are synthetic polymer containing repeating $$(R_2SiO)$$ units.

are natural occurring repeated $$(SiO_4)$$ units.

silicates with common $$SiO_4$$ unit.

ketones with silyl group $$(SiH_3)$$ similar to alkyl $$(SiH_3)_2CO$$

MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 12

Which of the following is incorrect regarding the structural features of borax?

0%
It has two tetrahedral and two triangular units
0%
All the boron atoms are not in same hybrid state
0%
Each boron atom of a pair is tetravalent
0%
Only one boron atom has a lone pair of electrons

Consider the following reactions:

$Cr_{2}O_{3}+2Al\rightarrow Al_{2}O_{3}+2Cr+heat$
ii)

$Al_{2}O_{3}+2Cr\rightarrow Cr_{2}O_{3}+2Al+heat$
iii)

$2Al+6NaOH\rightarrow 2Na_{3}AlO_{3}+3H_{2}$

Out of these, the possible reactions are:

0%
i and iii
0%
ii and iii
0%
i and ii
0%
i , ii and iii

Order of stability of the ions of Ge, Sn and Pb is:

0%
$Ge^{2+} < Sn^{2+} < Pb^{2+}$
0%
$Ge^{4+} < Sn^{4+} < Pb^{4+}$
0%
$Sn^{4+} < Sn^{2+}$
0%
$Pb^{2+} < Pb^{4+}$

Explanation

On moving down group IV A, the stability of the +2 oxidation state increases due to the inert pair effect which becomes more pronounced in heavier elements.

Hence, the correct order of the stability of oxidation states is

$Ge^{2+} < Sn^{2+} < Pb^{2+}$ .

So, the correct option is ' A'.

List -I List -2
A.

$Al_{2}O_{3}$ Dimer
B.

$AlCl_{3}$ Non-metal
C.

$B$ Acidic
D.

$B_{2}O_{3}$ Amphoteric

The correct match is:

0%
A-4, B-3, C-2, D-1
0%
A-1, B-4, C-3, D-2
0%
A-2, B-3, C-4, D-1
0%
A-4, B-1, C-2, D-3

Explanation

$Al_{2}O_{3}$ is amphoteric and

$AlCl_{3}$ forms a dimer.

Boron is a non-metal while

$B_2O_3$ is acidic in nature.

The correct option is 'D'

Select the correct order with respect to acidic nature of the given compounds.

0%
$Al_{2}O_{3}< In_{2}O_{3}< Tl_{2}O< B_{2}O_{3}$
0%
$Tl_{2}O< \, In_{2}O_{3}< Al_{2}O_{3}< B_{2}O_{3}$
0%
$Al_{2}O_{3}< B_{2}O_{3}< \: In_{2}O_{3}< Tl_{2}O$
0%
$Tl_{2}O< Al_{2}O_{3}< B_{2}O_{3}< \: In_{2}O_{3}$

Explanation

As we move from top to bottom in a group, metallic character increases and so the acidic nature of the oxides decreases.

Therefore,

$B_2O_3$ is acidic in nature,

$Al_2O_3$ and

$Ga_2O_3$ are amphoteric in nature, whereas

$In_2O_3$ and

$Tl_2O$ are basic in nature.

Hence, the correct order with respect to acidic nature of compounds is as follows:

$Tl_2O<In_2O_3<Al_2O_3<B_2O_3$

Hence, option

$B$ is correct.

Which of the following statement is/are correct?

0%
$B_{2}H_{6}$ is not an electron-deficient compound
0%
Boron has an extremely high melting point because of the strong binding forces in the covalent polymer
0%
$BF_{3}$ and $BrF_{3}$ molecules have different shapes
0%
Ortho boric acid contains pyramidal $BO_{3}^{-3}$ units

Explanation

$B_2H_6$ , diborane is an electron deficient compound.

B) Boron has a very strong crystalline lattice. So, it has an extremely high melting point.

$BrF_{3}$ has a bent structure whereas

$BF_{3}$ has a trigonal planar structure.

$BF_3$ has coplanar geometry due to

$sp^2$ hybridisation on boron atom whereas

$BrF_3$ has

$sp^3d$ hybridisation and possesses two lone pairs of electron on Br atom which leads to T shaped geometry.

D) Ortho boric acid contains triangular

$BO_3^{-3}$ units.

Option B and C are correct.

In representative elements the ionization energy values generally decrease smoothly down the group. But in boron group the values show an erratic trend. This is mainly due to :

0%
increase in metallic nature down the group
0%
inert pair effect in the heavier elements of the group
0%
d-block contraction and lanthanide contraction affecting the values of later elements due to poor shielding by d and f electrons
0%
the different type of packing in their crystalline structure

Which of the following have a three dimensional network structure?

0%
$\displaystyle\:SiO_{2}$
0%
$\displaystyle\:(BN)_{5}$
0%
$\displaystyle\:P_{4}(white)$
0%
$\displaystyle\:CCl_{4}$

Explanation

$P_4$ has tetragonal structure and other

Consider the following compound

If hybridization of 'C' is

$sp$ (

$s + px$ ) and nodal plane of

$\pi$ -bond present between C and S is 'XY' plane then which of the following statement(s) is/are CORRECT regarding given information for above compound.

0%
$\pi$ - bond between N and C atom is formed by sideways overlapping of $p_y$ orbitals
0%
$(2p_{\pi}$ – $3d_{\pi}$ ) back bond between N and Si atom can be formed by ( $d_{xz} – p_z$ ) orbital overlapping
0%

$\sigma$ - bond between N and Si atom is formed by $sp^3$ hybrid orbital of Si and $sp_{x}$ hybrid orbital of N atom.
0%
Six atoms of given compound are lying in same plane.

Which one of the following statements is incorrect about the structural properties of

$BF_{3}$ ?

0%
The unsual shortness of $B-F$ bonds is due to $p\pi - p\pi$ interaction between $B$ and $F$ atoms
0%
All the three $B-F$ bond lengths are equal
0%
Double bond is delocalized giving three resonance structures.
0%
Due to back bonding, $BF_{3}$ and $BCl_{3}$ are not isostructural molecules

Explanation

$B{ F }_{ 3 }$ has a trigonal planar structure, all three

$B-F$ bonds lie in the same plane. Thus, the p orbitals of Boron and Fluorine become parallel.
p-orbital of Boron is empty while fluorine has a lone pair. Fluorine donates its lone pair and this is called back bonding, or

$p\pi -p\pi$ bond. Resonating structures can be obtained.
Since the

$B{ F }_{ 3 }$ molecule is highly symmetrical, it has no dipole moment. All bonds are of equal length, thus option (B) is correct.
In Boron trihalides, the length of

$B-F$ bond would be shorter than expected for a single bond, due to

$p\pi -p\pi$ bonding.

$B{ Cl }_{ 3 }$ is also a trigonal planar molecule. Back bonding does not change the structure of the molecule. It simply decreases bond length

$B-X$ and decreases Lewis acidity of

$B{ X }_{ 3 }$ .
Thus, option

$(D)$ is incorrect.

The aqueous solution of 'A' is alkaline due to

0%
the presence of $Ca^{2+}$ ions
0%
the presence of $H_{3}BO_{3}$
0%
hydrolysis of $B_{4}O_{7}^{2-}$
0%
hydrolysis of $CO_{3}^{2-}$

Explanation

A is borax or sodium tetraborate

$(Na_2B_4O_7.10H_2O)$ .

One mole of A contains 10 moles of water of crystallization.

It is a white crystalline solid and is used in water softening. Its aqueous solution is alkaline due to the formation of sodium hydroxide upon hydrolysis of

$Na_2B_4O_7$ . The reaction is as follows:

$Na_2B_4O_7+7H_2O \rightleftharpoons 2NaOH + 4 H_3BO_3$

Select the correct statements.

0%
Oxides of boron $(B_{2}O_3)$ and silicon $(SiO_{2})$ are acidic in nature.
0%
Oxides of aluminium $(Al_{2}O_{3})$ and gallium $(Ga_{2}O_{3})$ are amphoteric in nature.
0%
Oxides of indium $(In_{2}O_{3})$ and thallium $(Tl_{2}O_{3})$ are basic in nature.
0%
Oxides of germanium $(GeO_{2})$ and tin $(SnO_{2})$ are acidic in nature.

Explanation

A, B and C are correct statements.

D :

$GeO_2$ is acidic while

$SnO_{2}$ is amphoteric in nature.

$SnO_2$ reacts with

$HCl$ to form

$SnCl_2$ and reacts with

$NaOH$ to form

$Na_2SnO_3$ .

Alkyl substituted chlorosilane,

$R_{n}SiCl_{(4-n)}$ , on hydrolysis and then condensation polymerisation yields silicones. What is the value of n in the alkyl substituted chlorosilane used for the preparation of cross-linked silicone?

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

Explanation

$RSiCl_{3}$ on hydrolysis gives a cross linked silicone.

How many oxygen atoms are shared per

$SiO_{4}$ tetrahedral unit in mineral wollastonite

$[Ca_{3}(Si_{3}O_{9})]$ ?

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

Explanation

It is cyclic silicate which has anion

$Si_{3}O_{9}^{6-}$ .
Two atoms of oxygen are shared per

$SiO_4$ tetrahedral unit.

Silica reacts with

$Mg$ to form a magnesium compound

$A$ .

$A$ reacts with dilute

$HCl$ and forms

$B$ . The compound

$B$ is:

0%
$MgSiO$
0%
$MgO$
0%
$Mg{ Cl }_{ 2 }$
0%
$Si{ Cl }_{ 4 }$

Explanation

It is given that silica reacts with

$Mg$ to form a magnesium compound

$A$ .

$A$ reacts with dilute

$HCl$ to form

$B$ .

The reactions are as follows:

$2Mg + SiO_2 \rightarrow 2MgO + Si$

$MgO + 2HCl \rightarrow MgCl_2 + H_2O$
Hence, the compound

$B$ is

$MgCl_2$ .

The chain length of silicone polymer can be controlled by adding:

0%
$MeSiCl_{3}$
0%
$Me_{2}SiCl_{2}$
0%
$Me_{3}SiCl$
0%
$Me_{4}Si$

Explanation

The chain length of silicone polymer can be controlled by adding

$Me_{3}SiCl$ .

$R \cdot + Me_{3}SiCl \rightarrow Me_{3}Si-R+ Cl \cdot$

$2Cl \cdot \rightarrow Cl_2$

Which of the following oxides is acidic in nature?

0%
$B_{2}O_{3}$
0%
$Al_{2}O_{3}$
0%
$Ga_{2}O_{3}$
0%
$In_{2}O_{3}$

Explanation

$B_{2}O_{3}$ is acidic and reacts with basic (metallic oxides forming metal borates.

Aluminum and galium oxides are amphoteric and those of indium and thallium are basic in their properties.

Hence, option A is correct.

Ammonia gas is not prepared by :

0%
reduction of sodium nitrite or sodium nitrate by the reaction of zinc dust and sodium hydroxide.
0%
hydrolysis of calcium cyanamide.
0%
heating ammoniumchloridewith slaked lime.
0%
heating of $(NH_{4})_{2}Cr_{2}O_{7}$

Explanation

(1)

$NO_{2}^{-}+7[H]^{-}\xrightarrow[]{Zn/NaOH}NH_{3}+2H_{2}O; NO_{3}^{-}+9[H]^{-}\xrightarrow[]{Zn/NaOH}NH_{3}+3H_{2}O$
(2)

$CaCN_{2} + 3H_{2}O \rightarrow CaCO_{3} + 2NH_{3}$
(3)

$NH_{4}Cl + Ca(OH)_{2} \rightarrow 2NH_{3} + CaCl_{2} + 2H_{2}O$
(4)

$(NH_{4})_{2}Cr_{2}O_{7} \rightarrow N_{2} + Cr_{2}O_{3} + 4H_{2}O$

In silicones, the maximum number of silicon atoms directly linked to an oxygen atom is :

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

Which among the following substances is not a component of the mixture used for glazing pottery?

0%
Feldspar
0%
Silica
0%
Borax
0%
Alumina

Which of the following does not react with cold water?

0%
$SiC$
0%
$AIN$
0%
$CaC_{2}$
0%
$Al_{4}C_{3}$

Explanation

$SiC$ is a covalent compound, it requires some amount of energy to react with water. So, it doesn't react with cold water.

Option A is correct.

$Me_3SiCl$ is mixed with

$Me_2SiCl_2$ to get a polymers called silicones of the type:

0%
0%
0%
Both A and B
0%
None of these

Which of following stability order is /are correct due to inert pair effect?

0%
$\displaystyle\:Hg^o >Hg^{2+}$
0%
$\displaystyle\:Bi^{3+} < Bi^{5}$
0%
$\displaystyle\:Pb^{2+} > Pb^{4+}$
0%
$\displaystyle\:Fe^{2+} < Fe^{3+}$

Explanation

The inertness of s subshell electrons towards the bond formation is called inert pair effect or it can be said as the inactiveness of electrons present in outermost shell (i.e. $ns^2$ ) to get unpaired and involve in bond formation is called inert pair effect.

for example:

1) In 13th group, thallium can exhibit $+1$ and $+3$ oxidation states but it is stable in $+1$ oxidation state only due to inert pair effect

2) In 14th group , lead shows both $+2$ and $+4$ oxidation states but it is stable in $+2$ oxidation state due to inert pair effect.

3)In group 15 elements, the stability of $+3$ oxidation state increases down the group and that of $+5$ oxidation state decreases down the group, due to inert pair effect."

Among the following, the element which show inert-pair effect are:

0%
$\displaystyle\:Bi$
0%
$\displaystyle\:Sn$
0%
$\displaystyle\:Pb$
0%
$\displaystyle\:C$

Explanation

The inertness of s subshell electrons towards the bond formation is called inert pair effect or it can be said as the inactiveness of electrons present in the outermost shell (i.e. $ns^2$ ) to get unpaired and involve in bond formation is called inert pair effect.

for example:

1) In 13th group, thallium can exhibit $+1$ and $+3$ oxidation states but it is stable in $+1$ oxidation state only due to inert pair effect

2) In 14th group, lead shows both $+2$ and $+4$ oxidation states but it is stable in $+2$ oxidation state due to inert pair effect.

3)In group 15 elements, ( $Bi$ ) the stability of $+3$ oxidation state increases down the group and that of $+5$ oxidation state decreases down the group, due to inert pair effect.

Hence, the correct options are $A$ , $B$ and $C$

The correct B --- F bond length follows the sequence :

0%
${ BF }_{ 3 }\quad <\quad { BF }_{ 2 }{ OH }\quad <\quad { BF }_{ 2 }{ NH }_{ 2 }\quad <\quad { B{ F }_{ 4 }^{ - } }$
0%
${ BF }_{ 2 }{ NH }_{ 2 }\quad <\quad { BF }_{ 2 }{ OH }\quad <\quad { BF }_{ 3 }\quad <{ BF }_{ 4 }^{ - }$
0%
${ { BF }_{ 3 } }\quad <\quad { BF }_{ 4 }^{ - }\quad <\quad { BF }_{ 2 }{ OH }\quad <\quad { BF }_{ 2 }{ NH }_{ 2 }$
0%
${ BF }_{ 3 }\quad <\quad { BF }_{ 2 }{ NH }_{ 2 }\quad <{ BF }_{ 2 }{ OH }\quad <\quad { BF }_{ 4 }^{ - }$

Microcosmic salt and borax are used in the identification of cations by dry tests. They are respectively :

0%
$Na{ BO }_{ 2 }$ and $Na{ PO }_{ 3 }$
0%
$Na{ NH }_{ 4 }{ HPO }_{ 4 }.{ 4H }_{ 2 }O$ and ${ Na }_{ 2 }{ B }_{ 4 }{ O }_{ 7 }.10{ H }_{ 2 }O$
0%
$Na{ PO }_{ 3 }$ and $Na{ BO }_{ 2 }$
0%
${ Na }_{ 2 }{ B }_{ 4 }{ O }_{ 7 }$ and $Na{ NH }_{ 4 }{ HPO }_{ 4 }$

Explanation

Microcosmic salt bead test is useful to identify cations like in borax bead test. When the microcosmic salt is heated on platinum wire loop to get sodium meta phosphate,

$Na(NH_4)HPO_4.4H_2O \rightarrow Na(NH_$)HPO_4 + 4H_2O$

$Na(NH_4)HPO_4 \rightarrow NaPO_3 +NH_3 +H_2O$
(Glass mass)
Now, we have

$NaPO_3$ which reacts with metallic oxides to give coloured orthophosphates.
Thus, option (B) is the correct answer.

Read the following write-ups and answer the questions at the end of it.

Silicons are synthetic polymers containing repeated

$R_{2}SiO$ units. Since, the empirical formula is that of a ketone

$(R_{2}CO)$ , the name silicone has been given to these materials. Silicones can be made into oils, rubbery elastomers and resins. They find a variety of applications because of their chemical inertness, water repelling nature, heat- resistance and good electrical insulating property. Commercial silicon polymers are usually methyl derivatives and to a lesser extent phenyl derivatives and are synthesised by the hydrolysis of

$R_{2}SiCl_{2}$ [R=methyl (Me) or phenyl (Ph)]

If we mix

$Me_{3}SiCl$ with

$Me_{2}SiCl_{2}$ , we get silicones of the type:

0%
0%
0%
both of the above
0%
none of the above

The silicate anion in the mineral kinoite is a chain of three

$SiO_{4}$ tetrahedra that share corners with adjacent tetrahedra. The mineral also contains

$Ca^{2+}$ ions,

$Cu^{2+}$ ions and water molecules in a 1:1:1 ratio. The mineral is represented as :

0%
$CaCuSi_{3}O_{10}H_{2}O$
0%
$Ca_{2}Cu_{2}Si_{3}O_{10}H_{2}O$
0%
$Ca_{2}Cu_{2}Si_{3}O_{10}2H_{2}O$
0%
none of these

Explanation

The mineral is represented as

$Ca_{2}Cu_{2}Si_{3}O_{10}2H_{2}O$ .
The charge on

$Si_3O_{10}$ unit is

$-8$ . It is balanced by two calcium ions each with

$+2$ charge and two copper ions, each with

$+2$ charge.
Also, the ratio of

$Ca^{2+}$ ions,

$Cu^{2+}$ ions and water molecules is

$1:1:1$ ratio.

(

${Si_2O_5}$ )

${\displaystyle^{2n-}_{n}}$ anion is obtained when :

0%
no oxygen of a ${SiO^{4-}_{4}}$ tetrahedron is shared with another ${SiO^{4-}_{4}}$ tetrahedron
0%
one oxygen of a ${SiO^{4-}_{4}}$ tetrahedron is shared with another ${SiO^{4-}_{4}}$ tetrahedron
0%
two oxygen of a ${SiO^{4-}_{4}}$ tetrahedron is shared with another ${SiO^{4-}_{4}}$ tetrahedron
0%
three oxygen of a ${SiO^{4-}_{4}}$ tetrahedron is shared with another ${SiO^{4-}_{4}}$ tetrahedron

Explanation

It can be seen from figure that three oxygen of a

${SiO^{4-}_{4}}$ tetrahedron is shared with another

${SiO^{4-}_{4}}$ tetrahedron.

$Al + N_2 \xrightarrow {\triangle} (A) \xrightarrow {H_2O} (B) \xrightarrow {K_2HgI_4/OH^-} (D)$ (brown ppt.)

What is

$(D)$ ?

0%
$H_2N - Hg - Hg - I$
0%
$H_2N - Hg - O - Hg - I$
0%
$H_2N - O -Hg- Hg - I$
0%
$H_2N - Hg - I - Hg - O$

Explanation

$Al + N_2 \xrightarrow {\Delta} AlN \xrightarrow {H_2O}NH_3 \xrightarrow {(K_2HgI_4/OH^-)}\, \, \, (D) : H_2N - Hg - O - Hg - I$ (brown ppt.) (Iodide of Millon's base)

Hence option

$B$ is correct.

One natural source of atmospheric

$CO_2$ is precipitation reaction such as the precipitation of silicates in the oceans:

$Mg^{2+}(aq)+SiO_2(dispersed)+2HCO_3(aq)\rightarrow MgSiO_3(s)+2CO_2(g)+H_2O(l)$

How many grams of magnesium silicate would be precipitated during the formation of

$100$ L of

$CO_2$ at

$30^oC$ and

$775$ torr?

$(Mg=24, Si=28, C=12, O=16)$

0%
$102\ g$
0%
$410 \ g$
0%
$205\ g$
0%
$51\ g$

Explanation

Moles of

$CO_2$ formed

$=n=\displaystyle\frac{PV}{RT}$

$PV=\displaystyle\frac{775}{760} atm\times 100L$

$RT=0.0821\ L \ atm$

$mol^{-1}$

$K^{-1}\times 303K$

$n = 4.0992$

$2$ moles

$CO_2$ are formed then

$MgSiO_3$ precipitated is

$1mol$ and when

$4.0992$ moles

$CO_2$ and formed

$MgSiO_3$ would be

$=\displaystyle\frac{4.0992}{2}$

$=2.0496\ mol$

(molar mass of

$MgSiO_3=100\ g$

$mol^{-1}$

$=2.0496\times 100 \ g$

$=204.96\ g$ is nothing but

$205\ g$

Hence, option

$C$ is correct.

Borax in its crystal possess:

0%
3 tetrahedral unit
0%
2 tetrahedral and 2 planar triangular units
0%
3 tetrahedral and 2 planar triangular units
0%
all tetrahedral units

Explanation

Borax

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

$Sp^2 \to$ trigonal planar

$Sp^3 \to$ tetrahedral

In borax

$2(Sp^2)$ trigonal planar and

$2(Sp^3)$ tetrahedral

1470646_151743_ans_210b24001aea4053b84c25c3c81b2fdf.PNG

Borax is converted into crystalline boron by the following steps:

${Borax \overset{X}{\rightarrow} H_3BO_3\overset{\Delta }{\rightarrow} B_2O_3\xrightarrow[\Delta ]{Y} B}$
Then

$X$ and

$Y$ are respectively:

0%
$HCl, Mg$
0%
$HCl, C$
0%
$C, Al$
0%
$HCl , Al$

Explanation

In first step,Boric acid is obtained by treating a hot concentrated solution of borax with hydrochloric acid or sulphuric acid. The resulting solution on concentration and cooling gives crystals of boric acid. here acid used is HCl or

$H_2SO_4$ .

Hence, the correct option is $A$ .

Select the correct statement(s) :

0%
When Al is added to potassium hydroxide solution,hydrogen gas is evolved
0%
${H_2 Si F_6}$ is formed when silica reacts with hydrogen fluoride followed by hydrolysis
0%
Phosphine gas is formed when red phosphorus is heated with NaOH
0%
( ${NH_4})_2 SO_4.FeSO_4.6H_2O$ is called alums

Explanation

(A) When Al is added to potassium hydroxide solution, hydrogen gas is evolved.

$2Al +2 KOH+2H_2O \rightarrow 2KAlO_2 + 3H_2$

Thus, option A is correct.

(B)

${H_2 Si F_6}$ is formed when silica reacts with hydrogen fluoride followed by hydrolysis.

$SiO_2 \xrightarrow [2.hydrolysis] {1. HF} {H_2 Si F_6}$

Thus, option B is correct.

(C) Phosphine is obtained by boiling white phosphorus with an aqueous solution of sodium hydroxide.
Thus, option C is correct.

(D)

${NH_4})_2 SO_4.FeSO_4.6H_2O$ is called Mohr's salt. It is a double salt.
Thus, option D is incorrect.

One type of silicon adhesive uses a polymer that has two

$OH$ groups left on the ends of the chain:

$HO-Si(CH_3)_2[O-Si(CH_3)_2]_n-O-Si(CH_3)_2-OH$
It has been found that this kind of polymer reacts easily with an organic-silicon molecule that contains acetate

$(CH_3COO)$ group. Which of these three molecules

$CH_3COOSi(CH_3)_3$ ,

$(CH_3COO)_2Si(CH_3)_2$ ,

$(CH_3COO)_3SiCH_3$ would you choose to add to the hydroxy polymer in order to make a strong adhesive?

0%
$CH_3COOSi(CH_3)_3$ ,
0%
$(CH_3COO)_2Si(CH_3)_2$
0%
$(CH_3COO)_3SiCH_3$
0%
None of the above

Explanation

${ \left( C{ H }_{ 3 }COO \right) }_{ 3 }SiC{ H }_{ 3 }-$ triacetoxy (methyl) silane.
This should give the cross-linking necessary for strength of adhesives.

Select correct statements:

0%
Hydrides of $B$ and $Si$ are volatile and catch fire on exposure to air.
0%
Oxides of $B$ and $Si(B_2O_3$ and $SiO_2)$ are acidic in nature.
0%
Borates and silicates have tetrahedral $BO_4$ and $SiO_4$ structural units.
0%
All the above are correct statements.

Explanation

Hydrides of silicon and boron are less stable because in these, hydrogen withdraw electrons from M-H bond and due to this they are volatile and catch fire on exposure to air.
Oxides of

$B$ and Si

$(B_2O_3$ and

$SiO_2)$ are electron deficient due to high electronegativity difference and shows acidic nature.
Borates and silicates have tetrahedral

$BO_4$ and

$SiO_4$ structural units.

Select correct statement about stability of cations.

0%
$Ge^{4+} > Sn^{4+} > Pb^{4+}$
0%
$Ge^{2+} < Sn^{2+} < Pb^{2+}$
0%
$Pb^{2+} > Pb^{4+}, Sn^{4+} > Sn^{2+}$
0%
All the above are correct statements.

Explanation

Common oxidation state of IVA group elements are +2, +4.
+2 state becomes gradually more stable down the group & +4 state stability decreases from carbon to Lead due to inert pair effect.
So order is

$Ge^{4+} > Sn^{4+} > Pb^{4+}$

$Ge^{2+} < Sn^{2+} < Pb^{2+}$
And for lower element Pb II oxidation state is more stable than IV oxidatio state.

$Pb^{2+} > Pb^{4+}, Sn^{4+} > Sn^{2+}$

Silicone's :

0%
are synthetic polymer containing repeating $(R_2SiO)$ units.
0%
are formed by hydrolysis of $R_2SiCl_2$ .
0%
are natural occurring repeated $(SiO_4)$ units.
0%
single $R_2SiO$ unit.

Explanation

SILICONES: Organic polymers containing silicon in them are called as silicones. These are organosilicon polymers containing

$R_2SiO$ repeating units and empirical formula analogous to ketone

$(R_2CO)$ .
Alkyl substituted : These are formed by hydrolysis of alkyl chlorosilanes and their polymerisation. The alkyl or aryl substituted chlorosilanes are prepared by the reaction of Grignard reagent and silicon tetrachlorides.

Silicone's are :

0%
synthetic polymers containing repeated $R_2SiO$ units.
0%
silicates with common $SiO_4$ unit.
0%
ketones with silyl group $(SiH_3)$ similar to alkyl $(SiH_3)_2CO$
0%
Zircon (neso silicates)

Explanation

These are organosilicon polymers containing

$R_2SiO$ repeating units and empirical formula analogous to ketone

$(R_2CO)$

$-\underset{|}{\overset{|}{S}}i-O-\underset{|}{\overset{|}{S}}i-linkage$

Organic polymers containing silicon in them are called as silicones.
These are formed by hydrolysis of alkyl chlorosilanes and their polymerisation. The alkyl or aryl substituted chlorosilanes are prepared by the reaction of Grignard reagent and silicon tetrachlorides.

The color imparted by Co(Il) compounds to glass is :

0%
deep blue.
0%
green.
0%
yellow.
0%
red.

Which is incorrect statement about silicons?

0%
They are repeating units $(SiO_4)$ in silicates
0%
They are synthetic polymers containing repeated $R_2SiO$ units
0%
They are formed by hydrolysis of $R_2SiCl_2$
0%
All the above are incorrect statements

Explanation

Organic polymers containing silicon in them are called as silicones. These are organosilicon polymers containing

$R_2SiO$ repeating units and empirical formula analogous to ketone

$(R_2CO)$ . They are formed by hydrolysis of

$R_2SiCl_2$ .

Select correct statement(s) :

0%
$Si$ uses all of its valence electrons in an $sp^3$ -hybridization and crystallizes in an fcc structure similar to diamond.
0%
Diamond is an electrical insulator, graphite is an electrical conductor but silicon is a semiconductor.
0%
Most common silica is quartz.
0%
All of the above are correct statements.

If we mix

$SiMe_3Cl$ with

$SiMe_2Cl_2$ , we get silicone's of the type:

0%
0%
0%
both (a) and (b)
0%
none of the above

Explanation

When

$Me_3SiCl$ is mixed with

$Me_2SiCl_2$ silicones of the type

$Me_3Si-O-Si(Me_2)-O-Si(Me_2)-OSiMe_3$ are obtained.
Each of the terminal silicon atom carries three methyl groups and each of the middle silicon atom carries two methyl groups.

$Ca_2B_6O_{11}\xrightarrow {\Delta}[X]+CaCO_3+NaBO_2$ (unbalanced equation) Which of the following is/are correct choice(s) for [X]?

0%
The structure of anion of crystalline [ $X$ ] has one boron atom $sp^3$ hybridized and other three boron atoms $sp^2$ hybridized
0%
$X$ with $NaOH_{(aq)}$ gives a compound which on reaction with hydrogen peroxide in alkaline medium yields a compound used as the brightener in soaps
0%
Hydrolysis of [ $X$ ] with $HCl$ or $H_2SO_4$ yields a compound which on reaction with $HF$ gives fluoroboric acid
0%
[ $X$ ] on heating with chromium salts in oxidizing flame gives green colored bead

Explanation

The compound X is borax

$(Na_2B_4O_7 )$ . It is prepared by boiling finely powdered colemanite mineral

$(Ca_2B_6O_{11})$ with sodium carbonate solution.

$Ca_2B_6O_{11} + 2 Na_2CO_3 \rightarrow 2 CaCO_3 + Na_2B_4O_7 + 2 NaBO_2$

The statements (B), (C) and (D) are correct. The statement (A) is incorrect.

(A) The structure of anion of crystalline [X] has two boron atoms

$sp^3$ hybridized and other two boron atoms

$sp^2$ hybridized.

(B) X with NaOH(aq), gives sodium metaborate

$NaBO_2$ which on reaction with hydrogen peroxide in alkaline medium yields hexahydrate sodium tetraaquadiperoxoborate (III).

$2NaBO_2 + 2H_2O_2 + 6H_2O \rightarrow Na_2[B_2(O_2)_2(OH)_4]•6H_2O$

$HCl$ or

$H_2SO_4$ yields boric acid which on reaction with

$HF$ gives fluoroboric acid.

$Na_2B_4O_7 + H_2SO_4 \rightarrow Na_2SO_4 + H_2B_4O_7$

$H_2B_4O_7 + 5H_2O \rightarrow 4H_3BO_3$

$H_3BO_3 + 4HF \rightarrow H[BF_4] + 3H_2O$

(D) [X] on heating with chromium slats in oxidizing flame gives green colored bead.

$Cr_2O_)3 + 3B_2O_3 \rightarrow 2Cr(BO_2)_3 \text {(chromium metaborate green)}$
This is borax bead test.

$B^{3+}$ ion can not exist in aqueous solution because of its :

0%
strong reducing ability
0%
strong oxidizing ability
0%
small size and large charge
0%
large size and small charge

Explanation

On account of higher charge density i.e. charge/size ratio it gets hydrolysed forming

$H_3BO_3$ .

$2B^{3+} + 6H_2O \longrightarrow 2H_3BO_3 + 6H^+$

Aqueous solution of borax reacts with 2 mol of acids. This is because of?

0%
Formation of 2 mol of $B(OH)_{3}$ only.
0%
Formation of 2 mol of $\left [B(OH)_{4} \right ]^{-}$ only.
0%
Formation of 1 mol each of $B(OH)_{3}$ and $\left [B(OH)_{4} \right ]^{-}$ only.
0%
Formation of 2 mol each of $\left [B(OH)_{4} \right ]^{-}$ and $B(OH)_{3}$ , of which $\left [B(OH)_{4} \right ]^{-}$ reacts with acid.

Borax is actually made of two tetrahedral and two triangular units joined together and should be
written as :

$Na_{2}\:[B_{4}O_{5}(OH)_{4}].8H_{2}O$ .
Consider the following statements about borax:
Each boron atom has four B-O bonds
EAch boron atom has three B-O bonds
Two boron atoms have four B-O bonds while other two have three B-O bonds
Each boron atom has one-OH groups
Select correct statement(s) -

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

Explanation

Borax

$Na_{2}\:[B_{4}O_{5}(OH)_{4}].8H_{2}O$

Identify the only 'INCORRECT' statement of the following.

0%
B - O - B bonds in sodium peroxoborate are similar
0%
B-O-B bonds in potassium trioborate are 3
0%
Ortho borate ion is triangular plannar
0%
B -O-B bonds in diborate ion is 1.

Identify the correct order of boiling point :

0%
$HF\,<\,HCl\,<\,HBr\,<\,HI$
0%
$NH_3\,<\,PH_3\,<\,AsH_3\,<\,SbH_3$
0%
$CH_4\,<\,SiH_4\,<\,GeH_4\,<\,SnH_4$
0%
$H_2S\,<\,H_2Se\,<\,H_2Te\,<\,H_2O$

Explanation

boiling point increases down the group in the periodic table because of the vanderwaals forces except for the molecules showing hydrogen boding i.e.

$H$ covalently bound to

$N, F$ or

$O$ .

The incorrect statements among the following are:

0%
$NCl_5$ does not exist while $PCl_5$ does
0%
Pb prefers to form tetravalent compounds
0%
The three $C-O$ bond are not equal in the carbonate ion
0%
Both $O_2^+$ and $NO$ are paramagnetic

Explanation

Explanation:

In there is absence of $d-orbitals$ . has $d-orbitals$ , so it exists.

Due to inert pair effect $P{b^{2 + }}$ is more stable than $P{b^{4 + }}$ . So it doesn’t form tetravalent compounds.

In carbonate ion the bonds are equal due to the resonance.

The configuration of $O_2^ + (8 + 8 - 1 = 15) = \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2}$ $\sigma 2p_z^2,\pi 2p_x^2 \approx 2p_y^2, {{\pi 2p}^*_{x}}^1 \approx \pi {{2p}^*_{y}}^0$

The configuration of $NO (7 + 8 = 15) = \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2}$ $\sigma 2p_z^2,\pi 2p_x^2 \approx 2p_y^2, {{\pi 2p}^*_{x}}^1 \approx \pi {{2p}^*_{y}}^0$

Both have one unpaired electron so they are paramagnetic.

Final answer: The statements B and C are incorrect. So the correct answer is options $B$ and $C.$

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 12 - MCQExams.com

0:0:2

Practice Class 11 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 12 - MCQExams.com

0:0:2

Practice Class 11 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.