$$Na_{2}B_{4}O_{7}\cdot 10H_{2}O$$

Which one of the following statements about $$H_{3}BO_{3}$$ is not correct?

It is prepared by acidifying an aqueous solution of borax.

It has a layer structure in which planar BO, units are joined by hydrogen bonds.

It does not act as proton donor as it acts as a Lewis acid by accepting hydroxyl ions.

Group $$13$$ elements show $$+1$$ and $$+3$$ oxidation states. Relative stability of $$+3$$ oxidation state may be given as :

$$Tl^{3+}>In^{3+}>Ga^{3+}>Al^{3+}>B^{3+}$$

$$Al^{3+}>Ga^{3+}>Tl^{3+}>In^{3+}>B^{3+}$$

$$Al^{3+}>B^{3+}>Ga^{3+}>Tl^{3+}>In^{3+}$$

Boron is unable to form $${ BF }_{ 6 }^{ 3- }$$ ions due to:

less reactivity towards halogens.

MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 8

3^{r d} A

$3^{rd}\ A$ group

T l

$Tl$ shows +

1

$\mathrm{1}$ oxidation state while other member shows +

3

$3$ oxidation state due to :

0%
presence of lone pair of electron in $Tl$
0%
Inert pair effect
0%
Large Ionic radius of $Tl$ ion
0%
None of above

Explanation

In 3rd A group Tl shows +1 oxidation state while other member shows +3 oxidation state due to inert pair effect. The inert pair effect is the tendency of the electrons in the outermost atomic S orbital to remain unionized or unshared in compounds of post-transition metals.

So in $Tl$ only outer electron i.e $np^{1}$ participate in chemical bonding whereas the $ns^{2}$ do not participate in chemical bonding in a chemical reaction but other elements in that group do not show inert pair effect.

Hence option B is correct.

What is hybridisation of boron in

[B (O H)_{4}]^{-}

$[B(OH)_4]^-$ ?

0%
$sp^2$
0%
$sp^3$
0%
$sp^3d^2$
0%
None of these

Explanation

Hybridisation of boron in

[B (O H)_{4}]^{-}

$[B(OH)_4]^{-}$ is

s p^{3}

$sp^{3}$

Hybridisation is

s p^{3}

$sp^{3}$ (Refer to Image 1)

electronic configuration of boron $$\longrightarrow1s^{2}2s^{2}2p^1$$

(Refer to Image 2)

electronic configuration of boron in first excited state

(Refer to Image 3)

Boron forms

4 s p^{3}

$4sp^3$ hybrid orbitals

The unpaired orbitals forms covalent bond with

3 - O H

$3-OH$ groups

The empty orbital forms co-ordinate covalent with

- O H^{-}

$-OH^{-}$ group

The boron atom in $B(OH)_3$ is electron deficient: it has 6 rather than 8 electrons in its valence shell. It acts as a Lewis acid by readily accepting the lone pair from the oxygen in a water molecule to go from sp2 to sp3 hybridisation.

Metals react with silicons to form:

0%
silicates
0%
silanes
0%
silicides
0%
mixture of silicates and silanes

The no. of unpaired electrons in carbon is :

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Which of the following statement is correct with respect to the property of elements in carbon family?With an increase in their atomic number their:

0%
atomic size decreases
0%
ionization energy increases
0%
metallic character decreases
0%
stability of $+2$ oxidation state increases

Explanation

As the atomic number of the element increases the stability of the compounds in +2 oxidation state increases down the group due to inert pair effect.
Hence option

D

$D$ is correct.

The hydrides formed by germanium are:

0%
$GeH_2$
0%
$GeH_4$
0%
$GeH_6$
0%
$GeH_3$

Explanation

Germanium(II) hydrides, also called germylene hydrides, are a class of Group $14$ compounds consisting of low-valent germanium and a terminal hydride. They are also typically stabilized by an electron donor-acceptor interaction between the germanium atom and a large, bulky ligand. Germanium is the chemical compound with the formula $GeH_4$ , and the germanium analogue of methane. It is the simplest germanium hydride and one of the most useful compounds of germanium.

Similarly $GeH_6$ is also one of the hydrides

The metalloid in the carbon family is:

0%
carbon
0%
tin
0%
silicon
0%
lead

Which of the following oxide is formed by carbon?

0%
Sesquioxide
0%
Suboxide
0%
Superoxide
0%
All

Explanation

Carbon suboxide

O = C = C = C = O

$O = C = C = C = O$

Which of the following characteristic property increases as we go down the carbon group?

0%
Acidic nature of oxides
0%
Electronegativity
0%
Metallic characters
0%
Non-metallic characters

The naturally occuring radioactive isotope of carbon is:

0%
$C-12$
0%
$C-13$
0%
$C-14$
0%
none of the above

Explanation

The longest-lived radioisotope is

^{14} C

$^{14}C$ , with a half life of 5,700 years.

The number of oxides formed by carbon is:

0%
$2$
0%
$3$
0%
$4$
0%
$5$

The only hydride formed by lead has the formula:

0%
$PbH_2$
0%
$PbH_4$
0%
$PbH_6$
0%
$PbH_3$

Explanation

A hydride is the anion of hydrogen, $H^{ - }$ , more commonly, it is a compound in which one or more hydrogen centres have nucleophilic, reducing, or basic properties. In compounds that are regarded as hydrides, the hydrogen atom is bonded to a more electropositive element or group. Compounds containing hydrogen bonded to metals or metalloid may also be referred to as hydrides. Almost all of the elements form binary compounds with hydrogen, the exceptions being $He, Ne, Ar, Kr, Pm, Os, Ir, Rn, Fr,$ and $Ra$

Plumbane, ${ PbH }_{ 4 }$ , is a metal hydride and group 14 hydride composed of lead and hydrogen.

$\left[ 1 \right]$ Plumbane is not well characterized or well known, and it is thermodynamically unstable with respect to the loss of a hydrogen atom.

$\left[ 2 \right]$ Derivatives of plumbane include lead tetrafluoride, ( ${ PbF }_{ 4 }$ ), and tetraethyllead, $\left( { \left( { CH }_{ 3 }{ CH }_{ 2 } \right) }_{ 4 }Pb \right)$

Physical properties of a substance are given.
P

\to

$\rightarrow$ It is colour less
Q

\to

$\rightarrow$ has intense smell
R

\to

$\rightarrow$ It's aqueous solution possesses basic property.
Which of the above physical properties are correct for sulphur dioxide?

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

The relative strength of trichlorides of boron group to accept a pair of electron is given by:

0%
$GaCl_3 > AlCl_3 > BCl_3$
0%
$AlCl_3 < BCI_3 < GaCl_3$
0%
$AlCl_3 < GaCl_3 < BCl_3$
0%
$BCl_3 < AlCl_3 < GaCl_ 3$
0%
$GaCl_3 < BCl_3 < AlCl_3$

Explanation

Lewis acidic strength is in the order

G a X_{3} < A l X_{3} < B X_{3}

$GaX_3 < AlX_3 < BX_3$
So, acidic strength of trichlorides is as follows:

G a C l_{3} < A l C l_{3} < B C l_{3}

$GaCl_3 < AlCl_3 < BCl_3$
Hence, strength of accepting electron is as follows:

G a C l_{3} > A l C l_{3} > B C l_{3}

$GaCl_3 > AlCl_3 > BCl_3$

The correct Lewis acid order for boron halides is:

0%
$BBr_3 > BCl_3 > BI_3 > BF_3$
0%
$BI_3 > BF_3 > BBr_3 > BCl_3$
0%
$BF_3 > BCl_3 > BBr_3 > BI_3$
0%
$BI_3 > BBr_3 > BCl_3 > BF_3$

Explanation

(i) On the basis of back bonding, smaller be the halide atom, more effective be the back bonding and show less tendency to accept a pair of electrons.

(ii) Size of halides increase down the group, hence the ability of back bonding decreases making the molecule more acidic.

Hence, correct order of Lewis acid for boron-halides is

B l_{3} > B B r_{3} > B C l_{3} > B F_{3}

$Bl_3 > BBr_3 > BCl_3 > BF_3$ .

An ionic compound is:

0%
$Sn{ Cl }_{ 2 }$
0%
$C{ Cl }_{ 4 }$
0%
$Ge{ Cl }_{ 4 }$
0%
$Si{ Cl }_{ 4 }$

Explanation

An ionic compound is

S n {C l}_{2}

$Sn{ Cl }_{ 2 }$ .
All tetrahalides of carbon family are covalent except

S n F_{4}

$\displaystyle SnF_4$ which is ionic.

In boron atom screening on outermost electron is due to:

0%
electrons of K shell only
0%
all the electrons of K and L shell
0%
two electrons of 1s and 2s each
0%
only by electrons of L shell

Explanation

Boron

= 1 S^{2} 2 S^{2} 2 P^{1}

$=1S^2 2S^2 2P^1$

Screening effect :- Higher is the screening electron causes less attraction from the nucleus and can be easily removed, which leads to lower value of ionisation potential.

To Screening effect in Boron in Boron is due to two electrons of

1 S

$1S$ and

2 S

$2S$ subshells.

The hybridised state of

{A l}^{3 +}

${ Al }^{ 3+ }$ in the complex ion formed when

A l {C l}_{3}

$Al{ Cl }_{ 3 }$ is treated with aqueous acid is:

0%
${ sp }^{ 3 }$
0%
${ dsp }^{ 2 }$
0%
${ sp }^{ 3 }{ d }^{ 2 }$
0%
${ sp }^{ 2 }d$
0%
${ sp }^{ 2 }$

Explanation

$Al^{ 3+ }$ aqueous acid $\rightarrow { \left[ Al{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 3+ }$ .

${ Al }^{ 3+ }$ ions give a complex ion in which ${ Al }^{ 3+ }$ ion form six coordinate covalent bonds with six ${ H }_{ 2 }O$ molecules.

Hence, it shows ${ sp }^{ 3 }{ d }^{ 2 }$ hybridisation.

(Due to availability of vacant $3d$ -orbitals.)

$Al=3{ s }^{ 2 }3{ p }^{ 1 }$

${ Al }^{ 3+ }=3{ s }^{ 0 }$

Four electron pairs denoted by $6{ H }_{ 2 }O$ molecules.

Addition of mineral acid to an aqueous solution of Borax, the compound formed is:

0%
Puroboric acid
0%
Boron hydride
0%
Meta boric acid
0%
Orthoboric acid

Explanation

Correct Option: $D$

Hint: The chemical formula of Borax is $Na_2B_4O_7$ .

Explanation:

When a solution of Borax reacts with a mineral acid such as $HCl$ orthoboric acid is formed.
The reaction is as shown below;

$Na_2B_4O_7+5H_2O+2HCl\rightarrow 4H_3BO_3+2NaCl$

Orthoboric acid is a weak, monobasic acid of boron.

Boron nitride has the structure of the type:

0%
Graphite type
0%
Diamond type
0%
Both diamond and graphite type
0%
$NaCl$ type

Explanation

Hexagonal boron nitride

(h - B N)

$(h-BN)$ is the equivalent in the structure of graphite.

Like graphite, its plate-like microstructure and layered lattice structure give it good lubricating properties.

Hence, option

A

$A$ is correct.

What is the chief constituent of Pyrex glass ?

0%
$B_2O_3$
0%
$SiO_2$
0%
$Al_2O_3$
0%
$NaO_2$

Explanation

Pyrex glass is obtained by fusing toghether 60 to 80%

S i O_{2}

$\displaystyle SiO_2$ , 10 to 25 %

B_{2} O_{3}

$\displaystyle B_2O_3$ and remaining amount of

A l_{2} O_{3}

$\displaystyle Al_2O_3$ .

S i O_{2}

$\displaystyle SiO_2$ is the chief constituent of Pyrex glass.

Hence, the correct answer is option

B

$B$ .

The element

Z = 114

$Z = 114$ has been discovered recently. It will belong to which of the following family/ group and electronic configuration?

0%
Nitrogen family, $[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{6}$
0%
Halogen family, $[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{5}$
0%
Carbon family, $[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{2}$
0%
Oxygen family, $[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{4}$

Explanation

The element

Z = 114

$Z=114$ has been discovered recently.

It will belong to carbon family and its electronic configuration is

[R n] 5 f^{14} 6 d^{10} 7 s^{2} 7 p^{2}

$\displaystyle [Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{2}$ . It belongs to seventh period and fourteenth group.

Hence, the correct option is

C

$C$ .

Colemanite is

0%
$Na_{2}B_{4}O_{7}\cdot 10H_{2}O$
0%
$Ca_{2}B_{6}O_{11} \cdot 5H_{2}O$
0%
$NaBO_{2}$
0%
$H_{3}BO_{3}$

Explanation

Colemanite, an ore of boron is

C a_{2} B_{6} O_{11} \cdot 5 H_{2} O

$Ca_{2}B_{6}O_{11}\cdot 5H_{2}O$ is a borate mineral found in evaporite deposits of alkaline lacustrine environments.

Colemanite is a secondary mineral that forms by alteration of borax and ulexite.

Hence, option B is correct.

Boron has an extremely high melting point because of:

0%
the strong van der Waals forces between its atoms
0%
the strong binding forces in the covalent polymer
0%
its ionic crystal structure
0%
allotropy

Which of the following is not silicate?

0%
Zeolite
0%
Feldspar
0%
Ultramarine
0%
Silica

Explanation

Except silica, all the other three are silicates.

Zeolite

- N a_{2} A l_{2} S i_{2} O_{8} \cdot x H_{2} O

$-\ Na_2Al_2Si_2O_8\cdot xH_2O$

Feldspar

- K A l S i_{3} O_{8}

$-\ KAlSi_3O_8$ ,

Ultramarine

- N a_{8} (A l O_{2})_{6} \cdot (S i O_{2})_{6} S_{2}

$-\ Na_8(AlO_2)_6 \cdot (SiO_2)_6S_2$

Elements of 14 group :

0%
Exhibit oxidation state of + 4 only
0%
Exhibit oxidation states of +2 and + 4
0%
form $M^{2-}$ and $M^{4+}$ ions
0%
form $M^{2+}$ and $M^{4+}$ ions

Explanation

Due to inert pair effect, elements of group 14 exhibit oxidation states of

+ 2

$+ 2$ and

+ 4

$+ 4$ .

Addition of HCl to an aqueous solution of

P b (N O_{3})_{2}

$Pb(NO_3)_2$ gives a

0%
Yellow Precipitate
0%
Baum Precipitate
0%
White Precipitate
0%
Black Precipitate

Explanation

P b (N O_{2})_{2} (a q) + 2 H C l (a q) \to P b C l_{2} (s) + 2 H N O_{3} (a q)

$Pb (NO_2)_2(aq)+ 2HCl (aq)\rightarrow PbCl_2 (s)+ 2 HNO_3 (aq)$ gives white precipitate

Which of the following is the most basic oxide?

0%
${ Sb }_{ 2 }{ O }_{ 3 }$
0%
${ Bi }_{ 2 }{ O }_{ 3 }$
0%
$Se{ O }_{ 2 }$
0%
${ Al }_{ 2 }{ O }_{ 3 }$

Explanation

{B i}_{2} O_{3}

${ Bi }_{ 2 }{ O }_{ 3 }$ is the most basic oxide.
More the oxidation state of the central metal atom, more is its acidity. Hence,

S e O_{2}

$Se{ O }_{ 2 }$ (O.S of Se

= + 4

$=+4$ ) is acidic. Further, for a given oxidation state, the basic character of the oxides increases with increasing size of the central atom.

Thus,

{A l}_{2} O_{3}

${ Al }_{ 2 }{ O }_{ 3 }$ and

{S b}_{2} O_{3}

${ Sb }_{ 2 }{ O }_{ 3 }$ are amphoteric and

{B i}_{2} O_{3}

${ Bi }_{ 2 }{ O }_{ 3 }$ is basic.

When borax is dissolved in water:

0%
$B(OH)_3$ is formed only
0%
$[B(OH)_4]^-$ formed only
0%
both $B(OH)_3$ and $[B(OH)_4]^-$ are formed
0%
$[B_3O_3(OH)_4]^-$ is formed only

Explanation

Hint: Borax reacts to give an alkaline solution.

Correct Answer: Option C

Explanation:

The reaction of borax in water is as follows:

{[B_{4} O_{7}]}^{2 -} + 7 H_{2} O \to 2 B (OH)_{3} + 2 {[B {(OH)}_{4}]}^{-}

${\left[ {{{\text{B}}_4}{{\text{O}}_7}} \right]^{2 - }} +\ 7{{\text{H}}_2}{\text{O}} \to 2\;{\text{B}}{({\text{OH}})_3} + 2{\left[ {\;{\text{B}}{{({\text{OH}})}_4}} \right]^ - }$

Hence both

B {(O H)}_{3}

$B{(OH)}_3$ and

[B {(O H)}_{4}]^{-}

$[B{(OH)}_4]^-$ are formed.

Which of the following molecule is /are chiral ?

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

The outermost electronic configuration of the most electronegative element is:

0%
${ ns }^{ 2 }{ np }^{ 3 }$
0%
${ ns }^{ 2 }{ np }^{ 4 }$
0%
${ ns }^{ 2 }{ np }^{ 5 }$
0%
${ ns }^{ 2 }{ np }^{ 6 }\quad$

Explanation

{n s}^{2} {n p}^{5}

${ ns }^{ 2 }{ np }^{ 5 }$ configuration represents the most electronegative element as after gaining one electron it becomes more stable (inert gas configuration) [electronegativity is the tendency of attracting electron].

Fulorine (atomic number 9) is the most electronegative element. Its electronic configuration is

1 s^{2} {2 s}^{2} {2 p}^{5}

$1s^2{ 2s }^{ 2 }{ 2p }^{ 5 }$ . It is a halogen.

Which one of the following statements about

H_{3} B O_{3}

$H_{3}BO_{3}$ is not correct?

0%
It is a strong tribasic acid.
0%
It is prepared by acidifying an aqueous solution of borax.
0%
It has a layer structure in which planar BO, units are joined by hydrogen bonds.
0%
It does not act as proton donor as it acts as a Lewis acid by accepting hydroxyl ions.

Explanation

1. It is not a tribasic acid.

Option A is incorrect for $H_3BO_3$ .

2. Aqueous solution, instead of donating the $-OH$ group it breaks the $H-OH$ bond of water and accepts the $-OH$ group from it leaving in the solution hence it is mono basic acid.

Elements of group 13 shows :

0%
only +1 oxidation state
0%
only +3 oxidation state
0%
+1 and +3 oxidation states
0%
+1, +2 and +3 oxidation states

S i F_{4}

$SiF_4$ gets hydrolysed then which of the following is not expected to from?

0%
$SiO_2$
0%
$Si(OH)_2F_2$
0%
$H_2SiF_6$
0%
$Si(OH)_4$

Explanation

Partial hydrolysis is not possible for

S i F_{4}

$SiF_{4}$

4 H_{2} O + S i F_{4} ⟶ S i (O H)_{4} + H F

$4H_{2}O+SiF_{4}\longrightarrow Si(OH)_{4}+HF$

H_{2} O + S i F_{4} ⟶ H_{2} S i O_{3} + H_{2} S i F_{6}

$H_{2}O+SiF_{4}\longrightarrow H_{2}SiO_{3}+H_{2}SiF_{6}$

2 H_{2} O + S i F_{4} ⟶ S i O_{2} + 4 H F

$2H_{2}O+SiF_{4}\longrightarrow SiO_{2}+4HF$

Hence

S i (O H)_{2} F_{2}

$Si(OH)_{2}F_{2}$ is not possible

In group 13, electronegativity first decreases from B to Al and then increase marginally down the group. This is because of :

0%
non- metallic nature of B
0%
discrepancies in atomic size of element
0%
ability of B and Al to form $p\pi$ - $p\pi$ multiple bonds
0%
irregular trend in electronegativity throughout the periodic table

Which of the following metals does not show inert pair effect?

0%
Thallium
0%
Gallium
0%
Indium
0%
Aluminium

The reason behind the lower atomic radius of Ga as compared to Al is ?

0%
poor screening effect of d-electrons for the outer electrons from the increased nuclear charge
0%
increased force of attraction of increased nuclear charge on electrons
0%
increased ionisation enthalpy of Ga as compared to Al
0%
Anomalous behaviour of Ga

Aluminium exhibits +3 oxidation state. As we move down the group, +1 oxidation states get more stable. This is a consequence of :

0%
increasing size of the atom
0%
inert pair effect
0%
electron deficient nature
0%
$p\pi$ - $p\pi$ bonding

Explanation

Answer:- (B) inert pair effect

The atoms of group 13 elements have 3 valence electrons, two in s-subshell and one in p subshell.

The +1 oxidation state becomes more stable as we move down to the group due to inert pair effect. After the removal of 1 electron from p orbital, the remaining

n s^{2}

$ns^2$ electrons behave like stable noble gases due to inert pair effect and do not take part in compound formation. The two electrons present in the s-shell are strongly attracted by the nucleus and do not participate in bonding. This inert pair effect becomes more and more prominent on moving down the group.

Hence, the +1 oxidation state gets more stable as we move down the group.

Group

13

$13$ elements show

+ 1

$+1$ and

+ 3

$+3$ oxidation states. Relative stability of

+ 3

$+3$ oxidation state may be given as :

0%
$Tl^{3+}>In^{3+}>Ga^{3+}>Al^{3+}>B^{3+}$
0%
$B^{3+}>Al^{3+}>Ga^{3+}>In^{3+}>Tl^{3+}$
0%
$Al^{3+}>Ga^{3+}>Tl^{3+}>In^{3+}>B^{3+}$
0%
$Al^{3+}>B^{3+}>Ga^{3+}>Tl^{3+}>In^{3+}$

Explanation

Answer:- (B)

B^{+ 3} > {A l}^{+ 3} > {G a}^{+ 3} > {I n}^{+ 3} > {T l}^{+ 3}

$B^{+3} > {Al}^{+3}> {Ga}^{+3}> {In}^{+3} > {Tl}^{+3}$

Reason:-

As we move down the group the inert pair effect becomes more and more prominent which causes the decrease in stability of elements in +3 oxidation state.

Hence, (B) is correct answer.

Which is the most electropositive element?

0%
Na
0%
Cu
0%
Cs
0%
Ca

Chemically borax is ?

0%
sodium metaborate
0%
sodium orthoborate
0%
sodium tetraborate decahydrate
0%
sodium hexaborate.

Explanation

Answer:- (C) Sodium tetraborate decahydrate

Borax, also known as sodium perborate is an important boron compound, a mineral, and the salt of boric acid. It is usually a white powder consisting of soft colorless crystals that dissolve easily in water.

Chemical formula of borax is -

{N a}_{2} B_{4} O_{7} .10 H_{2} O

${Na}_2B_4O_7.10H_2O$ and called as sodium tetraborate decahydrate.

Aluminium oxide is not reduced by chemical reactions due to ?

0%
its highly stable nature
0%
its highly unstable nature
0%
its amphoteric nature
0%
its highly explosive nature

Explanation

Answer:- its highly stable nature

Aluminium oxide is highly stable so extraction of aluminium is done by electrolytic process.

In this process, alumina

{A l}_{2} O_{3}

${Al}_2O_3$ is used along with flourspar and cryolite.

Boron is unable to form

{B F}_{6}^{3 -}

${ BF }_{ 6 }^{ 3- }$ ions due to:

0%
non-availability of d-orbitals
0%
small size of boron atom
0%
non-metallic nature
0%
less reactivity towards halogens.

Explanation

Boron does not have d orbitals. Because of this, it is not possible for the element to expand on its octet. This causes the maximum covalence of boron to be restricted to 4. Beyond this value, boron will become unstable and thus, cannot form

{B F_{6}}^{3 -}

${BF_6}^{3-}$ ions.

Hence, the correct answer is option

A

$\text{A}$ .

The tendency of group 14 elements to show +2 oxidation state increases in the order of:

0%
$C < Si < Sn < Pb < Ge$
0%
$C < Si < Ge < Sn < Pb$
0%
$Ge < Sn < Pb < C < Si$
0%
$Pb < Sn < Ge < C < Si$

Explanation

The tendency to show

+ 2

$+2$ oxidation state increases as we move down the group. This is due to inert pair effect which causes the inability of

n s^{2}

$ns^2$ electrons of valence shell to participate in bonding.

Thus, the stability of elements in

+ 2

$+2$ oxidation state increases as we move down the group in the order

C < S i < G e < S n < P b

$C < Si < Ge < Sn < Pb$ .

Hence, the correct answer is option

B

$\text{B}$ .

The first member of the p-block elements differs from the remaining members of their corresponding groups due to :

0%
small size and absence of d-orbitals
0%
diagonal relationship with other elements
0%
difference in ability to form double and triple bonds
0%
high ionisation enthalpy

Formula for the following compound Thallium (I) sulphate is

T l_{2} S O_{4}

$Tl_{2}SO_4$ .

0%
True
0%
False

Explanation

Thallium(I) sulfate

(T l_{2} S O_{4})

$(Tl_2SO_4)$ or thallous sulfate is the sulfate salt of thallium in the common +1 oxidation state, as indicated by the Roman numeral I. It is often referred to as simply thallium sulfate. Thallium(I) sulfate is colourless, odourless, tasteless, and highly toxic.

Which of the following properties correctly explain the compound

S i O_{2}

$SiO_2$ ?

0%
Linear, basic
0%
Tetrahedral, acidic
0%
Tetrahedral, basic
0%
Linear, acidic

Explanation

Silicon dioxide,

S i O_{2}

$SiO_2$ is one of the most common minerals in the earth. It has a tetrahedral arrangement with one silicon-bonded to four oxygen atoms. Some oxygen atoms will be bonded to two silicon atoms so that two tetrahedral are joined at a corner (bridging atoms). The orientation can be random, leading to an amorphous structure.

Silicon dioxide has no basic properties as it doesn't contain oxide ions and it doesn't react with acids. Instead, it is very weakly acidic, reacting with strong bases.

Thus,

S i O_{2}

$Si{O}_{2}$ has a tetrahedral structure and is acidic in nature.

Hence, the correct answer is option

B

$\text{B}$ .

Electropositive character for the elements of group 13 follows the order ?

0%
B > Al > Ga > In > Tl
0%
B < Al < Ga < In < Tl
0%
B < AL > Ga < In > Tl
0%
B < Al > Ga > In > Tl

Explanation

Answer:- (D)

B < A l > G a > I n > T l

$B<Al>Ga>In>Tl$

Reason:-

The electropositive character of elements increases from boron to aluminium due to smaller size and high ionisation potential of B than Al.

However, the electropositive character decreases slowly from Gallium to Thallium due to extra d-electrons which exerts a very little shielding effect on outer electrons. Therefore, these electrons are more firmly held by the nucleus and hence there is a decrease in electropositive character.

Hence, (D) is correct answer.

Which of the following ions is the most stable?

0%
$Sn^{2+}$
0%
$Ge^{2+}$
0%
$Si^{2+}$
0%
$Pb^{2+}$

Explanation

Hint: These compounds belong to the carbon family having a covalency of $4$ .

Correct Answer: Option D

Explanation:

The common oxidation states exhibited by this family are $+4$ and $+2$ .
The tendency to show $+2$ oxidation state increases in down the group.
Due to the inert pair effect, down the group the tendency to form ions with $+2$ oxidation state increases from $Ge$ to $Pb$ .
The stability order is in increasing order of atomic number.

Hence, the correct answer is

P b^{2 +} .

$Pb^{2+}.$

Silicon is an important constituent of:

0%
sand
0%
atmosphere
0%
plants
0%
water bodies

Explanation

Answer:- (A) sand

Silicon is the second most abundant element in the Earth's crust, making up

25.7 %

$25.7\%$ of it by weight. It occurs in clay, feldspar, granite, quartz and sand, mainly in the form of silicon dioxide (also known as silica) and silicates (compounds containing silicon, oxygen and metals).

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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