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

Sulphuryl chloride

$(SO_2Cl_2)$ reacts with white phosphorus

$(P_4)$ to give:

0%
$PCl_5, SO_2$
0%
$OPCl_3, SOCl_2$
0%
$PCl_5, SO_2, S_2Cl_2$
0%
$OPCl_3, SO_2, S_2Cl_2$

Explanation

Sulphuryl chloride (

$SO_2Cl_2$ ) reacts with white phosphorus (

$P_4$ ) to give phosphorous pentachloride and sulphur dioxide.

$\displaystyle 10SO_2Cl_2 + P_4 \rightarrow 4PCl_5 + 10SO_2$

Hence, option A is correct.

Bond energy of

$F_2, Cl_2$ and

$Br_2$ follow the order:

0%
$F - F > Cl - Cl > Br - Br$
0%
$Cl - Cl >Br - Br > F - F$
0%
$Br - Br > Cl - Cl > F - F$
0%
$Cl - Cl > F - F > Br - Br$

Explanation

The order of bond energy is :

$\displaystyle Cl-Cl (57 \: kcal/mol) > Br-Br (45.5 \: kcal/mol) > F-F (38 \: kcal/mol)$

Why is argon used in lamps?

0%
Argon forms molecules when electricity is passed through it
0%
Argon is inert and so does not react with the hot filament
0%
Argon is less dense than air
0%
Argon produces light when it burns

Ammonia gas can be dried over by:

0%
$Ca{ Cl }_{ 2 }$
0%
conc. ${ H }_{ 2 }S{ O }_{ 4 }$
0%
${ P }_{ 2 }{ O }_{ 5 }$
0%
quick lime

Explanation

Ammonia is prepared by heating a mixture of calcium hydroxide and ammonium chloride.

$Ca{(OH)}_2(s) + 2NH_4Cl(s) \longrightarrow Ca{Cl}_2(s) + 2H_2O(l) + 2NH_3(g)$

Concentrated sulphuric acid conc.

$H_2SO_4$ and anhydrous calcium chloride

$Ca{Cl}_2$ are not used to dry ammonia because they react with it. The gas is passed through fresh quicklime (solid calcium oxide lumps) to effectively dry it. Ammonia is collected by upward delivery as it is lighter than air.

Haber process is used for the production of:

0%
$NH_{3}$
0%
$HNO_{3}$
0%
$H_{2}SO_{4}$
0%
$O_{3}$

Explanation

Haber process is used for the production of ammonia (

$NH_3$ ) gas.

$N_{2}(g) + 3H_{2}(g) \leftrightharpoons 2NH_{3}(g)$

The table shows some information about elements in Group VII of the Periodic Table.

Name	State at room temperature	Colour
Chlorine	Gas	yellow-green
Bromine	liquid	brown
Iodine	?	?
Astaine	solid	black

Which information about iodine completes the table?

0%
State - liquid, colour - black
0%
State - liquid, colour - green
0%
State - solid, colour - grey
0%
State - solid, colour - yellow

Explanation

$\text{D) State-solid,Colour-yellow}$

$\text{Iodine is solid at room temperature and appears yellow in low concentrat}$

Which oxide is insoluble in aqueous sodium hydroxide?

0%
MgO
0%
$Al_2O_3$
0%
$P_4O_{10}$
0%
$SO_2$

Explanation

$\text{MgO is insoluble in aqueous sodium hydroxide.}$

Which of the following inert gas participate in chemical reaction?

0%
$Xe$
0%
$He$
0%
$Ne$
0%
None

Explanation

Xenon participate in chemical reactions because ionising power of

$Xe$ and oxygen are almost same and atomic size

$Xe$ is large.

Nitric acid can be obtained from ammonia via the formations of the intermediate compounds:

0%
nitric oxide and nitrogen dioxide
0%
nitrogen and nitric oxide
0%
nitric oxide and dinitrogen pentoxide
0%
nitrogen and nitrous oxide

Explanation

Ostwald process for the manufacture of nitric acid:

$\underset { \text{Ammonia} }{ 4{ NH }_{ 3 } } +{ 5O }_{ 2 }\xrightarrow [ 700-{ 800 }^{ o }C ]{ Pt\;\text{gauge }} \underset { \text{Nitric oxide} }{ 4NO } +6{ H }_{ 2 }O\quad ;\Delta H=-21.5\ kcal$

$2NO+{ O }_{ 2 }\xrightarrow [ ]{ { 50 }^{ o }C } \underset { \text{Nitrogen dioxide} }{ 2{ NO }_{ 2 } }$

$4{ NO }_{ 2 }+2H_2O+{ O }_{ 2 }\longrightarrow \underset {\text{ Nitric acid} }{ 4H{ NO }_{ 3 } } \quad$

Which of the following elements can be involved in p

$\pi-d\pi$ bonding?

0%
Carbon
0%
Nitrogen
0%
Phosphorus
0%
Boron

Explanation

If there is bonding between two atoms where one atom is having one vacant orbital and another is having one lone pair of electrons, then if this electron pair is donated to that respective vacant orbital then the bonding is called

$p\pi -p\pi$ or

$p\pi -d\pi$ depending on the orbital to which the electron pair is donated and from which the electron pair is donated.

Nitrogen, Carbon, and Boron can not form

$p\pi -d\pi$ bond because of the absence of d-orbitals in its valence shell. But Phosphorus can form

$p\pi -d\pi$ bonds.

Which among the following is inert gas and not very soluble in water too?

0%
$\displaystyle { O }_{ 2 }$
0%
$\displaystyle { CO }_{ 2 }$
0%
$\displaystyle { N }_{ 2 }$
0%
$\displaystyle { H }_{ 2 }{ SO }_{ 4 }$
0%
$\displaystyle He$

Explanation

$He$ is a n inert or noble gas.Noble/Inert gases are any of the seven chemical elements that make up Group

$18$ (VIIIa) of the periodic table. The elements are helium

$(He)$ , neon

$(Ne)$ , argon

$(Ar)$ , krypton

$(Kr)$ , xenon

$(Xe)$ , radon

$(Rn)$ , and element

$118$ ununoctium

$(Uuo)$ . The noble gases are colourless, odourless, tasteless, nonflammable gases.

Pressure is increased in the Haber process for making ammonia. It favors :

0%
the reverse reaction
0%
the forward reaction
0%
neither reaction
0%
none

Explanation

Pressure is increased in the Haber process for making ammonia. It favors the forward reaction.

$\displaystyle N_2 + 3H_2 \rightarrow 2NH_3$
The reaction proceeds with decrease in the number of moles of gaseous species. With increase in pressure, the equilibrium shifts in forward direction with decrease in the number of moles of gaseous species so that, the effect of increased pressure is nullified.

Which one of the following order is correct for the bond dissociation enthalpy of halogen molecules?

0%
$I_{2} > Br_{2} > Cl_{2} > F_{2}$
0%
$Cl_{2} > Br_{2} > F_{2} > I_{2}$
0%
$Br_{2} > I_{2} > F_{2} > Cl_{2}$
0%
$F_{2} > Cl_{2} > Br_{2} > I_{2}$

Explanation

Bond dissociation energy of halogen family decreases down the group as the size of atom increases.

The decreasing order for the bond dissociation enthalpy of halogen molecules is

$\displaystyle Cl_{2} \text { 57 kcal/mol} > Br_{2} \text { 45.5 kcal/mol} > F_{2} \text { 38 kcal/mol} > \text { 35.6 kcal/mol} I_{2}$ .

A halogen molecule having larger atoms should have low dissociation energy and vice versa. Fluorine is an exception because of interelectronic repulsion is present in small atom fluorine.

Hence, the correct option is

$B$ .

Which of the following is most electronegative element?

0%
Al
0%
F
0%
Ca
0%
Li
0%
C

Which element is more likely to exist in many different forms?

0%
Barium
0%
Phosphorous
0%
Rubidium
0%
Fluorine
0%
Manganese

Find the gases which is the densest at standard temperature and pressure.

0%
Helium
0%
Argon
0%
Carbon dioxide
0%
Xenon
0%
Nitrogen

Explanation

The gas which is the densest at standard temperature and pressure is xenon.
At standard temperature and pressure, pure xenon gas has a density of 5.761 kg/m

$^3$ , about 4.5 times the surface density of the earth's atmosphere, 1.217 kg/m

$^3$ . As a liquid, xenon has a density of up to 3.100 g/mL, with the maximum density occurring at the triple point.

In commercial industry, ammonia is prepared by :

0%
decomposition of salts
0%
contact process
0%
arc process
0%
combining hydrogen and nitrogen gases (Haber process)

Explanation

In commercial industry, ammonia is prepared by combining hydrogen and nitrogen gases (Haber process).

$N_2 + 3H_2 \rightleftharpoons 2NH_3+24\ kcal$

Read the two statements given below and answer whether they are true or false.

Statement I:

${He}_{2}$ is not a commonly known form.

Statement II:

$He$ is lighter than air.

0%
Both the statements are false
0%
Statement $I$ is true and Statement $II$ is false
0%
Statement $I$ is false and Statement $II$ is true
0%
Both the statements are true

Monatomic gas among the following at STP is :

0%
bromine
0%
hydrogen
0%
calcium
0%
neon
0%
nitrogen

Explanation

Noble gases are the only chemical elements which are stable single atom molecules at standard temperature and pressure. The noble gases are almost entirely unreactive or inert because, they all have a full outer shell of electrons. These are helium

$(He)$ , neon

$(Ne)$ , argon

$(Ar)$ , krypton

$(Kr)$ , xenon

$(Xe)$ and radon

$(Rn)$ . The heavier noble gases, like

$Xe$ , are exceptions as they can form chemical compounds by expending octet.

What are the missing products of the following reaction?

$\displaystyle { NH }_{ 4 }Cl+{ Ca\left( OH \right) }_{ 2 }\rightarrow \_ \_ \_ +Ca{ Cl }_{ 2 }$

0%
$\displaystyle { N }_{ 2 }$
0%
$\displaystyle { NH }_{ 3 }$
0%
$\displaystyle { H }_{ 2 }O$
0%
$\displaystyle { NH }_{ 3 }+{ N }_{ 2 }$
0%
$\displaystyle { NH }_{ 3 }+{ H }_{ 2 }O$

Explanation

Calcium hydroxide reacts with ammonium chloride to give ammonia, water and calcuim chloride.

${ 2NH }_{ 4 }Cl+Ca{ (OH) }_{ 2 }\rightarrow { 2NH }_{ 3 }+{ 2H }_{ 2 }O+{ CaCl }_{ 2 }$

The number of valence electrons present in noble gases is:

0%
$6$
0%
$18$
0%
$8$
0%
$4$

Explanation

The number of valence electrons present in noble gases is 8.
For example the noble gas with atomic number 18 has electronic configuration

$\displaystyle 1s^2 \: 2s^22p^6 \: 3s^23p^6$ . Since it has completed its octet, it is stable and least reactive.

The noble gases which form clathrate compounds:

0%
$Ar$
0%
$Xe$
0%
$Kr$
0%
$Ne$

Identify the true statement regarding nitrates of all metals.

0%
All nitrates decompose to give $N{ O }_{ 2 }$
0%
All nitrates are soluble in water
0%
All nitrates are insoluble in water
0%
All nitrates are stable to heat

Formation of

${O}^{-}$ is ________, whereas formation of

${O}^{-2}$ is __________.

0%
exothermic, endothermic
0%
endothermic, exothermic
0%
endothermic, endothermic
0%
exothermic, exothermic

Explanation

The formation of an

$O^-$ ion is exothermic, which means energy is released in the process. This is because adding an electron to

$O$ is much easier as it is electropositive and has a high electron affinity. The electron gain enthalpy in this case is

$-142 {\text{ kjmol}}^{-1}$ .

$O + e^- \rightarrow O^-$

$[\Delta r^\circ = -142 {\text { kjmol}^{-1}}]$

To form an

$O^{2-}$ ion, another electron has to be added to

$O^-$ ion. However, since

$O^-$ ion is already negatively charged, it resists the addition of a negatively charged electron. Thus, because of the repulsive coulumbic forces between the ion and the electron, more energy is required to overcome these forces. So, the formation of

$O^{2-}$ is an endothermic process as energy is absorbed. The electron gain enthalpy in this case is

$+844 {\text { kjmol}^{-1}}$ .

$O^- + e^- \rightarrow O^{2-}$

$[\Delta r^\circ = +844 {\text{ kjmol}^{-1}}]$

Nitrogen reacts with a metal

$M$ to produce a compound

$X$ . This compound

$X$ when dissolved in water forms pungent gas

$Y$ which can be identified easily by the formation of white fumes in the presence of

$HCl$ .

Identify Y:

0%
Nitrogen dioxide
0%
Ammonia
0%
Nitrous oxide
0%
Nitric oxide

Explanation

$3Ca+{ N }_{ 2 }\xrightarrow [ \Delta ]{ } { Ca }_{ 3 }{ N }_{ 2 }$

${ Ca }_{ 3 }{ N }_{ 2 }+6{ H }_{ 2 }O\xrightarrow [ ]{ } 3Ca{ (OH) }_{ 2 }+2N{ H }_{ 3 }$

From the reaction we can see that the metal

$M$ is

$Ca$ and pungent gas

$Y$ is Ammonia.

Industrial preparations of

$HN{ O }_{ 3 }$ by Ostwald's process involves:

0%
reduction of $N{ H }_{ 3 }$
0%
oxidation of $N{ H }_{ 3 }$
0%
hydrogenation of $N{ H }_{ 3 }$
0%
hydrolysis of $N{ H }_{ 3 }$

Explanation

Ammonia is converted to nitric acid in 2 stages. It is oxidized by heating with oxygen in the presence of a catalyst such as platinum with 10% rhodium, to form nitric oxide and water. This reaction is strongly exothermic, making it a useful heat source once initiated.

Stage two encompasses two reactions and is carried out in an absorption apparatus containing water. Initially nitric oxide is oxidized again to yield nitrogen dioxide. This gas is then readily absorbed by the water, yielding the desired product (nitric acid, albeit in a dilute form), while reducing a portion of it back to nitric oxide.

Hence, industrial preparations of

${ HNO }_{ 3 }$ by Ostwald's process involves oxidation of

$NH_{ 3 }$ .

Which of the following compounds do not give ammonia on reaction with water?

0%
${ Mg }_{ 3 }{ N }_{ 2 }$
0%
$AlN$
0%
$CaC{ N }_{ 2 }$
0%
$Ca{ \left( CN \right) }_{ 2 }$

Explanation

$\text{Let's see, what happens:}$

$\text{when the given, compounds react with water}$

$\mathrm{Mg}_{2}\mathrm{~N}_{2}+6\mathrm{H}_{2}\mathrm{O}\longrightarrow 2 \mathrm{Mg}(\mathrm{OH})_{2}+2 \mathrm{NH}_{3}$

$AlN+3\mathrm{H}_{2}\mathrm{O}\longrightarrow\mathrm{NH}_{3}+\mathrm{Al}(\mathrm{OH})_{3}$

$\begin{array}{l}\mathrm{Ca}(\mathrm{CN})_{2}+2\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}+2\mathrm{HCN}\\\mathrm{CaCN}_{2}+3\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{CaCO}_{3}+2 \mathrm{NH}_{3}\end{array}$

$\begin{array}{l}\text { Herce, only } \mathrm{Ca}(\mathrm{CN})_{2}\text { doesn't give } \mathrm{NH}_{3} \text { , on reaction with water. } \\\qquad \text { So, option(D) is correct. }\end{array}$

The allotrope of this element is the primary absorber of UV solar radiation in Earth's atmosphere. The element is:

0%
carbon
0%
nitrogen
0%
oxygen
0%
neon
0%
argon

__________ is a greenish yellow gas having an irritating choking and unpleasant odour.

0%
Fluorine
0%
Bromine
0%
Chlorine
0%
Hydrogen sulphide

$I$ . The ideal gas law can be applied to a sample of xenon at a temperature of

$-{273}^{o}C$ .

$II$ . Xenon is a noble gas.

0%
Statement $I$ is true, Statement $II$ is true
0%
Statement $I$ is true, Statement $II$ is false
0%
Statement $I$ is false, Statement $II$ is true
0%
Statement $I$ is true, Statement $II$ is true and is a correct explanation of the phenomena described in $I$

Explanation

Since,

$-273°C\ =\ 0\ K$ .

At absolute zero temperature i.e. 0 K, a gas will no longer be a gas, but rather it becomes a solid. So, the gas law will not be applicable in this case.

So, correct answer is option

$C$ .

Statement 1: By mass, oxygen is the most abundant element in the human body.
Statement 2: It is principally found as

$O_2$ in the bloodstream.

0%
Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
0%
Both Statement 1 and Statement 2 are correct, but Statement 2 is NOT the correct explanation of Statement 1.
0%
Statement 1 is correct, but Statement 2 is not correct.
0%
Statement 1 is not correct, but Statement 2 is correct.

Explanation

By mass, oxygen is the most abundant element in the human body.

Oxygen is found in the human body as oxy-haemoglobinin the bloodstream but not as $O_2$

So statement 1 is correct but not 2.

Hence option C is correct.

Regardless of its electron configuration, the given element must always be paramagnetic when it's a single, neutrally charged atom.

0%
Carbon
0%
Nitrogen
0%
Oxygen
0%
Neon
0%
Argon

In which of the following choices are the noble gases arranged correctly in order of increasing boiling point?

0%
$Xe, Kr, Ar, Ne$
0%
$Ne, Ar, Kr, Xe$
0%
$Kr, Xe, Ne, Ar$
0%
$Ar, Ne, Kr, Xe$

What occurs in the minerals of the apatite family?

0%
phosphorous
0%
arsenic
0%
carbon
0%
nitrogen

Explanation

$\text {Apatite is the group of phosphate ion. }\\\text {phosphorous occurs in minerals of the apatite family.Hence,A is correct option.}$

Which selection is the correct formula for the chlorate ion?

0%
${Cl}^{-}$
0%
$Cl{O}^{-}$
0%
$Cl{O}_{2}^{-}$
0%
$Cl{O}_{3}^{-}$
0%
$Cl{O}_{4}^{-}$

Explanation

The correct formula for the chlorate ion is

$Cl{O}_{3}^{-}$
The formula

$Cl{O}^{-}$ represents hypochlorite ion.
The formula

$Cl{O}_{2}^{-}$ represents chlorite ion.
The formula

$Cl{O}_{4}^{-}$ represents perchlorate ion.

Which is true regarding nitrogen?

0%
Less electronegative
0%
Has low ionisation enthalpy
0%
d-orbitals are available
0%
Ability to form $p\pi -p\pi$ bonds with itself

Explanation

p-pi-p-pi bond is when bonding occurs in elements having their vacant place and vacant electron in the p orbital. Example: Nitrogen (N

$_2$ )

Hence, the correct option is

$D$

Nitrogen and phosphorus are both in Group 15 of the Periodic Table. Phosphorus forms a chloride with the formula

$P{Cl}_{5}$ but nitrogen does not form

$N{Cl}_{5}$ .
Which statements help to explain this?
Nitrogen's outer shell cannot contain more than eight electrons.
Nitrogen cannot have an oxidation state of

$+5$
Nitrogen is less electronegative than phosphorus.

0%
1, 2 and 3 are correct
0%
1 and 2 only are correct
0%
2 and 3 only are correct
0%
1 only is correct

Which of the following order of acidic strength is correct?

0%
$HClO > HClO_{2} > HClO_{3} > HClO_{4}$
0%
$HClO_{4} > HClO_{2} > HClO_{3} > HClO$
0%
$HClO_{2} > HClO > HClO_{4} > HClO_{3}$
0%
$HClO_{4} > HClO_{3} > HClO_{2} > HClO$

Explanation

Acidic strength will increase with an increase in the oxidation state of

$Cl$ in its oxyacid.

Since the oxidation state of

$Cl$ in

$HClO_4$ is maximum followed by

$HClO_3,HClO_2,$ and

$HClO$ .

THus option

$D$ is correct.

$N_2(g) + 3H_2(g)\overset{Fe+Mo}{\rightleftharpoons}2NH_3(g)$ ; Haber's process.

Mo is used as :

0%
a catalyst
0%
a catalytic promoter
0%
an oxidising agent
0%
a catalytic poison

Explanation

$NH_3$ is manufactured by Haber's process.

$N_2(g) + 3H_2(g)\overset{Fe+Mb}{\rightleftharpoons}2NH_3(g)$
In this process, pure nitrogen and hydrogen gases react in the ratio of

$1:3$ by volume in

$400^0C-500^0C$ temperature,

$200-900$ atm pressure in presence of iron as catalyst and Mo as promoter.

Sea weeds are an important source of :

0%
Iron
0%
Chlorine
0%
Iodine
0%
Bromine

For the elements:

$Cl, Br, F, Al, C, Li, Cs$ &

$Xe$ .
Choose the best option. The element having 8 electrons in the outermost shell is:

0%
$Xe$
0%
$Cs$
0%
$Br$
0%
$C$

The proof of oxidizing action of hydrogen peroxide in acid solution is in the formation of

0%
$O_2$
0%
$H_2O$
0%
both $H_2O$ and $O_2$
0%
both $H_3O^+$ and $O_2$

Explanation

$2H_2O_2\rightarrow 2H_2O+O_2$

Which of the following molecule/ion does not exist?

0%
$PCl_6{^-}$
0%
$ICl_3$
0%
$BF_4{^-}$
0%
$XeF_3{^-}$

Which one of the following noble gases is least polarizable?

0%
$Xe$
0%
$Ar$
0%
$Ne$
0%
$He$
0%
All have equal polarizablilty

Explanation

$He$ is least polarizable. Among noble gases,

$He$ has smallest size. lower is the size, more strongly are the valence electrons held with nucleus and lower is the polarizability.

The sublimation of iodine is due to

0%
Low melting point
0%
Weak $I - I$ Bond
0%
Weak van der Waals forces among $I_2$ molecules
0%
Lone pair - lone pair repulsion in $I_2$ molecule

Explanation

Iodine is a solid which is easily sublimable. This is due to the weak vanderwaals forces of attraction among

$I_2$ molecules.

Which of the following bond has highest bond energy?

0%
F-F
0%
Cl-Cl
0%
Br-Br
0%
I-I

Explanation

Correct Option: $B$

Explanation:

•

$Flourine$ ,

$Chlorine$ ,

$Bromine$ and

$Iodine$ are the group

$17$ elements.

• Bond energy is the amount of energy needed to break the atoms involved in a molecular bond into free atoms.

•

$F-F$ bond is the weak bond because the repulsion between the lone electron pairs of the two fluorine atoms is huge, Also the ionization energy of the

$Flourine$ is very low and hence it can react easily with other elements and break the

$F-F$ bond. The bond length of two

$F$ atoms is more compared to the other group

$17$ elements.

•

$Cl-Cl$ bond is the strongest bond because of the smallest bond length compared to all the other halogen elements, Due to stronger bond and lesser repulsion, the bond length gets small.

• As we know that, The more the electrons are attracted, the stronger is the bond formed.

• The bond dissociation energy of

$Flourine$ is however , lower than

$Chlorine$ and

$Bromine$ due to its small size its inter electronic repulsion is very high, so the bond dissociation enthalpy order as follows

$Cl_2$ >

$Br_2$ >

$F_2$ >

$I_2$ .

•

$I_2$ has the smallest bond dissociation energy due to the largest size.

Thus, the correct option is B

2170135_221294_ans_97f42bdbece8481caa141bec6e897de1.jpeg

Fluorine differs from the other halogens due to :

0%
Small size
0%
Very high electronegativity
0%
Non-availability of d-orbitals
0%
All of these

Atoms in

$P_4$ molecule of white phosphorus are arranged regularly in the following way :

0%
at the corners of a cube.
0%
at the corners of an octahedron.
0%
at the corners of a tetrahedron.
0%
at the center and corners of a tetrahedron.

Explanation

Atoms in

$P_4$ molecule of white phosphorus are arranged regularly at the corners of a tetrahedron such that each atom is single-bonded to other three.The

$P-P-P$ bond angle is

$60^{0}$ .

Option (C) is correct.

Which of the following is/are 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 is paramagnetic

Explanation

$NCl_3$ is hydrolyzed by hot water to release ammonia and hypochlorous acid.

$NCl_3 + 3H_2O\rightarrow NH_3 + 3HOCl$

$NH_3$ is less stable than

$PH_3$ , because

$NH_3$ is more stable than

$PH_3$ due to inter molecular hydrogen bonding.

$NH_3$ is a weak reducing agent compared to

$PH_3$ , because from

$NH_3$ to

$PH_3$ reducing nature increases down the group.

D) Nitric oxide is paramagnetic, because it is having unpaired electrons.
Hence, options A,C & D are correct.

White phosphorus

$(P_4)$ has :

0%
six P-P single bonds
0%
four P-P single bonds
0%
four lone pairs of electrons
0%
PPP angle of $60^o$

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

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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