Colemanite $$: Ca_2B_6O_{11} \cdot 5H_2O$$

Ordinary form of borax $$: Na_2B_4O_7 \cdot 5H_2O$$

Boronatrocalcite $$: 2Mg_3B_8O_{15} \cdot MgCl_2$$

Octahedral form of borax $$: Na_2B_4O_7 \cdot 10H_2O$$

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

Correct match is:

0%
Ordinary form of borax $: Na_2B_4O_7 \cdot 5H_2O$
0%
Colemanite $: Ca_2B_6O_{11} \cdot 5H_2O$
0%
Boronatrocalcite $: 2Mg_3B_8O_{15} \cdot MgCl_2$
0%
Octahedral form of borax $: Na_2B_4O_7 \cdot 10H_2O$

Aqueous solutions of borax acts as a buffer because:

0%
it contains weak acid and its salt with strong base
0%
it contains tribasic acid and strong base
0%
it contains number of neutral water molecules
0%
none of the above

Boron does not form

${B}^{3+}$ cation easily. It is due to

0%
energy required to form ${B}^{3+}$ ion is very high
0%
Boron is non-metal
0%
Boron is semi-metal
0%
none of the above

Electron affinity of:

0%
carbon is greater than oxygen
0%
sulphur is higher than oxygen
0%
chlorine is higher than bromine
0%
chlorine is higher than fluorine

Which of the following is most stable?

0%
$Pb^{2+}$
0%
$Ge^{2+}$
0%
$Si^{2+}$
0%
$Sn^{2+}$

Oxidation number of carbon in

${H}_{2}{C}_{2}{O}_{4}$ is

0%
$+4$
0%
$+3$
0%
$+2$
0%
$-2$

Explanation

${H}_{2}\overset { \ast }{ {C}_{2} } {O}_{4}$

Let,

$x$ be the oxidation number of carbon in

${H}_{2}{C}_{2}{O}_{4}.$

Now,

$2\times (+1)+2\times x+4\times(-2) =0;\ \Rightarrow x=+3$

The stable compound in

$+1$ oxidation state are formed by:

0%
$B$
0%
$Al$
0%
$Tl$
0%
$Ga$

$Tl I_3$ is an ionic compound which furnishes the following ions:

0%
$Tl^{3+}$ and $I_3^-$
0%
$Tl^{3+}$ and $I^-$
0%
$Tl^{+}$ and $I^-$
0%
$Tl^{+}$ and $I_3^-$

Explanation

$Tl I_3$ is an ionic compound which furnishes the following ions:

$TlI_{3}\rightleftharpoons Tl^{+}+I_{3}^{-}$

$Tl$ is in

$+1$ oxidation state because of inert pair effect.

Which of the following halides is most acidic?

0%
$BCl_3$
0%
$SbCl_3$
0%
$BiCl_3$
0%
$CCl_4$

Explanation

A species which is more electron deficient will attract electron easily and hence will be more acidic. Boron atom in $BCl_3$ , has six electrons in the outermost orbit and thus it can accept a pair of electrons form a donor molecule to complete its octet. The effective overlap between orbital's decreases due to large size of p orbital of halides. Hence $BCl_3$ is most acidic than other options.

Outermost configuration of two

$p-$ block elements

$A$ and

$B$ are

$ns^{2}np^{1}$ and

$ns^{2}np^{6}$ respectively. The elements are if

$n=4$ ;

0%
$A=Ga, B=Xe$
0%
$A=In, B=Xe$
0%
$A=Ga, B=Kr$
0%
$A=Zn, B=Kr$

While reviewing the third row elements is periodic table, point out the correct statement:

0%
sodium has the largest size
0%
chlorine has the maximum electronegativity
0%
first ionisation potential of $Mg >$ first ionisation potential aluminium
0%
chlorine is the strongest reducing agent

Oxidation number of

$C$ in

${C}_{6}{H}_{12}{O}_{6}$ is

0%
$+6$
0%
$-6$
0%
$0$
0%
$+4$

Explanation

$C_6H_{12}O_6$

$6x+12-12=0$

$C$ has oxidation number

$=0$

The oxidation number of carbon in

${CH}_{2}O$ is

0%
$-2$
0%
$+2$
0%
$0$
0%
$+4$

Explanation

$\overset { \ast }{ C } {H}_{2}O$

$x+2-2=0$

$x=0$

The chemical name of borax is

0%
Sodium orthoborate
0%
Sodium metaborate
0%
Sodium tetraborate
0%
Sodium tetraborate decahydrate

Explanation

The chemical name of borax is Sodium tetraborate decahydrate

$(Na_2B_4 O_7 . 10 H_2 O) .$

Oxidation number of carbon in

${CH}_{3}-Cl$ is

0%
$-3$
0%
$-2$
0%
$-1$
0%
$0$

Explanation

$\overset { \ast }{ C } {H}_{3}-Cl$

$x+3(+1)+(-1)\times 1=0;x=-2$

The oxidation number of carbon in

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

0%
$0$
0%
$+2$
0%
$-2$
0%
$+4$

Explanation

$\overset { \ast }{ C } {H}_{2}{Cl}_{2}$

Let,

$x$ be the oxidation number of carbon in

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

Now,

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

Oxidation state of elemental carbon is

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Thallium shows different oxidation states because

0%
it is a transition metal
0%
of inert-pair effect
0%
of its high reactivity
0%
of its amphoteric character

How many ionisation energies can carbon have

0%
$1$
0%
$2$
0%
$4$
0%
$6$

The tetrahedral nature of carbon was first given by

0%
Kekule
0%
Le Bell and Vant Hoff
0%
Pauling
0%
Armstrong and Bayer

Explanation

The discovery of stereoisomerism is due to

$\text{Vant Hoff and Le Bel,}$ who proposed tetrahedral geometry to a carbon atom.

Carbon containing molecules present in all life forms includes:

0%
Carbon dioxide, carbonates and hydrogen carbonates
0%
Nucleic acids, carbon dioxied, carbonates and hydrogen carbonates
0%
Proteins, carbohydrates, fats, nucleic acids and vitamins
0%
Hormones, carbohydrates, Carbonates and hydrogen carbonates

In nature, carbon is found in elemental forms as:

0%
Carbon dioxide and carbonates
0%
Diamond and graphite
0%
Hydrogen carbonates and graphite
0%
Carbon dioxide and graphite

The correct statement is :

0%
$+5$ oxidation state of H is more stable than its $+3$ oxidation state.
0%
$G^{2+}$ is the better oxidizing agent than $G^{4+}$ .
0%
$+3$ oxidation state of H is more stable than its $+5$ oxidation state.
0%
The ionization energy of G is greater than that of Tin.

Which of the following compound is most acidic?

0%
$Cl_2O_7$
0%
$P_4O_{10}$
0%
$SO_3$
0%
$B_2O_3$

Explanation

The oxidation state of non-metal increases the acidic nature of oxide increase

$Cl_2O_7$ is more acidic.

Carbon forms four covalent bonds by sharing its four valence electrons with four univalent atoms, e.g. hydrogen. After the formation of four bonds, carbon attains the electronic configuration of

0%
helium
0%
neon
0%
argon
0%
krypton

Select correct statement(s) :

0%
Thermodynamic stability of graphite > diamond, but reverse order is true for their kinetic stability
0%
Melting point of $NaCI > LiCI$ , but reverse order is true for their thermal stability
0%
Ionisation energy of $N_2 > O_2$ , but reverse order of ionisation energy is true for their corresponding atoms
0%
Lewis acidic strength of $Bel_2 > BeF_2$ , but reverse order is true for their melting point

If P to T are second period p-block elements then which of the following graph show correct relation between valence electrons in

$P_2$ to

$T_2$ (corresponding molecules) and their bond order is :

Which of the following statement is incorrect?

0%
$B_2C_5$ is acidic in nature
0%
$AC_2$ is acidic in nature
0%
$FC_3$ is basic in nature
0%
$GC_2$ is amphoteric in nature

The oxidation number of carbon in diamond is:

0%
$zero$
0%
$+4$
0%
$-4$
0%
$+2$

The nature of aqueous solution of borax is :

0%
Acidic
0%
Neutral
0%
Amphoteric
0%
Basic

Explanation

When borax is dissolved in water it ionizes and hydrolysis of borate ion gives OH (the hydroxyl ion) and boric acid. Since boric acid is a weak acid and

$NaOH$ is strong a alkali an aqueous solution of borax is alkaline in nature.

In which of the following elements +3 oxidation state is most stable?

0%
In
0%
Ga
0%
Al
0%
TI

Explanation

$Al$ is the most stable element in +3 oxidation state.

Which of the following is an organo silicon polymer ?

0%
Silica
0%
Silicone
0%
Silicone carbide
0%
Silicic acid

$SiCl_{4} \overset{H_2O}{\longrightarrow} X \overset{Heat}{\longrightarrow} Y \overset{NaOH}{\longrightarrow} Z$
X, Y and Z in the above reaction are

0%
$X \rightarrow SiO_2 , Y \rightarrow Si , Z \rightarrow NaSi$
0%
$X \rightarrow Si(OH)_4, Y \rightarrow SiO_2, Z \rightarrow Na_2SiO_3$
0%
$X \rightarrow Si(OH)_4, Y \rightarrow Si, Z \rightarrow SiO_2$
0%
$X \rightarrow SiO_2, Y \rightarrow SiCl_{4}, Z \rightarrow Na_2SiO_3$

Group

$13$ elements show

0%
only $+1$ oxidation state
0%
only $+3$ oxidation state
0%
$+1$ and $+3$ oxidation states
0%
$+1, +2$ and $+3$ oxidation states.

Explanation

Group -

$13$ elements show

$+1$ and

$+3$ oxidation states. Tendency to show

$+3$ oxidation state decreases down the group.

Carborumdum is a:

0%
molecular solid
0%
covalent solid
0%
ionic solid
0%
amorphous solid

Which substance is having molecular solid :

0%
graphite
0%
$C_{60}$
0%
gold
0%
$Ca_{3}(PO_{4})_{2}$

Choose the correct order of

$C - C$ bond length in the given compounds:

0%
Acetylene < ethylene < graphite < benzene < ethane
0%
Acetylene < ethylene < benzene < graphite < ethane
0%
Acetylene < graphite < ethylene < benzene < ethane
0%
Acetylene < benzene < graphite < ethylene < ethane

Explanation

thane contains carbon-carbon single bond.

Benzene is aromatic in nature. It has resonance. All the carbon-carbon bonds in benzene have partial double bond character.

Ethylene contains carbon-carbon double bond.

Acetylene contains carbon-carbon triple bond.

Hence, the correct order of carbon-carbon bond length is

$C_2H_6>C_6H_6>C_2H_4>C_2H_2$ .

Acetylene < benzene < graphite < ethylene < ethane

The incorrect stability order of

$+3$ and

$+1$ states of

$13th$ group elements (boron family) is:

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

Correct order of bond angle in given species is:

0%
$SiO_{4}^{4-} > PCl_{3} > NCl_{3} > SbH_{3} > H_{2}Te$
0%
$SiO_{4}^{4-} > NCl_{3} > PCl_{3} > SbH_{3} > H_{2}Te$
0%
$SiO_{4}^{4-} > H_{2}Te > SbH_{3} > PCl_{3} > NCl_{3}$
0%
$NCl_{3} > PCl_{3} > SiO_{4}^{4-} > SbH_{3} > H_{2}Te$

Explanation

$\angle Cl - P - Cl$ in

$PCl_{3} > \angle H - Sb - H$ in

$SbH_{3},$ as in

$SbH_{3},$ bond pairs are formed by overlapping of almost pure p-orbitals.

$\angle H - Sb - H$ in

$SbH_{3} < \angle H - Te - H$ in

$H_{2}Te$ because two lone pairs are present on

$Te$ while at

$Sb$ there is one lone pair.

Hence, Option "B" is the correct answer.

Which is the hardest compound of boron ?

0%
$B_{2}O_{3}$
0%
$BN$
0%
$B_{4}C$
0%
$B_{2}H_{6}$

Explanation

$B_{4}C$ is the hardest compound of boron .

Boron carbide

$B_{4}C$ is a very hard substance like

$SiC.$

Which is not true about borax ?

0%
It is a useful primary standard for titrating against acids
0%
One mole of borax contains $4B - O - B$ bonds
0%
Aqueous solution of borax can be used as buffer
0%
It is made up of two triangular $BO_{3}$ units and two tetrahedral $BO_{4}$ units

Explanation

In borax

$(Na_{2}B_{4}O_{7} . 10H_{2}O),$ among

$10$ water molecules

$2$ molecules are part of structure, i.e., exists

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

$Na_{2}[B_{4}O_{5}(OH)_{4}] . 8H_{2}O + 2HCl \rightarrow 2NaCl + 4H_{3}BO_{3} + 5H_{2}O$
Methyl orange

$(pH = 3.7)$ is used to detect end point. Aq. solution of borax acts as buffer, as borax is salt of strong base

$NaOH$ and weak acid

$H_{3}BO_{3}.$

Hence, Option "B" is the correct answer.

1831989_1860839_ans_9ed11d562aa0461a92555a5e59cfd111.png

Which of the following order is incorrect ?

0%
Lewis basic character : $NH_{3} > PH_{3} > AsH_{3} > SbH_{3}$
0%
Bond dissociation energy : $HF > HCl > HB r > HI$
0%
Thermal stability : $H_{2}O > H_{2}S > H_{2}Se > H_{2}Te$
0%
Bond angle : $CH_{4} > SiH_{4} > GH_{4} > SnH_{4}$

Explanation

Bond angle :

$CH_{4} > SiH_{4} > GeH_{4} > SnH_{4}$

this is wrond order because of SNH4 show highest bond angle

$SiO_{2}$ reacts with :

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

Explanation

$SiO_2 + Na_2CO_3= Na_2SiO_3$

$SiO_2 + HF= SiF_6$

so A and C is correct

List -I List -2
A. IVA group   Crystalline form
B. Onyx   2 Amorphous form of silica
C. Lubricant Graphite
D. Hardest material   4.ns2 np2
Diamond
The correct match with A B C D is ?

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

Explanation

IV A Group has 4 valence electrons and are arranged as

$ns^2$

$np^2$ .

Onyx is the amorphous form of silica.

Diamond is one of the hardest materials due to the strong covalent bonds which connect various atoms in the lattice.

Graphite is used as a lubricant due to the weak bonding between its layers, which allows the layers to slide on one another.

An alkyl halide reacts with a group 14 element

$Y$ at 570K with

$Cu$ as a catalyst and produces a dialkyl chloro compound

$Z$ . The compound

$Z$ on hydrolysis gives another compound which is a strong water repellent and quite inert chemically. The dioxide of

$Y$ is acidic in nature. The alkyl halide can also be obtained from methane after monosubstitution. Identify

$Y$ and

$Z$ .

0%
$Y = Si$ and $Z = (CH _{3})_{2} SiCl_{2}$
0%
$Y = Si$ and $Z = CH_{3}Cl$
0%
$Y = C$ and $Z = SiCl_{4}$
0%
$Y = Si$ and $Z = CCl_{4}$

Explanation

Monosubstitution of methane gives methyl chloride. It reacts at

$570K$ with

$Si$ having

$Cu$ as catalyst to produce

$(CH _{3})_{2} SiCl_{2}$ , which is the compound

$Z$ .

$(CH _{3})_{2} SiCl_{2}$ on hydrolysis gives

$(CH _{3})_{2} Si(OH)_{2}$ , which is a strong water repellent and quite inert chemically. Also

$SiO_2$ is acidic in nature.

Option 'A' is correct.

A metal

$X$ produces an oxide and nitride on burning in air, but does not liberate hydrogen with alkali. Another metal

$Y$ produces an oxide and nitride on burning in air and also liberates hydrogen with alkalies. The metals

$X$ and

$Y$ respectively are:

0%
Na, Mg
0%
Mg, Al
0%
Na, Al
0%
Al, Mg

Explanation

When magnesium is burnt in air, it forms magnesium oxide and magnesium nitride.

$2Mg+O_2 \rightarrow 2MgO$

$3Mg+N_2 \rightarrow Mg_3N_2$

Magnesium does not liberate hydrogen with alkali.

When aluminium is burnt in air, it forms aluminium oxide and aluminium nitride.

$4Al+3O_2 \rightarrow 2Al_2O_3$

$2Al+N_2 \rightarrow 2AlN$

Aluminum also reacts with alkali to liberate hydrogen.

$2Al+2NaOH+2H_2O \rightarrow 2NaAlO_2 + 3H_2$

The substance 'B' is

0%
$Na_2B_{4}O_{7}$
0%
$B_{2}O_{3}$
0%
$H_{3}BO_{3}$
0%
$HBO_{2}$

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:

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

When borax is heated, it loses water and swells into a white mass which is 'B'. The formula is

$Na_2B_4O_7$ . The reaction is as follows:

$Na_2B_4O_7.10H_2O \xrightarrow {heat} \underset {B} {Na_2B_4O_7} + 10H_2O$

On further heating, 'B' forms glossy solid called borax glass and borax bead (compound C or

$B_2O_3$ ). The reaction is as follows:

$Na_2B_4O_7 \xrightarrow {heat} 2NaBO_2 + \underset {C} {B_2O_3}$

When 'C' is heated with nickel oxide, a brown bead 'D' is obtained which is nickel metaborate

$(Ni(BO_2)_2)$ . The reaction is as follows:

$B_2O_3 + NiO \rightarrow \underset {\text{brown bead} \ (D)} {(Ni(BO_2)_2)}$

Match the ores from List-I with their corresponding formulae from List-II.

List -I List -II
A. Borax 1)

$Al_{2}O_{3}.2H_{2}O$
B. Razorite 2)

$Ca_{2}B_{6}O_{11}.5H_{2}O$
C. Bauxite 3)

$Na_{2}B_{4}O_{7}.4H_{2}O$
D.Colemanite 4)

$Na_{2}B_{4}O_{7}.10H_{2}O$

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

Explanation

Borax:

$Na_{2} B_{4} O_{7} .10H_{2} O$

Razorite:

$Na_{2} B_{4} O_{7} .4H_{2} O$

Bauxite:

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

Colemanite:

$Ca_{2} B_{6} O_{11} .5H_{2} O$

So, the correct option is 'C'.

The number of

$-OH$ units directly bonded to boron atoms in borax is:

0%
10
0%
4
0%
5
0%
3

The tendency to show inert pair effect follows the order :

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

Explanation

Down the group inert pair effect increases so the correct order is

$B < Al < Ga < In < Tl$ .

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

0:0:1

Practice Class 11 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.