When $${ SO }_{ 2 }$$ is passed through acidified $${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$$ solution:

green $${ Cr }_{ 2 }({ SO }_{ 4 }{ ) }_{ 3 }$$ is formed.

Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms.

Each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms.

Silicon atom is bonded two oxygen atoms

There are double bonds to two oxygen atoms

a good conductor of electricity

each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms

each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms

silicon atoms is bonded to two oxygen atoms

there are double bonds between silicon and oxygen atoms

MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 10

Which of the following acid has the same molecular weight and equivalent weight ?

0%
$H_2SO_4$
0%
$H_3PO_2$
0%
$H_3PO_3$
0%
$H_3PO_4$

Explanation

Equivalent weight equal to molecular weight divided by n- factor.
$n- factor$ in case of $H_3PO_2 = 1$
Then Equivalent weight equal to molecular weight in case of $H_3 PO_2$ .

Covalency of sulphur in

S O_{4}^{2 -}

$SO_{4}^{2-}$ ion is:

0%
4
0%
6
0%
2
0%
5

Explanation

The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion. The oxidation number of the sulfur atom in the

S O_{4}^{- 2}

$SO_4^{-2}$ ion must be

+ 6

$+6$ . For example, because the sum of the oxidation numbers of the atoms in this ion must equal -2.

Colour of the solution when

K I

$KI$ reacts with

B r_{2}

$Br_2$ is:

0%
pale yellow
0%
reddish-brown
0%
black
0%
none of the above

Explanation

They undergo a displacement reaction where a more reactive halogen in this case bromine displaces iodine from its salt.

{B r}_{2_{(l)}} + {2 K I}_{(a q)} \to {2 K B r}_{(a q)} + I_{2}

${ Br }_{ { 2 }_{ \left( l \right) } }+{ 2KI }_{ \left( aq \right) }\rightarrow { 2KBr }_{ \left( aq \right) }+{ I }_{ 2 }$

In this reaction, the solution will have a slight change in colour to a different shade of brown since the bromine and iodine solution are both in a different shade of brown. Since it's reddish-brown.

Hence, option B is correct.

When

{S O}_{2}

${ SO }_{ 2 }$ is passed through acidified

K_{2} {C r}_{2} O_{7}

${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$ solution:

0%
the solution truns blue.
0%
${ SO }_{ 2 }$ is reduced.
0%
green ${ Cr }_{ 2 }({ SO }_{ 4 }{ ) }_{ 3 }$ is formed.
0%
the solution is decolourised.

Explanation

Green

C r_{2} (S O_{4})_{3}

$Cr_2(SO_4)_3$ is formed When

S O_{2}

$SO_2$ is bubbled through acidified solution of

K_{2} C r_{2} O_{7}

$K_2Cr_2O_7$ , the orange colour of

K_{2} C r_{2} O_{7}

$K_2Cr_2O_7$ solution turns to clean green due to formation of chromium sulphate.

Hence option C is correct answer.

H C l

$HCl$ cn be oxidised by:

0%
$KMnO_4$
0%
$Cl_2$
0%
Conc. $H_2SO_4$
0%
$KMnO_4$ and Conc. $H_2SO_4$

Explanation

$H^+$ in $HCl$ is the reactant reduced, and $HCl$ is the oxidizing agent. $Zn$ is the reactant oxidized and the reducing agent. $Cu^{2+}$ in $CuCl_2$ is the reactant reduced, and $CuCl_2$ is the oxidizing agent. $Cl_2$ is the reactant reduced and the oxidizing agent.

Select the correct statements:

0%
First IP of Mg is less than that of Al
0%
$O^+$ has lower electronegativity than that of $O^-$
0%
Electron affinity of Cl is lesser than that of F
0%
Second IP of oxygen is greater than that of N

Explanation

solution:

The second ionisation energy is the energy required to remove the electron from the corresponding mono-valent cation of the respective atom.

Oxygen has greater second ionisation energy than fluorine.

This is because oxygen atom has stable electronic configuration,

2 s^{2}, 2 p^{3}

$2s^2, 2p^3$ after removing one electron, the O+ shows greater ionisation energy than

F^{+}

$F^+$

hence the correct option: D

Blue vitriol has:

0%
Ionic bond
0%
Coordinate bond
0%
Hydrogen bond
0%
All the above

Explanation

Option D is the correct answer.

Blue vitriol or copper sulphate pentahydrate

$\text{Blue vitriol$$ or $$copper sulphate pentahydrate}$ has a molecular formula of

C u S O_{4} .5 H_{2} O .

$CuSO_4.5H_2O.$ of the five water molecules associated with this structure, only four are directly replaceable by

N H_{3}

$NH_3$ , which suggest that water molecule is somehow differently attached.

Blue vitriol is

C u S O_{4} .5 H_{2} O

$CuSO_4.5H_2O$ and it has all types of bonds.

The structure of

C u S O_{4} .5 H_{2} O

$CuSO_4.5H_2O$ in the solid-state is shown above.

The

C u^{2 +}

$Cu^{2+}$ ions are attracted towards

S O_{4}^{2 -}

$SO_4^{2−}$ ions not only by ionic interactions (electrovalent) but also by coordinate covalent bonds.

The

C u^{2 +}

$Cu^{2+}$ ions form coordinate covalent bonds with water as well as sulfate ions.

There are covalent bonds in water and sulfate ions.

1646951_1220913_ans_c6c66bfabfa04ec58cb78eae660d7594.jpg

The process requiring absorption of energy is :

0%
$N\rightarrow { N }^{ - }$
0%
$F\rightarrow { F }^{ - }$
0%
$Cl\rightarrow { Cl }^{ - }$
0%
$H\rightarrow { H }^{ - }$

The correct order of melting point of 14th group element is:

0%
$C > Si > Ge > Pb > Sn$
0%
$C < Si < Ge < Sn < Pb$
0%
$Sn > Pb > Ge > Si >C$
0%
$Si > C > Ge > Sn > Pb$

Explanation

The melting point of Group-14 decreases down the group with the exception of

P b

$Pb$ whose melting point is slightly higher than that of

S n

$Sn$ .

Hence the correct order is :

C > S i > G e > S n > P b

$C > Si > Ge > Sn >Pb$

Maximum covalency shown by sulphur is:

0%
$2$
0%
$4$
0%
$6$
0%
none

Explanation

S = 1 s^{2} 2 s^{2} 2 P^{6} 3 s^{2} 3 p^{4} = 6

$S=1s^22s^22P^63s^23p^4=6$

Sulpher is the member of

16

$16$ group element so it's maximum covalency is

6

$6$ also in it's excited state it has

6

$6$ unpaired electron thus it has

6

$6$ maximum covalency for example in

S F_{6}

$SF_{6}$ the maximum valency is

6

$6$ .

What the covalency of phosphorous atom in excited state ?

0%
3
0%
4
0%
5
0%
1

Explanation

Nitrogen atoms can form up to four covalent bonds.In the ammonium ion( $NH_4^+$ ), the nitrogen atom forms four covalent bonds, again filling the outermost orbital with eight electrons: Phosphorus can form up to five covalent bonds, as in phosphoric acid $(H_3PO_4)$ .

In the following which has highest boiling point:

0%
$H I$
0%
$H F$
0%
$H Br$
0%
$HCl$

Explanation

The order of boiling is

H F > H I > H B r > H C l

$HF>HI>HBr>HCl$
Due to strong hydrogen bonding in HF. The boiling point of HF is very high. So, HF has highest boiling point.

The most abundant IV A group element in the earth's crust is :

0%
Germanium
0%
Carbon
0%
Silicon
0%
Tin

Oxidation state of carbon in HCN and HNC is respectively:-

0%
$2, 4$
0%
$4,2$
0%
$2, 2$
0%
$4, 4$

Average atomic weight of boron is

10.10

$10.10$ and boron exists in two isotopic forms

B^{10}

$B^{10}$ and

B^{11}

$B^{11}$ . The percentage abundance of

B^{10}

$B^{10}$ is?

0%
$10\%$
0%
$90\%$
0%
$50\%$
0%
$20\%$

Explanation

Let the percentage in the form of

B^{10}

$B^{10}$ is x %

Therefore, the percentage in the form of

B^{11}

$B^{11}$ is (100-x) %

Average atomic mass =

x \times \frac{10}{100} + (100 - x) \times \frac{11}{100}

$x \times \dfrac{10}{100} + (100-x) \times \dfrac{11}{100}$

10.10 = \frac{10 \times x}{100} + \frac{1100 - 11 x}{100}

$10.10 = \dfrac{10 \times x}{100} + \dfrac{1100-11x}{100}$

1010 = 1100 - x

$1010 = 1100 - x$

x = 90

$x = 90$

Therefore, the percentage in which

B^{10}

$B^{10}$ exists is 90 %

B F_{3}

$BF_{3}$ on hydrolysis forms

0%
$H_{3}BO_{3}$
0%
$HBF_{4}$
0%
Both $(A)$ and $(B)$
0%
None of these

Boron exists in +3 state where as thallium exist in both +1 & +3 oxidation state, due to:

0%
inter pair effect
0%
lanthanoid contraction
0%
Diagonal relationship
0%
none of these

What is the colour of a solution of

{A l}_{2} {({S O}_{4})}_{3}

$\mathrm { Al } _ { 2 } \left( \mathrm { SO } _ { 4 } \right) _ { 3 }$ in water?

0%
blue
0%
pink
0%
green
0%
colourless

Explanation

The colour of solution

A l_{2} (S O_{4})_{3}

$Al_2(SO_4)_3$ in water is colourless. Hence Aluminium and sulphate reacts equally and dissolves

Which of the following order regarding the melting point is correct?

0%
$Si>Ge>Pb>Sn$
0%
$Si>Pb>Ge>Sn$
0%
$Ge>Si>Pb>Sn$
0%
none of these

Boron when heated with carbon powder at high-temperature forms

0%
$BC_2$
0%
$B_4C_3$
0%
$B_2C_3$
0%
$B_4C$

Eka-boron was
subsequently named as

0%
Gallium
0%
Germanium
0%
Scandium
0%
Molybdenum

Aluminium becomes passive when treated with______

0%
$Conc.H_{2}SO_{4}$
0%
$Conc.HNO_{3}$
0%
$Conc.HCl$
0%
$dil.HCl$

Explanation

Aluminium becomes passive when treated with Conc

H N O_{3}

$HN{O}_{3}$ due to formation of

{A l}_{2} O_{3}

${Al}_{2}{O}_{3}$ . It forms a protective coating on the surface of the metal.

Which of the following compound is amphoteric?

0%
$Al_2O_3$
0%
$CrO_3$
0%
$SO_2$
0%
$CO_2$

Explanation

An amphoteric oxide is an oxide that can act as either an acid or base in a reaction to produce a salt and water.

A l_{2} O_{3}

$Al_2O_3$ reacts with acid as well as base. Hence, it is amphoteric in nature.

Hence, option A is correct.

Maximum covalency shown by boron is:

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

In silicon dioxide:

0%
Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms.
0%
Each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms.
0%
Silicon atom is bonded two oxygen atoms
0%
There are double bonds to two oxygen atoms

Which of the following is amorphous form of silica?

0%
Quartz
0%
Kieselguhr
0%
Tridymite
0%
Cristobalite

The oxidation state of P in

H_{3} P O_{5}

$H_3PO_5$ is :

0%
+3
0%
+5
0%
+7
0%
None of these

S i_{6} O_{18}^{12 -}

$Si_6O^{12-}_{18}$ unit is an example of

0%
3D Silicate
0%
Double chain silicate
0%
Cyclic silicate
0%
2D silicate

More than four bonds are made by how many elements in carbon family?

0%
$1$
0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

The elements belong to carbon family are:

C a r b o n - (C)

$Carbon-(C)$

S i l i c o n - (S i)

$Silicon-(Si)$

G e r m a n i u m - (G e)

$Germanium-(Ge)$

T i n - (S n)

$Tin-(Sn)$

L e a d - (P b)

$Lead-(Pb)$

F l e r o v i u m - (F l)

$Flerovium-(Fl)$

Except carbon all other elements contain empty d-orbitals in their outermost shell, thus can form more than four-bonds.

Therefore 5 is the correct answer.

Dry ice is

0%
Solid $CO$
0%
Solid $SO_{2}$
0%
Solid $CO_{2}$
0%
Solid $O_{2}$

Carbon shows tetravalency due to:

0%
$s^2 p^2$ hybridization
0%
$d^3 p$ hybridization
0%
$dsp^2$ hybridization
0%
$sp^3$ hybridization

The basic nature of the hydroxides of group

13 (I I I A)

$13 (III A)$ decreases progressively down the group.

0%
True
0%
False

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 last elements of the

p

$p$ -block in 6th period is represented by the outer most electronic configuration.

0%
$7s^27p^6$
0%
$5f^{14}6d^{10}7s^27p^0$
0%
$4f^{14}5d^{10}6s^26p^6$
0%
$4f^{14}5d^{10}6s^26p^4$

The element with atomic number 35 belongs to :

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

Explanation

The electronic configuration of the element with atomic number 35 is

[A r] 3 d^{10} 4 s^{2} 4 p^{5}

$[Ar]3d^{10}4s^24p^5$ . The last electron enters to p-subshell. Therefore, it is a p-block element.

In view of the signs of

Δ_{r} G^{\circ}

$\Delta_r G^{\circ}$ for the following reaction

P b O_{2} + P b \to 2 P b O

$PbO_2 + Pb \rightarrow 2 PbO$ ,

Δ_{r} G^{\circ}

$\Delta_r G^{\circ}$ < 0

S n O_{2} + S n \to 2 S n O

$SnO_2 + Sn \rightarrow 2 SnO$ ,

Δ_{r} G^{\circ}

$\Delta_r G^{\circ}$ > 0 which oxidation states are more characteristic for lead and tin?

0%
For lead +4, for tin +2
0%
For lead +2, for tin +2
0%
For lead +4, for tin +4
0%
For lead +2, for tin +4

Explanation

P b O_{2} + P b \to 2 P b O

$PbO_{2} + Pb \rightarrow 2PbO$

+4 0 +2

Hence ,

Δ G

$\Delta G$ > 0 Spontaneous +2 is more stable

S n O_{2} + S n \to 2 S n O

$SnO_{2} + Sn \rightarrow 2SnO$

+4 0 +2

Hence ,

Δ G

$\Delta G$ > 0 Non - Spontaneous +2 is less stable

So for lead +2 and for Sn +4

Hence , option D is correct.

What is the general electronic configuration of group

14

$14$ or

I V A

$IVA$ elements:

0%
$ns^2\ np^4$
0%
$ns^2\ np^2$
0%
$ns^2\ np^6$
0%
$ns^2$

Which of the following statements is correct?

0%
Elements of group 15 form electron deficient hydrides
0%
All elements of group 14 form electron precise hydrides
0%
Electron precise hydrides have tetrahedral geometries
0%
Electron-rich hydrides can act as Lewis acids

Colemanite is a mineral of:

0%
$Ca$
0%
$B$
0%
$Al$
0%
$Mg$

B^{3 +}

$B^{3+}$ cannot exist in aqueous solution because of its

0%
Strong reducing ability
0%
Strong oxidizing ability
0%
Small size and large charge
0%
Large size and small charge

Explanation

Being covalent in nature because of small size and large charge it gets hydrolysed in an aqueous solution. In other words, on account of higher charge density i.e. charge/size ratio, it gets hydrolysed forming

H_{3} B O_{3}

$H_{3}BO_{3}$

2 B^{3 +} + 6 H_{2} O \to 2 H_{3} B O_{3} + 6 H^{+}

$2B^{3+} + 6H_{2}O \rightarrow 2 H_{3}BO_{3} + 6H^{+}$

Hence, option C is correct.

Carbon atoms in diamond are bonded with each other in a configuration:

0%
linear
0%
planar
0%
octahedral
0%
tetrahedral

Explanation

Diamond is the hardest material known on earth. Each carbon atom is surrounded by four other carbon atoms, which lie at the corner of a regular tetrahedron. Valence bond connects each carbon atom with four others. Carbon atoms in diamond are

s p^{3}

$sp^3$ hybridized.

Hence the correct option is

(D)

$(D)$ .

1615195_1727452_ans_e3cc3c884ba64e14b82f25076356b37a.png

Which one is borax?

0%
$Na_2B_4O_7 \cdot 10H_2O$
0%
$Na_2B_4O_7 \cdot 4H_2O$
0%
$Ca_2B_6O{11} \cdot 5H_2O$
0%
$CaB_4O_7$

Alumina is _______

0%
a bad conductor of electricity
0%
a good conductor of electricity
0%
a dehydrating agent
0%
soluble in water

Which of the following is a semiconductor?

0%
$C$
0%
$Pb$
0%
$Ge$
0%
$Sn$

Which of the following oxides has three dimensional structure?

0%
$CO$
0%
$CO_2$
0%
$SiO_2$
0%
$SO_2$

Explanation

$CO, CO_2$ are linearly shaped molecules whereas $SO_2$ is a $bent$ shaped molecule.

$SiO_2$ has a $three-dimensional$ structure as $Si$ in $SiO_2$ is $sp^3$ hybridised with no lone pairs. Silicon dioxide is a covalent, network solid in which each silicon atom is tetrahedrally surrounded by 4 oxygen atoms

Option C is correct.

Which of the following is a covalent carbide?

0%
$CaCO_3$
0%
$Al_4C_3$
0%
$WC$
0%
$SiC$

Which silicon compounds is used as lubricant?

0%
Asbestos
0%
Silicone
0%
Zeolite
0%
Mica

In silicon oxide:

0%
each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms
0%
each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms
0%
silicon atoms is bonded to two oxygen atoms
0%
there are double bonds between silicon and oxygen atoms

Explanation

The tetravalency of

S i

$Si$ is surrounded by four oxygen atoms. Much of the silicon and oxygen in the Earth’s crust is present as the compound silicon dioxide also known as silica.
1642112_1728454_ans_06accee530c4419198ed0e89ef500aa1.png

1642112_1728454_ans_06accee530c4419198ed0e89ef500aa1.png

Possible oxidation states of boron family elements are

0%
$+1$
0%
$+2$
0%
$+3$
0%
$+4$

Which of the following statements is incorrect?

0%
Boron shows only $+3$ oxidation state.
0%
In $Ga+3$ oxidation state is more stable than $+1$ oxidation state.
0%
In $Sn+2$ oxidation state is more stable than $+4$ oxidation state.
0%
In $Tl+1$ oxidation state is more stable than $+3$ oxidation state.

Explanation

In the group 13, 14, 15 the stability of lower oxidation state increases down in the group.

But,

S n^{+ 4}

$Sn^{+4}$ is more stable than

S n^{+ 2}

$Sn^{+2}$ due to its configuration.

The electronic configuration of Sn is

5 s^{2} 5 p^{4}

$5s^{2} 5p^{4}$ .

S n^{4}

$Sn^{4}$ , it lose four electrons and achieve full filled electronic configuration.

Due to its stability ,

S n^{+ 4}

$Sn^{+4}$ is more stable than

S n^{2}

$Sn^{2}$ .

Among the all statements mentioned,

(C) is incorrect

$\textbf{(C) is incorrect}$ .

Hence, option will be

(C)

$(C)$ .

Inert pair effect is not exhibited by:

0%
$Pb$
0%
$Bi$
0%
$Tl$
0%
$B$

Explanation

Inert pair effect

$\textbf{Inert pair effect}$ is the tendency of the two electrons in the outermost atomic s-orbital to remain unshared in compounds of post-transition metals.

Inert pair effect is not exihibited by

Boron

$\textbf{Boron}$ due to some relativistic effects caused by the enormous speed of electrons on lower energy-levels in heavier atoms.It shows only single state i.e. +3.

Hence,

$\textit{Hence,}$

correct option is (D) B

$\textbf{correct option is (D) B}$ .

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 10 - MCQExams.com

0:0:3

Practice Class 11 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 10 - MCQExams.com

0:0:3

Practice Class 11 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.