MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 14

Borax structure contains?

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two $BO_{4}$ groups and two $BO_{3}$ groups
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four $BO_{4}$ groups only
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four $BO_{3}$ groups only
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three $BO_{4}$ groups and one $BO_{3}$ groups

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

	Column-I		Column-II
P	Borax	i	$Na_3AlF_6$
Q	Inorganic benzene	ii	$Na_2B_4O_7.10H_2O$
R	Cryolite	iii	$Al_2O_3.2H_2O$
S	Bauxite	iv	$B_3N_3H_6$

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P-ii, Q-iv, R-i, S-iii
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P-i, Q-ii, R-iii, S-iv
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P-ii, Q-iii, R-i, S-iv
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P-ii, Q-i, R-iv, S-iii

The

$M.P.$ of

$SnCl_4$ is less than that of

$SnCl_{2}$ . The suitable reason for the given observation is:

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There is more charge on $Sn^{4+}$
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The size of $Sn^{4+}$ is small
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Ionic potential of $Sn^{4+}$ is high
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The shape of $SnCl_{4}$ is tetrahedral

The correct polymerization tendency order of oxoanions of third period non-metal is:

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$Si>P>S>Cl$
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$Cl>S>P>Si$
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$P>Si>Cl>S$
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$S>P>Cl>Si$

$\mathrm { SiO } _ { 2 }$ can react with:

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$NaOH$ only
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$HF$ only
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Neither $HF$ nor $NaOH$
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Both $HF$ and $NaOH$

Explanation

Silicon dioxide reacts with sodium hydroxide to product sodium silicon and water. This reaction takes place at a temperature of

$900-1000\ ^oC$

$2NaOH+S:O_{2}\longrightarrow Na_{2}S:O_{3}+H_{2}O$

Silicon dioxide reacts with hydrogen fluroide to produce hydrogen hexafluro silicate & water

$S:O_{2}+6HF\longrightarrow H_{2}[S:F_{6}]+2H_{2}O$

$S:O_{2}$ definitely reacts with strong bases like

$NaOH$

$S:O_{2}$ is a very stable compound, and therefore it can also react with

$HF$ .

Graphite is soft while diamond is hard because:

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graphite is in powder form.
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diamond has $sp^{2}$ - hybridisation but graphite has $sp^{3}$ -hybridisation.
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graphite is in planar form while diamond is in tetrahedral form.
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graphite is covalent and diamond is ionic.

The correct order of atomic radii of group 13 elements is:

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$B < Al < Ga < In$
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$B < Ga < Al < In$
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$B < Al < Ga$ $\sim$ $In$
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$B < Al$ $\sim$ $Ga > In$

Select correct order:-

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$F_2 > Cl_2 > Br_2 > I_2$ (order of bond energy)
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$O > S > Se > Te$ (Order of tendency of catenation)
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$BF_3 > BCl_3 > BBr_3 > BI_3$ ( Order of lewis acidic strength)
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$p-Nitrophenol < o-Nitrophenol$ (Order of vapour pressure)

The hydridisations of boron in borax are

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Two borons are $sp^2$ and other two are $sp^3$ hybridised
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One boron is $sp^3$ and three borons are $sp^3$ hybridised
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Three borons are $sp^2$ and one boron is $sp^3$ hybridised.
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All are in $sp^3$ hybridisation.

What is correct statement among the following about the structure of borax molecule?

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all the four boron atoms are $s{p^3}$ -hybridized
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all the four boron atoms are $s{p^2}$ - hybridized
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two of the four boron atoms are $s{p^3}$ -hybridized and the other two are $s{p^2}$ -hybridized
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Out of the 10 water molecules generally shown as water of crystallization in its MF,4 molecules of ${H_2}O$ are really involved in the structural formation

Pick the correct statement :

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Borax is white crystalline solid containing $[B_{4}O_{5}(OH)_{4}]^{2-}$
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Aqueous solution of borax is acidic in nature.
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Cobalt gives green colour in borax bead test.
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All are incorrect.

Which of the following oxidation states are the most characteristic for lead and tin?

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$+2,\,+2$
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$+4,\,+2$
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$+2,\,+4$
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$+4,\,+4$

Which of the following statement is wrong regarding borax?

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Alkaline solution is formed when barox is dissolved in water
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Each boron atom is borax ,olecular is bonded to four oxygen atoms
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Borax is used a buffer
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Geometry around two B atoms is triangular while it is tetrahedral around the other two units

$(CH_3)_3CCH_2COOH$ is more acidic than

$(CH_3)_3SiCH_2COOH$ because :

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Size of Si is more than that of carbon
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Electronegativity of carbon is less than that of silicon
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Silicon is more electropositive than carbon due to which $(CH_3)_3SiCH_2COO^-$ because
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None of the above

Stability of hydrides of carbon family is in order

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${\text{C}}{{\text{H}}_{\text{4}}}{\text{ > Si}}{{\text{H}}_{\text{4}}}{\text{ > Ge}}{{\text{H}}_{\text{4}}}{\text{ > Sn}}{{\text{H}}_{\text{4}}}{\text{ > Pb}}{{\text{H}}_{\text{4}}}$
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${\text{C}}{{\text{H}}_{\text{4}}}{\text{ < Si}}{{\text{H}}_{\text{4}}}{\text{ < Ge}}{{\text{H}}_{\text{4}}}{\text{ < Sn}}{{\text{H}}_{\text{4}}}{\text{ < Pb}}{{\text{H}}_{\text{4}}}$
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${\text{C}}{{\text{H}}_{\text{4}}}{\text{ > Sn}}{{\text{H}}_{\text{4}}}{\text{ > Ge}}{{\text{H}}_{\text{4}}}{\text{ > Si}}{{\text{H}}_{\text{4}}}{\text{ > Pb}}{{\text{H}}_{\text{4}}}$
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None of these

The most abundant

$IVA$ group elements in the earth's crust is

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Germanium
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Carbon
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Silicon
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Tin

Which is incorrect statement about silicones?

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They are repeating units ( $Si{O}_{4}$ ) in silicates
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They are synthetic polymers containing repeated ${R}_{2}SiO$ units
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They are formed by hydrolysis and on subsequent polymerisation of ${R}_{2}Si{Cl}_{2}$
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Silicones are used as a lubricants

Which one is non aromatic compound

The number of naturally occurring

$p-block$ elements, that are diamagnetic is

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$7$
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$6$
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$11$
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$5$

Which of the following does not exist ?

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$BiCl_{5 }$
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Xe ${ H }_{ 4 }$
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$BiF_{ 5 }$
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Both (A) and (B)

Which of the following is more stable?

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$Pb^{4+}$
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$Sn^{2+}$
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$Ge^{4+}$
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$Ge^{2+}$

For which of the following change in Gibbs free energy will be negative?

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$Ge^{+2} \longrightarrow Ge^{+4}$
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$Pb^{+2} \longrightarrow Pb^{+4}$
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$Sn^{+4} \longrightarrow Sn^{+2}$
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$Ga^{+3} \longrightarrow Ga^{+4}$

Total number of sigma bonds formed by all the boron atoms in borax is

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14
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4
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9
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7

In which of the following element +1 oxidation state is more stable ?

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$B$
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$Al$
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$Ga$
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$Tl$

The number of

$OH$ units directly linked to boron atoms in

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

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2
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3
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4
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10

Choose the wrong statement:

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$Ge, Sn$ and $Pb$ from $+2$ as well as $+4$ oxidation state
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$Ge^{2+}$ and $Sn^{2+}$ compounds act as reducing agents
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$Pb^{4+}$ compounds act as oxidizing agents
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Stability increases from $Ge^{2+}$ to $Pb^{2+}$

Aluminium exist in +3 state where as thallium exist in both +1 & +3 oxidation state. Reason for this is

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inert pair effect
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Lanthanoid contraction
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Diagonal relationship
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none of these

Which of the following silicate is called disilicate?

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Orthosilicate
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Pyrosilicate
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Single -chain silicate
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None of these

Select the correct statements :

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The $+1$ oxidation compared to the $+3$ oxidation state gets gradually more stabilized as we move from top to bottom in III group
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$Tl (III)$ being unstable, acts as a good oxidizing agent to get reduced to $Tl (I)$
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$SnCl_2$ is good oxidizing agent
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All are correct

Which of the following compounds are the common products obtained in the hydrolysis of

$XeF_6$ and

$XeF_4$ ?

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$XeO_2F_2$
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$HF$
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$XeO_3$
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Both (A) and (B)

The solution of borax in water is:

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Neutral
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Acidic
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Alkaline
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Amphoteric

Which of the following is/are example of orthosilicate?

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$ZrSiO_4$
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$Mg_2SiO_4$
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$Zn_3Si_2O_7$
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Asbsetos

Carbon black is obtained when methane is :

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heated in absense of air
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heated in presence of nitrogen
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heated in presence of ammonia
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heated with steam

Single Correct Answer Type
Consider the given sequence of reaction and identify
P and Q:
Q (Green colored)

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$P=Na_3,BO_3,Q=Cr(BO_2)_3$
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$P=B_2O_3,Q=CrBO_3$
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$P=Na_2B_4O_7,Q=Cr(BO_2)_3$
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$P=B_2O_3,Q=Cr(BO_2)_3$

Which of the following elements do not form carbide?

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$B$
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$Al$
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$In$
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$Ga$

$Me_3SiCl$ is used during polymerisation of organo silicones because:

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The chain length of organo silicone polymers can be controlled by adding $Me_3SiCl$
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$Me_3SiCl$ blocks the end terminal of silicone polymer
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$Me_3SiCl$ improves the quality and yield of the polymer
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$Me_3SiCl$ acts as a catalyst during polymerisation

Borax on heating strongly above its melting point melts to a liquid, which then solidifies to a transparent mass commonly known as borax-bead. The transparent glassy mass consists of:

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Sodium pyroborate
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Boric anhydride
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Sodium meta-berate
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Boric anhydride and sodium metaborate

Explanation

Borax (

$Na_{2}B_{4}O_{7}.10H_{2}O$ ) when dissolved in water forms orthoboric acid and an alkaline solution. On heating strongly above its melting point, borax first looses its water and melts to a liquid, which then solidifies to a transparent mass, commonly known as borax-bead. This borax-bead is made of boric anhydride and sodium metaborate.

$Na_{2}B_{4}O_{7}.10H_{2}O \overset{Heat}{\rightarrow} B_{4}O_{7} \overset{Heat}{\rightarrow} 2NaBO_{2} + B_{2}O_{3}$

Hence , option D is correct.

An element belongs to 3rd period and group -13 of the periodic table. Which of the following properties will be shown by the elements?

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Good conductor of electricity
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Liquid, Metallic
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Solid, metallic
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Solid, Non - metallic

Aqueous solution of a salt (Y) is alkaline to litmus. On strong heating, it swells-up to give a glassy material. When conc.

$H_2SO_4$ is added to a hot concentrated solution of (Y), white crystals of a weak acid separates out. Hence, the compound (Y) is:

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$Na_2SO_4.10H_2O$
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$Ca_2P_6O_{11}.10H_2O$
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$Na_2B_6O_{11}$
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$Na_2B_4O_7.10H_2O$

Explanation

$H_3BO_3$ is weak acid.

Option D is correct.
1548972_1726168_ans_bf0c75c092594e508b86c9fdc83a47e2.JPG

The incorrect statement/s among the following is/are:

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$NCl_5$ does not exists while $PCl_5$ does
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Lead prefers to form tetravalent compound
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The three $C-O$ bonds are not equal in the carbonate ion
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Both ${O_2}^{+}$ and NO are paramagnetic

Explanation

$NCl_5$ does not exist while

$PCl_5$ does because there are no vacant d-orbitals in

$N.$

Lead prefers to form divalent compounds due to inert pair effects.

In carbonate ion, all the three

$C-O$ bonds are equal due to resonance.

Both

$O_2^+$ and

$NO$ are paramagnetic as there is one unpaired electron.

Among

$Al_2O_3, SiO_2, P_2O_3$ , and

$SO_2$ , the correct order of acid strength is:

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$Al_2O_3 < SiO_2 < SO_2 < P_2O_3$
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$SiO_2 < SO_2 < Al_2O_3 < P_2O_3$
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$SO_2 < P_2O_3 < SiO_2 < Al_2O_3$
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$Al_2O_3 < SiO_2 < P_2O_3 < SO_2$

Explanation

Acidinc strength depends upon non-metallic character of atom. As we move left to right in a period non-metallic character increases and hence acidity increases.

So correct order is:

$Al_2O_3 < SiO_2 < P_2O_3 < SO_2$

Which element exhibits both

$+1$ and

$+3$ oxidation numbers in its compounds?

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$B$
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$Be$
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$Sn$
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$Tl$

Select the incorrect statement(s):

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$p$ -orbital can accommodate $6e^-$
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$Na \xrightarrow{I.E_1} Na^+ \xrightarrow{I.E._2} Na^{2+} \xrightarrow{I.E._3} Na^{3+}$ ; the correct order of ionization energy is $I.E._2 > I.E._3 > I.E._1$
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$Xe$ is a noble gas
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$B$ is a non-metal

Explanation

(a)

$\to$ p-orbital can accommodate a maximum of 2 electrons.

(b)

$\to Na:1s^2 2s^22p^63s^1$

$Na^+ : 1s^2 2s^2 2p^6$

$Na^{2+} : 1s^2 2 s^2 2p^5$

$Na^{3+} : 1s^2 2s^2 2p^4$

$I.E. _3 > I.E._2 > I.E._1$

$I.E._3$ is greater than

$I.E._2$ because to convert

$Na^{2+} \to Na^{3+}$ , electron is removed from a cation having more positive charge.

The correct order of atomic radii of group

$13$ elements is

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$B < Al \sim Ga > In$
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$B < Al < Ga<In$
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$B < Al < Ga \sim In$
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$B < Ga < Al \sim In$

Explanation

Hint: There is some exception in the atomic radius of group 13 elements.

Step 1: Atomic radii of group 13 elements

$B$ : Boron has the smallest atomic radius among group 13 elements.

$Ga$ : In $Ga$ , the $d$ orbital does not shield the nucleus effectively. So the effective nuclear charge is higher for the outermost electrons. So, the atomic radius is smaller.

$Al$ : Down the group, the effective nuclear charge decreases as the number of shells increases, So the atomic radius of $Al$ is larger.

$In$ : In $In$ , the $d$ orbital does not shield the nucleus effectively. So the effective nuclear charge is higher for the outermost electrons. But down the group, the effective nuclear charge decreases as the number of shells increases. So, these two effects result in the atomic radius of $In$ similar to the atomic radius of $Al$ .

Step 2: Atomic radius order

$B<Ga<Al\mathtt{\sim}In$

Final Step: Correct option is (D) $B<Ga<Al\mathtt{\sim}In$

The number of

$OH$ units directly linked to boron atoms in

$Na_2B_4O_7 \cdot 10H_2O$ are:

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$2$
0%
$3$
0%
$4$
0%
$10$

Silica is reacted with sodium carbonate. Which gas liberated?

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

The blue coloured mineral 'Lapis Lazuli' which is used as a semi-precious stone is a mineral of the following class:

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sodium alumino silicate
0%
zinc cobaltate
0%
basic copper carbonate
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prussian blue

In silica

$SiO_2$ , each silicon atom is bonded to:

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two oxygen atoms
0%
four oxygen atoms
0%
one silicon and two oxygen atoms
0%
one silicon and four oxygen atoms

Which is likely to show inert pair effect?

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$K$
0%
$Mg$
0%
$Al$
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$Pb$

${ SiCl }_{ 4 }\xrightarrow [ ]{ { H }_{ 2 }O } X\xrightarrow [ ]{ 1000^0C } Y$
In the above reaction,

$X$ and

$Y$ respectively are:

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$SiO_2$ and $Si$
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$H_4SiO_4$ and $SiO_2$
0%
$H_2SiCl_6$ and $SiO_2$
0%
$H_4SiO_4$ and $Si$

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 14 - MCQExams.com

Zircon $(Z r S i O_{4})$ and forsterite $(M g_{2} S i O_{4})$ are examples of orthosilicates. They contain discrete tetrahedral of [SiO₄]⁴⁻.

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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 14 - MCQExams.com

Zircon (ZrSiO4​) and forsterite (Mg2​SiO4​) are examples of orthosilicates. They contain discrete tetrahedral of [SiO₄]⁴⁻.

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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Zircon $(Z r S i O_{4})$ and forsterite $(M g_{2} S i O_{4})$ are examples of orthosilicates. They contain discrete tetrahedral of [SiO₄]⁴⁻.