partially soluble in cold water & fairly soluble in hot water.

MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 1

What is the general electronic configuration of

$p-block$ elements?

0%
$\left( n \right) { s }^{ 2 }\left( n-1 \right) { p }^{ 1-6 }$
0%
${ ns }^{ 2 }{ np }^{ 1-6 }$
0%
Both $a$ and $b$
0%
None of these

What is the n factor or valancy factor of ozone during the change

$2$ O

$_3$

$\rightarrow$

$3$ O

$_2$ ?

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Which is correct order of stability:

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

Kernite is a mineral containing:

0%
aluminium
0%
gallium
0%
boron
0%
silicon

Explanation

The chemical formula for Kernite is

$Na_2B_4O_7.4H_2O$ . It contains boron.

______ can be found in bauxite and sphalerite in traces.

0%
Boron
0%
Aluminium
0%
Gallium
0%
Indium

Lorandite is a mineral of:

0%
boron
0%
aluminium
0%
gallium
0%
thallium

Explanation

Lorndite is a thallium arsenic sulfosalt with the chemical formula:

$TlAsS_2$ . Though rare, it is the most common thallium-bearing mineral.

Beryl, cryolite are minerals of:

0%
boron
0%
aluminium
0%
thallium
0%
indium

Explanation

Aluminium also occurs in the minerals beryl, cryolite, garnet, spinel, and turquoise.
Beryl -

${Be}_3{Al}_2{(SiO_3)}_6$ .
Cryolite -

${Na}_3AlF_6$

Aluminium is usually found in

$+3$ oxidation state. In contrast, thallium exists in

$+1$ and

$+3$ oxidation states. This is due to :

0%
lanthanoid contraction
0%
lattice effect
0%
diagonal relationship
0%
inert pair effect

Explanation

Inert pair effect is the prominent character of the p-block element.

In this, the high molecular weight element of the group show a lower oxidation state.

This is because on going down the group, the shielding effect increases, but d-subshell and f-subshell show poor shielding effect.

The high molecular weight members of p-block groups contain d-subshell and f-subshell.

Because of this, they show a lower oxidation state.

Hence, option

$D$ is correct.

The element that shows greater ability to form

$p\pi -p\pi$ multiple bonds, is:

0%
$Si$
0%
$Ge$
0%
$Sn$
0%
$C$

Explanation

Carbon atom has

$2$ p orbitals able to form strongest

$p\pi -p\pi$ bonds due to its size.

Correct statements among a to d regarding silicones are:

(a) They are polymers with hydrophobic character.

(b) They are biocompatible.

(d) Usually, they are resistant to oxidation and used as greases.

0%
(a), (b) and (c) only.
0%
(a), and (b) only.
0%
(a), (b), (c) and (d).
0%
(a), (b) and (d) only.

Explanation

Silicones are the polymer of silicon-containing

$[-(R)_3Si-O-]$ linkage.

It is a polymer with hydrophobic character thus used in making water-resistant seals.

They are biocompatible.

They have high thermal stability and low dielectric strength which is not correct.

They are also resistant to oxidation. Because of there wax-like texture, they are also used in greases.

Hence, the correct option is D.

Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is:

0%
$\mathrm{R}_{2}\mathrm{S}\mathrm{i}\mathrm{C}1_{2}$
0%
$\mathrm{R}_{3}\mathrm{S}\mathrm{i}\mathrm{C}1$
0%
$\mathrm{R}_{4}\mathrm{S}\mathrm{i}$
0%
$\mathrm{R}\mathrm{S}\mathrm{i}\mathrm{C}1_{3}$

Explanation

Trichlorosilane is the basic ingredient used in the production of purified cross linked silicon polymer. In water, it rapidly decomposes to produce a silicone polymer while giving off hydrochloric acid.

$RSiCl_3 + H_2O \rightarrow RSi(OH)_3 + HCl$

Linear and cyclic silicones are produced by the reaction of water with organochlorosilanes of the general formula,

$R_2SiCl_2$ followed by a polymerization reaction that occurs by the elimination of a molecule of water from two hydroxyl groups of adjacent

$R_2Si(OH)_2$ molecules.

The amorphous form of silica is:

0%
Quartz
0%
Kieselguhr
0%
Cristobalite
0%
Tridymite

The correct statements among I to III regarding group

$13$ elements oxides are:
(I) Boron trioxide is acidic
(II) Oxides of aluminium and gallium are amphoteric.
(III) Oxides of indium and thallium are basic.

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

Explanation

All statements are correct.

$B_{2}O_{3}\rightarrow acidic$ .

$Al_{2}O_{3}$ and

$Ga_{2}O_{3}$ are amphoteric oxides.

$In$ and

$Tl$ oxides are basic in nature.

As we go down the group non-metallic character decreases and metallic character increases.

0%
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
0%
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
0%
Statement-1 is True, Statement-2 is False
0%
Statement-1 is False, Statement-2 is True

The crystalline form of borax has :

0%
Tetranuclear $\displaystyle { \left[ { B }_{ 4 }{ O }_{ 5 }{ \left( OH \right) }_{ 4 } \right] }^{ 2- }$ unit
0%
All boron atoms in the same plane
0%
Equal number of $\displaystyle { sp }^{ 2 }$ and $\displaystyle { sp }^{ 3 }$ hybridized boron atoms
0%
One terminal hydroxide per boron atom

Explanation

Correct formula of borax is

$Na_2[B_4O_5(OH)_4].8H_2O$ .

Borax has tetranuclear.

$[B_4O_5(OH)_4]^{2-}$ unit.

Only two B atom lies in the same plane.

Two Boron are

$sp^2$ & two are

$sp^3$ hybridised.

One terminal hydroxide is present per boron atom.

Hence, options

$A$ ,

$C$ and

$D$ are correct.

What is the change in oxidation number of carbon in the following reaction?

$CH_4 (g) + 4Cl_2 (g)\longrightarrow CCl_4 (l) + 4HCl(g)$

0%
$0$ to $+ 4$
0%
$-4$ to $+ 4$
0%
$0$ to $-4$
0%
$+4$ to $+ 4$

Explanation

$\overset {-4} CH_4 (g) + 4Cl_2 (g)\longrightarrow \overset {+4} CCl_4 (l) + 4HCl(g)$

The oxidation state of

$H$ and

$Cl$ are +1 and -1 respectively. Therefore the oxidation state of C in methane is

$-4$ and that in carbon tetrachloride is

$+4$ .

Which of the following is incorrect statement ?

0%
$PbF_4$ is covalent in nature
0%
$SiCl_4$ is easily hydrolysed
0%
$GeX_4 (X=F,Cl,Br,I)$ is more stable than $GeX_2$
0%
$SnF_4$ is ionic in nature

Explanation

$PbF_4$ and

$SnF_4$ are ionic in nature.

Hence option A is the correct answer.

Which one of the following elements is unable to form

$MF_{6}^{3-}$ ion?

0%
$B$
0%
$Ga$
0%
$In$
0%
$Al$

Explanation

B is unable to form

$MF_{6}^{3-}$ ion.

This is because

$\displaystyle B^{3+}$ has no vacant d orbitals. It has only on s and 3 p orbitals in the valence shell.

Hence, it can accept 4 electron pairs and can form

$\displaystyle BF_4^-$ .

therefore option A is correct.

The stability of

$+1$ oxidation state among

$Al$ ,

$Ga$ ,

$In$ and

$Tl$ increases in the sequence:

0%
$Tl < In < Ga < Al$
0%
$In < Tl < Ga < Al$
0%
$Ga < In < Al < Tl$
0%
$Al < Ga < In < Tl$

Explanation

The stability of

$+1$ oxidation state among

$Al$ ,

$Ga$ ,

$In$ and

$Tl$ increases in the sequence.

$\displaystyle Al < Ga < In < Tl$ .
This is due to inert pair effect due to which, on moving down the group, the stability of

$+3$ oxidation state decreases while the stability of

$+1$ oxidation state increases.

The species, having bond angles of

$120^{\circ}$ is:

0%
$BCl_3$
0%
$PH_{3}$
0%
$ClF_{3}$
0%
$NCl_{3}$

Explanation

Correct Option: $A$

Hint: The species which have trigonal planar shape possess $120^◦$ bond angle.

Explanation:

Compound	Hybridization	Shape	Bond angle	Reason
$BCl_3$	$sp^2$	Trigonal planner	$120^◦$	In $BCl_3$ $B$ have three bonded pair electron but no lone pair so, it posses trigonal planner structure.
$PH_3$	$sp^3$	Pyramidal	$93^◦$	In $PH_3$ $P$ have three bonded pair electron and two lone pairs so, it posses pyramidal structure.
$ClF_3$	$sp^3d$	T-shaped	$90^◦$	$ClF_3$ posses two lone pair which create repulsion on bonded pair electrons and posses T-shape.
$NCl_3$	$sp^3$	Pyramidal	$107^◦$	In $NCl_3$ nitrogen have two lone pairs and posses $sp^3$ hybridization and have pyramidal shape.

Final Answer: $BCl_3$ have $120^◦$ bond angle.

Glass is soluble in :

0%
HF
0%
$\mathrm{H}_{2}\mathrm{S}\mathrm{O}_{4}$
0%
$\mathrm{H}\mathrm{Cl}\mathrm{O}_{4}$
0%
Aqua regia

Explanation

$4\ HF+SiO_2\xrightarrow \ SiF_4+2H_2O$
This reaction is known as
itching of glass

$SiO_2$ present in glass reacts
with HF to form

$SiF_4$ . Therefore
glass is soluble in HF

Solid silicones are stable up to:

0%
$100^{\circ}$
0%
$250^{\circ}$
0%
$300^{\circ}$
0%
$400^{\circ}$

Explanation

Solid silicones rubbers retain their quality from -100

$^{o}$ C to 250

$^{o}$ C.

Which of the following elements has a limited co-ordination number of four?

0%
$Sn$
0%
$C$
0%
$Si$
0%
$Ge$

The correct order of atomic radii in group

$13$ elements is _________.

0%
$B < Ga < Al < Tl < In$
0%
$B < Al < In < Ga < Tl$
0%
$B < Ga < Al < In < Tl$
0%
$B < Al < Ga < In < Tl$

Explanation

The correct order of atomic radii in group

$13$ elements is:

$B (\text { 85 pm }) < Ga (\text {135 pm }) < Al (\text {143 pm }) < In (\text { 167 pm }) < Tl (\text {170 pm })$ .

On moving down the group, the atomic radius increases from boron to thallium. The atomic radius of Ga is less than that of Al. Ga has 10 d electrons which do not screen the nuclear charge in an effective manner and hence electrons in the outermost shell of Ga experience more force of attraction towards the nucleus than Al. This results in a decrease in atomic radius.

Hence, option C is correct.

Which of these is not a monomer for a high molecular mass silicone polymer?

0%
$MeSiCl_{3}$
0%
$Me_{2}SiCl_{2}$
0%
$Me_{3}SiCl$
0%
$PhSiCl_{3}$

The element which, forms oxides in all oxidation states varying from +1 to +5 is:

0%
$N$
0%
$P$
0%
$As$
0%
$Sb$

Explanation

Nitrogen atom can form oxides in all oxidation states varying from +1 to +5.

$N_2O$ Nitrogen oxidation state is +1

$NO$ Nitrogen oxidation state is +2

$N_2O_3$ Nitrogen oxidation state is +3

$NO_2$ Nitrogen oxidation state is +4

$N_2O_5$ Nitrogen oxidation state is +5.

So answer is A.

Among the following substituted silanes, the one which will give rise to cross-linked silicon polymer on hydrolysis is:

0%
$R_3SiCl$
0%
$R_4Si$
0%
$RSiCl_3$
0%
$R_2SiCl_2$

Explanation

Among the following substituted silanes, the one which will give rise to cross-linked silicone polymer on hydrolysis is trichloroalkylsilane

$RSiCl_3$ .

On hydrolysis,

$RSiCl_3$ will give

$RSi(OH)_3$ which will polymerize to cross-linked silicone polymer.

$\displaystyle RSiCl_3 \xrightarrow [-3HCl]{3H_2O} RSi(OH)_3 \xrightarrow {\text {polymerisation}}$ cross-linked silicone polymer.

Which of the element shows

$+4$ oxidation state?

0%
$Sn$
0%
$Ra$
0%
$Fr$
0%
$Sc$

Explanation

Only tin is capable of exhibiting stable +4 oxidation state while others cannot.

$Sn^{+4}$ state is stable and forms

$SnCl_4$ compound.

Which one of the following pentafluorides cannot be formed?

0%
$PF_5$
0%
$AsF_5$
0%
$SbF_5$
0%
$BiF_5$

Explanation

Bismuth due to inert pair effect exists with only +3 oxidation state. +5 oxidation state is uncommon for Bi. So, bismuth does not form pentahalide.

Hence, option

$D$ is correct.

What type of oxide would Eka-Aluminium(Gallium) form?

0%
$GaO_3$
0%
$Ga_3O_2$
0%
$Ga_2O_3$
0%
$GaO$

Explanation

Eka-aluminium (Gallium) would form the oxide

$\displaystyle Ga_2O_3$ .
Ga has 3 valence electrons and

$O$ has 6 valence electrons. Ga shows a

$+3$ oxidation state and

$O$ shows

$-2$ oxidation state. So, 2 Ga atoms will share 6 electrons and

$3-O$ atoms will share 6 electrons to form

$\displaystyle Ga_2O_3$ .

Silica is heated with carbon in electric furnace to form :

0%
silicon carbonate
0%
silicon carbide or carborundum
0%
silicon monoxide
0%
none

Explanation

When quartz or sand are heated with an excess of coke in an electric arc furnace at

$2000-2500$

$^{o}$ C silicon carbide or carborundum is formed.

$SiO_2 + 3C \rightarrow SiC +2CO$

Quartz is the form ?

0%
$CO_{2}$
0%
$SiO_{2}$
0%
$SO_{2}$
0%
$NO_{2}$

The structure of

$SiO_{2}$ ?

0%
trigonal
0%
linear
0%
tetrahedral
0%
angular

Explanation

Si in Silica is

$sp_3$ hybridised giving it a tetrahedral structure.

Which of the following is used for making optical
instruments :

0%
$SiO_{2}$
0%
Si
0%
$SiH_{4}$
0%
SiC

Density is highest for :

0%
Si
0%
Ge
0%
Sn
0%
Pb

Explanation

Explanation:

The density of elements increases as we move from top to bottom in a group.

In group 14 elements, as we move from $C$ to $Pb$ the number of orbits increases.

So the size of the atom increases and its density also increases.

Thus $Pb$ has the highest density.

Final Answer: Correct option (D).

Silicones are:

0%
toxic
0%
non-toxic
0%
bitter
0%
sour

Silica dissolves in NaOH solution to form:

0%
quartz
0%
sodium silicate
0%
carborundum
0%
jaspar

Explanation

Silica is almost non-reactive due to the presence of very strong Si-C bonds. They are unaffected by dihydrogen, halogen and acids except

$HF$ and

$NaOH$ to form

$SiF_4$ and

$Na_2SiO_3$ respectively. The reaction is as follow:

$SiO_{2} + 2NaOH \rightarrow Na_{2}SiO_{3} +H_{2}O$

In silicones, silicon is strongly linked to:

0%
oxygen
0%
nitrogen
0%
sulphur
0%
carbon.

Explanation

$Si-O-Si-O-Si$ bonds are found in silicones.

Which is the crystalline form of silica?

0%
Agate
0%
Jaspar
0%
Onyx
0%
Crystobalite

Explanation

Silica (

$SiO_2$ ) occurs in various crystallographic forms such as quartz, crystobalite and tridymite that are interconvertible at suitable temperatures.

Rock crystal is:

0%
NaCl
0%
Sand
0%
Quartz
0%
Agate

Glass reacts with HF forming :

0%
$\mathrm{F}_{2}$
0%
$\mathrm{H}_{2}\mathrm{S}\mathrm{i}\mathrm{F}_{6}$
0%
$\mathrm{C}\mathrm{a}\mathrm{F}_{2}$
0%
$\mathrm{N}\mathrm{a}\mathrm{F}$

Explanation

$SiO_2\ + 6HF\rightarrow H_2SiF_6+2H_2O$

When glass reacts with HF,

$H_2SiF_6$ is formed,

Self protective metal among the following is:

0%
Na
0%
Al
0%
Fe
0%
K

Explanation

$Al$ when comes in contact with air to form

$Al_{2}O_{3}$ which forms a protective passive layer on the metal. This layer cannot be removed by chemical means.

Hence, option B is correct.

Borax is:

0%
miscible in water.
0%
immiscible in oil.
0%
partially soluble in cold water & fairly soluble in hot water.
0%
insoluble in water.

Which metal powder suspended in oil is used as a paint for mirrors?

0%
Fe
0%
Sn
0%
Ag
0%
Al

Explanation

The reflective property of aluminium is being used here. Powdered aluminium suspended in oil is used as paint. The low cost and easy availability make it useful in this.

This compound is also known as Diaspore having formula

$Al_{2}O_{3}.H_{2}O$

The maxium covalency of aluminium is:

0%
4
0%
6
0%
8
0%
3

Explanation

The maximum covalency of Aluminium is 6. Since Aluminium belongs to the 3rd period, it possesses a 3d subshell as well. As a cation it can use

$sp^{3}d^{2}$ hybrid orbitals for a covalency of 6.

$Na_{2}B_{4}O_{7.}\, 10\, H_{2}O$ is:

0%
Borax
0%
Kernite
0%
Glauber salt
0%
Colemanite

Explanation

$Na_{2}B_{4}O_{7.}\, 10\, H_{2}O$ is borax..

Option A is correct.

Boric acid is prepared from borax by the action of

0%
$HCl$
0%
$NaOH$
0%
$CO_{2}$
0%
$Na_{2}CO_{3}$

Explanation

Boric acid is prepared by the action of

$HCl$ on borax. The reaction is as follows:

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

Which of the following molecules is nonpolar?
(i)

$PbCl_4$
(ii)

$BF_3$
(iii)

$SnCl_2$
(iv)

$CS_2$

0%
(i), (ii), (iii)
0%
(i), (ii), (iii), (iv)
0%
(i), (ii), (iv)
0%
(ii), (iii), (iv)

The reagent that has action on glass is:

0%
aqua regia
0%
hydrofluoric acid
0%
oleum
0%
fuming nitric acid

Explanation

HF reacts with glass

$6\ HF+SiO_2\rightarrow H_2SiF_6+2\ H_2O$
etching of glass is done by HF.

The compound used for etching of glass is:

0%
HI
0%
HBr
0%
HCl
0%
HF

Explanation

HF is used in etching of glass because it gets attached to the silica glass.

$SiO_2+6HF\rightarrow H_2\ SiF_6+2\ H_2O$

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 1 - 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 1 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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