MCQ Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 6

Number of electrons present in

$^{26} Fe$ is ___________.

0%
26
0%
27
0%
28
0%
29

Explanation

Number of electron present in the

$Fe^{26}$ is 26 because its atomic number is 26 which represent no. of electrons or protrons. Option 'A' is correct.

Which one of the following statements is correct?

0%
The elements like F, Cl, Br etc having high values of electron affinity act as strong oxidising agent
0%
The elements having low values of ionisation energies act as strong reducing agent
0%
The formation of $Be^{-}$ (g) from $Be$ (g) is an endothermic process
0%
All of these

Explanation

Halogens have high electron affinity and are strong oxidizing agents. The elements with low IE readily give electrons for reduction. Be is a metal. Thus, converting it to

${Be}^{-}$ is endothermic.

The pair of amphoteric hydroxides is:

0%
$Al(OH)_3, LiOH$
0%
$Be(OH)_2, Mg(OH)_2$
0%
$B(OH)_3, Be(OH)_2$
0%
$Be(OH)_2, Zn(OH)_2$

Explanation

The pair of amphoteric hydroxides is

$Be(OH)_2, Zn(OH)_2$ .
Amphoteric oxides reacts with acids as well as bases.
Beryllium hydroxide is amphoteric in nature due to small size, high electronegativity and high ionisation energy of beryllium.

$\displaystyle Be(OH)_2 + 2HCl \rightarrow BeCl_2 + 2H_2O$

$\displaystyle Be(OH)_2 + 2NaOH \rightarrow Na_2BeO_2 +2H_2O$
The amphoteric behaviour of zinc hydroxide is as shown.

$\displaystyle Zn(OH)_2 + 2HCl \rightarrow ZnCl_2 + 2H_2O$

$\displaystyle Zn(OH)_2 + 2NaOH \rightarrow \underset { \displaystyle \text {sodium zincate} } {Na_2ZnO_2} + H_2O$

Which group of elements does not show diagonal relationship?

0%
$Li,\,Mg$
0%
$Be,\,Al$
0%
$B,\,Si$
0%
$C,\,P$

Explanation

In addition to horizontal and vertical trends, there is a diagonal relationship between elements such as Li and Mg, Be and Al, B and Si.

A diagonal relationship is said to exist between certain pairs of diagonally adjacent elements in the second and third periods of the periodic table.

But

$C,\,P$ doesn't show diagonal relationship.

Consider the

$M(OH)_3$ formed by all the group 13 elements. The correct sequence of acidic strength of hydroxides

$[M(OH)_3]$ is:-

0%
$B(OH)_3 < Al(OH)_3 > Ga(OH)_3 > In(OH)_3 > Tl(OH)_3$
0%
$B(OH)_3 > Tl(OH)_3 > Al(OH)_3 > In(OH)_3 > Ga(OH)_3$
0%
$Al(OH)_3 > Ga(OH)_3 > B(OH)_3 > In(OH)_3 > Tl(OH)_3$
0%
$B(OH)_3 > Al(OH)_3 > Ga(OH)_3 > In(OH)_3 > Tl(OH)_3$

Explanation

As we know,

I.P

$\propto$ acidic strength of hydroxides

and

$\displaystyle \xleftarrow [ I.P. ]{B\:>\:Al\:>\:Ga\:>\:In\:>\:Ti}$

So order of acidic strength is

$B(OH)_3 > Al(OH)_3 > Ga(OH)_3 > Ti(OH)_3$

Many properties of an element and its compounds can be predicted from the position of that an element in the periodic table. What property could not be predicted this way?

0%
The acidic or basic nature of its oxide
0%
The formula of its oxide
0%
The number of isotopes it has
0%
Its metallic or non-metallic properties

Explanation

$\left( \ast \right)$ Many properties of an element from position of element in periodic table some properties.

(i) Acidic or basic nature of oxide

(ii) Oxidationstate

(iii) Formula of oxide

(iv) Metallic or non-metallic oxide.

But it cannot tell number of isotopes it has.

Ans :- C

Which of the following properties of the elements of group

$2^{nd}$ (alkaline earth metals) does not increase with increasing atomic number?

0%
Stability of carbonate
0%
Solubility of hydroxide
0%
Reactivity with water
0%
First ionization energy

The given diagram represents _________.

0%
$S$
0%
$S^-$
0%
$Cl$
0%
$Cl^-$

Explanation

The electronic configuration is defined as a distribution of electrons of an atom or molecule in the atomic or molecular orbital.

A diagram shows the distribution of an electron in shells of the atom. There are total

$16$ protons and

$16$ neutrons. A number of protons are equal to an atomic number. So, the given diagram is of sulphur having one negative charge. In neutral atom Number of protons is equal to a number of electrons and this diagram has one electron more i.e.

$17$ electron. Hence, it is

$S^-$ represented by the diagram.

An element M belongs to the group 1 and period 3 of the modern periodic table. Out of the following statement, which is the true about 'M'?

0%
It forms an acidic oxide having formula $M_2 O_3$
0%
It is a non-metal
0%
It is highly electro positive in nature
0%
It has a valency of three

Poor shielding of nuclear charge by d or f-orbital electrons is responsible for which of the following facts?

0%
Atomic radius of Nb (4-1 series) is comparable to that of Ta (5-1 series)
0%
The $1^{st}$ ionisation energy of copper is less than that of zinc
0%
The value of electron gain enthalpy is more negative for sulphur than for oxygen.
0%
The $1^{st}$ ionisation energy for gold is greater than that of silver.

Explanation

Poor shielding of nuclear charge by d or f-orbital electrons is responsible for the main reason why

$1^{st}$ ionization energy of copper is less than that of zinc.

Which of the following statements are true about the modern form of the periodic table?

0%
Properties of the elements are the periodic functions of their atomic number
0%
There are 7 periods
0%

There are 18 numbered groups
0%
All of the above

Which of the following element has maximum first ionisation potential?

0%
V
0%
Ti
0%
Cr
0%
Mn

Explanation

Mn has maximum first ionisation potential due to stable half filled d-orbitals. Electronic configuration of

$Mn$ is

$1s^22s^22p^63s^23p^64s^23d^5$ .

Sodium has ___________ as compared to potassium.

0%
less electronegativity
0%
more ionization potential
0%
larger atomic radius
0%
lower melting point

Explanation

Alkali metals can readily lose their one electron to form +1 oxidation state and can easily ionize this electron and this

$ns_1$ configuration results in alkali metals to have large ionic and atomic radii and they have high second ionization energy. This ionization energy decreases down the group.

An element having the highest first ionization energy in a given period is :

0%
inert gas
0%
transition element
0%
representative element
0%
inner transition element

Among Li, Be, N and F, the element having the largest atomic radius is

0%
Li
0%
Be
0%
N
0%
F

The electronic configuration of the element which is just above the element with atomic number 43 in the same periodic group is :

0%
$1s^2,\, 2s^2\, 2p^6,\, 3s^2\, 3p^6\, 3d^5,\, 4s^2$
0%
$1s^2,\, 2s^2\, 2p^6,\, 3s^2\, 3p^6\, 3d^{10}\, 4s^2\,4p^5$
0%
$1s^2,\, 2s^2\, 2p^6,\, 3s^2\, 3p^6\, 3d^6,\, 4s^1$
0%
$1s^2,\, 2s^2\, 2p^6,\, 3s^2\, 3p^6\, 3d^{10},\, 4s^1\,4p^6$

Explanation

The element with atomic number 43 is Tc and the element just above this element is 43-18= 25 which is Mn

$[Ar]{ 3d }^{ 5 }{ 4s }^{ 2 }$

The 1st I.E. of N as compared to C is:

0%
greater
0%
smaller
0%
equal
0%
none of these

According to Mendleev's periodic law which properties of the elements are the periodic function of their atomic masses?

0%
Physical properties
0%
Chemical properties
0%
Physical and chemical properties both
0%
None of these

Which of the following statements is incorrect about the trends in periodic properties in the periodic table ?

0%
Metallic character increases from Li to Cs
0%
The acidity of oxides increases from left to right in a period
0%
Noble gases have the highest electron affinity among all the elements in the periodic table
0%
Ionisation energy of aluminium is lower than that of magnesium

Which of the following has the highest value of ionisation energy?

0%
Be
0%
He
0%
Ne
0%
Xe

Number of electrons in

$M$ shell of an element with atomic number

$28$ is _______.

0%
18
0%
16
0%
8
0%
2

The element having electronic configuration

$[Kr]4d^{10}\, 4f^{14}\, 5s^2\, 5p^6\, 5d^1\, 6s^2$ belongs to :

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

Electronic configuration of

$Si$ in ground state is _____________.

0%
$2, 8, 1$
0%
$2, 8, 2$
0%
$2, 8, 4$
0%
$2, 8, 3$

Which of the following elements has the least ionisation energy?

0%
$Li$
0%
$Cs$
0%
$Mg$
0%
$Ca$

Explanation

On moving down the group

$I.E$ decreases. So from

$Li$ to

$Cs$ I.E. decreases. Calcium and magnesium belong to alkaline earth metals i.e. group 2 so they have high I.E as it increases along a period when electrons are tightly held to nucleus in the same shell. In moving down new shell is added to the atom and valence electron is loosely held to nucleus so

$Cs$ has least I.E.

What are the electronic configuration of

$Na^{+}$ and

$Cl^{-}$ ions?

0%
$Na^{+}=2, 8, 2$ and $Cl^{-}=2, 8, 7$
0%
$Na^{+}=2, 8$ and $Cl^{-}=2, 8, 8$
0%
$Na^{+}=2, 8, 2$ and $Cl^{-}=2, 8, 6$
0%
$Na^{+}=2, 8$ and $Cl^{-}=2, 8, 7$

Explanation

Atomic number of

$Na = 11$ and thus atomic no. of

$Na^{+}$

$=$ 10 and E.C

$= 2, 8$
Atomic number of Cl

$=$ 17 and thus atomic no. of

$Cl^{-}$

$=$ 18 and E.C

$=$ 2, 8, 8

Which element of 2nd period forms most acidic oxide?

0%
Flourine
0%
Nitrogen
0%
Boron
0%
Carbon

Explanation

The acidity of the oxides increases along the

$II$ period from left to right. Thus the oxide of Florine is most acidic in nature

The electronic configuration of a dipositive metal ion

$M^{2+}$ is

$2, 8, 14$ and its ionic weight is

$56$ . The number of neutrons in its nucleus would be:

0%
$30$
0%
$32$
0%
$34$
0%
$42$

Explanation

Dipositive ion has

$24$ electrons and hence atom of the metal will have

$26$ electrons and

$26$ protons.

No. of neutrons = Atomic wt. - No. of protons

$= 56-26$

$= 30$

Hence, option

$A$ is correct.

When Li, Na, and K are placed in decreasing order of their ionisation energy, the order is :

0%
Li, K, Na
0%
K, Na, Li
0%
Na, Li, K
0%
Li, Na, K

Who was the first to classify the elements into metals and non-metals?

0%
Meyer
0%
Mendeleev
0%
Lavoisier
0%
None of the above

Ionisation energy in the alkali metals group :

0%
increases from Li to Cs
0%
decreases from Li to Cs
0%
remains the same
0%
first increases then decreases

Which is the correct electronic configuration of

$Cr$ (Chromium)?

0%
$1s^2 2s^22p^63s^23p^64s^1 3d^5$
0%
$1s^22s^22p^63s^23p^64s^23d^4$
0%
$1s^22s^22p^63s^23p^64s^23d^6$
0%
$1s^22s^22p^63s^23p^64s^13d^8$

Explanation

Atomic number of

$Cr = 24$
Electronic configuration of

$Cr= 1s^2, 2s^2,2p^6,3s^2,3p^6,4s^1,3d^5$
Exactly half-filled and d orbitals give extra stability to

$Cr$ atom.

The element with the highest first ionisation energy among the following is:

0%
Oxygen
0%
Nitrogen
0%
Boron
0%
Carbon

Which of the following has minimum ionisation energy?

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

Which of the following metal forms amphoteric oxide?

0%
Copper
0%
Silver
0%
Aluminium
0%
Iron

Explanation

An amphoteric oxide is a substance that can chemically react as either acid or base. Aluminium oxide shows acid and basic properties of oxide so it is amphoteric. Thus,

$Al_2O_3$ entails the marking point at which a change over from a basic oxide to acidic oxide occurs.

Which of the following element would lose an electron easily?

0%
$Mg$
0%
$Na$
0%
$K$
0%
$Ca$

Which one of the following elements has the highest first ionization potential?

0%
$C$
0%
$N$
0%
$B$
0%
$O$

Explanation

Nitrogen has the highest first ionization energy, though it should have less than that of oxygen while going across a period. The electronic configuration of nitrogen is

${ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }_{ x }^{ 1 },{ 2p }_{ y }^{ 1 },{ 2p }_{ z }^{ 1 }$ . Nitrogen has three electrons in

$2p$ subshell which are singly filled. So, by Hund's Rule, half filled and fully filled orbitals are most stable. So, nitrogen is in a stable state and does not want to lose electrons. Hence, it has the highest first ionization energy.

A newly discovered element

$X$ is placed in group

$IA$ of the periodic table since it forms :

0%
an oxide which is acidic
0%
a volatile chloride having a formula $XCl$
0%
an ionic chloride having the formula $XCl$
0%
an insoluble $X{SO}_{4}$

Explanation

The in group I of the periodic table are called the alkali metals. They form ionic compounds when they react with non-metals. Their ions have a single positive charge. For example, sodium forms sodium ions, $Na^+$ .

The elements in Group 7 of the Periodic Table are called the halogens. They form ionic compounds when they react with metals. Their ions have a single negative charge. For example, chlorine forms chloride ions, $Cl^-$ .

Group $1A$ elements form $XCl$ type chloride.

The electronic configuration of the element with Atomic number 19 is___________.

0%
2,8,7
0%
2,9,8
0%
2,8,8,1
0%
2,10,7

Explanation

The electronic configuration of the element with Atomic number 19 is 2,8,8,1 or

$1s^2 2s^22p^6 3s^23p^6 4s^1$ or

$[Ar] 4s^1$ . The element is potassium. It is an alkali metal with one valence electron.

Which of the following burns to form an oxide, which is gaseous at room temperature?

0%
Hydrogen
0%
Phosphorus
0%
Calcium
0%
Sulphur

Explanation

$H_2+O_2 \rightarrow H_2O(l)$

$P+O_2 \rightarrow P_4O_{10}(s)$

$Ca+O_2 \rightarrow CaO(s)$

$S+O_2 \rightarrow SO_2(g)$

What type of oxide would Eka-aluminium form?

0%
$EO_3$
0%
$E_3O_2$
0%
$E_2O_3$
0%
$EO$

Explanation

Eka- Aluminium (E) forms oxide with the same chemical formula as Aluminium oxide. Thus, the formula can be represented as

${ E }_{ 2 }{ O }_{ 3 }$ .

Hence, option C is correct.

The element with the atomic number

$9$ resembles with the element having the atomic number:

0%
$8$
0%
$17$
0%
$36$
0%
$27$

Ionization potential in a period is lowest for :

0%
halogens
0%
inert gases
0%
alkali metals
0%
alkaline earth elements

Metals and non-metals are separated by a zig-zag line.

0%
True
0%
False

Which of the following sets of elements has strongest tendency to form anions ?

0%
$As, Se, Br$
0%
$P, S, Cl$
0%
$N, O, F$
0%
$Sb, Te , I$

Explanation

Electronegativity is the tendency to attract a shared pair of electrons and this property increases moving across from left to right in a period and decreases on moving down the group. So

$N, O, F$ are highly electronegative in nature.

Arrange the following metal in the increasing order of their reactivity towards water: Zinc, Iron, Magnesium, Sodium.

0%
Iron < Magnesium < Sodium < Zinc
0%
Iron < Zinc < Magnesium < Sodium
0%
Magnesium < Iron < Sodium < Zinc
0%
Sodium < Iron < Magnesium < Zinc

Electronic configuration of vanadium is:

0%
$[Ar] 3d^{10} 4s^1$
0%
$[Ar] 3d^{9} 4s^2$
0%
$[Ar] 3d^{5} 4s^1$
0%
$[Ar] 3d^{3} 4s^2$

Explanation

Electronic configuration is the number of electrons distributed per shell.

Example:- Electronic configuration of Vanadium

$[Ar]3d^34s^2$

Electronic configuration of Scandium

$[Ar]3d^14s^2$ .

The first and second ionization energies (KJ mol

$^{-1}$ ) of three elements A, B, and C are given below.

	A	B	C
IE $_1$	402	650	1140
IE $_2$	2642	1060	2070

Identify the element which is likely to be an alkali-metal.

0%
A
0%
B
0%
C
0%
Can't be predicted

A particular atom has the 4th shell as its valence shell. If the difference between the number of electrons between K and N shell and L and M shell is zero, find the electronic configuration of its stable ion.

0%
2, 8, 8, 2
0%
2, 8, 18, 8
0%
2, 8, 8
0%
2, 8

Explanation

Number of electrons in shell is given by

$2n^2$

where

$n$ = energy level

Maximum number of electrons in

$K$ shell

$\longrightarrow 2n^2=2\times (1)^2=2$

$[n=1]$

Maximum number of electrons in

$L$ shell

$\longrightarrow 2n^2=2\times (2)^2=2\times 4=8$

$[n=2]$

Maximum number of electrons in

$M$ shell

$\longrightarrow 2n^2=2\times (3)^2=2\times 9=18$

$[n=3]$

Maximum number of electrons in

$N$ shell

$\longrightarrow 2n^2=2\times (4)^2=2\times 16=32$

$[n=4]$

Difference between

$K$ and

$N$ shell is zero.

$K$ can accommodate maximum

$2$ electron. Therefore, for difference to be zero

$N$ should also have

$2$ electrons.

$\therefore K=2$ electrons

$N=2$ electrons

Difference between

$L$ and

$M$ is also zero. Maximum electron in

$L=8$ . So, for difference to be zero

$M$ should also have

$8$ electron in it.

$\therefore L=8e^-$

$M=8e^-$

Electrons per shell in

$K$

$L$

$M$

$N$ [

$N$ is the

$4^{th}$ valence shell]

$2$

$8$

$2$

Stable ion is formed by loss of

$2$ electrons from

$N$ shell.

$\therefore$ Electronic configuration will be

$2,8,8$ .

The element with the highest ionization energy is:

0%
lithium
0%
hydrogen
0%
helium
0%
neon

Explanation

He contains fully filled

$1s^{2}$ orbital which has more penetrating effect and is very close to the nucleus and hence has higher value of ionisation energy.

Hence , option C is correct .

Element

$A$ has electronic configuration 2,

$B$ has configuration 2,8,

$C$ has configuration 2,8 while

$D$ has 2,8,Which element will show similar chemical properties?

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

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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