the valence shell of the beryllium atom and would be electron deficient and unstable. It polymerizes to remedy the electron deficiency. The bonds formed cannot be explained by the classical theories of bonding. They are “banana-shaped” molecular orbitals holding three atoms

$Be -- H -- Be$ together and are called 3 centred – 2 electrons (

$3c – 2e^-$ ) bonds.

The metal oxides of alkaline earth metal can be formed by heating corresponding carbonates except:

0%
$BeO$
0%
$MgO$
0%
$CaO$
0%
none of these

Explanation

Option

$(A)$ is correct.

The metal oxides of alkaline earth metal can be formed by heating corresponding carbonates except

$BeO$ due to small size of beryllium ion.

$BeO$ is covalent and has polymeric nature.

All alkaline earth metals combine directly with

$H_2$ and form ionic hydrides as

$MH_2$ except:

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

Explanation

Option

$(C)$ is correct.

$Be$ does not combine directly with

$H_2$ but

$BeH_2$ is prepared from preformed beryllium(II) compounds and

$BeH_2$ is covalently bonded.

All other alkaline earth metals react directly with hydrogen to produce the binary hydrides (

$MH_2$ ). The hydrides of the heavier alkaline earth metals are ionic, but

$BeH_2$ has polymeric structures that reflect covalent character.

Which of the following arrangements is correct in respect to solubility in water?

0%
$CaSO_{4} > BaSO_{4} > BeSO_{4} > MgSO_{4} > SrSO_{4}$
0%
$BeSO_{4} > MgSO_{4} > CaSO_{4} > SrSO_{4} > BaSO_{4}$
0%
$BaSO_{4} > SrSO_{4} > CasO_{4} > MgSO_{4} > BeSO_{4}$
0%
$BeSO_{4} > CaSO_{4} > MgSO_{4} > SrSO_{4} > BaSO_{4}$

Explanation

The correct order of solubility in water is

$BeSO_{4} > MgSO_{4} > CaSO_{4} > SrSO_{4} > BaSO_{4}$

The increasing order of the ionic size is-

$Be^{2+}<Mg^{2+}<Ca^{2+}<Sr^{2+}<Ba^{2+}$ .

The hydration energy decreases more rapidly than lattice energy, hence, solubility decreases down the group. (Hydration energy

$\propto \dfrac {charge}{size}$ ).

Oxidation number of calcium in calcium phosphate :

0%
$1^+$
0%
$1^-$
0%
$0$
0%
$2^+$
0%
$3^+$

Explanation

$Ca$ is the Alkaline earth metal hence it has electrovalency to be +2, only. Hence, its oxidation state , can not change, from 2.

$Ca_3(PO_4)_2$ :
As charge of phosphate

$(PO_4^{3-})$ group is

$-2$ , Let the oxidation state of

$Ca$ be

$x$ . and whole unit is neutral.

$3 \times x +2\times (-3) = 0$

$3x -6 =0$

$3x = +6$

$x = \dfrac {+6}{3} = +2$
So ,

$Ca$ has

$+2$ oxidation state.

Beryllium differs in properties from other elements of its own group but shows resemblance with aluminium because of:

0%
relatively bigger ionic radius and high polarising power of $Be$
0%
relatively smaller ionic radius and high polarising power of $Be$
0%
relatively bigger ionic radius is the only reason behind this
0%
none of the above

Explanation

Due to relatively smaller ionic radius and high polarising power, Be differs in properties from other elements of its own group. Polarising powers of

$Be^{2+}$ and

$Al^{3+}$ . ions are almost the same, thus they show similarities in their properties.

Sodium ions helps in:

0%
transmission of nerve signals
0%
transport of sugar and amino acids into cells
0%
activation of different enzymes
0%
oxidation of glucose to produce ATP

Explanation

Option

$(A),\ (B)$ are correct.

Sodium ions help in transmission of nerve signals by sodium/potassium pump.

Sodium ions also help in transport of sugar and amino acids into cells through indirect active transport via an extracellular medium.

Alkaline earth's metals are denser than alkali metals because metallic bonding in alkaline earth's metal is:

0%
stronger
0%
weaker
0%
volatile
0%
none of the above

Explanation

Alkaline earth metals (

${ns}^{2}$ ) are denser than alkali metals (

${ns}^{1}$ ) because metallic bonding in alkaline earth metals is stronger than alkali metals due to presence of two electrons in valence shell as compared to one electron in alkali metals.

Berly is an important ore of:

0%
$Be$
0%
$Ba$
0%
$Mg$
0%
$Sr$

Explanation

Option

$(A)$ is correct.

Berly is an important ore of

$Be$ . It is composed of beryllium aluminium cyclosilicate with the chemical formula of

$Be_3Al_2Si_6O_{18}$ . Beryl is a mineral that contains a significant amount of beryllium. Beryllium is a very rare metal, and that limits the occurrence of Beryl to a few geological situations where beryllium is present in sufficient amounts to form minerals. It mainly occurs in igneous and metamorphic rocks

The correct increasing order of solubility of the sulphates in water is:

0%
$BeSO_4>MgSO_4>CaSO_4>SrSO_4>BaSO_4$
0%
$BaSO_4>SrSO_4>CaSO_4>MgSO_4>BeSO_4$
0%
$BaSO_4>CaSO_4>SrSO_4>BeSO_4MgSO_4$
0%
$BaSO_4>SrSO_4>MgSO_4>CaSO_4> BeSO_4$

Explanation

The solubility of the sulphates in water decreases down the group:

$\underset { Soluble }{ \underbrace { BeSO_{ 4 } > MgSO_4 } } > \underset { Sparingly \quad soluble}{ \underbrace { CaSO_{ 4 } } } > \underset { Insoluble }{ \underbrace { SrSO_{ 4 } > BaSO_4 } }$

The high solubilities of

$BeSO_4$ and

$MgSO_4$ are due to the high enthalpy of hydration of the smaller

$Be^{2+}$ and

$Mg^{2+}$ ions, which overcomes the lattice energy factor.

Hence, option

$A$ is correct.

$BeF_2$ is soluble in water, whereas the fluorides of other alkaline earth metals are insoluble because of:

0%
greater hydration energy of $Be^{2+}$ ion as compared to crystal lattice
0%
covalent nature of $BeF_2$
0%
ionic nature of $BeF_2$
0%
none of these

Explanation

Option

$(A)$ is correct.

$BeF_2$ is soluble in water due to greater hydration energy of

$Be^{2+}$ ion as compared to the crystal lattice.The other fluorides are

almost insoluble in water. Since on descending the group lattice energy decreases more rapidly than the hydration energy.

Which of the following alkaline earth metal can never have more than four molecules of water of crystallisation?

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

Explanation

Option

$(C)$ is correct.

$Be$ can never have more than four molecules of water of crystallisation as it has only four available orbitals in its valence shell. (

$Be = 2s^2 2p_x^02p_y^0 2p_z^0$ ).

The other elements in II A group may have more than four molecules of water of crystallisation as they have

$d$ - orbitals as well.

Which of the following statement is true regarding alkaline earth metals?

0%
Beryllium and barium compounds are poisonous in nature
0%
Beryllium shows diagonal relationship with aluminium
0%
Beryllium salts impart yellow colour to flame
0%
Both $A$ and $B$

Explanation

Due to comparatively same size as aluminium, Beryllium show diagonal relationship with aluminium. Same charge/size ratio.

Due to the high reactivity of metals, water-soluble compounds of

$Be$ and

$Ba$ are poisonous. So option

$D$ is correct.

Which of the given halide does not conduct electricity in fused state and in solutions?

0%
$BeCl_2$
0%
$MgCl_2$
0%
$CaCl_2$
0%
none of these

Explanation

$Be{ Cl }_{ 2 }$ cannot conduct electricity in fused state because charged particles cannot move within the crystal.

The thermal stability of alkaline earth metal carbonates-

$MgCO_3, CaCO_3, SrCO_3$ and

$BaCO_3$ , follows the order:

0%
$BaCO_3 > SrCO_3 > CaCO_3 > MgCO_3$
0%
$CaCO_3 > SrCO_3 > MgCO_3 > BaCO_3$
0%
$MgCO_3 > CaCO_3 > SrCO_3 > BaCO_3$
0%
$BaCO_3 > SrCO_3 > MgCO_3 > CaCO_3$

Consider the metals

$B, Al, Mg$ and

$K$ . What will be the correct order of their metallic character?

0%
$K> Mg> Al> B$
0%
$Al> Mg> B> K$
0%
$B> Al> Mg> K$
0%
$Mg> Al> K> B$

$LiOH$ reacts with

$CO_{2}$ to form

$Li_{2}CO_{3}$ (atomic mass of

$Li = 7)$ . The amount of

$CO_{2}$ (in g) consumed by

$1\ g$ of

$LiOH$ is closest to _______.

0%
$0.916$
0%
$1.832$
0%
$0.544$
0%
$1.088$

Explanation

Balanced chemical equation is:

$2LiOH+C{ O }_{ 2 }\rightarrow { Li }_{ 2 }C{ O }_{ 3 }+{ H }_{ 2 }O$

Molecular weight of each specie is:

$LiOH$ =24 g/mol;

$C{ O }_{ 2 }$ =44 g/mol;

${ Li }_{ 2 }C{ O }_{ 3 }$ = 74 g/mol;

${ H }_{ 2 }O$ =18 g/mol

From the reaction:

2 mol of

$LiOH$ (

$24\times2=48\ g$ ) consumes 1 mol of

$CO_2$ (44 g)

Therefore, 1 g of

$LiOH$ will consume

$\frac { 1 }{ 48 } \times 44\quad g\quad C{ O }_{ 2 }=0.916\ g\ CO_2$

Option A is the correct answer.

Both lithium and magnesium display similar properties due to the diagonal relationship; however, the one which is incorrect, is:

0%
both form nitrides
0%
nitrates of both $Li$ and $Mg$ yield $NO_{2}$ and $O_{2}$ on heating
0%
both from basic carbonates
0%
both form soluble bicarbonates

Elements showing diagonal relationship are:

0%
$Li$ and $Be$
0%
$Li$ and $Mg$
0%
$Li$ and $Na$
0%
$Al$ and $Si$

Explanation

Elements showing diagonal relationship are

$Li$ and

$Mg$ . This diagonal relationship is due to similarity in ionic sizes and charge/ radius ratio of elements. Polarizing power of ions of

$Li^+$ and

$Mg^{2+}$ and electronegativity of

$Li$ and

$Mg$ are similar.

Select the incorrect choice:

0%
The solubility of alkaline earth metal carbonates, sulfates, and chromates increases from $Be$ to $Ba$ .
0%
The solubility of alkaline earth metal hydroxides is less than that of alkali metal hydroxides.
0%
Solubility of alkaline earth metal oxides increases from $Be$ to $Ba$ .
0%
$SO_2$ on passing in lime water turns it milky.

The most abundant metal ion present in the human body is

0%
$Zn^{2+}$
0%
$Ca^{2+}$
0%
$Na^+$
0%
$Fe^{2+}$

Explanation

Most abundant metal ion in human body is

$Ca^{2+}$ .

Identify the elements having same valence shell.

0%
$P$ and $Q$
0%
$Q$ and $R$
0%
$P$ and $R$
0%
$P$ , $Q$ and $R$

When a solution of alkaline earth metal in liquid ammonia is evaporated:

0%
they form hexammoniates , $M(NH_{3})_{6}$
0%
The ammoniates decomposed at high temperatures
0%
The tendency to form ammoniates decreses with increases in the size of the metal
0%
The ammoniates are good conductors of electricity

Why is

$LiF$ almost insoluble in water whereas

$LiCl$ is soluble not only in water but also in acetone?

0%
For a substance to dissolve in water, its hydration energy must be greater than its lattice energy.
0%
Due to small size of $Li^+$ ion and $F^-$ ions, lattice energy of $LiF$ is very high. $LiF$ is covalent in nature. Therefore, $LiF$ is insoluble in water.
0%
Due to larger size of $Cl^-$ ion, the lattice energy of $LiCl$ is less than its hydration energy. Hence, $LiCl$ is soluble in water and in acetone.
0%
Due to the large size of $F^-$ ion and high hydration energy, the $LiF$ is insoluble in water.

Explanation

$\rightarrow$ Lattice energy is defined as the amount of energy required to separate one mole of ionic compound into its gaseous ions.

$\rightarrow$ Lattice energy of

$LiF\rightarrow$

$1009$ KJ/mol

$LiCl\rightarrow$

$845$ KJ/mol

$\rightarrow$ HIgher the lattice point, higher is the melting point of the alkali metal which halide and hence lower is its solubility in water.

$\rightarrow$ Hence,

$LiF$ is almost insoluble in water, while

$LiCl$ is soluble in water and acetone.

Choose the correct answer form the alternatives given.
When alkaline earth metals dissolve in ammonia, they form coloured solution like alkali metals . Which of the following observations regarding the reaction are correct?
(i) Dilute solution are bright blue in colour due to solvated electrons.
(ii) These solutions decompose to form amides and hydrogen.
(iii) From this solution the ammoniates

$[M(NH_3)_6]^{2+}$ can be recovered by evaporation.

0%
Only (i) and (ii)
0%
Only (i) and (ii) and (iii)
0%
Only (ii) and (iii)
0%
Only (i)

Explanation

All alkali metal dissolve in liquid ammonia giving deep blue solutions which are conducting in nature. These solutions contain ammoniated cations and ammoniated electrons as shown below:

$M + ( x + y ) NH_3 \rightarrow M+(NH_3 )_x + e^-(NH_3)_y$

The blue colour of the solution is considered to be due to ammoniated electrons which absorb energy corresponding to red region of the visible light for the their excitation to higher energy levels. The transmitted light is blue which imparts blue colour to the solutions.

From this solution the ammoniates $[M(NH_3)_6]^{2+}$ can be recovered by evaporation.
These solutions decompose to form amides and hydrogen.

$Be$ and

$Al$ exhibit a diagonal relationship. Which of the following statements about them is/are not true?
(i) Both react with HCI to liberate

$H_2$ .
(ii) They are made passive by

$HNO_3$ .
(iii) Their carbides give acetylene on treatment with water.
(iv) Their oxides are amphoteric.

0%
(iii) and (iv)
0%
(i) and (iii)
0%
(i) only
0%
(iii) only

Explanation

The carbides of both Be and Al liberate methane with water.

$Be_2C + 2H_2O \rightarrow 2BeO + CH_4$

$Al_4C_3 + 6H_2O \rightarrow 2Al_2O_3 + 3CH_4$

Beryllium shows a diagonal relationship with aluminium. Which of the following similarity is incorrect?

0%
$Be_2C$ like $Al_4C_3$ yields methane on hydrolysis .
0%
Be like Al is rendered passive by $HNO_3$ .
0%
$Be(OH)_2$ like $Al(OH)_3$ is basic.
0%
Be forms beryllates A1 forms aluminates.

Explanation

The two structures involve the only movement of electrons and not of atoms or groups, hence these are resonating structures. Both metals have the tendency to form covalent compounds Diagonal relationship of Be with Al Because of its small size. Be differs from other earth alkaline earth metals but resembles in many of its properties with Al on account of the diagonal relationship.

The correct option is that $Be (OH)_{2}$ like $Al (OH)_{3}$ both are basic in nature.

The correct answer is C.

What is the biological importance of

$Na^+$ and

$K^+$ ions in cell fluids like blood plasma?

0%
They participate in transmission of nerve signals .
0%
They regulate the number of red and white blood corpuscles in the cell.
0%
They can be present in any amount in the blood since they are absorbed by the cells.
0%
They regulate the viscosity and colour of the blood.

Explanation

The process of moving sodium and potassium ions across the cell membrane is an active transport process involving the hydrolysis of ATP to provide the necessary energy. It involves an enzyme refer to as

${ Na }^{ + }/{ K }^{ + }-ATPase$ . This process is responsible for maintaining the large excess of

${ Na }^{ + }$ outside the cell and the large excess of

${ K }^{ + }$ ions on the inside.

It accomplishes the transport of three

${ Na }^{ + }$ to the outside of the cell and the transport of two

${ K }^{ + }$ ions to the inside. This unbalanced charge transfer contributes to the separation of charge across the membrane. The sodium-potassium pump is an importance contributor to action potential produced by nerve cells.

The

${ Na }^{ + }$ and

${ K }^{ + }$ ions participate in the transmission of nerve signals.

Which among the following elements of group-

$2$ exhibit anomalous properties?

0%
Be
0%
Mg
0%
Ca
0%
Ba

A metal M reacts with nitrogen to give nitride which on reaction with water produces ammonia gas. Metal M can be:

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

Explanation

The metal Lithium shows an exceptional behaviour by reacting in atmosphere of nitrogen to give lithium nitride

${Li}_{3}N$ which upon reaction with water produces ammonia gas.

$6Li+ {N}_{2}\rightarrow3{Li}_{3}N$

$3Li_{3}N+H_{2}\rightarrow 3LiOH + NH_{3}$

Sulphates of Be and Mg are readily soluble in water but sulphates of Ca, Sr and Ba are insoluble. This is due to the fact that:

0%
the greater hydration enthalpies of $Be^{2+} and Mg^{2+}$ overcome the lattice enthalpy
0%
high lattice enthalpy of $Be^{2+}$ and $Mg^{2+}$ makes them soluble in water
0%
solubility decreases from $BeSO_4$ to $BaSO_4$ due to increase in ionic size
0%
$BeSO_4$ and $MgSO_4$ are ionic in nature while other sulphates are covalent

Choose the correct answer form the alternatives given.
Which of the following compounds are not arranged in correct order as indicated?

0%
$SrCl_2 < CaCl_2 < MgCl_2 < BeCl_2$ (increasing extent of hydrolysis)
0%
$SrCl_2 < CaCl_2 < MgCl_2 < BeCl_2$ (increasing lattice energy)
0%
$CaSO_4 < MgSO_4 < BeSO_4$ (increasing stability)
0%
$Be(OH)_2 < Mg(OH)_2 < Ca(OH)_2$ (increasing solubility)

Explanation

$BeSO_4 < MgSO_4 < CaSO_4$ (increasing stability)

Temp.of Decomposition of $BeSO_4 , MgSO_4 , CaSO_4$ are 773K ,11681 K and 1422K respectively.

Which of the following statement is not correct.

0%
$LiOH$ is amphoteric in nature.
0%
$LiCl$ is soluble in pyridine.
0%
$Li_3N$ is stable while $Na_3N$ doesn't exist even at room temperature.
0%
$BeO$ is amphoteric in nature.

Beryllium and aluminium exhibit many properties which are similar. But, the two elements differ in:

0%
Forming covalent halides
0%
Forming polymeric hybrids
0%
Exhibiting maximum covalency in compounds
0%
Exhibiting amphoteric nature in their oxides

Explanation

Due to diagonal relationship, Be and Al exhibits similar properties But covalency of

$Be$ is 4 because it has an absence of d-orbital whereas, covalency of

$Al$ is up to 6 due to the presence of vacant d-orbital.

Hence they differ in (C) Exhibiting the maximum covalency in compounds.

The reducing power of metal depends on various factors. Suggest the factors which make

$Li$ , the strongest reducing agent in aqueous solution.

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Sublimation enthalpy
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Ionisation enthalpy
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Hydration enthalpy
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Electron-gain enthalpy

Explanation

With the small size of

$Li^+$ ion, lithium has the highest hydration enthalpy which accounts for its high negative

$E^{\circ}$ value and its high reducing power.

Ionic mobility of

$L{ i }^{ + }$ is less than that of

$N{ a }^{ + }$ and

${ K }^{ + }$ ions both, because:

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ionisation potential of Li is small
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charge density of $L{ i }^{ + }$ ion is high
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hydration tendency of $L{ i }^{ + }$ ion is high
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$L{ i }^{ + }$ has two electron

Explanation

The alkali metal ion exist as hydrated ions

$M+(H_2O)_n$ in the aqueous solution. The degree of hydration, decreases with ionic size as we go down the group. Hence

$Li^+$ ion is mostly hydrated e.g.

$[Li(H_2O)_6]^+$ . Since the mobility of ions is inversely proportional to the size of the their hydrated ions, hence the increasing order of ionic mobility is

$Li^+$ .

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

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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