MCQ Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 3

Addition of halogen

$(C{l_2} \ or \ B{r_2})$ to alkenes goes through formation of :

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Carbocation
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Carbanion
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Cyclic halonium ion
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Carbon free radical

Alkali metals dissolve in liquid ammonia and produces _______.

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deep blue solution
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a solution which are conducting in nature
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red coloured solution
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Both $1$ and $2$

Explanation

Alkali metal when reacts with liquid ammonia reaction is given by,

$M +(x+y) NH_3\rightarrow M^+(NH_3)x + e^{-}(NH_3)_y$

this electrons is free and gives very good conductivity as well blue colour.

When

$PbO_{2}$ reacts with conc.

$HNO_{3}$ , the gas evolved is

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$NO_{2}$
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$O_{2}$
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$N_{2}$
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$N_{2}O$

Which one of the halogen is prepared by only electrolysis method?

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${ Cl }_{ 2 }$
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${ Br }_{ 2 }$
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${ F }_{ 2 }$
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${ I }_{ 2 }$

which of the following is not characteristic of interhalogen compounds?

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They are more reactive than halogens.
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They are covalent in nature.
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They have low boiling point and are non-volatile.
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they are diamagnetic.

The basic radial which is detected by the reagent which is also used to detect phosphate is:

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$Cd^{2+}$
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$Sn^{2+}$
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$Sb^{3+}$
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$As^{3+}$

Which of the following product (s) is/are obtained when

$C{I_2}O$ reacts with

$N{H_3}$ ?

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$N{O_2}$
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${N_2}$
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$NC{I_3}$
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$N{H_4}CI$

Explanation

$C{l_2}O + N{H_3} \to {N_2} + N{H_4}Cl + {H_2}O$

Which one does not liberate

$NH_3$ when undergoes hydrolysis?

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Acetanliide
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Acetonitrile
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Acetamide
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Phenyl isocyanide

Which is not correct a reaction?

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$XeF_{6}+NaF\rightarrow \left [ XeF_5 \right ]^{+} (2F^{-}) (Na^{+})$
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$Xef6+PF5\rightarrow \left [ XeF_5^{+} \right ]\left [ PF^{-}_6\right ]$
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$XeF_{6}+SbF_{5}\rightarrow \left [ XeF_5^{+} \right ]\left [ SbF_6^{-} \right ]$
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$XeF_{2}+PF_{5}\rightarrow \left [XeF^{+} \right ]\left [ PF_6^{-} \right ]$

Most reactive form of phosphorus is

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White phosphorus
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Red phosphorus
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Black phosphorus
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All allotropes are equally reactive

Which of the following statements is correct ?

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In trisilylamine $p\pi - d\pi$ back bonding takes place.
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These are two $p\pi - d\pi$ bonds in molecule.
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Fraction of s-character in hydride around the central atom in molecule is 0.5
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${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{COH}}$ is more acidic than ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{SiOH}}$

The correct statement(s) about the oxoacids,

$HCl{O}_{4}$ and

$HClO$ , is(are):

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the central atom in both $HCl{O}_{4}$ and $HClO$ is $s{p}^{3}$ hybridized
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the conjugate base of $HCl{O}_{4}$ is weaker base than ${H}_{2}O$
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$HCl{O}_{4}$ is formed in the reaction between ${Cl}_{2}$ and ${H}_{2}O$
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$HCl{O}_{4}$ is more acidic than $HClO$ because of the resonance stabilization of its anion

Explanation

The correct statement(s) about the oxoacids,

$\displaystyle HClO_4$ and

$\displaystyle HClO$ , are the following

(A) The central atom in both

$\displaystyle HClO_4$ and

$\displaystyle HClO$ is

$\displaystyle sp^3$ hybridized. The central

$Cl$ atom in

$\displaystyle HClO_4$ has 4 bonding domains and zero lone pairs of electrons.
The central

$O$ atom in

$\displaystyle HClO$ has 2 bonding domains and 2 lone pairs of electrons.

(B) The conjugate base of

$\displaystyle HClO_4$ is weaker base than

$\displaystyle H_2O$ . This is because

$\displaystyle HClO_4$ is stronger acid than water. A stronger acid has weaker conjugate base and vice versa.

(D)

$\displaystyle HClO_4$ is more acidic than

$\displaystyle HClO$ because of the resonance stabilization of its anion. Greater is the extent of delocalisation, greater is the stabilisation of negative charge and higher is the acidity. In

$\displaystyle HClO_4$ , negative charge is delocalised over

$4 O$ atoms.
The following statement is incorrect.

(C)

$\displaystyle HClO_4$ is formed in the reaction between

$\displaystyle Cl_2$ and

$\displaystyle H_2O$ ."
Instead, the following methods are used.

$\displaystyle NaClO_4+HCl \rightarrow NaCl + HClO_4$
Laboratory preparations of

$\displaystyle HClO_4$ include the following
Treatment of barium perchlorate with sulfuric acid to give barium sulfate and perchloric acid.
Nitric acid is mixed with ammonium perchlorate to form nitrous oxide and perchloric acid

The correct structure of tribromooctaoxide is:

Explanation

The correct structure is A

$O=\overset{O}{\overset{||}{\underset{O}{\underset{||}{Br}}}}-\overset{O}{\overset{||}{\underset{O}{\underset{||}{Br}}}}-\overset{O}{\overset{||}{\underset{O}{\underset{||}{Br}}}}=O$
Tribromooctaoxide

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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Both Assertion and Reason are incorrect

Explanation

It is the most electronegative element, given that it is the top element in the halogen group, and therefore is very reactive.

It has low bond dissociation enthalpy making it more reactive and form noble gas configuration.

Hence, option

$A$ is correct.

Assertion: F is more electronegative than

$Cl$ .
Reason: F has high electron affinity than

$Cl$ .

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If both assertion and reason are true and the reason is a correct explanation of the assertion
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If both assertion and reason are true but reason is not a correct explanation of the assertion
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If the assertion is true but reason is false
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If both the assertion and reason are false
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If the assertion is false but the reason is true

Explanation

Here assertion is true but reason is false. F is more electronegative than chlorine, in fact it is most electronegative element of the periodic table but its electron affinity, the tendency to accommodate electrons, is less than the

$Cl$ due to non availability of

$d$ -electrons.

Assertion: The fluorine has lower reactivity.
Reason :

$F - F$ bond has low bond dissociation energy.

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Both assertion and reason are true and the reason is a correct explanation of the assertion
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Both assertion and reason are true but reason is not a correct explanation of the assertion
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Assertion is true but reason is false
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Both the assertion and reason are false
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Assertion is false but the reason is true

Explanation

Fluorine have the highest electronegativity of all the elements because it has the largest nuclear charge and the least atomic shell, so it has the largest nuclear pull force. In other words, fluoride has the strongest ability to pull or catch an electron to form negative ion (that is, electronegativity).

Bond dissociation energy of $F_2$ is fairly low due to the lone pair - lone pair repulsions of the two flourine atoms, which are placed so close to each other due to small size of the atom.

Inert gases are mixed in iodine vapours. Then there are ________ between them.

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H-bonding
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Vander Waal's forces
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Electrostatic forces
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Metallic bonds

Explanation

Noble gases are least reactive. Their inertness to chemical reactivity is beacuse of the completely filled

$ns^2np^6$

electronic configuration in their valence shell except helium.

Hence, their only exists the short range force i.e. Vander waal's force between them.

The strongest oxidising agent is :

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$F_2$
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$Cl_2$
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$I_2$
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$Br_2$

Explanation

An element which can readily accept an electron is a good oxidising agent. Halogens are good oxidising agents because they have strong tendency to accept an electron due to

$(i)$ small atomic size,

$(ii)$ high electronegativity and

$(iii)$ high electron affinity. Fluorine, being smallest, is best oxidising agent.

The correct order of solubility in water for He, Ne, Ar, Kr, Xe is :

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$He > Ne > Ar > Kr > Xe$
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$Xe > Kr > Ar > Ne > He$
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$Ne > Ar > Kr > He > Xe$
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$Ar > Ne > He > Kr > Xe$

Explanation

As the molecular weight of noble gas atoms increases down the group its polarity increases due to which van-der-waal's force between them increases. Due to increased polarity of heavier inert gas, its solubility in water also increases. So, most soluble gas will be

$Xe$ and least soluble will be

$He$ . So correct order is

$Xe > Kr > Ar > Ne He$ .

The reaction of

${P}_{4}$ with

$X$ leads selectively to

${P}_{4}{O}_{6}$ . The

$X$ is:

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dry ${O}_{2}$
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a mixture of ${O}_{2}$ and ${N}_{2}$
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moist ${O}_{2}$
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${O}_{2}$ in presence of aq. $NaOH$

Explanation

Phosphorus trioxide (

$P_4O_6$ ) is prepared by heating phosphorus in the limited supply of oxygen and nitrogen.

${ P }_{ 4 }+3{ { O }_{ 2 } }\xrightarrow [ ]{ { O }_{ 2 }+{ N }_{ 2 } } { P }_{ 4 }{ O }_{ 6 }$

${N}_{2}$ prevents further oxidation of

${ P }_{ 4 }{ O }_{ 6 }$ to

${ P }_{ 4 }{ O }_{ 10 }$ .

For electron affinity of halogens which of the following is correct?

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$Br > F$
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$F > Cl$
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$Br < Cl$
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$F < I$

Explanation

Electron affinity decreases in a group from top to bottom due to increase in size, so the expected order of electron affinity for halogens is

$F > Cl > Br > I$
However, it is not so. The electron affinity of elements of

$III$ period is higher than that of

$II$ period because elements of

$II$ period have a small size and greater electron density so the incoming electrons suffer a repulsive force. Thus the order of electron affinity is

$Cl > F > Br > I$ .

The electronic configuration of halogens in their valence shell is:

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$\mathrm{n}\mathrm{s}^{2}$
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$\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{6}$
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$\mathrm{n}\mathrm{s}^{0}$
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$\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{5}$

Explanation

$\underline{Halogens}$
Electronic configuration of halogens in their valence shells =

$ns^{2} np^{5}$

Fluorine does not show variable oxidation states because_________

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it is a halogen
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of absence of d-orbitals
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absence of s-orbital
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absence of p-orbitals

Assertion (A): Oxygen exhibits anomalous behavior.
Reason (R): Absence of d-orbitals

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Both Assertion and Reason are true and Reason is the correct explanation of Assertion
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Both Assertion and Reason are true and Reason is not the correct explanation of Assertion
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Assertion is true but Reason is false
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Assertion is false but Reason is true

Anomalous behaviour of oxygen, compared to other VIA group elements is due to:

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its high electronegativity
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its small atomic size
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non availability of d-orbitals
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all

The oxidation number of chlorine in KClO is:

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+ 1
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+ 3
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+ 5
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+ 7

Explanation

$\overset{+1}{K}Cl\overset{-2}{O}$
oxidation state of alrali metals = +1
oxygen = -2

Let oxidation state of chlorine is x.

$\therefore$ x+1+(-2) = 0

$\therefore$ x = +1

Thus, oxidation state of chlorine in potassium chlorate is +1.

The halogen which can form both cations and anions is:

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F
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Cl
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Br
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I

Explanation

Iodine can form both cations, as well as anions. Iodine can form cations due to the poor shielding effect of d orbitals where electrons can be removed from the valence shell and the cation is formed.

The cations and anions of Iodine is represented as follow:

$I^{-}$ and

$I^{+}$

The number of p-electrons in bromine atom is:

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17
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7
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15
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13

Explanation

$Br - Z=35$

Its electronic configuration is given as:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^5$

So, total p-electrons

$=6+6+5= 17$

Which of the following is not formed in the reaction of Xe with

$F_2$ ?

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$\mathrm{X}\mathrm{e}\mathrm{O}\mathrm{F}_{4}$
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$\mathrm{X}\mathrm{e}\mathrm{F}_{2}$
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$\mathrm{X}\mathrm{e}\mathrm{F}_{4}$
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$\mathrm{X}\mathrm{e}\mathrm{F}_{6}$

Explanation

$Xe+F_2\xrightarrow \ XeF_2$

$Xe+2F_2\xrightarrow \ XeF_4$

$Xe+3F_2\xrightarrow \ XeF_6$

$XeOF_4$ is never formed in reaction of Xe with

$F_2$

The halogen which can displace three halogens from their compounds is:

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$F$
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$Cl$
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$Br$
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$I$

Fluorine shows anomalous behavior in VIIA group due to:

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its small size
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high electronegativity
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absence of d-orbitals
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all the above

Halogen elements are those which are:

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diatomic and form $X^{--}$ ions
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monatomic and form $\mathrm{X}^{-}$ ions
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diatomic and form $\mathrm{X}^{-}$ ions
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monoatomic and form $X^{--}$ ions

Explanation

Halogen atoms have 7 electrons in their valence shell. So, they require only one electron to complete its octet. Thus, halogens are present in a diatomic form in order to complete their octet by sharing of electrons. Also,

$x+e^- \rightarrow x^-\ (stable\ for\ Halogens)$

Which is called lazy gas?

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Kr
0%
Ar
0%
He
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Ne

What is the atomic number (Z) of the noble gas that reacts with fluorine?

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54
0%
10
0%
18
0%
2

Explanation

Atomic number 54 is for Xe.

It reacts with fluorine and forms compounds, like

$XeF_2$ ,

$XeF_4$

Option A is correct.

The word Argon represents the term_______

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Sun
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Hidden
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Lazy
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Stranger

Halogens means:

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ore forming elements
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sea salt products
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inert gases
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rare gases

Explanation

Sea salt

$\to$ like

$Mgcl_{2},\ MgBr_{2}$ have halogens.

$\therefore$ Halogens are sea salt products

The element that oxidizes water to

$\mathrm{O}_{2}$ with a large evolution of heat is:

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$\mathrm{F}_{2}$
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$\mathrm{Cl}_{2}$
0%
$\mathrm{B}\mathrm{r}_{2}$
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$I_{2}$

Explanation

$F_2$ is the most powerful oxidizing
agent and evolves large amount
of heat due to the formation of H - bond
with

$H_2O$ ( It is also the most electronegative element )

The oxidation state of chlorine in bleaching powder is:

0%
-2
0%
- 1
0%
+ 1
0%
-1 and + 1

Explanation

The chemical formula of bleaching powder is

$CaOCl_{2}$ .

Here, calcium is bonded to

$Cl^{-}$ ion and

$OCl^{-}$ ion respectively.

$Cl^{-}$ the oxidation state of chlorine is -1 and in

$OCl^{-}$ ion the oxidation state of chlorine is +1.

Fluorine does not show positive oxidation states because:

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it is most electronegative element
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it forms only anions in ionic compounds
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it cannot form multiple bonds
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it shows non-bonded electron pair repulsion due to small size.

$\mathrm{F}_{2}$ is better oxidising agent than

$\mathrm{B}\mathrm{r}_{2}$ . It is due to:

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small size of flourine atom
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more electron repulsions in flourine
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higher electronegativity of flourine
0%
non-metallic nature of fluorine

Explanation

F is more electronegative than Br. Therefore it can easily accept

$e^-$ and oxidise others.
Therefore

$F_2$ is better oxidising agent than

$Br_2$

The element which can displace three other halogens from their compound is :

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$Cl$
0%
$Br$
0%
$I$
0%
$F$

Which is the lightest gas?

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Helium
0%
Oxygen
0%
Hydrogen
0%
Nitrogen

Inversion temperature of helium is very low. So when helium is allowed to expand into vacuum it gets:

0%
Cooled
0%
Heated
0%
Neither cooled, nor heated X
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Liquified

Explanation

$He$ gets heated at their inversion temperature on expansion into vacuum.

The noble gas which has abnormal behaviour in its liquid state is_______.

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He
0%
Ar
0%
Xe
0%
Ne

Which of the following are formed by xenon?

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$\mathrm{X}\mathrm{e}\mathrm{F}_{7}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{4}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{5}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{3}$

Explanation

$XeF_4 -$ This compound is formed by xenon .

Xe compounds can contain only even no. of fluorine atoms.

Xe has a complete filled 5p configuration.As a result when it undergoes bonding with an odd number (3 or 5) of F atoms it leaves behind one unpaired electron. This causes the molecule to become unstable.

Which of the following fluoride(s) of xenon cannot be prepared?

0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{2}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{3}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{6}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{7}$

Viscosity is very low for?

0%
Ar
0%
Xe
0%
He
0%
Kr

Explanation

Noble gases have low melting and boiling point and lowest boiling point among them is of

$He$ (

$4.2K$ ). This is because they have only type of inter-atomic interaction that is weak dispersion forces.

During fractional evaporation of liquid air, nitrogen comes out as a fraction along with:

0%
Ar, Kr, Xe
0%
Only Ar
0%
He and Ne
0%
Kr, Xe

Which of the following compound cannot be prepared?

0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{2}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{3}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{4}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{6}$

Explanation

Xe has a complete filled 5p configuration. As a result when it undergoes bonding with an odd number (3 or 5) of F atoms it leaves behind one unpaired electron. This causes the molecule to become unstable. As a result

$XeF_3$ ,

$XeF_5$ not possible.

Only even no. of fluorides can combine with Xe. Therefore

$XeF_3$ cannot be formed, Only even bonds can be formed.

The heat of vaporization is very high for ?

0%
He
0%
Ne
0%
Ar
0%
Xe

Explanation

Heat of vapourisation

$\times$ van der Waal's forces

$\times$ malecular mass
Xe has high malecular mass
Therefore heat of vapourisation
is very high for Xe

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 3 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 3 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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