Bauxite ore is generally contaminated with impurity of oxides of two elements $${E}_{1}$$ and $${E}_{2}$$. Which of the following statements is correct?

$${E}_{1}$$ is a non-metal and belongs to 3rd period while $${E}_{2}$$ is a metal and belong to fourth period

One of the two oxides has three dimensional polymeric structure.

MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 15

An inorganic compound

(A)

$(A)$ made of two most occurring elements in the earth's crust and used in building construction when made to react with carbon, forms a poisonous gas

(B)

$(B)$ which is most stable diatomic molecule. Compound

(A)

$(A)$ and

(B)

$(B)$ are:

0%
$SiO_2, CO_2$
0%
$SiO_2, CO$
0%
$SiO_2, N_2$
0%
$CaO, CO_2$

Explanation

S i O_{2}

$SiO_2$ is a compound of oxygen and silicon, the two most abundant elements of earth's crust and is used in building construction.

S i O_{2} + 2 C ⟶ S i + 2 C O

$SiO_2 +2C\longrightarrow Si+2CO$ (piosonous gas and a stable diatomic molecule)

Option B is correct.

Silica is reacted with sodium carbonate. What is the gas liberated?

0%
$CO$
0%
$O_2$
0%
$CO_2$
0%
$O_3$

Explanation

When silica is reacted with sodium carbonate, it gives sodium silicate and carbon dioxide gas is liberated.The addition of

S i O_{2}

$SiO_{2}$ greatly enhanced the decomposition of

N a_{2} C O_{3}

$Na_{2}CO_{3}$ .

The reaction is given as:

S i O_{2} + N a_{2} C O_{3} \to N a_{2} S i O_{3} + C O_{2}

$SiO_{2} + Na_{2}CO_{3} \rightarrow Na_{2}SiO_{3} + CO_{2}$

As the gas liberated is

C O_{2}

$CO_{2}$ .

Hence, the correct option is (C).

2 R C l + S i C u p o w d e r - ----- \to 570 K R_{2} {S i C l}_{2} H_{2} O - - \to R_{2} S i {(O H)}_{2} P o l y m e r i s a t i o n - -------- \to

$2RCl+Si\xrightarrow [ 570\ K ]{ Cu\ powder } { R }_{ 2 }{ SiCl }_{ 2 }\xrightarrow [ ]{ { H }_{ 2 }O } { R }_{ 2 }Si{ (OH) }_{ 2 }\xrightarrow [ ]{ Polymerisation }$

(A) . (A)

$(A). (A)$ is:

0%
cyclic silicone
0%
cross linked silicone
0%
linear silicone
0%
none of these

Which one of the following is wrong?

0%
Oxygen and sulphur belong to the same group of periodic table
0%
Oxygen is a gas while sulphur is solid
0%
Both show $+2,+4$ and $+6$ oxidation states
0%
$H_{2}S$ has no hydrogen bonding

Which of the following is not correct?

0%
Hydrolysis of $NCl_3$ gives $NH_3$ and $HClO$
0%
$NH_3$ is less stable than $PH_3$
0%
$NH_3$ is a weak reducing agent compared to $PH_3$
0%
Nitric oxide in solid state exhibits diamagnetic property

Xenon hexafluoride reacts with silica to form a xenon compound X. The oxidation state of Xe in X is :

0%
+2
0%
+4
0%
+6
0%
0

Explanation

S i O_{2} + 2 X e F_{6} \to 2 X e O F_{4} + S i F_{4}

$SiO_2 + 2XeF_6 \rightarrow 2XeOF_4 + SiF_4$

Oxidation state of Xe in (X) is +6.

Option C is correct.

{N a}_{2} B_{4} O_{7} \cdot {10 H}_{2} O

${Na}_{2}{B}_{4}{O}_{7}\cdot{10H}_{2}O$ is correctly represented as:

0%
${2NaBO}_{2}\cdot{Na}_{2}{B}_{2}{O}_{3}\cdot{10H}_{2}O$
0%
${Na}_{2}\left[{B}_{4}{O}_{5}\left(OH\right)_{4}\right]\cdot{8H}_{2}O$
0%
${Na}_{2}\left[{B}_{4}\left(H_{2}O\right)_{4}\right){O}_{7}]\cdot{6H}_{2}O$
0%
all of the above

Bauxite ore is generally contaminated with impurity of oxides of two elements

E_{1}

${E}_{1}$ and

E_{2}

${E}_{2}$ . Which of the following statements is correct?

0%
${E}_{1}$ is a non-metal and belongs to 3rd period while ${E}_{2}$ is a metal and belong to fourth period
0%
One of the two oxides has three dimensional polymeric structure.
0%
Both (A) and (B) are correct
0%
None of the above

Inert pair effect is shown by:

0%
$Ra$
0%
$Bi$
0%
$Pb$
0%
$Tl$

Which of the following elements of matter would best convey that there is life on earth

0%
Oxygen
0%
Hydrogen
0%
Carbon
0%
Iron

Explanation

Life on Earth is based on carbon,

$\text{Life on Earth is based on carbon,}$ likely because each carbon atom can form bonds with up to four other atoms simultaneously. This quality makes carbon well-suited to form the long chains of molecules that serve as the basis for life as we know it, such as proteins and DNA.

Which of the following compounds of boron does not exist in the free form?

0%
$BCl_3$
0%
$BF_3$
0%
$BBr_3$
0%
$BH_3$

Explanation

Hydride of boron does not exist in

B H_{3}

$BH_3$ form. It is stable as its dimer diborane (

B_{2} H_{6}

$B_2H_6$ )

Option D is correct.

Carbon has zero oxidation number in________

0%
$CO$
0%
${CH}_{4}$
0%
${CH}_{2}{Cl}_{2}$
0%
${CH}_{3}Cl$

Explanation

x + 2 \times (+ 1) + 2 (- 1) = 0; x = 0

$x+2\times (+1)+2(-1)=0;x=0$ in

{C H}_{2} {C l}_{2}

${CH}_{2}{Cl}_{2}$

Option C is correct.

C a_{2} B_{6} O_{11} + N a_{2} C O_{3} Δ \to [X] + C a C O_{3} + N a B O_{2}

$Ca_{2}B_{6}O_{11}+Na_{2}CO_{3}\overset{\Delta }{\rightarrow} \left [ X \right ]+CaCO_{3}+NaBO_{2}$ (unabalanced equation)
correct choice (s) for [X] is/are

0%
Structure of anion of crystalline [X] has one born atom $sp^{3}$ hybridised and other three boron atoms $sp^{2}$ hybridised
0%
X with NaOH (aq) gives a compound which on reaction with hydrogen peroxide in alkaline medium yields a compound used as brighter in soaps
0%
hydrolysis of [X] with HCl or $H_{2}SO_{4}$ yields a compound which on reaction with HF gives fluroboric acid.
0%
[X] on heating with chromium salts in oxidising flame gives green coloured bead in cold.

Explanation

Compound

[X]

$\left [ X \right ]$ is

N a_{2} B_{4} O_{7}

$Na_{2}B_{4}O_{7}$

(A)two boron atoms are

s p^{2}

$sp^{2}$ while other two boron atoms are

s p^{3}

$sp^{3}$ hybridised

(B)Product is sodium peroxoborate which is used as brightner in soaps

(C)

H_{3} B O_{3}

$H_{3}BO_{3}$ is formed which with HF gives

H B F_{4}

$HBF_{4}$

(D) Green baed in oxidising as well as in reducing flame in cold

The decrease stability of higher oxidation state in p-block with increasing atomic number is due to

0%
Increase in bond energy as going down the group
0%
Inert pair effect.
0%
Both (A) and (B)
0%
None is correct

On the addition of mineral acid to an aqueous solution of borax, the compound formed is

0%
Boro dihydride
0%
Orthoboric acid
0%
Metaboric acid
0%
Pyroboric acid

Explanation

Addition of mineral acid to an aqueous solution of Borax, orthoboric acid is formed.

N a_{2} B_{4} O_{7} + 2 H C l + 5 H_{2} O \to 4 H_{3} B O_{3} + 2 N a C l

$Na_{2}B_{4}O_{7} + 2HCl + 5H_{2}O \rightarrow 4H_{3}BO_{3} + 2NaCl$

Hence, the correct answer is option B.

Column-I	Column-II
$A: He$	$i:$ High electron affinity
$B: Cl$	$ii:$ Most electropositive element
$C: Ca$	$iii:$ Strongest reducing agent
$D: Li$	$iv:$ Highest ionization energy

The correct match of contents in Column -I with those in Column -II is:

0%
$A-iii, B-i, C-ii, D-iv$
0%
$A-iv, B-iii, C-ii, D-i$
0%
$A-i, B-ii, C-iii, D-iv$
0%
$A-iv, B-i, C-ii, D-iii$
0%
$A-ii, B-iv, C-i, D-iii$

Which reagent is used to remove

S O_{4}^{2 -}

$SO_{4}^{2-}$ and

C l^{-}

$Cl^{-}$ ?

0%
$BaSO_{4}$
0%
$NaOH$
0%
$Pb(NO_{3})_{2}$
0%
$KOH$

Explanation

$PbSO_{4}$ and

$PbCl_{2}$ are insoluble in cold water hence the reagent

$Pb(NO_{3})_{2}$ is used to remove

$SO_{4}^{2-}$ and

$Cl^{-}.$

In the electrothermal process, the compound displaced by silica from calcium phosphate is:

0%
Calcium
0%
Phosphine
0%
Phosphorus
0%
Phosphorus pentoxide

Explanation

In the electrothermal process, the compound displaced by silica from calcium phosphate is phosphorus pentoxide. It is then reduced by coke in electric furnace to get white phosphorus.

$Ca_3(PO_4)_2 +3SiO_2 \to 3CaSiO_3 +P_2O_5$

In laboratory silicon can be prepared by the reaction :

0%
By heating Carbon in electric furnace
0%
By heating Potassium with Potassium dichromate
0%
Silica with Magnesium
0%
None of these

Explanation

In laboratory, Silicon is prepared by the reaction of Silica with Magnesium.

Reaction :

$\mathrm{SiO_2 + 2Mg \rightarrow Si +2MgO}$

Hence, Option "C" is the correct answer.

Silica is added to roasted copper ore during smelting in order to remove

0%
Cuprous sulphide
0%
Cuprous oxide
0%
Ferrous oxide
0%
Ferrous sulphide

Explanation

Silica is added during smelting in order to remove iron oxides from copper sulphide and oxides as slag.

$\underset{\text{Impurity}}{FeO}+\underset{\text{Flux}}{SiO_2} \rightarrow \underset{\text{Slag}}{FeSiO_3}$

Which of the following compounds does not exist?

0%
$SbCl_3$
0%
$BiCl_5$
0%
$SbCl_5$
0%
$AsCl_5$

Explanation

$BiCl_5$ does not not exist because

$+3$ oxidation state of

$Bi\$ is more stable than

$+5$ due to inert pair effect.

Which element froms maximum compound in chemistry?

0%
O
0%
H
0%
Si
0%
C

Composition of borax is

0%
$Na_2B_4O_7.4H_2O$
0%
$Na_2B_4O_7.10H_2O$
0%
$NaBO_2$
0%
$Na_2BO_3$

Explanation

Borax is

$\mathrm{Na_2B_4O_7.10H_2O}$

Hence, Option "B" is the correct answer.

Silicon dioxide is formed by the reaction of

0%
$SiCl_{4}+H_{2}O$
0%
$SiO_{2}+HF$
0%
$SiO_{2}+NaOH$
0%
$SiCl_{4}+NaOH$

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

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

In which of the following the inert pair effect is most prominent ?

0%
$C$
0%
$Ge$
0%
$Si$
0%
$Pb$

Explanation

Due to poor or ineffective shielding of the

$ns^{2}$ electrons of the valence shell by intervening d- and f-electrons, the s-electrons are reluctant to participate in bond formation.

Increasing order of stability of the +2 oxidation state of the ions ?

0%
$Ca^{2+} < Ba^{2+} < Sr^{2+}$
0%
$Pb^{2+} < Ge^{2+} < Sn^{2+}$
0%
$Ge^{2+} < Sn^{2+} < Pb^{2+}$
0%
$Cu^{2+} < Au^{2+} < Ag^{2+}$

Which of the following halides is inert towards hydrolysis at room temperature?

0%
$SiCl_4$
0%
$PCl_3$
0%
$NCl_3$
0%
$NF_3$

A cyclic skeleton of silicon and oxygen can constructed by the silicate ion composition :

0%
$Si_{2}O_{7}^{4-}$
0%
$Si_{2}O_{5}^{2-}$
0%
$SiO_{3}^{2-}$
0%
$SiO_{4}^{4-}$

Correct stability order of metal cation is/are :

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

There are three elements P, Q and R which belong to the p-block of periodic table they all form trifluorides with

$F_2$ , such that

$PF_3$ is lewis acid but

$QF_3$ is weak lewis base (dipole moment 0.23D). These two compounds react with each other in presence of

$F_2$ to produce

$QF^+_4$ ,

$PF^-_4$ The compound

$RF_3$ is a T-shape interhalogen molecule. Which of the following is an incorrect statement with reference to the above information?

0%
All the R-F bond lengths are equal in $RF_3$ molecules
0%
In the structure the angle $A \neq 180^0$
0%
The Q and P both are $sp^3$ hybridised in $QF^+_4$ , $PF^-_4$
0%
The elements P, Q, R can be $B$ , $N$ and $Cl$ respectively

The incorrect order is:

0%
$N > P > As :$ Strength of $\pi$ -bond with oxygen atom
0%
$SiF_{4} > SiCl_{4} > SiBr_{4} > SiI_{4} :$ Thermal stability
0%
$H_{2}S > H_{2}Se > H_{2}Te :$ Arrhenious acid character
0%
$CaBr_{2} > MgBr_{2} > BeBr_{2} :$ Melting point

Explanation

Correct order of Arrhenious acid character:

$H_{2}S < H_{2}Se < H_{2}Te$

The value of

$K_{a}$ is maximum for

$H_{2}Te,$ due to least bond energy among given hydrides.

Hence, Option "C" is the correct answer.

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 15 - 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 15 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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