MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 9

The shapes and hybridisation of

$BF_3 \,\,and \,\,{ BH }_{4 }^{- }$ respectively are:

0%
$BF_3$ - Trigonal, $sp^2$ hybridisation ;

${ BH }_{4 }^{- }$ - Square planar, $sp^3$ hybridisation
0%
$BF_3$ - Triangular, $sp^3$ d hybridisation ;

${ BH }_{4 }^{- }$ - Hexagonal, $sp^3$ d hybridisation
0%
$BF_3$ - Trigonal, $sp^2$ hybridisation ;

${ BH }_{4 }^{- }$ - Tetrahedral, $sp^3$ hybridisation
0%
$BF_3$ - Tetrahedral, $sp^3$ hybridisation;

${ BH }_{4 }^{- }$ - Tetrahedral, $sp^3$ hybridisation.

Explanation

$BF_3 \; \rightarrow$ As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three

$sp^2$ hybridised orbitals of boron with the sp orbitals of three halogen atoms. Boron is

$sp^2$ hybridised in

$BF_3$ .

${BH_{4}}^{-} \; \rightarrow$ Boron-hydride ion

$({BH{_4}}^{–})$ is formed by

$sp^3$ hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.

Hence, the correct answer is option

$\text{C}$ .

An oxide X in its normal form is almost non-reactive due to very high X - O bond enthalpy. It resists the attack by halogens, hydrogen and most of acids and metals even at elevated temperatures. It is only attacked by HF and NaOH. The oxide X is :

0%
$SiO_2$
0%
$CO_2$
0%
$SnO_2$
0%
$PbO_2$

Explanation

Since X is non-reactive and it cannot be attacked by the halogens, hydrogen, and acids as well as metals. Even at high-temperature reaction is not happening. But exceptionally it reacts with HF and NaOH. Thus, X will be $SiO_{2}$ .

It doesn't react because of the difficulty of breaking up its giant covalent structure. It is insoluble in the water and organic solvents.

The correct option is A.

The bond dissociation energy of

$B - F$ in

$BF_3$ is 646 kJ

$mol^{-1}$ whereas that of

$C - F$ in

$CF_4$ is 515 kJ

$mol{-1}$ . Comment on the above statement.

0%
stronger $\sigma$ bond between $B$ and $F$ in $BF_3$ as compared to that between $C$ and $F$ in $CF_4$
0%
significant $p\pi - p\pi$ interaction between $B$ and $F$ in $BF_3$ whereas there is no possibility of such interaction between $B$ and $F$ in $CF_4$
0%
lower degree of $p\pi -p\pi$ interaction between $B$ and $F$ in $BF_3$ than that between $C$ and $F$ in $CF_4$
0%
smaller size of $B-$ atom as compared to that of $C-$ atom.

Explanation

$BF_3$ is a Lewis acid due to incomplete octet of Boron. Among

$BF_3, BCl_3, BBr_3$ and so on,

$BF_3$ is the weakest Lewis acid due to back bonding of lone pair electron of F to p-orbital of Boron.

B has vacant available p-orbital in

$BF_3$ and hence it involves pπ−pπ back bonding which is not possible in

$CF_4$ as C does not have any vacant orbital.

Hence, option

$B$ is correct.

Topaz, opal, agate, and onyx are precious and beautiful stones found in nature. All these are different forms of:

0%
silica
0%
soda
0%
carbon
0%
lime

Explanation

Topaz is a rare silicate mineral with a chemical composition of $Al_2SiO_4(FOH)_2$ . It usually forms in fractures and cavities of igneous rocks such as pegmatite and rhyolite, late in their cooling history.
Opal is a non-crystalline form of the mineral silica and is formed from amorphous "balls" or "lumps" of silica. So the opal is a very unique gem compared with others, like diamonds or a ruby which is formed crystalline. This is the reason why variation of colours are in opals.
Agate is a banded microcrystalline form of the mineral Quartz, and does not occur in visible crystals. It occurs in nodules, in massive form, as botryoidal, mammilary, and stalactitic formations, as smooth rounded pebbles, as amygdules, and as the linings of geodes.
Onyx is formed of bands of chalcedony in alternating colors. It is cryptocrystalline, consisting of fine intergrowths of the silica minerals quartz and moganite. Its bands are parallel to one another, as opposed to the more chaotic banding that often occurs in agates.
Hence option A is correct answer.

Which of the following is not matched correctly with its use?

0%
piezoelectric material - Quartz
0%
Ion-exchangers - Graphite
0%
Filtration plants - silica
0%
Electrical insulators - Silicones

Explanation

Ion exchange is an exchange of ions between two electrolytes or between an electrolyte solution and a complex. Zeolites are used as ion-exchangers in softening of hard water.

Hence, option

$B$ is not correctly matched.

Borax (sodium tetraborate) has many uses, some of which are mentioned here, except?

0%
For giving a glaze to linen, in laundering
0%
In bleaching of paper pulp
0%
Making heat-resistant
0%
As a flux in welding and soldering

Silica gel is a very useful:

0%
raw material for soap
0%
drying agent
0%
wetting agent
0%
solvent

Which of the following bonds is shown in silicones?

0%
$O-\!\!\!\!(\underset{|}{ \overset{|}Si}-O-\!\!\!\!)_n$
0%
0%
$O-\underset{O}{\underset{|}{\overset{O}{\overset{|}{Si}}}}-O$
0%

Explanation

Answer:-

Silicones are polymers, contains

${({R}_{2}Si{O})}_{n}$ units as shown in fig.

Thus, (A) is correct answer.

The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in

$[B(OH_4)]^-$ and the geometry of the complex are respectively:

0%
$sp^3$ , tetrahedral
0%
$sp^3$ , square planar
0%
$sp^3d^2$ , octahedral
0%
$dsp^2$ , square planar

Explanation

Boron has electronic configuration:

$1s^22s^22p_x^1 2p^0_y 2p^0_z$

In the excited state,2s orbital electrons are impaired and one electron is shifted to a p-orbital. Now, hybridisation occurs between one s-and three p-orbitals to give

$sp^3$ hybridisation and tetrahedral geometry.

Spaceships have to be air-tight because in space there is no air, only vaccum. Which of the following compounds has been used for air-tight closing of the door (hatch) in lunar modules?

0%
Chlorides
0%
Silicones
0%
Bakelite
0%
Hydrocarbons

Clay is used for making bricks and flower pots. If the clay article is not dried before 'firing' in a kiln, there will be evaporation of water from the surface and the :

0%
kiln fire will be put out by steam
0%
article will becomes soft
0%
article will become hard
0%
article will develop cracks

Elements of group 14 exhibit oxidation sate of:

0%
+4 only
0%
+2 and +4 only
0%
+1 and +3 only
0%
+2 only

Which is not correct?

0%
$GeO$ is acidic
0%
$Ge{ Cl }_{ 2 }$ is more stable than $Ge{ Cl }_{ 4 }$
0%
$Ge{ Cl }_{ 2 }$ is acidic
0%
$Ge{ Cl }_{ 4 }$ in $HCl$ forms ${ \left[ Ge{ Cl }_{ 6 } \right] }^{ 2- }$ ion

Explanation

The phenomena of the shielding effect of electrons explains this. Since there are fewer electrons in $Ge ^{4+}$ , the nucleus has a stronger pull on those electrons than in $Ge ^{+2}$ .
Both $Ge ^{4+}$ and $Ge ^{2+}$ $Ge ^{2+}$
Hence option B is not correct.

Which of the following statements is correct with respect to the property of elements with increase in atomic number in the carbon family (group 14)?

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

Assertion : The heavier p-block elements do not form strong

$\pi$ bonds.
Reason : The heavier elements of p-block form

$d\pi-d\pi$ bonds.

0%
Both A and R are true and R is the correct explanation of A.
0%
Both A and R are true but R is not the correct explanation of A.
0%
A is true but R is false.
0%
A is false but R is true.

Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding:

0%
$MeSiCl_3$
0%
$Me_2SiCl_2$
0%
$Me_3SiCl$
0%
$Me_4Si$

Explanation

${Me}_{3}SiCl$

The chain length of Silicone polymer can be controlled by adding

${(C{H}_{3})}_{3}SiCl$ , which blocks the ends as shown below:

${(C{H}_{3})}_{3}SiCl \xrightarrow[-HCl]{{H}_{2}O}{(C{H}_{3})}_{3}SiOH$

Element

$M+{ N }_{ 2 }\xrightarrow [ ]{ \Delta } \xrightarrow [ ]{ { H }_{ 2 }O } { NH }_{ 3 }$
Element

$M$ belonging to group

$13$ can be ?

0%
$B$ or $Al$
0%
$Ga$ or $Al$
0%
$B$ or $Ga$
0%
$In$ or $TI$

Explanation

$2B + N_2 \longrightarrow 2BN$

$2Al +N_2 + \longrightarrow 2AlN$

These 2 nitrites on heating with presence of

$H_2O$ releases

$NH_3$ . So, the group

$13$ elements are

$B$ and

$Al$

Ionisation enthalpy

$(\Delta _iH_1\ kJ \ mol^{-1})$ 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

On moving down the group from

$B$ to

$Tl$ , a regular decreasing trend in the I.E. is not observed.

Element	$B$	$Al$	$Ga$	$In$	$Tl$
I.E. (kJ/mol)	801	577	579	558	589

In Ga, there are ten d-electrons in the penultimate shell which screen the nuclear charge less effectively and thus, outer electron is held firmly. As a result, the ionisation energy of both Al and Ga is nearly the same. The increase in ionisation energy from In to Tl is due to poor screening effect of 14f electrons present in the inner shell.

Thus, the correct order is,

$B>Al<Ga>In<Tl$

The exhibition of the highest co-ordination number depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as a central atom in

$MF_6^{3-}$ ?

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

Explanation

The element

$M$ in the complex ion

${M{F}_{6}}^{3-}$ has a coordination number of six. Since boron has only s and p-orbitals and no d-orbitals, therefore, boron is restricted to a maximum covalence of four (using 2s and three 2p orbitals). Thus,

$B$ cannot form a complex of the type

${M{F}_{6}}^{3-}$ .

Which of the following is correct?

0%
$\%$ abundance of $B^{10}=$ % abundance of $B^{11}$
0%
$\%$ abundance of $B^{10}>$ % abundance of $B^{11}$
0%
$\%$ abundance of $B^{10}<$ % abundance of $B^{11}$
0%
none of the above

Explanation

Let the percentage existence of

$B10$ isotope be

$x$ . Therefore percentage existence of the other isotope is

$(100-x)$

mass of

$B-10$ isotope is

$10amu$

mass of

$B-11$ isotope is

$11 amu$

Therefore .

$\dfrac{10x}{100+11(100-x)/100}=10.2$

$10x+11(100-x)=1020$

$1100-x=1020$

$x=80$

Therefore the percentage of Boron-10 isotope is

$80$ %

and Boron-11 isotope is

$(100-x)=20$ %

Hence option B is the correct answer

Which among

${ CH }_{ 4 }, { SiH }_{ 4 }, { GeH }_{ 4 }, { SnH }_{ 4 }$ is the most volatile?

0%
${ CH }_{ 4 }$
0%
${ SiH }_{ 4 }$
0%
${ GeH }_{ 4 }$
0%
${ SnH }_{ 4 }$

Explanation

$CH_4$ is most volatile among all because methane has a very low molecular weight of 16. So as the molecular weight increases, the vander waals forces of attraction between the molecules also increases.

The element with atomic number

$33$ belongs to:

0%
group $13$
0%
group $14$
0%
group $15$
0%
group $16$

Explanation

Electronic configuration of atomic number 33 will be

${1s}^22s^22p^63s^23p^64s^2{3d}^{10}4p^3$

So, its configuration is similar to 15th group and it has 5 valence electrons .

Hence it belongs to 15th group of the periodic table.

Borax, disodium tetraborate is an a important boron compound, a mineral, and a salt of boric acid. The number of terminal

$B-----OH$ presents in borax is ?

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

The figure shows there are 4

$B-OH$ bonds present in Borax.
1074149_1048676_ans_c1d0e394e315403ab49bd973d32a10ee.gif

Which of the following statements is correct?

0%
$BCl_3$ and $AlCl_3$ are both Lewis acids and $BCl_3$ is stronger acid than $AlCl_3$
0%
$BCl_3$ and $AlCl_3$ both Lewis acids and $AlCl_3$ is stronger acid than $BCl_3$
0%
$BCl_3$ and $AlCl_3$ are both equally strong Lewis acids
0%
Both $BCl_3$ and $AlCl_3$ are not Lewis acids

Explanation

Both $BCl_3$ and $AlCl_3$ are Lewis acids.
$BCl_3$ is a stronger Lewis acid as the bond formed with the base will involve 2p orbital overlap which is stronger than the 3p orbital overlap in the case of $AlCl_3$ . Hence, the bond formed between $BCl_3$ and the Lewis base will be stronger.

Stability order of

$+3$ and

$+1$ states of boron family elements is:

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

Explanation

Due to Inert pair effect, in the elements which are at a lower position in a group, there are d and f orbitals which have a poor shielding effect. So outermost s electrons feel more attractions with the nucleus and become more difficult to lose. So as 2 electrons are not getting lost, oxidation state decreases from +3 to +1.
Hence, $Tl^{+}>In^{+}>Ga^{+}$ and $Ga^{3+}>In^{3+}>Tl^{3+}$

Which of the following statements is correct with respect to the property of elements with increase in atomic number in the carbon family (group 14)?

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

Which of the following is correct order of boiling point?

0%
${CCl}_{4}\ >{SiCl}_{4}$
0%
${SiCl}_{4}\ >{CCl}_{4}$
0%
${SiCl}_{4}={CCl}_{4}$
0%
None of these

Explanation

$CCl_4$ &

$SiCl_4$ ,

$C$ has small size than

$Si$ thus in

$CCl_4$ molecules cover large contact area which make the intermolecular forces more stronger between them.

Thus, boiling point of

$CCl_4$ is greater than

$SiCl_4$ .

Moissan boron is_____

0%
Amorphous boron of ultra purity
0%
Crystalline boron of ultra purity
0%
Amorphous boron of low purity
0%
Crystalline boron of low purity

Four hydrogenation reactions maximum catalytic activity is shown by:

0%
Group $5-11$ elements
0%
Group $5-9$ elements
0%
Group $7-9$ elements
0%
Group $9-11$ elements

Which of the following statement is incorrect?

0%
The hydroxide of aluminium is more acidic than that of boron
0%
The hydroxide of boron is basic whereas that of aluminium is amphoteric
0%
The hydroxide of boron is acidic whereas that of aluminium is amphoteric
0%
The hydroxide of both boron and aluminium are amphoteric

What the covalencey of phosphorus atoms in excited state?

0%
$3$
0%
$4$
0%
$5$
0%
$1$

Explanation

Phosphorus can have expanded octet, because it can shift it's lone pair electrons (

$3s$ orbital electrons) to empty

$3d$ obital during excited state and thus can form 5 bonds. So now there is 5 spaces along the singly filled orbitals to keep electrons or we may say that there can be 5 bonds now.

Borax when dissolved in water exhibts

0%
Alkaline nature
0%
Acidic nature
0%
Neutral nature
0%
Amphoteric nature

Explanation

when borox is dissolved in water it forms sodium borate and boric acid which forms a strong base hence it is alkaline in nature .

Borax dissociate into

$B_4{O_7}^{2-}$ ions

After dissociation

$B_4{O_7}^{2-} \rightarrow {2OH}^{-} +4H_3BO_3$

formation of

${OH}^-$ shows alkaline nature .

Which of the following is correct w.r.t. to the radii related to lead?

0%
$Pb^{4+} > Pb^{2+} > Pb$
0%
$Pb > Pb^{2+} > Pb^{4+}$
0%
$Pb^{2+} > Pb > Pb^{4+}$
0%
Depends upon conditions

The relative stability of +1 oxidation state of group 13 elements follows the order:

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

Explanation

Due to the inert pair effect as we move down the group in

$13^{th}$ group lower oxidation state become more stable.

So, the correct order is:

$Al < Ga < In < Tl$ .

$CO_2$ is a gas, while

$SiO_2$ is a solid but both are:

0%
covalent containing $\pi$ -bond
0%
molecules having $p\pi$ - $d\pi$ bonding
0%
acidic
0%
discrete molecules

Explanation

$CO_2$ consists of individual molecules with one central carbon atom double bonded to two oxygen atoms. Silicon does not form double bonds with oxygen. $CO_2$ is gas but $SiO_2$ is solid because the $SiO_2$ is in giant molecular structure but the $CO_2$ is simple molecular structure.because of this reason $CO_2$ is gas but $SiO_2$ is solid.
$CO_2$ is an acidic oxide can be explained by Lux-Flood concept & extended Lewis theory. $CO_2$ with $H_2O$ produces $H_2CO_3$ i.e. carbonic acid.
Silicon dioxide has no basic properties - it doesn't contain oxide ions and it doesn't react with acids. Instead, it is very weakly acidic, reacting with strong bases. Silicon dioxide doesn't react with water, because of the difficulty of breaking up the giant covalent structure.

In which of the following molecule back bonding is present?

0%
$BF_3$
0%
$H_2 O$
0%
$NH_3$
0%
$HF$

Boron exists in two isotopic form

$B^{10}$ and

$B^{11}$ . If the average atomic weight of

$B$ is

$10.20\ u$ , select the correct statement.

0%
$\%$ abundance of $B^{10}=\%$ abundance of $B^{11}$
0%
$\%$ abundance of $B^{10} > \%$ abundance of $B^{11}$
0%
$\%$ abundance of $B^{10} < \%$ abundance of $B^{11}$
0%
$\%$ abundance of $B^{10}=\dfrac{\%\ abundance\ of\ B^{11}}{2}$

Explanation

Let the percentage existence of

$B^{10}$ isotope be

$x$ . Therefore percentage existence of the

$B^{11}$ isotope is

$(100-x)$

mass of

$B^{10}$ isotope is

$10amu$ .

mass of

$B^{11}$ isotope is

$11 amu$ .

Therefore,

$\dfrac{10+11(100-x)}{100} = 10.2$

$10+11(100-x) = 1020$

$$x=80

Therefore the percentage of

$Boron-10$ isotope is

$80$ %

and

$Boron-11$ isotope is

$(100-x)=20$ %

Hence, option B is the correct answer.

The electropositive character is the maximum for which of the following elements in group 13?

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

Boric acid is used in carom boads for smooth gliding of pawns because

0%
$H_3BO_3$ molecules are loosely chemically bonded and hence soft
0%
its low density makes it fluffy
0%
it can be powdered to a very small grain size
0%
it is chemically inert with the plywood
0%
H-bonding in $H_3BO_3$ GIVES IT A LAYERED STRUCTURE.

Explanation

Boric acid is a solid lubricant and is used to reduce friction. It is used as a lubricant to make motion smooter. In carrom boards, boric acid is used for smooth gliding of porous because

$H_3BO_3$ molecules are loosely chemically bounded and hence soft and thus reduce friction.

Teflon polymer is formed by the polymerisation of:

0%
$CH_2CN-CN$
0%
$F_2C=CF_2$
0%
$Cl_2C=CH_2$
0%
$H_2C=CHCl$

Which of the following compounds of elements in group IV expected to be most ionic?

0%
$PbCl_{2}$
0%
$PbCl_{4}$
0%
$CCl_{4}$
0%
$SiCl_{4}$

Explanation

According to fajans rule greater the charge on cation,greater would be the polarising power and greater covalent character.

$PbCl_2\rightarrow Pb^{2+}$

$PbCl_4\rightarrow Pb^{+4}$

$PbCl_2$ is ionic and

$PbCl_4$ in covalent.

Similarly

$CCl_4,SiCl_4$ are also covalent.

Select the incorrect statement among the following:

0%
${BF}_{3}$ and $B{Cl}_{3}$ both have equal bond angles
0%
$Xe{O}_{2}{F}_{2}$ has two $p\pi-p\pi$ bonds
0%
${SF}_{6}$ has all equal bond lengths
0%
${CH}_{2}{Cl}_{2}$ is less polar than ${CH}_{3}Cl$

Explanation

1) Lesser the electronegativity of the attached atoms more is the bond angle.
2) Size of attached atoms (eg

$Br, \; Cl$ or

$F$ in this case)
The bond order will be-

$BBr_3>BCl_3>BF_3$

The element with lowest electronegativity in 13th group is:

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

Explanation

Down the

$13^{th}$ group, electronegativity first decreases from B to Al and then increases marginally. This is because of the discrepancies in the atomic size of the elements.

Hence option B is the correct answer.

Aluminium does not react with:

0%
NaOH
0%
HCI
0%
$N_2$
0%
$HNO_3$

Explanation

Before reaction with $HNO_3$ it forms an oxide layer of $Al_2O_3$ .

oxide layer does not allow reaction of $HNO_3$ with Aluminium.

Because of this reason $Al$ vessel is used to store $HNO_3. HNO_3$ does'nt react with $Al$ Or any other metal (except manganese and magnesium) because $HNO_3$ is a strong oxidising agent.

$H_{2}O_{2}$ is a:

0%
monobasic acid
0%
dibasic acid
0%
neutral compound
0%
weak alkali

Explanation

$1.$

$H_2O_2$ is a dibasic acid.

$2.$ A dibasic acid is a type of acid that has two oxygen ions to donate to a base in an acid base reaction.
Therefore a dibasic molecule has two replaceable hydrogen atoms.

$3.$ An acid is said to be monobasic, dibasic or tribasic according to the number of replaceable hydrogen
atoms,

$4.$ Since,

$H_2O_2$ has two replaceable hydrogen atoms, That's why it is a dibasic acid.

Compound A is produced by absorbing dinitrogen trioxide in

$Na_{2}CO_{3}$ solution. Compound A is:

0%
$Na_{2}O$
0%
$NaNO_{2}$
0%
$N_{2}O$
0%
$Na_{2}N_2O_{2}$

Which one of the following is the correct statement?

0%
Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase
0%
${B_2}{H_6}.2N{H_3}$ is known as 'inorganic' benzene
0%
Boric acid s a protoic acid
0%
Beryllium exhibits coordinatin number of six

Which of the following metal is highest electropositive ( metallic) in nature?

0%
Be
0%
Rb
0%
Mn
0%
Mg

Explanation

Generally all the alkali and alkaline-earth metal are electropositive in nature. Whereas the p-block elements like O, Cl, C, N are electronegative elements.

The most electropositive element among the given elements is

$Rb$

thus, Option B is correct.

Note: alkali metals are more electropositive than alkaline-earth metal

Which silicon compound is used in machinery (with moving parts) in a manner similar to the use of an allotrope of carbon (crystalline) which possesses properties of both a covalent network solid and a molecular solid?

0%
Silica gel
0%
Zeolite
0%
Silicone
0%
Asbestos

Explanation

Little packets of silica gel are found in all sorts of products because silica gel is a desiccant -- it adsorbs and holds water vapor. In leather products and foods like pepperoni, the lack of moisture can limit the growth of mold and reduce spoilage. In electronics it prevents condensation, which might damage the electronics. If a bottle of vitamins contained any moisture vapor and were cooled rapidly, the condensing moisture would ruin the pills. You will find little silica gel packets in anything that would be affected by excess moisture or condensation.

Silica gel is nearly harmless, which is why you find it in food products. Silica, or ( $SiO_2$ ), is the same material found in quartz. The gel form contains millions of tiny pores that can adsorb and hold moisture. Silica gel is essentially porous sand.

Hybridisation in

$Xe{O_3}$ molecule ?

0%
$s{p}$
0%
$s{p^3}$
0%
$s{p^2}$
0%
none

Explanation

$XeO_3$ has structural formula.
Number of lone pair =

$1$
Hybridisation in

$sp^3$
No. of bonds = 3
Lonepair+bondpair= 3+1=4

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.