MCQ Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 2

Industrial preparations of

$HNO_{3}$ by Ostwald's process involves:

0%
reduction of $NH_{3}$
0%
oxidation of $NH_{3}$
0%
hydrogenation of $NH_{3}$
0%
hydrolysis of $NH_{3}$

Explanation

$4NH_3 (g) + 5O_2 (g) \rightarrow 4NO (g) + 6H_2O (g)$

$2NO(g) +O_2(g) \xrightarrow {50^oC} 2NO_2(g)$

$4NO_2(g) + 2H_2O(l) +O_2(g) \rightarrow 4HNO_3(aq)$

Oxidation of Ammonia occurs in the preparation of ammonia.
Option B is correct.

Choose correct squence for the following species.

0%
$F_{2} > Cl_{2} > Br_{2} > I_{2}$ (Reducing power)
0%
$Cl_{2}< F_{2} < Br_{2} < I_{2}$ (Reactivity)
0%
$I_{2} > Br_{2} > F_{2}> Cl_{2}$ (Bond length)
0%
$Cl_{2} > Br_{2} >F_{2} >I_{2}$ (Bond dissociation enthalpy)

A solution of colorless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt(s) H is(are) :

0%
$NH_4NO_3$
0%
$NH_4NO_2$
0%
$NH_4Cl$
0%
$(NH_4)_2SO_4$

Explanation

Option (A), (B) are correct.

The colorless salt H is either

$NH_4NO_3$ or

$NH_4NO_2$ . The colorless salt produces

$NH_3$ gas (non-inflammable) on boiling with excess of

$NaOH$ . On addition of zinc dust (a reducing agent) to this solution, sodium nitrite or sodium nitrate will liberate

$NH_3$ gas again.

$NH_4NO_3+ NaOH \longrightarrow NH_3+NaNO_3+H_2O$

$7NaOH+NaNO_3+4Zn\longrightarrow 4Na_2ZnO_2+NH_3+2H_2O$

$NH_4NO_2+NaOH \longrightarrow NaNO_2+NH_3+H_2O$

$3Zn+5NaOH+NaNO_2 \longrightarrow 3Na_2ZnO_2+NH_3$

Which of the following reaction represents the balanced form of conversion of Ca to ammonia?

0%
${ 2NH }_{ 4 }Cl+Ca\left( OH \right) _{ 2 }\xrightarrow { \triangle } { CaCl }_{ 2 }+{ 2H }_{ 2 }O+{ 2NH }_{ 3 }$
0%
${ NH }_{ 4 }Cl+Ca\left( OH \right) _{ 2 }\xrightarrow { \triangle } { CaCl }_{ 2 }+{ 2H }_{ 2 }O+{ NH }_{ 3 }$
0%
${ 2NH }_{ 4 }Cl+3Ca\left( OH \right) _{ 2 }\xrightarrow { \triangle } 3{ CaCl }_{ 2 }+{ 2H }_{ 2 }O+{ 2NH }_{ 3 }$
0%
${ 4NH }_{ 4 }Cl+Ca\left( OH \right) _{ 2 }\xrightarrow { \triangle } 2{ CaCl }_{ 2 }+{ 2H }_{ 2 }O+{ 2NH }_{ 3 }$

The solubility of iodine in carbon tetrachloride is due to

0%
Ionic solvent-solute interaction
0%
Charge-transfer phenomenon
0%
Van der Waals interaction
0%
Dissociation of iodine

Explanation

Both

$I_2$ and

$CCl_4$ are non polar in nature. The interaction between them is van der waal.

Which one of the following arrangements does not give the correct picture of the trends indicated against it?

0%
$F_2 > Cl_2 > Br_2 > I_2$ : Bond dissociation energy
0%
$F_2 > Cl_2 > Br_2 > I_2$ : Electronegativity
0%
$F_2 > Cl_2 > Br_2 > I_2$ : Oxidizing nature
0%
$F_2 > Cl_2 > Br_2 > I_2$ : Electron gain enthalpy

Explanation

$Cl_2 > Br_2>F_2 > I_2$ : Bond dissociation energy.

As we from top to bottom as atomic size increases Bond dissociation energy decreases but exception for F due to high electronegativity. Incorrect order.

$F_2 > Cl_2 > Br_2 > I_2$ : Electronegativity.

As we from top to bottom as atomic size increases and electronegativity decreases. Correct order.

$F_2 > Cl_2 > Br_2 > I_2$ : Oxidizing nature.

As we from top to bottom as atomic size increases oxidizing nature decreases due to decrease in electronegativity. Correct order.

$Cl_2 >F_2 > Br_2 > I_2$ : Electron gain enthalpy.

As we from top to bottom as atomic size increases and effective nuclear charge decreases and hence electron gain enthalpy decreases except for F due to high electronic repulsion. Incorrect order.

Hence A & D are correct options.

The boiling points of neon and krypton are

$-245.{ 9 }^{ \circ }C$ and

$-152.{ 9 }^{ \circ }C$ respectively. Using these data estimate the boiling point of argon. (report answer as the negative of the closest integer)

0%
$-289^oC$
0%
$-299^oC$
0%
$-199^oC$
0%
$-189^oC$

Explanation

The boiling point of argon can be roughly approximated as the average of the boiling points of neon and krypton.

$\dfrac {-245.9 \: ^0C-152.9 \: ^0C}{2}=-199.4 \: ^0C$
Hence, the boiling point of argon is approximately

$-199.4 \: ^0C$ .

The high oxidising power of fluorine is due to its :

0%
high electron affinity.
0%
high heat of dissociation and low heat of hydration.
0%
low heat of dissociation and high heat of hydration.
0%
high heat of dissociation and high heat of hydration.

The Haber process can be represented by the following scheme:

Find out D?

0%
$NH_4Cl$
0%
$CaCl_2$
0%
$NH_4HCO_3$
0%
$Ca(OH)_2$

Explanation

In Haber process:

$(i)CaO + H_2O\rightarrow \underset{(A)}{Ca(OH)_2}$

$(ii)NH_3 + H_2O + CO_2 NH_4CO_3 \rightarrow \underset{(B)}{ NH_4CO_3}$

$(iii) 2NaHCO_3 \rightarrow Na_2CO_3 + \underset{(C)}{CO_2} + H_2O$

Thus,

$A = Ca(OH)_2, B = NH_4HCO_3, C=Na_2CO_3, D=NH_4Cl, and\ E=CaCl_2$

Thus,D is

$NH_4Cl$ .

Hence, option A is correct.

Which of the following gases has the greatest density at STP conditions?

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

Explanation

The highest density among the inert gases is of Neon (Ne). This a factual data. Thus the highest density among given options is of

$Ne$ , as all the options are of inert gases.

Statement 1: Oxidation state for fluorine is -1 in all compounds.
Statement 2: Fluorine is the most electronegative element.

0%

Statement 1 is correct but statement 2 is not correct.
0%

Statement 1 is not correct but statement 2 is correct.
0%

Both the statements 1 and 2 are not correct.
0%

Both the statements 1 and 2 are correct and statement 2 is not correct explanation of statement 1.
0%

Both the statements 1 and 2 are correct and statement 2 is correct explanation of statement 1.

The steps for the manufacture of nitric acid by Ostwald process are given. Arrange them in sequence.
(a) Recirculation of nitric oxide
(b) Addition of

${H}_{2}O$ to nitrogen dioxide
(c) Oxidation of nitric oxide
(d) Catalytic oxidation of ammonia

0%
d c b a
0%
d c a b
0%
c d b a
0%
c d a b

Explanation

In ostwald process Ammonia is converted to nitric acid in 2 stages. It is oxidized (in a sense "burnt") by heating with oxygen in the presence of a catalyst to form nitric oxide and water. Nitric oxide is oxidized again to yield nitrogen dioxide. This gas is then readily absorbed by the water and forms nitric acid. Thus the steps of process are :
(d) Catalytic oxidation of ammonia
(c) Oxidation of nitric oxide
(b) Addition of

${H}_{2}O$ to nitrogen dioxide
(a) Recirculation of nitric oxide

Atoms in a

${ P }_{ 4 }$ molecule of white phosphorus are arranged regularly in which of 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 centre and corners of a tetrahedron

Explanation

White phosphorus has the following structure
The structure shows that the four

$P$ atoms in white phosphorus lie at the corners of a regular tetrahedron.

Chlorine was discovered by:

0%
Priestley
0%
Davy
0%
Shceele
0%
Cavendish

Explanation

Chlorine was discovered by Shceele in 1774. He prepared it by heating

$\displaystyle MnO_2$ with

$\displaystyle HCl$ and named it oxymuriatic gas.

How many oxides of monomeric nature can nitrogen form?

0%
$1$
0%
$2$
0%
$4$
0%
$5$

Explanation

$N$ has 5 valence electron, therefore molecules with

$N_2O_x$ will have even number of electrons and will be non radical. For monomeric oxide,

$N_2O$ is the only monomeric nitrogen oxide.

Therefore option A is correct.

Which of the following does not contain chlorine?

0%
Rock salt
0%
Sylvine
0%
Cryolite
0%
Carnallite

Explanation

$Cl$ occurs commonly both in the earth's crust and seawater.

the most common compound of chlorine in seawater is

$NaCl$ and minimum amount of

$KCl$ is also occur in seawater.

the most mineral of

$Cl$ are (Rocksalt)

$NaCl$ ,(Sylvine)

$KCl$ and(carnallite)

$KClMgCl_2$ .

$\left( { Na }_{ 3 }Al{ F }_{ 6 } \right)$ Cryolite does not contain chlorine.

Nitrogen does not show pentacovalency because of:

0%
absence of s-orbitals
0%
absence of d-orbitals
0%
its small size
0%
none of these

Explanation

The electronic configuration of nitrogen is

$1s^22s^22p^3$ . Nitrogen don't have d-orbitals and thus don't show pentacovalency.

Ammonia gas can be collected by the displacement of:

0%
Conc. $H_2SO_4$
0%
Brine
0%
Water
0%
Mercury

Which catalyst is used in Haber process for manufacturing of

$NH_3$ ?

0%
$Pt$
0%
$Fe+Mo$
0%
$CuO$
0%
$Al_2O_3$

Explanation

Ammonia(

$NH_3$ ) is prepared from

$N_2,H_2$ in 1:3 ratio using

$Fe$ catalyst and

$Mo$ promoter.
Hence option B is correct.

Which of the following halide of nitrogen is stable in nature?

0%
$NF_ 3$
0%
$NCl _3$
0%
$NBr_ 3$
0%
$NI _3$

Explanation

Nitrogen has very small size as compared to the halogens, which have much larger sizes. Due to this, they can not remain bonded to the nitrogen atom and hence are highly unstable.

$NF_3$ is stable because fluorine and nitrogen have comparable sizes and electronegativity.

Which of the the following properties of oxygen differs from the rest of members of its family?

0%
High value of ionization energy
0%
Oxidation state (2,4,6)
0%
Polymorphism
0%
Formation of hydride

Explanation

Oygen shows a veriety of oxidation states like

$-2,-1,-\cfrac{1}{2}$ and

$2,4,6$ . Where as the other elements in the group do not show this many oxidation states. This makes the oxygen different from other elements of the group.

Hence option B is correct.

Which of the following chemicals contains chlorine?

0%
Fischer salt
0%
Epsom salt
0%
fremy's salt
0%
Spirit of salt

Explanation

Spirit of salt an old name for hydrochloric acid (

$HCl$ ).

Hence option D is correct,

Which of the following gas is heavier than air?

0%
$N_2$
0%
$O_2$
0%
$CO_2$
0%
$H_2$

Explanation

The molecular weights of the

$N_2,O_2,CO_2,H_2$ are 28,32,44,2 respectively

$CO_2$ is heavier than air (

$O_2$ )

Hence option

$C$ is correct.

Which of the following is a part of the catalyst system used in Haber's process?

0%
Copper oxide
0%
Iron
0%
Potassium oxide
0%
Aluminium oxide

Explanation

$\begin{array}{l}\text { Haber's Process - } \\\text { The haber process is used in } \\\text { the manufacturing of ammonia from nitrogen and } \\\text { hydrogen, and then goes on to explain the reasons } \\\text { for the conditions used in the process. } \\\text { In the process use the some catalyst Iron, } \\\text { potassium oxide , aluminium oxide }\end{array}$

$N_{2}+3 H_{2} \dfrac{Fe / K_{2} O}{Al_{2} O_{3}} \rightarrow 2 N H_{3}$

The noble gas mixture is cooled in a coconut bulb at

$173 K$ . The gases that are not adsorbed are:

0%
Helium and Neon
0%
Helium and Xenon
0%
Xenon and Neon
0%
none of these

Explanation

When the mixture of noble gas is cooled in a coconut bulb at

$173\ K$ then argon, krypton and xenon are adsorbed on charcoal while Helium and Neon are not adsorbed.

Hence option

$A$ is correct.

Select the correct statement for brown ring test.

0%
Nitrates are converted to $NO$
0%
$NO$ forms complex with $Fe$
0%
$Fe^{2+}$ acts as reducing agent here.
0%
Brown colour is due to the complex formation.

Explanation

$2HNO_3+ 3H_2SO_4 + 6FeSO_4 \rightarrow 3Fe_2(SO_4)_3 + 2NO + 4H_2O$

$[Fe(H_2O)_6]SO_4+NO \rightarrow [Fe(H_2O)_5(NO)]SO_4+H_2O$ Nitrates are converted into nitric oxide.

$NO$ forms complex with

$Fe$ .

$Fe^{2+}$ acts as reducing agent.

$[Fe(H_2O)_5(NO)]SO_4$ is brown coloured complex formed.

In the Ostwald's process for the manufacture of

$HNO_3$ , the catalyst used is:

0%
$Fe$
0%
$Pt$
0%
$V_2O_5$
0%
$Mo$

Explanation

Ostwal's process is a chemical process for the manifacture of nitric acid in large scale.In the ostwalds process, during the preparation of

$HNO_3$ , which can use for fertilisers, vanadium pentaoxide (

$V_2O_5$ ) is used as catalyst.

An atom has

$2K, 8L$ and

$5M$ electrons.
Choose the correct statement(s) regarding it.
A. Trivalent anion of this atom will have

$12$ protons in its nucleus.
B. Trivalent cation of this atom will have six p-electrons in it.
C. This atom form an amphoteric oxide of formula

$X_{2}O_{3}$ .
D. One of its allotrope is tetra atomic

$(X_{4})$

0%
$A$ and $B$
0%
$B$ only
0%
$B$ and $C$
0%
$B$ and $D$

Explanation

In given atom the structure is, K(2) L(8) M(5)

Configuration will be:

$X=1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$

$X – 3e \rightarrow X^{3+}$

So, it will be trivalent metal cation.

$X^{3+} = 1s^{2}2s^{2}2p^{6}3s^{2}$

So, p electrons will be 6.

As per the configuration, this element is the phosphorous (atomic number 15). It exists as allotrope in tetra atomic i.e. $X_{4}$ format.

The correct option is D.

Ammonia is not produced in the reaction of:

0%
$NH_4Cl$ and $KOH$
0%
$AlN$ with water
0%
$NH_4Cl$ with $NaNO_2$
0%
$NH_4Cl$ with $Ca(OH)_2$

Explanation

Reaction between

$NH_4Cl,NaNO_2$ do not produce

$NH_3$ instead it produces

$N_2$

$NH_4Cl+NaNO_2\rightarrow N_2+2H_2O+NaCl$

Which of the following pairs are correctly matched?

Haber process - Manufacture of

$NH_3$
Leblanc process - Manufacture of

$H_2SO_4$
Birkeland-Eyde process - Manufacture of

$HNO_3$
Solvay process - Manufacture of

$Na_2CO_3$

0%
$2, 3, 4$
0%
$1, 2, 3, 4$
0%
$1, 2$
0%
$1, 3, 4$

Nitrogen forms stable

$N_{2}$ molecule but phosphorus is converted to

$P_{4}$ from

$P_{2}$ because:

0%
p $\pi$ -p $\pi$ bonding is strong in phosphorus
0%
p $\pi$ -p $\pi$ bonding is weak in phosphorus
0%
triple bond is present in phosphorus
0%
single $P - P$ bond is weaker than $N -N$ bond

Explanation

$P$ has empty

$d$ orbital, that can be filled by an additional electron. And since its size is bigger it is least likely to handle triple bond, as size increases

$\pi$ overlapping decreases.

Ammonia is not produced in the reaction of ________.

0%
$NH_4Cl$ with KOH
0%
$AlN$ with water
0%
$NH_4Cl$ with $NaNO_2$
0%
$NH_4Cl$ with $Ca(OH)_2$

Explanation

$N{ H }_{ 4 }Cl+KOH\rightarrow KCl+{ H }_{ 2 }O+N{ H }_{ 3 }$

$AlN+3{ H }_{ 2 }O\rightarrow N{ H }_{ 3 }+Al{ (OH) }_{ 3 }$

$N{ H }_{ 4 }Cl+NaN{ O }_{ 2 }\rightarrow { N }_{ 2 }+2{ H }_{ 2 }O+NaCl$

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

Option C is correct.

Each of the following is true for white and red phosphorus except that they:

0%
are both soluble in $CS_{2}$
0%
can be oxidized by heating in air
0%
consist of the same kind of atoms
0%
can be converted into one another.

Explanation

White phosphorus, yellow phosphorus or simply tetraphosphorus ( $P_4$ ) exists as molecules made up of four atoms in a tetrahedral structure. As phosphorous molecule is non polar,so it dissolves in non polar solvent like carbon disulfide, $CS_2$ .

Hence option A is correct.

White phosphorus

$P_4$ has the following characteristics.

0%
$6$ P-P single bonds
0%
$4$ P-P single bonds
0%
$4$ lone pair of electrons
0%
P-P-P angle of $60^o$

Explanation

In white phosphorus

$(P_4),$

$6$ P-P single bonds

$4$ lone pair of electrons

Bond angle

$\angle$ P-P-P

$=60^o$

Hence, options A, C and D are correct.

$2$ moles of

$PCl_{5}$ were heated in a closed vessel of

$2$ litre capacity. At equilibrium,

$40\%$ of

$PCl_{5}$ is dissociated into

$PCl_{3}$ and

$Cl_{2}$ . The value of equilibrium constant is:

0%
$0.266$
0%
$0.53$
0%
$2.66$
0%
$5.3$

Explanation

$PCl_5(g)\leftrightharpoons PCl_3(g)+Cl_2(g)$

2 mol 0 0

-2

$\alpha$ mol 2

$\alpha$ mol

2 (1-

$\alpha$ )mol 2

$\alpha$ mol 2

$\alpha$ mol

When degree of dissociation,

$\alpha=40$ %=

$0.4$

At equilibrium , the molar concentration of the component of mixture

$[PCl_5(g)]=\cfrac {2(1-\alpha)}{V}=\cfrac {1(1-0.4)}{2}=0.6$ mol/L

$[PCl_3(g)]=\cfrac {2\alpha}{V}=\cfrac {2\times 0.4}{2}=0.4$ mol/L

$[Cl_2(g)]=\cfrac {2\alpha}{V}=\cfrac {2\times 0.4}{2}=0.4$ mol/L

So,

$Eq^m$ constant,

$K_c=\cfrac {[PCl_3(g)]\times [Cl_2(g)]}{[PCl_5(g)]}$

$=\cfrac {0.4\times 0.4}{0.6}$

$=0.27$ mol/L

Hence, the correct option is

$A$ .

In Haber's process

$50.0\ g$ of

$N_{2}(g)$ and

$10.0\ g$ of

$H_{2}(g)$ are mixed to produced

$NH_{3}(g)$ . What are the number of moles of

$NH_{3}(g)$ formed?

0%
$3.33$
0%
$2.36$
0%
$2.01$
0%
$5.36$

Explanation

$N_2 + 3H_2 \rightarrow2NH3$

Here.

28 g

$N_2$ reacts with 6 g of

$H_2$

So,

50 g of

$N_2$ react with

$\dfrac{6}{28}\times50 g -H_2 = 10.7 g$ but we have only 10g so

$H_2$ is limiting reagent..

6 g

$H_2$ gives 34g

$NH_3$

10g

$H_2$ will give

$\dfrac{34}{6}\times10=56.67g -NH_3$

So number of moles of

$NH_3$ formed is,

$=\dfrac{Given-mass}{Molar-mass}=\dfrac{56.67}{17}=3.33-moles$

Fluorine is the best oxidising agent because it has:

0%
highest electron affinity
0%
highest reduction potential
0%
highest oxidation potential
0%
lowest electron affinity.

Explanation

Correct Option: $B$

Hint: Fluorine has the least bond enthalpy

Explanation:
Fluorine is an element that has high electronegativity.

Fluorine has low bond enthalpy and highest reduction potential so more ability to loose the electrons.

Fluorine can also be easily reduced.

Fluorine is thus one of the best oxidizing agents.

Henceforth Correct option is the highest reduction potential (B)

Which among the following statements is incorrect?

0%
$XeF_{4}$ and $SbF_{5}$ combine to form salt.
0%
$He$ and $Ne$ do not form clathrates.
0%
$He$ has highest boiling point in its group.
0%
$He$ diffuses through rubber and polyvinyl chloride.

Explanation

As we move from top to bottom in g group the boiling point of the elements increases as the atomic mass increases.So

$He$ is the first element in the group it should have less boiling point than the other elements.

So the given statement

$He$ has highest boiling point is the wrong.

Hence option C is correct.

Potassium permanganate

$\rightarrow$ potassium manganate +

$A$ + oxygen.
What is

$A$ in the given reaction?

0%
Manganese
0%
Manganese dioxide
0%
Manganese trioxide
0%
Manganese tetraoxide

Explanation

The compound

$A$ in the given reaction is Manganese dioxide. It is obtained by heating of potassium permanganate to form oxygen.

$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$

While preparing oxygen from potassium permanganate a plug of cotton must be put in a hard glass test tube.

0%
True
0%
False

Chlorine has two naturally occuring isotopes

$_{ 17 }^{ 37 }{ Cl }$ and

$_{ 17 }^{ 35 }{ Cl }$ and average atomic mass of chlorine is

$35.5$ . The percentage of

$_{ 17 }^{ 37 }{ Cl }$ is:

0%
25%
0%
75%
0%
40%
0%
60%

Explanation

Let ,

The percentage of

$Cl^{37}$ be

$x$ and The percentage of

$Cl^{37}$ is

$y = (100-x)$

Average atomic mass =

$\frac{Atomic-mass \times abundance}{100}$

$35.5 = \frac{37x + 35(100-x)}{100}$

$37x + 35(100-x) = 3550$

$2x = 50$

$\therefore x = 25 %$

$y = 100 - x = 75 %$

The percentage of

$Cl^{37}$ = 25%

The percentage of

$Cl^{35}$ = 75%

Hence , option A is correct .

Which of the following is the polar molecules ?

0%
$XeF{l_4}$
0%
$S{F_4}\;$
0%
$Si{F_4}$
0%
none

Phosphours sulphide

${ P }_{ 4 }{ S }_{ 3 }$ , a well known chemical used in match industry has many

$P-S$ bonds ?

0%
4
0%
2
0%
6
0%
8

Which of the following interhalogens cannot exist?

0%
$ClF_5$
0%
$ClI_5$
0%
$ICl_2^-$
0%
$IF_4^-$

Chlorine is prepared in the laboratory by the action of:

0%
Zn and HCl
0%
CaCO $_3$ and HCl
0%
Pb $($ NO $_3)_2$ and HCl
0%
MnO $_2$ and HCl

Select reaction in which at least one compound is formed as product, in which Xe is in +6 oxidation state?

0%
$XeF_6+CsF \xrightarrow[thermal \, decomposition]{\Delta} product$
0%
$XeF_6 \overset{Hydrolysis} {\rightarrow} \quad ? \xrightarrow[disproportion]{OH^-} product$
0%
$Xe+F_2\overset{873\, K, 7 bar}{\rightarrow} \quad ? \quad \overset{SF_6}{\rightarrow}product$
0%
$Xe_{excess}+F_2 \quad \overset{400^0C,1 bar}{\rightarrow}\xrightarrow[of water]{Excess} product$

The atomic number of

$N$ isThe atomic number of 3rd member of nitrogen family will be:

0%
23
0%
15
0%
33
0%
43

An aqueous solution of a halogen salt of potassium reacts with the same halogen

$X_2$ to give

$KX_3$ , a brown-colored solution, in which halogen exists as

$X^-_3$ ion,

$X_2$ as a Lewis acid, and

$X^-$ as a Lewis base, halogen base is:

0%
Chlorine
0%
Bromine
0%
Iodine
0%
Fluorine

Element which never has positive oxidation state in any of its compounds is?

0%
Chlorine
0%
Bromine
0%
Oxygen
0%
Fluorine

Which of the following statements is true?
(i) The order of acidic strength of haloacid is
HF>HCl>HBr>HI.
(ii) The order of stability of compounds is
HF<HCI<HBr<HI.

0%
(i) is correct and (ii) is incorrect
0%
(i) is incorrect and (ii) is incorrect
0%
Both are incorrect
0%
Both are correct

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 2 - 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 2 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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