Match the column I with column II and mark the appropriate choice Column I Column II $$(A)$$ Boron fibres $$(i)$$ Heat resistant glasses $$(B)$$ Borax $$(ii)$$ Bulletproof vest $$(C)$$ Aluminium $$(iii)$$ Filler in automobile tyre $$(D)$$ Carbon black $$(iv)$$ Transport industry

$$(A)\rightarrow (ii),(B)\rightarrow (i),(C)\rightarrow (iv),(D)\rightarrow (iii)$$

$$(A)\rightarrow (i),(B)\rightarrow (ii),(C)\rightarrow (iii),(D)\rightarrow (iv)$$

$$(A)\rightarrow (ii),(B)\rightarrow (iii),(C)\rightarrow (i),(D)\rightarrow (iv)$$

$$(A)\rightarrow (iii),(B)\rightarrow (ii),(C)\rightarrow (i),(D)\rightarrow (iv)$$

each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms.

there are double bonds between silicon and oxygen atoms

silicon atom is bonded to two oxygen atoms

each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms

MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 13

Borax is used as a buffer since-

0%
Its aqueous solution contains equal amount of weak acid and its salt.
0%
It is easily available.
0%
Its aqueous solution contains equal amount of strong acid and its salt.
0%
Statement that borax is a buffer, is wrong.

When borax is dissolved in water :

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

Electronic configuration of only one p-block element is exceptional. One molecule of that element consists of how many atoms of it?

0%
One
0%
Two
0%
Three
0%
Four

Explanation

The electronic configuration of an only p-block element is exceptional. One molecule of that element consists of only one atom of it.

As the exception is found in only He, as helium has the electronic configuration to be

$1s^2$ . But other member of the same group has

$ns^2\; np^6$ configuration.

So, this is an exceptional case where He does not resemble its group in its configuration.
And as He is a noble gas element so it is monoatomic gas.

The following element forms a molecule with eight of its own atoms:

0%
$Si$
0%
$S$
0%
$Cl$
0%
$P$

Which of the following product is formed when

$SiF_4$ reacts with water?

0%
$SiF_3$
0%
$H_4SiO_4$
0%
$H_2SO_4$
0%
$H_2SiF_4$

Explanation

In reaction with water,

$SiF_4$ (like

$SiCI_4$ ) gets hydrolysed to form

$H_4SiO_4$ (silicic acid).

$SiF_4+4H_2O\longrightarrow Si(OH)_4$ or

$H_4SiO_4+ 4HF$
(Silicic acid)

The ground glass stopper is often seen to get stuck in the neck of glass bottles containing aq

$NaOH$ solution This is due to:

0%
formation of solid $\displaystyle Na_{2}CO_{3}$ around the neck of the bottle due to NaoH reacting with atmospheric $\displaystyle CO_{2}$
0%
formation of solid $\displaystyle Na_{2}SiO_{3}$ around the neck of the bottle
0%
creation of partial vacuum due to highly hygroscopic nature of $NaOH$
0%
high coefficient of thermal expansion of glass

Explanation

The ground glass stopper is often seen to get stuck in the neck of glass bottles containing aq

$NaOH$ solution This is due to formation of solid

$Na_2SiO_3$ around the neck of the bottle

$\displaystyle 2NaOH(aq)+SiO_{2}\rightarrow \underset{fuses\: \: with\: \: glass}{Na_{2}SiO_{3}(s)+H_{2}O}$

Hence option

$B$ is correct.

Stannite is an ore of __________.

0%
tin
0%
lead
0%
germanium
0%
carbon

Explanation

Stannite is a mineral, a sulfide of copper, iron, and tin. The chemical formula

${Cu}_2FeSnS_4$ .

Which of the following electrolytes is least effective in causing flocculation of ferric hydroxide sol?

0%
$K_{3}[Fe(CN)_{6}]$
0%
$K_{2}CrO_{4}$
0%
$KBr$
0%
$K_{2}SO_{4}$

Explanation

(i) Sols are coagulated by oppositely charged ions.
(ii) Greater the charge on ion causing coagulation, higher will be its coagulating power.

$\therefore Fe(OH)_{3}$ is positively charged sol.

$\therefore$ Anion will coagulate this sol.
(a)

$K_{3} [Fe(CN)_{6}]\rightarrow 3K^{+} + [Fe(CN)_{6}]^{3-}$
(b)

$K_{2}Cr_{2}O_{7}\rightarrow 2K^{+} + Cr_{2}O_{7}^{2-}$
(c)

$K Br\rightarrow K^{+} + Br^{-}$
(d)

$K_{2}SO_{4} \rightarrow 2K^{+} + SO_{4}^{2-}$

$\because$ In

$KBr, Br^{-}$ has smallest negative charge among given choices.

$\therefore$ It will be least effective in coagulating

$Fe(OH)_{3}$ .

The hydrides of the first elements in groups 15 - 17, namely

$N{H}_{3}, {H}_{2}O$ and

$HF$ respectively, show abnormally high values for melting and boiling points. This is due to:

0%
small size of $N, O$ and $F$
0%
the ability to form extensive intermolecular $H$ -bonding
0%
the ability to form extensive intramolecular $H$ -bonding
0%
effective van der Waal's interaction

Explanation

The hydrides of the first elements in groups 15 - 17, namely

$NH_3$ ,

$H_2O$ and HF respectively, show abnormally high values for melting and boiling points. This is due to the ability to form extensive inter-molecular H-bonding. In these hydrides, H atom is attached to highly electronegative N, O or F atoms resulting in stronger hydrogen bonds which lead to molecular associations. Large amount of energy is needed to break these hydrogen bonds.

Quartz,

$\displaystyle Si{ O }_{ 2 }$ , which is insoluble in most of the solvent with a high melting point. Which type of force best describe the physical properties of quartz?

0%
Hydrogen bonding
0%
Covalent bonding
0%
Ionic bonding
0%
$P$ bonding
0%
Dispersion forces

Explanation

Option

$(A)$ and

$(B)$ are correct.

The physical properties of quartz are due to covalent bonding and hydrogen bonding.

Each silicon atom is covalently bonded to four oxygen atoms. Each oxygen atom is covalently bonded to two silicon atoms.Thus all atoms in the entire crystal are linked together by covalent bonding which makes quartz hard, brittle, and have high melting points.

When quartz molecules are built up to form crystals and silicates, several linked

$SiO_4$ units with attached hydrogen by hydrogen bonding.

Group 13 elements are attacked by non-oxidising acids except:

0%
aluminium
0%
gallium
0%
indium
0%
boron

Explanation

Option

$(D)$ is correct.

Boron is not attacked by non-oxidising acids such as

$HCl$ but reacts with oxidising acids such as conc.

$HNO_3$ and conc.

$H_2SO_4$ forming boric acid.

Boron, being chemically a non-metal, is resistant to attack by non-oxidising acids but the other members of the group react with non-oxidising acids as typical metals and evolve hydrogen.

$Me_2SiCl_2$ on hydrolysis will produce :

0%
$(Me)_2Si(OH)_2$
0%
$(Me)_2Si=O$
0%
$[-O-(Me)_2Si-O-]_n$
0%
$Me_2 SiCl(OH)$

Explanation

Dimethyldichlorosilane

$\displaystyle Me_2SiCl_2$ on hydrolysis will produce

$\displaystyle [-O-(Me)_2Si-O-]_n$ , which is a silicone, a thermoplastic polymer.

$\displaystyle Me_2SiCl_2 \xrightarrow [\text{hydrolysis}]{H_2O} Me_2Si(OH)_2$

$\displaystyle n Me_2Si(OH)_2 \xrightarrow {\text{polymerisation}}[-O-(Me)_2Si-O-]_n$

In borax the number of

$B-O-B$ links and

$B-OH$ bonds present are respectively:

0%
five and four
0%
four and five
0%
three and four
0%
five and five

Explanation

There are

$5B-O-B$ bonds and

$4B-O-H$ bonds.

The correct increasing order of the stability of

$Al^+,\,Ga^+,\,In^+,\,Tl^+$ ions are:

0%
$Al^+\, <\,Ga^+\, <\,Tl^+\, <\,In^+$
0%
$Al^+\, <\,Ga^+\, <\,In^+\, <\,Tl^+$
0%
$Tl^+\, <\,In^+\, <\,Ga^+\, <\,Al^+$
0%
$Tl^+\, <\,Al^+\, <\,Ga^+\, <\,In^+$

Explanation

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. The term inert pair effect is often used in relation to the increasing stability of oxidation states that are two less than the group valency for the heavier elements of groups 13, 14, 15 and 16.
As an example in group 13 the +1 oxidation state of

$Tl$ is the most stable and

$Tl^{3+}$ compounds are comparatively rare. The stability of the +1 oxidation state increases in the following sequence:

$Al^+<Ga^+<In^+<Tl^+$

The compound that will react most readily with gaseous bromine has the formula is:

0%
$C_4H_{10}$
0%
$C_2H_4$
0%
$C_3H_6$
0%
$C_2H_2$

Explanation

The compound that will react most readily with gaseous bromine has the formula

$C_3H_6$ .

Unsymmetrical alkenes generally are more reactive than the symmetrical alkenes, alkynes and alkanes. That is why, propene is more reactive than the other given compounds.

Hence, option C is correct.

Kieselguhr contains :

0%
carbon
0%
silicon
0%
boron
0%
iodine

Explanation

Keisulguhr is a naturally occurring soft silicons sedimentary rock that is easily crumbled into fine white-to-off-white powder. It contains

$80-90$ % silica, with

$2-4$ % alumina. So it has silicon.

The number of oxygen atoms shared per

$SiO_{4}^{4-}$ tetrahedron in the following are:

I. Two-dimensional sheet structured silicates
II. Cyclic silicates and
III. Single strand chain silicates

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

Which of the following compounds undergo disproportionation in aqueous solution?

0%
$TlCl_3$
0%
$GaCl$
0%
$InCl$
0%
$TlCl$

Explanation

A reaction in which the same species is simultaneously oxidized as well as reduced is called a disproportionation reaction.

$GaCl$ and

$InCl$ undergo disproportionation in aqueous solution.

$(i)$ Its aqueous solution is alkaline to litmus.

$(ii)$ It swells up to glassy material

$Y$ on strong heating.

$(iii)$ When the concentrated solution of sulphuric acid is added to a hot solution of

$X$ .white crystal of an acid

$Z$ separates out.

The Compound

$Y$ is :

0%
sodium borate & boric oxide
0%
flower of sulphur
0%
bleaching powder
0%
none of these

Explanation

When heated borax undergoes various transitions it first loses water molecules and swells. Then resulting in the formation of glassy material.

$Na_2B_4O_7.10H_2O\xrightarrow {\quad \triangle\quad }Na_2B_4O_7\xrightarrow {\quad \triangle\quad } \underset{sodium\ metaborate}{2NaBO_2}+\underset{boric\ oxide}{B_2O_3}$

Sodium metaborate and boric oxide form the glassy material.

Y is sodium metaborate and boric oxide.

The reason(s) for small radius of

$Ga$ compared to

$Al$ is/are :

0%
poor screening effect of $d$ and $f$ orbitals.
0%
increase in nuclear charge
0%
presence of higher orbital
0%
higher atomic number

Explanation

$Ga\left( 31 \right) =\left[ Ar \right] 4{ s }^{ 2 }3{ d }^{ 10 }4{ p }^{ 1 }$

$Al\left( 13 \right) =1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 }3{ s }^{ 2 }3{ p }^{ 1 }$
Small radius of

$Ga$ to poor screening effect of d, f orbital and increase in nuclear charge.

Read the following statements and identify compound

$Z$

$(i)$ Its aqueous solution is alkaline to litmus.

$(ii)$ It swells up to glassy material

$Y$ on strong heating.

$(iii)$ When the concentrated solution of sulphuric acid is added to a hot solution of

$X$ .white crystal of an acid

$Z$ separates out.

0%
hydrochloric acid
0%
boric acid
0%
sulphuric acid
0%
none of these

Element showing the phenomenon of allotropy is:

0%
$Al$
0%
$Pb$
0%
$Sn$
0%
$Cu$

Explanation

Among the given elements tin (

$Sn$ ) shows the allotropy.
β-tin (the metallic form, or white tin), which is stable at and above room temperature, is malleable. In contrast, α-tin (nonmetallic form, or gray tin), which is stable below 13.2 °C (55.8 °F), is brittle.
Hence option

$C$ is correct.

The element that forms stable compounds in low oxidation state is:

0%
$Mg$
0%
$Al$
0%
$Ga$
0%
$Tl$

Explanation

The element that forms stable compounds in low oxidation state is

$Tl$ . The two electrons in the outer shell tend to remain paired and do not participate in compound formation (inert pair effect).

Due to this,

$Tl(I)$ compounds are more stable than

$Tl(III)$ compounds.

$Tl(III)$ ions act as oxidising agents and changes to

$Tl(I)$ ions.

$\displaystyle Tl^{3+}+2e^- \rightarrow Tl^+$

Hence, option

$D$ is correct.

$CH_3CH=\overset{\overset{\displaystyle CH_3}{|}}{C}CH_3$ changes to __________ in hydroboration-oxidation reaction.

0%
$CH_3\underset{\underset{\displaystyle OH}{|}}{C}H\underset{\underset{\displaystyle CH_3}{|}}{C}HCH_3$
0%
$CH_3CH_2\overset{\overset{\displaystyle OH}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}CH_3$
0%
$CH_3CH_2\underset{\underset{\displaystyle CH_3}{|}}{C}HCH_2OH$
0%
none of these

Pick out the wrong statement.

0%
Nitrogen has the ability to form $p\pi -p\pi$ bonds with itself
0%
Bismuth forms metallic bonds in elemental state
0%
Catenation tendency is higher in nitrogen when compared with other elements of the same group
0%
Nitrogen has higher first ionization enthalpy when compared with other elements of the same group
0%
Arsenic forms $d\pi -d\pi$ bonds with transition metals

Explanation

Catenation tendency of phosphorus is higher than that of nitrogen due to high bond energy of

$P - P$ single bond as compared to

$N - N$ single bond.

$(Me)_2SiCl_2$ on hydrolysis will produce.

0%
$(Me)_2Si(OH)_2$
0%
$(Me)_2Si=O$
0%
$[-O-Me_2(Si-O-)]_n$
0%
$Me_2Si(Cl)(OH)$

In the structure of borax, the numbers of boron atoms and B-O-B units, respectively, are :

0%
4 and 5
0%
4 and 3
0%
5 and 4
0%
5 and 3

Explanation

The structure of borax is shown. It contains 4 B atoms and 5 B-O-B bonds.

Hence option

$A$ is correct.

The anion which cannot be formed by boron is:

0%
$B{ F }_{ 6 }^{ 3- }$
0%
$B{ H }_{ 4 }^{ - }$
0%
$B{ \left( OH \right) }_{ 4 }^{ - }$
0%
$B{ O }_{ 2 }^{ - }$

Explanation

Boron cannot form

$B{ F }_{ 6 }^{ 3- }$ anions due to absence of

$d$ orbitals in boron. Since

$d$ orbitals are not present in boron, it cannot expand its octet beyond 8.

$B{ F }_{ 6 }^{ 3- }$ anion will have 12 valence electrons around B.

$1^\circ$ alcohols preferably undergo dehydration in presence of

$Al_{2}O_{3}$ :

0%
$E_{1}$
0%
$E_{2}$
0%
$S_{N}1$
0%
$S_{N}2$

Explanation

Primary alcohol dehydrates through the $E_2$ mechanism

Oxygen donates two electrons to a proton from sulfuric acid $H_2SO_4$ ,

forming an alkyloxonium ion.

Then the nucleophile $H_2SO_4$ back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.

Match the column I with column II and mark the appropriate choice

Column I	Column II
$(A)$ Boron fibres	$(i)$ Heat resistant glasses
$(B)$ Borax	$(ii)$ Bulletproof vest
$(C)$ Aluminium	$(iii)$ Filler in automobile tyre
$(D)$ Carbon black	$(iv)$ Transport industry

0%
$(A)\rightarrow (i),(B)\rightarrow (ii),(C)\rightarrow (iii),(D)\rightarrow (iv)$
0%
$(A)\rightarrow (ii),(B)\rightarrow (i),(C)\rightarrow (iv),(D)\rightarrow (iii)$
0%
$(A)\rightarrow (ii),(B)\rightarrow (iii),(C)\rightarrow (i),(D)\rightarrow (iv)$
0%
$(A)\rightarrow (iii),(B)\rightarrow (ii),(C)\rightarrow (i),(D)\rightarrow (iv)$

Explanation

(A) Boron fibers

$\rightarrow$ Boron carbide fibers and fiber composites may be used in the manufacture of ballistic materials for protective, durable and lightweight body armor.

$\rightarrow$ (ii) Bulletproof vest

(B) Borax

$\rightarrow$ used to produce various types of high strength and heat-resistant ceramic products.

$\rightarrow$ (i) Heat-resistant glasses

$\rightarrow$ More than a quarter of all aluminium is used in the transport sector.

$\rightarrow$ (iv) Transport industry

(D) Carbon black

$\rightarrow$ Carbon black is mainly used as a reinforcing filler in tires and other rubber products.

$\rightarrow$ (iii) Filler in automobile tyre

Which of the following statements is incorrect for following species:

$BO_3^{3-}, \, CO_3^{2-}, \, NO_3^-$

0%
They are isoelectronic and isostructural.
0%
They have same bond order for E-O bond, where E is the central atom.
0%
The extent of $\pi$ -bonding is in the order of $CO_3^{2-} \, < \, NO_3^- \, < \, BO_3^{3-}$
0%
Bond angles are same in every cases.

Explanation

$\begin{array}{l} \text { (A) All of them have total } 32 \text { electrons are isoelectronic. } \\ \text { (B) Bond order }=\frac{\text { Total no. of bond }}{\text { Total } \sigma \text { -bonds. }} \text { It is } \frac{4}{3} \text { in } \mathrm{CO}_{3}^{2-} \text { and } \mathrm{NO}_{3} \text { - } \\ \text { but } 1 \text { in } \mathrm{BO}_{3}^{3-} \end{array}$

$\text { (c) Extent of } \pi \text { -bonding : } \mathrm{NO}_{3}^{-}>\mathrm{CO}_{3}^{2-}>\mathrm{BO}_{3}^{3-}$

$\begin{array}{l} \text { (2) Bond angles are nearly } 120^{\circ} \text { in each compound. } \\ \text { so, correct answer is (C)} \end{array}$

2002183_992448_ans_a967ac35cb41402ba7035a7de841306a.jpeg

Hydrolysis of

$SiCl_4$ gives ?

0%
$Si(OH)_4$
0%
$SiOCI_2$
0%
$SiO_2$
0%
$H_2SiO_4$

Explanation

Hydrolysis of

$SiCl_4$ gives

$SiO_2$ (silica or silicon dioxide).

$SiCI_ {4(l)}+2H_2O_((l)\rightarrow SiO_{2(s)}+ 4HCI_{aq}$

	List-I (ppt)		List-II (colour)
(P)	${ Ag }_{ 2 }Cr{ O }_{ 4 }$	(1)	Yellow
(Q)	${ Cu }_{ 2 }\left[ Fe{ \left( CN \right) }_{ 6 } \right]$	(2)	White
(R)	$Ba{ SO }_{ 4 }$	(3)	Chocolate brown
(S)	$Pb{ I }_{ 2 }$	(4)	Scarlet Red

0%
$P-4, Q-2, R-1, S-3$
0%
$P-2, Q-3, R-4, S-1$
0%
$P-4, Q-3, R-2, S-1$
0%
$P-3, Q-4, R-1, S-2$

Explanation

$Ag_2CrO_4:$ Silver chromate : It is scarlet red in color.

$Cu_2[Fe{(CN)}_6]:$ It has a octahedral geometry of iron. and it is chocolate brown in color.

$BaSO_4:$ The precipitate of barium sulfate is white in color.

$PbI_2:$ Lead II iodide forms a bright yellow odorless crystalline solid , that becomes orange and red when heated.

Product (A) and (B) in above reaction is :

0%
$O^--\overset{O}{\overset{||}{S}}-O-H, \, O^--\overset{O}{\overset{||}{S}}-O-CH_3$
0%
$O^--\overset{O}{\overset{||}{S}}-O-H, \,\, O^--\underset{O^-}{\underset{|}{\overset{O}{\overset{||}{S_{\bigoplus}}}}}-O-CH_3$
0%
$O^--\overset{O}{\overset{||}{S}}-O-CH_3, \,\, O^\ominus-\underset{CH_3}{\underset{|}{\overset{O}{\overset{||}{S_{\bigoplus}}}}}-H$
0%
d $O^--\underset{H}{\underset{|}{\overset{O}{\overset{||}{S_{\bigoplus}}}}}-O, \,\, O^--\underset{CH_3}{\underset{||}{\overset{O}{\overset{||}{S}}}}-O^-$

Explanation

$O^-$ act as a strong nucleophilic. Upon reaction with methyl iodide,it gives substitution product.And it get proportional in acidic condition.

A is the correct answer.

1077947_991887_ans_d925fc9b4c754f06b9991c6b9930aa6b.PNG

What happens when silicon is heated with methyl chloride in presence of copper as a catalyst at 573K?

0%
Methyl substituted chlorosilanes are formed
0%
Only $Me_4$ Si is formed
0%
Polymerised chains of $(CH_3)_3SiCl$ are formed
0%
Silicones are formed

Explanation

Heating of Silicon with methyl chloride at high temperature in the Presence of copper produces a mixture of mono, di and tri methylchlorosilanes and tetramethylsilane.

$C{H}_{3}Cl + Si \; \xrightarrow[570 \; K]{\text{Cu Powder}} \; C{H}_{3}Si{Cl}_{3} + {(C{H}_{3})}_{2}Si{Cl}_{2} + {(C{H}_{3})}_{3}SiCl + {(C{H}_{3})}_{4}Si$

Methyl substituted chlorosilanes are formed.

Hence, the correct answer is option

$\text{A}$ .

Under hydrolytic condition, the compounds used for preparation of linear polymer and for chain termination, respectively, are:

0%
$CH_3SiCl_3$ and $Si(CH_3)_4$
0%
$(CH_3)_2SiCl_2$ and $(CH_3)_3SiCl$
0%
$(CH_3)_2SiCl_2$ and $CH_3SiCl_3$
0%
$SiCl_4$ and $(CH_3)_3SiCl$

Explanation

Answer:-

${(C{H}_{3})}_{2}Si{Cl}_{2}$ and

${(C{H}_{3})}_{3}SiCl$

Under hydrolytic conditions, the compound used for the preparation of linear polymer and for chain termination are

${(C{H}_{3})}_{2}Si{Cl}_{2}$ and

${(C{H}_{3})}_{3}SiCl$ respectively.

The hydrolysis of dimethyldichlorosilane produces dimethyl silane diol which on condensation polymerization forms a linear polymer.

The hydrolysis of trimethylchlorosilane produces trimethyl silanol which cannot undergo polymerization as it contains only one -OH group. Hence, it is used for chain termination.

Which of the following is the correct statement about silicones?

0%
They are made up of $SiO_4^{4-}$ units.
0%
They are polymers made up of $R_2SiO$ units.
0%
They are water soluble compounds.
0%
They are hydrophillic in nature.

Explanation

Answer:- (B) They are polymers made up of

${({R}_{2}Si{O})}_{n}$ units.

Silicones are polymers which contains

${({R}_{2}Si{O})}_n$ chains, where R is any alkyl or phenyl group, present with each silicon atom. They are hydrophobic in nature.

Which of the following compounds have zero dipole moment ?

0%
$BF_3$
0%
$SnCl_2$
0%
$H_2O$
0%
$NH_3$

Which order of boiling point is correct?

0%
$He < Ne < Ar < O_{2} < N_{2}$
0%
$He < Ne < N_{2} < Ar < O_{2}$
0%
$He < Ne < N_{2} < O_{2} < Ar$
0%
$He < Ar < N_{2} < Ne < O_{2}$

How many derivate structures of

${B}_{3}{N}_{3}{H}_{4}{X}_{2}$ can be derived from the basic structure by the replacement of two hydrogen atoms?

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

Diborane on reaction with oxygen gives a boron compound A.'A' can also be obtained by strongly heating:

0%
${ H }_{ 2 }{ BO }_{ 2 }$
0%
${ BCl }_{ 3 }$
0%
${ NaBH }_{ 4 }$
0%
${ B }_{ 3 }{ N }_{ 3 }{ H }_{ 6 }$

In silicon dioxide

0%
there are double bonds between silicon and oxygen atoms
0%
silicon atom is bonded to two oxygen atoms
0%
each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms
0%
each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms.

Explanation

$SiO_2$ is a a 3D network solid in which each silicon atom is surrounded by 4 oxygen atoms, and each oxygen atoms is surrounded by two silicon atoms.
1584745_1192585_ans_70049469609f4f4b9ce4e4b1823a9ada.png

1584745_1192585_ans_70049469609f4f4b9ce4e4b1823a9ada.png

Super conductors are derived from compounds of:

0%
p-block elements
0%
lanthanides
0%
actinides
0%
transition elements

Which of the following anion is present in chain structure of silicate?

0%
${ [{ Si }_{ 2 }{ O }_{ 5 }^{ 2- }] }_{ n }$
0%
${ [{ Si }{ O }_{ 3 }^{ 2- }] }_{ n }$
0%
${ { Si }{ O }_{ 4 }^{ 4- } }$
0%
${ { { Si }_{ 2 } }{ O }_{ 7 }^{ 6- } }$

Most stable oxidation state of gold is:

0%
+1
0%
+3
0%
+2
0%
0

Heating of ammonium dichromate produces all these compounds except:

0%
NH $_{3}$
0%
H $_{2}$ O
0%
N $_{2}$
0%
Cr $_{2}$ O $_{3}$

Explanation

Solid ammonium carbonate decomposes as it is heated.

Balanced equation:

$(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3+N_2+H_2 O$

Thus, ammonia $(NH_3)$ is not obtained as a product int the given reaction.

When Borax is heated with solid salt metal metaborate is produced. Which of the following reaction actually involve the metal metaborate formation?

0%
Acid base reaction between metal oxide and $B_2O_3$
0%
Oxidation reduction reaction between metal oxide and $B_2O_3$
0%
Substitution reaction between $NaBO_2$ & metal oxide
0%
Complex formation reaction between metal oxide and $B_2O_3$

The oxidation state of the most electronegative element in the products of the reaction

${ BaO }_{ 2 }$ with

${ H }_{ 2 }{ SO }_{ 4 }$ are :

0%
0 and -1
0%
-2 and +1
0%
-1 and -2
0%
-2 and 0

In which of the following compounds

$B-F$ bond length is shortest?

0%
${ BF }_{ 4 }^{ - }$
0%
$BF_3$ $\leftarrow$ $NH_3$
0%
$BF_3$
0%
$BF_3$ $\leftarrow$ $N(CH_3)_{3}$

Which is/are true statements among the following

0%
$PH_5, SCl_6$ and $FCl_3$ do not exist.
0%
$p\pi - d\pi$ bonds are present in $SO_2$ molecule.
0%
$12$ $P-O$ bonds are present in $P_4O_6$ molecule.
0%
Cation of $PCl_5(s)$ has tetrahedral shape.

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 13 - 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 13 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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