has a saturated outermost shell

has one electron in outermost shell

has eight electrons is outer shell

has two electrons in outermost shell

MCQ Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 1

Which of the following elements belongs to halogen group?

0%
$1s^2 , 2s^2 , 2p^6 ,3s^2, 3p^5$
0%
$1s^2 , 2s^2 , 2p^6, 3s^2 , 3p^6 , 4s^2$
0%
$[Ar], 3d^{10},4s^2 ,4p^5$
0%
$[Kr], 4d^{10} , 5s^2, 5p^5$

Explanation

The halogens are the

$17^{th}$ group of the periodic table. They contain 7 electrons in the valence shell and have general valence shell electronic configuration of

$ns ^2np^5$ .

Elements of halogen group are:

Elements	Atomic number	Electronic configuration	Group member	Period
Flourine	9	$[He]2s^22p^5$	17	2
Chlorine	17	$[Ne]3s^2 3p^5$	17	3
Bromine	35	$[Ar]3d^{10}4s^2 4p^5$	17	4
Iodine	53	$[Kr]4d^{10}5s^2 5p^5$	17	5
Astatine	85	$[Xe]4f^{14}5d^{10}6s^2 6p^5$	17	6

Hence option

$A,\ C$ and

$D$ are correct .

In the modern periodic table, elements are arranged in order of increasing atomic number which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz s, p, d and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle filling of a new energy shell. In accordance with the Aufbau principle, the seven period (1 to 7) have 2,8,8,18,18,32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f - block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table.
The elements with atomic number

$35,53$ and

$85$ are all ________.

0%
Noble gases
0%
Halogens
0%
Heavy metals
0%
Light metals

The elements with atomic numbers 10, 18, 36, 54 and 86 are all:

0%
light metals
0%
inert gases
0%
halogens
0%
rare-earths

Which of the block of elements does belong to inert gases?

0%
s-block
0%
p-block
0%
d-block
0%
f-block

A certain element forms a solid oxide which dissolves in water to form an acidic solution. The element is:

0%
$S$
0%
$Mg$
0%
$Na$
0%
$P$

Which of the following statements is correct?

0%
All halogens form oxyacids
0%
Only chlorine and bromine form oxyacids
0%
All halogens except fluorine form oxyacids
0%
Only iodine forms oxyacid

A radioactive halogen is:

0%
polonium
0%
radon
0%
astatine
0%
iodine

How many electrons are present in the outermost orbit of halogens?

0%
1
0%
3
0%
5
0%
7

Explanation

The general electronic configuration of halogens is

$ns^2np^5$ . Therefore, halogens have seven electrons in their outermost shell.

Therefore, option (D) is correct.

Halogens show usually:

0%
Electro-valency
0%
Covalency
0%
Both electrovalency and covalency
0%
Coordinate valency

The outermost electronic configuration of the most electronegative element is:

0%
$ns^2np^3$
0%
$ns^2 np^4$
0%
$ns^2 np^5$
0%
$ns^2 np^6$

Radon is a:

0%
drug
0%
explosive
0%
trade name of nylon
0%
noble gas

Noble gases are chemically inactive because of:

0%
mono-atomic nature
0%
low boiling points
0%
closed valency shell
0%
rare occurrence

Which of the following represents the outermost shell electronic configuration of halogens?

0%
$ns^2np^3$
0%
$ns^2np^6$
0%
$ns^2 np^4$
0%
$ns^2np^5$

Explanation

The general electronic configuration of valence shell of Halogens is

$ns^2np^5$ .

Hence, Option "D" is the correct answer.

White Phosphorus contains

0%
$P_5$ molecules
0%
$P_4$ molecules
0%
$P_6$ molecules
0%
$P_2$ molecules

Explanation

White Phosphorus contains

$P_4$ molecules.

Hence, Option "B" is the correct answer.

Wilhem Ostwald redefined action of

0%
Anamers
0%
Isomers
0%
Catalyst
0%
Geometry of monomers

Which of the following are correct?

0%
$XeO_3$ has four $\sigma$ and four $\pi$ bonds
0%
The hybridization of $Xe$ in $sp^3d^2$
0%
Among the noble gases, the occurrence of argon is highest in the air
0%
Liquid helium is used in cryogenic liquids

Explanation

$A)$

$XeO_3$ has three double bonds and each double bond has

$1\sigma$ and

$1\pi$ bonds. Hence, the total number of bonds are

$3\sigma$ and

$3\pi$ bonds.

$B)$ In

$XeO_3$ , Xe undergoes

$sp^3$ hybridization.

$C)$ The noble gases are present in the atmosphere in small amounts

$:\ 0.934\%$ Argon.

$0.0018\%$ Neon.

$0.00052%$ Helium. Hence, argon is the most abundant noble gas in the atmosphere.

$D)$ Cryogenic liquids are liquefied gases that are kept in their liquid state at very low temperatures. Liquefied gases, such as liquid nitrogen and liquid helium, are used in many cryogenic applications. Liquid hydrogen and liquid helium are both so cold that they can liquefy the air they contact.

Hence, the correct answer is option C and D.

1565597_1734112_ans_1ec90399684a4bfcb42852d448b04fe4.png

Which of the following pairs are correctly matched?

0%
A halogen which is liquid at room temperature - Bromine
0%
The most electronegative element- Fluorine
0%
The halogen having highest electron affinity- Chlorine
0%
The strongest oxidising agent -Iodine

An interhalogen compound is :

0%
$I_3^-$
0%
$(CN)_2$
0%
$BrF_3$
0%
$N_3H$

Ammonia gas is not prepared by

0%
reduction of sodium nitrite or sodium nitrate by the reaction of Zinc dust and sodium hydroxide.
0%
hydrolysis of calcium cyanamide.
0%
heating ammonium chloride with slaked lime
0%
heating of $(NH_{4})_{2}Cr_{2}O_{7}$

Explanation

(A)

$NO_{2}^{-}+7\left [ H \right ]^{-}\xrightarrow{Zn/NaOH} NH_{3}+2H_{2}O$ ;

$NO_{3}^{-}+9\left [ H \right ]^{-}\xrightarrow{Zn/NaOH}NH_{3}+3H_{2}O$

(B)

$CaCN_{2}+3H_{2}O\rightarrow CaCO_{3}+2NH_{3}$

(C)

$NH_{4}Cl+Ca(OH)_{2}\rightarrow 2NH_{3}+CaCl_{2}+2H_{2}O$

(D)

$\left ( NH_{4} \right )_{2}Cr_{2}O_{7}\rightarrow N_{2}+Cr_{2}O_{3}+4H_{2}O$

Option D is correct.

Atoms in a

$P_4$ molecules of white phosphorus are arranged regularly in space in which of the following way

0%
at the corners of tetrahedron
0%
at the corners of a cube
0%
at the corners of a four membered ring
0%
at the centre and corners of an equilateral triangle

Explanation

Atoms in a

$P_4$ molecules of white phosphorus are arranged regularly in space at the corners of the tetrahedron.
1647331_1758755_ans_feed78c66ed94edc9e5afc06c5917a17.png

1647331_1758755_ans_feed78c66ed94edc9e5afc06c5917a17.png

Which inert gas having highest boiling point?

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

Explanation

$Xe$ gas is having the highest boiling point since it has the highest mass and size.

Every noble gas atom

0%
has a saturated outermost shell
0%
has one electron in outermost shell
0%
has eight electrons is outer shell
0%
has two electrons in outermost shell

Which statement is not correct for nitrogen?

0%
It has a small size
0%
It does not radily react with $O_2$
0%
It is a typical non-metal
0%
$d$ - orbitals are available for bonding

Explanation

$N_7 \to 1s^2, 2s^2, 2p^3$

$d-$ orbitals are absent in nitrogen.

Which of the following substances when boiled with caustic sodasolution will evolve ammonia

0%
Ethylamine
0%
Aniline
0%
Acetamide
0%
Acetoxime

Explanation

$CH_3CONH_2+NaOH\xrightarrow[]{Boil}CH_3COONa+NH_3$

Acetamide

Option C is correct.

The last member of Group 18 is

0%
Helium
0%
Neon
0%
Argon
0%
Radon

Which of the following inert gas liquifies easily?

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

Explanation

The maximum temperature at which gas can be liquified is called its critical temperature. The gas which has a high boiling point will change into liquid first and so critical temperature of the gas will be more.

$Kr$ gas will liquify easily among the given gas since it has highest boiling point because of its mass.

Oxygen molecule exists in the form of three molecules as

$O_2, O_3$ and

$O_4$ .

0%
True
0%
False

Explanation

Oxygen molecule do not exists in the form of

$O_4$ it exist in two forms which are

$O_2$ (oxygen gas ) and

$O_3$ (ozone).

Hence, the statement is false.

The atoms of which of the following pair of element are most likely to exist in free state ?

0%
Hydrogen and Helium
0%
Argon and Carbon
0%
Neon and Nitrogen
0%
Helium and Neon

Explanation

$\textbf{Correct option: D}$

$\textbf{Explanation for correct option:}$

Helium and Neon element are most likely to exist in free state.

Among these pair hydrogen, helium, argon, carbon, neon, nitrogen, constitute the noble metals thereby these elements are not meant for bonding with other elements. Hence their existence is achieved in their free state.

$\textbf{Conclusion:}$

So the correct option is $D$

Which compounds is prepared by the following reaction:

$\underset { (2:1\ volume \ ratio) }{ Xe+F_{2} }\xrightarrow [ 673\ K ]{ Ni }$

0%
$XeF_{4}$
0%
$XeF_{2}$
0%
$XeF_{6}$
0%
None of these.

Explanation

$\underset{(2:1)}{Xe+F_{2}}\xrightarrow [ 673\ K ]{ Ni } XeF_{2}$

The incorrect statement regarding

$15^{th}$ group hydrides

$(EH_{3}) . [E = N, P, As, Sb, Bi]$

0%
$NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3} :$ Thermal stability
0%
$N - H > P - H > As - H > Sb - H > Bi - H : E - H$ bond dissociation enthalpy
0%
$NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3} :$ Reducing character
0%
$NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3} :$ Basicity

Explanation

Reducing character

$\propto \dfrac{1}{'E - H' \,bond \,dissociation \,enthalpy}$

Hence, correct reducing character is

$NH_{3} < PH_{3} < AsH_{3} < SbH_{3} < BiH_{3}.$

Hence, Option "C" is the correct answer.

$X - O - X$ bond (where X = central atom) is not present in species.

0%
$Cl_{2}O_{7}$
0%
$H_{2}N_{2}O_{2}$
0%
$N_{2}O_{5}$
0%
$H_{2}S_{2}O_{7}$

Explanation

$H_2N_2O_2$ is a compound which have not present in species X-O-X bond

Which of the following oxy anion(s) contain

$P - O$ bond order equal to

$1.5$ ?

0%
$H_{2}P_{2}O_{6}^{2-}$
0%
$H_{2}PO_{3}^{-}$
0%
$H_{2}PO_{4}^{-}$
0%
$H_{2}PO_{2}^{-}$

$Xe+F_2 \xrightarrow{1:20} X \xrightarrow{H_2O} Y \xrightarrow{H_2O} Z \xrightarrow{H_2O} XeO_3$
Select correct option(s) for X, Y, Z and given chemical change :

0%
X, Y and Z are in same oxidation state
0%
All have equal number of lone pair on central atom
0%
All are non-planar
0%
All have equal number of covalent bonds

In the synthesis of sodium carbonate by solvay (or ammonia-soda) process, the recovery of ammonia is done by treating

$NH_4Cl$ with

$Ca(OH)_2$ . The by-product obtained in this process is:

0%
$CaCl_2$
0%
$NaCl$
0%
$NaOH$
0%
$NaHCO_3$

Explanation

$2NH_4Cl+Ca(OH)_2 \rightarrow 2NH_3+H_2O+CaCl_2$

$CaCl_2$ is obtained as by-product in this process.

In spite of having less electron affinity than chlorine, fluorine is the strongest oxidizing agent among the halogens. The most important reason for this phenomenon is :

0%
lower bond dissociation energy of $F_{2}$ than $Cl_{2}$
0%
much lower heat of hydration of $F^-$ than $Cl^-$
0%
much greater heat of hydration of $F^-$ than $Cl^-$
0%
greater electronegativity of $F$ than $Cl$

Explanation

In spite of having less electron affinity than chlorine, fluorine is the strongest oxidising agent among the halogens. The most important reason for this phenomenon is much greater heat of hydration of

$F^-$ than

$Cl^-$ which compensates for the lower electron affinity of

$F$ than of

$Cl$ . Hence, fluorine is the strongest oxidizing agent.

Which of the following arrangement of bond dissociation enthalpy of halogens is correct?

0%
$F-F>Cl-Cl>Br-Br$
0%
$F-F< Cl-Cl < Br-Br$
0%
$F-F > Cl-C l< Br-Br$
0%
$F-F < Cl-Cl > Br-Br$

Explanation

Bond energy is inversely dependent on bond length. The order of bond length is

$Br_2 > Cl_2 > F_2.$ .

Hence, the order of bond energy should be

$Br_2 < Cl_2 < F_2.$ . But, bond dissociation enthalpy of

$F_{2}$ is less than

$Cl_{2}$ because of small size of

$F_{2}$ and large inter-electronic repulsions.

Thus, order of bond energy is

$Cl_{2} > F_{2}$ and

$Br_2 < Cl_2$ .

Bleaching powder and bleach solution are produced on a large scale and is used in several house-hold products. The effectiveness of bleach solution is often measured by iodometry. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is:

0%
$Cl_{2}O$
0%
$Cl_{2}O_{7}$
0%
$ClO_{2}$
0%
$Cl_{2}O_{6}$
0%
$Cl_2O_3$
0%
$Cl_2O_5$
0%
$Cl_3O_9$

Explanation

Bleaching powder is

$Ca(OCl)Cl$ .

So, bleaching powder contains a salt of

$HOCl$ .

$Cl_2O$ is an anhydride of

$HOCl$ .

$Cl_2O+H_2O\Rightarrow 2HOCl$ .

Hence, option A is correct.

The products of complete hydrolysis of

$XeF_{4}$ are :

0%
$XeO_{3}$
0%
$Xe$
0%
$HF$
0%
$XeO_{4}$

Explanation

$6XeF_{4}+12H_{2}O\rightarrow 4Xe+2XeO_{3}+24HF+3O_{2}$

In the molecule

$XeO_{2}F_{2}$ , there exists one

$n$ -fold axis of symmetry and in the molecule

$XeO_{2}F_{4}$ , the number of

$2$ -fold axis of symmetry is

$m$ . The value of

$m + n$ is :

0%
$6$
0%
$7$
0%
$8$
0%
$5$

Explanation

$XeO_2F_2$ , there exists one

$2\ (n)$ -fold axis of symmetry, while in

$XeO_2F_4$ , there exists four

$(m)$

$2$ -fold axis of symmetry. The structures are given in the figure. Hence,

$n = 2$ and

$m = 4$ .

Therefore,

$n+m=2+4=6$

Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of :

0%
$NO$
0%
$NO_{2}$
0%
$N_{2}O$
0%
$N_{2}O_{4}$

Explanation

The decomposition of

$HNO_3$ is represented by the equation:

$4HNO_{3}\rightarrow 2H_{2}O+4NO_{2}+O_{2}$

Due to the formation of nitrogen dioxide, it turns yellow-brown.

Hence, option B is correct.

$XeF_2$ reacts with

$PF_5$ to give:

0%
$XeF_6$
0%
$[XeF]^+ [PF_6]^-$
0%
$XeF_4$
0%
$[PF_4]^+ [XeF_3]^-$

Explanation

$PF_5$ is a fluoride ion acceptor so it yields cationic species with xenon fluorides.

$[XeF_2]+ [PF_5]\rightarrow [XeF]^+ [PF_6]^-$

So, the correct answer is

$B$

Which of the following reactions does not produce ammonia as one of the products?

0%
$Mg_3N_2 + H_2O\rightarrow$ Products
0%
$(NH_4)_2SO_4 + CaO \rightarrow$ Products
0%
$NaNO_2 + Zn\xrightarrow{NaOH}$ Products
0%
$NH_4NO_3 \xrightarrow[ ]{\Delta }$ Products

Ammonia can be dried by :

0%
conc. $H_2SO_4$
0%
$P_4O_{10}$
0%
$CaO$
0%
$CaCl_2$

Explanation

Correct option: $C$

Explanation:
Method of preparation of Ammonia from Ammonium Chloride

By slowly heating $NH_4Cl$ and $Ca(OH)_2$

$2NH_4Cl+Ca\left ( OH \right )_2 \overset{\Delta}{\rightarrow} CaCl_2 +2NH_4OH \rightarrow 2NH_3+2H_2O$

$CaO$ dries ammonia because it is basic in nature and doesn't react with ammonia.

Ammonia cannot be dried by using concentrated sulphuric acid or phosphorus pentoxide because it reacts to form ammonium sulfate or ammonium phosphate respectively.

$2NH_3+H_2SO_4\rightarrow \left ( NH_4 \right )_2SO_4$

$P_2O_5+4NH_3+3H_2O\rightarrow 2\left ( NH_4 \right )_3PO_4$

Neither calcium chloride can be used for drying ammonia gas as it forms ammoniates.

Hence, ammonia can be easily dried by passing over soda lime.

Which of the following set of elements have the strongest tendency to form anions?

0%
N, O and F
0%
P, S and Cl
0%
N, P and Cl
0%
N, P and S

Ammonia gas is not prepared by :

0%
reduction of sodium nitrite or sodium nitrate by the reaction of zinc dust and sodium hydroxide
0%
hydrolysis of calcium cyanamide
0%
heating ammonium chloride with slaked lime
0%
heating of $(NH_4)_2Cr_2O_7$

Explanation

(A)

$NO_2^-+7[H]^-\xrightarrow {Zn/NaOH} NH_3+2H_2O; NO_3^-+9[H]^-\xrightarrow {Zn/NaOH} NH_3+3H_2O$

(B)

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

(C)

$NH_4Cl+Ca(OH)_2\xrightarrow {\Delta} 2NH_3+CaCl_2+2H_2O$

(D)

$(NH_4)_2Cr_2O_7\xrightarrow {\Delta} N_2+Cr_2O_3+4H_2O$

Option D is correct

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 the metal

$M$ .

0%
Calcium
0%
Sodium
0%
Cobalt
0%
Iron

Explanation

Calcium metal reacts with nitrogen to form calcium nitride.

If we consider metal

$M$ as calcium.

$3\underset{M}{Ca}+N_2 \rightarrow Ca_3N_2.$

The calcium nitride would react with water to form

$NH_3$ , which will react with

$HCl$ to give white fumes of

$NH_4Cl$ salt.

Thus, the metal

$M$ is calcium.

Hence, option (A) is the right answer.

The reaction of

$XeF_6$ with silica

$(SiO_2)$ gives:

0%
$XeO_3$ and $SiF_4$
0%
$XeF_4$ and $SiF_4$
0%
$XeOF_2$ and $SiF_4$
0%
$XeOF_4$ and $SiF_4$

Explanation

The product of the given reaction is as follows:

$2XeF_6 + SiO_2 \rightarrow 2XeOF_4+SiF_4$

Hence, option D is correct.

Which of the following molecule doesn't have

$d\pi -p\pi$ bonding?

0%
$SO_2$
0%
$P_4O_{10}$
0%
$H_3PO_4$
0%
$B_3N_3H_6$

Explanation

The structural study of

$B_3N_3H_6$ shows that it has hexagonal ring with alternating B and N atom along with H atom. Each B and N are

$sp^2$ hybridized and form three sigma bond one to terminal H atom and two to the adjacent N or B atom. There is a p

$\pi$ -p

$\pi$ donation of filled nitrogen p

$_z$ atomic orbital to adjacent Boron p

$_z$ atomic orbitals. So, borazine contain 3 pi and 12 sigma bonds. Hence, there is no involvement of any d-orbital in hybridization of

$B_3N_3H_6$ .

Which of the following do not exist?

0%
$\displaystyle\:SH_{6}$
0%
$\displaystyle\:HFO_{4}$
0%
$\displaystyle\:FeI_{3}$
0%
$\displaystyle\:HClO_{3}$

Explanation

only

$HClO_3$ exists as

The pair(s) that yield(s) the same gaseous product on reaction with water is/are :

0%
$AlN$ and $CaCN_2$
0%
$Ca$ and $CaH_2$
0%
$F_2$ and $XeF_4$
0%
$K$ and $KO_2$

Explanation

(A)

$AlN+3H_2O\rightarrow Al(OH)_3+NH_3; CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

(B)

$Ca+2H_2O\rightarrow Ca(OH)_2+H_2; CaH_2+H_2O\rightarrow Ca(OH)_2+H_2$

(C)

$2F_2+2H_2O\rightarrow 4HF+O_2; 6XeF_4+12H_2O\rightarrow 2XeO_3+24HF+4Xe+3O_2$

(D)

$2K+2H_2O\rightarrow 2KOH+H_2; 2KO_2+2H_2O\rightarrow+H_2O_2+O_2$

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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