specific atomic number and mass number

MCQ Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 9

Select the group where

$EN$ increases down the group.

0%
$Zn, Cd, Hg$
0%
$F, Cl, Br$
0%
$Li, Na, K$
0%
$Ca, Sr, Ba$

What is the correct order of ionisation energy?

0%
B > Al < Ga > In < Tl
0%
B < Al > Ga > In > Tl
0%
B > Al > Ga > In >Tl
0%
B > Al < Ga < In < Tl

Explanation

In the group of Boron, as we move down ionic radii increase thereby decreasing ionisation potential/energy, as larger the radii more loosely held is the outermost

$e^-$ .

$\therefore B>Al>Ga>In>Tl$

Hence, the correct option is

$\text{C}$

Which of the following gases is most predominating in the Sun?

0%
Hydrogen
0%
Nitrogen
0%
Ozone
0%
Helium

Explanation

The most predominat gas is sum is hydrogen. The percentage of abundance of elements in sum is as follows:

Hydrogen :

$91.2\%$

Helium :

$8.7\%$

Oxygen :

$0.078\%$

In hydrogen atom the energy difference between the states

$n=2$ and

$n=3$ is

$EeV$ . The ionisation potential of

$H$ atom is:

0%
$3.2E$
0%
$5.6E$
0%
$7.2E$
0%
$13.2E$

Explanation

Energy difference,

$E=13.6\left[\cfrac {1}{(2)^2}-\cfrac {1}{(3)^2}\right]$

$\implies E=\cfrac{5}{9\times 4}\times 13.6$

$eV$

$\implies E=\left(\cfrac {5}{36}\right)\times 13.6$

$eV$

Ionization potential

$=13.6$

$V=\cfrac {36}{4}E$

$=7.2E$

The first ionisation enthalpy of the elements are in the order of:

0%
$C < Si <N < P$
0%
$N < Si < C < P$
0%
$Si < N < C < P$
0%
none of the above

Which is correct increasing order of their tendency of the given elements to form

$M^{3-}$ ion?

0%
$Bi > Sb > As > P > N$
0%
$Bi < Sb < As < P < N$
0%
$N < P < Sb < Bi < As$ $
0%
$Bi > Sb \sim N \sim P > As$

Explanation

$Bi < Sb < As < P < N$ This is the nitrogen family.

$N$ have highest electronegativity due to which it is having high tendency of forming

$M^{3-}$ . As we move down electronegativity decreases thereby introducing new

$e^-$ is less feasible.

Match the column:-

(A)	${ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }3p^{ 4 }$	(P)	s-block , II A, 2nd period
(B)	$[Rn]6d^27s^2$	(Q)	d -block, II B, 4th period
(C)	${ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }{ 3d }^{ 10 }{ 4s }^{ 2 }$	(R)	f-block , III B, 7th period
(D)	$1s^22s^2$	(S)	p-block VI A, 3rd period

0%
$A-Q, B-P, C-S, D-R$
0%
$A-R, B-P , C-S , D-Q$
0%
$A-S, B-R, C-Q,D-P$
0%
None of the above

The first ionization enthalpies for three elements are 1314, 1680, and 2080 kJ

$mol^{-1}$ , respectively. The correct sequence of the element is ______.

0%
O,F and Ne
0%
F,O and Ne
0%
Ne,F and O
0%
F,Ne and O

Explanation

Electronic configuration of

$O$ ,

$F$ ,

$Ne$ is

$1s^22s^22p^4$ ,

$1s^22s^22p^5$ and

$1s^22s^22p^8$ respectively

and its ionization gives monocation (

$O^+$ ,

$F^+$ ,

$Ne^+$ ) that has configuration as:

$1s^22s^22p^3$ ,

$1s^22s^22p^4$ and

$1s^22s^22p^7$

Since

$O$ attains stable half-filled configuration after ionization, will be more stable with least ionization enthalpy.

$Ne$ having completely filled configuration before ionization, so will have high ionization enthalpy and

$F$ will have higher ionization than

$O$ but lower than

$Ne$ .

Therefore order of first ionization enthalpyis:

$O$ <

$F$ <

$Ne$ .

Option A is correct.

Choose the correct answer form the alternatives given.

$Ca^{2+}$ is isoelectronic with:

0%
$Na$
0%
$4Mg^{2+}$
0%
$Ba^{2+}$
0%
$Ar$

Explanation

The electronic configuration of Ca(Z=20) is

${ Ca } =\ { 1s }^{ 2 }{ 2s }^{ 2 }2p^{ 6 }3s^{ 2 }{ 3p }^{ 6 }4{ s }^{ 2 }$

while The electronic configuration of

${ { Ca }^{ 2+ } }$ is

${ { Ca }^{ 2+ } }=\ { 1s }^{ 2 }{ 2s }^{ 2 }2p^{ 6 }3s^{ 2 }{ 3p }^{ 6 }$ = Ar

The no.of electrons in

${ { Ca }^{ 2+ } }$ is 18 after the loss of two electrons. So it has noble gas configuration of Ar.

The main reason for showing anomalous properties of the first member of a group in s or p-block is :

0%
maximum chemical reactivity
0%
maximum electronegativity and different configurations
0%
small size, large charge/radius ratio
0%
tendency to form multiple bonds

Two elements

$X$ and

$Y$ contain only one electron in the outer level. Element

$X$ is reactive and loses electron easily while element

$Y$ is relatively unreactive and non-corrosive. The elements

$X$ and

$Y$ respectively are:

0%
$Cs$ and $Li$
0%
$Rb$ and $Na$
0%
$Li$ and $Cu$
0%
$Ag$ and $Au$

Explanation

Alkali metals (such as

$Li$ ) and coinage metals (such as

$Cu$ ), have one valence electron each and belong to groups IA and IB respectively. Alkali metals are highly reactive while coinage metals are very less reactive.

Which of the following elements will have highest ionisation energy?

0%
$1s^{2}2s^{2}2p^{6}3s^{1}$
0%
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$
0%
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$
0%
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}$

Explanation

Element with electronic configuration

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ will have highest ionisation energy. 3p orbital is half filled and stable. Removal of electron from stable, half filled p orbital requires more energy. Also, on moving from left to right in a period, the ionisation energy increases.

Which of the following is not a periodic property for the elements?

0%
Electronegativity
0%
Atomic size
0%
Occurrence in nature
0%
Ionization energy

The electronic states

$X$ and

$Y$ of an atom are depeicted below:

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

$Y: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1}$
Which of the following statements is not correct?

0%
$X$ represents an alkali metal.
0%
Energy is required to change $X$ into $Y$
0%
$Y$ represents ground state of the element.
0%
Less energy is required to remove an electron from $X$ than from $Y$

Explanation

Electronic configuration for given elements

$X$

$= 1s^22s^22p^63s^1, Y=1s^22s^22p^63s^23p^64s^1$

shows that they belong to the same group (group 11, alkali metals) and

$Y$ is bigger in size than

$X$ . Thus, the distance between the nucleus and valence shell in

$Y$ is greater as compared to

$X$ and therefore

$Y$ experience less effective nuclear charge and needs less energy to remove the valence electron than that from

$X$ atom.

Beryllium has higher ionization enthalpy than boron. This can be explained as :

0%
beryllium has higher size than boron hence its ionisation enthalpy is higher
0%
penetration of $2p-$ electrons to the nucleus is more than the $2s-$ electrons
0%
it is easier to remove an electron from $2p-$ orbital as compared to $2s-$ orbital due to more penetration of $s-$ electrons
0%
ionisation energy increases in a period

Which of the following can most easily form unipositive gaseous ion?

0%
$1s^{2} 2s^{2} 2p^{6} 3s^{2}$
0%
$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}$
0%
$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{2}$
0%
$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{3}$

Explanation

The element with electronic configuration

$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}$ can most easily form unipositive gaseous ion.

$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}$ readily loses an electron to form

$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{0}$ which is stable electronic configuration as

$3p$ subshell is empty and

$3s$ subshell is completely filled.

An energy of

$24.6\ eV$ is required to remove one of the electrons from a neutral helium atom. The energy

$(in\ eV)$ required to remove both the electrons from a neutral helium atom is:

0%
$2$
0%
$49.2$
0%
$51.8$
0%
$79.0$

Explanation

Energy to remove one electron is,

$E_1=24.6$

$eV$

The energy required to remove the second electron is

$E_2=Z^2Rhc\left(\cfrac{1}{1^2}-\cfrac{1}{\infty^2}\right)=2^2\times 13.6$

$eV=54.4$

$eV$

Therefore total energy required

$E=E_1+E_2=24.6+54.4=79.0$

$eV$

The first ionisation energies of the elements of the first transition series (

$Ti$ to

$Cu$ ):

0%
increases as the atomic number increases
0%
decreases as the atomic number increases
0%
do not show any change as the addition of electrons takes place in the inner $(n - 1) d-$ orbitals
0%
increases from $Ti$ to $Mn$ and then decreases from $Mn$ to $Cu$

The highest ionization energy is exhibited by:

0%
halogens
0%
alkaline earth metals
0%
transition metals
0%
noble gas

Who is father of the Periodic Table of Elements?

0%
Johann Baeyer
0%
Dmitri Mendeleev
0%
Alfred Nobel
0%
Johannes Van der Waals

Which of the following is not correctly matched with the given example?

0%
An element of first transition series which has highest second ionization enthalpy- $Cu$
0%
An element of first transition series with highest third ionisation enthalpy- $Zn$
0%
An element of first transition series with lowest enthalpy of atomisation- $Zn$
0%
Last element of third transition series- $Cd$

Explanation

$H$ extends from lanthanum to mercury .last element of

$3rd$ transition sense is molecule

$Cd(z=48)$ belongs to

$2nd$ series of d-block.

The first ionization energy for

$Li$ is

$5.4$ eV and electron affinity of

$Cl$ is

$3.61$ eV. The

$\Delta H$ (in kJ/mol) for the reaction

$Li(g)+Cl(g)\rightarrow Li^+(g)+Cl^-(g)$ is:

(If the resulting ions do not combine with each other)

0%
$70$
0%
$100$
0%
$172$
0%
$270$

Explanation

Given

$Li(g)\rightarrow{ Li }^{ + }$ +

${ e }^{ - }$ IE=

$5.4$ eV/atom

$Cl(g)+{ e }^{ - }\rightarrow { Cl }^{ - }$ (g)

${ E }_{ z }=-3.61$ eV/atom

Add both reaction:

$\Delta H=\left( 5.4-3.61 \right)$ eV/atom

$=\left( \left( 5.4-3.61 \right) \times 1.6\times { 10 }^{ -19 }\times 6.023\times { 10 }^{ 23 }\times { 10 }^{ -3 } \right) \\ =172.4kJ{ mol }^{ - }$

Which reference to concept of ionization potential, which one of the following sets are correct

0%
$U > K > Cs$
0%
$B > U > K$
0%
$Cs > U > B$
0%
$Cs > U > K$

In each of the following pairs,state which atom/ion has higher ionization enthalpy and more negative electron gain enthalpy?

0%
$I, { I }^{ - }$
0%
$Br, { Br }^{ - }$
0%
$Li, { Li }^{ - }$
0%
$Br, I$

Explanation

$IE=I>{I}^{-} , EA=I>{I}^{-}$

$IE=Br>{Br}^{-} , EA= Br>{Br}^{-}$

$IE={Li}^{-}>Li , EA=Li>{Li}^{-}$

$IE=Br>I , EA=Br>I$

Negatively charged species gives

${e}^{-}$ easily due to more electron density on it

$,$ and in group

$IE$ or

$EA$ decreases down the group due to less

${Z}_{eff}.$

The successive ionization energies for element X is given below

$IE_1$ :

$250 kJ mol^{-1}$

$IE_2$ :

$820 kJ mol^{-1}$

$IE_3$ :

$1100 kJ mol^{-1}$

$IE_4$ :

$1400 kJ mol^{-1}$
Find out the number of valence electrons for the element X.

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

General electronic configuration of outermost and penultimate shell is $\left( {n - 1} \right)\,\,{s^2}\,\,\left( {n - 1} \right){p^{6\,\,}}\left( {n - 1} \right)\,{d^x}\,\,n{s^2}$ . If $n = 4$ and $x = 5$ then number of protons in the nuleus will be :-

0%
$> \,\,25$
0%
$< \,\,24$
0%
$25$
0%
$30$

Explanation

Given

$n = 4 \times = 5$

$\begin{matrix}(4-1)s^2 \\ 3s^2 \end{matrix}\begin{matrix} (4-1)p^6 \\ 3p^6 \end{matrix}\begin{matrix} (4-1)d^5 \\ 3d^5 \end{matrix}\begin{matrix} 4s^2 \\ 4s^2 \end{matrix}$

Total electrons

$= 2 + 6 + 5 + 2 = 15$

Electrons in 1st and 2nd orbit

$= 2 + 8 = 10$

Total electrons

$= 10 + 15 = 25$

No. of electrons

$=$ no. of protons

So, total protons

$= 25$

The number of electrons having

$n + s = 2.5$ in sulphide ion is:

0%
$3$
0%
$4$
0%
$6$
0%
$2$

Explanation

Electronic configuration of sulphide ion.

${ 1s }^{ 2 }{ 2s }^{ 2 }{ 3s}{ 2 }{ 3p }^{ 3 }$

Given that

$n+s=2.5$

$n=2$ ,

$s=0.5=\dfrac { 1 }{ 2 }$

$n=2$ means

${ 2s }^{ 2 }$ .

So, Two electrons having

$n+s=2.5$ .

Ionic potential

$\left( \phi \right)$ of electropositive element will be highest in which of the following compound:

0%
$CsCl$
0%
$Mg{ CI }_{ 2 }$
0%
$AI{ F }_{ 3 }$
0%
${ SF }_{ 6 }$

Explanation

Ionic potential is the ratio of electric charge to radius of ion

$.$ If measures charge density on surface of ion

$.$ Smaller radii ion have more ionic potential

$.$

So ionic potential

$={SF}_{6}>{AlF}_{3}>{MgCl}_{2}>CsCl$

Strong electropositive property shown by the following:

0%
$Mg$
0%
$P$
0%
$Si$
0%
$Cl$

Explanation

Electropositivity is a result of less

${Z}_{eff}$ on atom

$,$ and since metals have less

${Z}_{eff}$ thus more electropositivity

$.$

Which compound is mainly responsible for the depletion of ozone layer?

0%
$CO_2$
0%
$CH_4$
0%
$CH_3OH$
0%
$CF_2Cl_2$

Explanation

$CF_2Cl_2$ is the main reason behind depletion of ozone. It is produced by Home Refrigerator etc.

Elements of which of the following group

$\left ( s \right )$ of periodic table do not from hydrides.

0%
group 7,8,9
0%
group 13
0%
group 15,16,17
0%
group 14

Explanation

The metals from group

$7,8,9$ do not form hydrides and the element

$Cr$ ,the only element in group

$6$ which forms hydride.

Which of the following sulphates has the highest solubility in water?

0%
$Mg{SO}_{4}$
0%
$Ba{SO}_{4}$
0%
$Ca{SO}_{4}$
0%
$Be{SO}_{4}$

Explanation

Answer (D)

$BeSO_4$

Solubility depends upon the resulting enthalpy of lattice energy and hydration enthalpy. For

$MgSO_4$ , hydration enthaply is more hence it is soluble in water. Whereas, for

$BaSO_4$ it is not enough that's why it is not soluble in water.

As sulphate ion has a much larger size , the relative increase in size from Be to Ba doesn't matter and the compounds have more or less the same lattice enthalpy. But hydration enthalpy steadily decreases from Be to Ba.. This solubility decreases.

Which of the following pair has electron deficient compounds?

0%
$B_2H_6, \ AlCl_3$
0%
$C_2H_6, \ Al_2Cl_6$
0%
$SF_2, \ Cl_2O$
0%
$NaBH_4, \ ICl$

Explanation

(Refer to Image 1)

In Diborane structure, each Boron bonds with

$4$ hydrogen atoms. But Boron valency is

$3$ . Each Boron gives

$3$ valence electrons but for the

$4^{th}$ bond, there is no electron from Boron side which makes it electron deficient.

(Refer to Image 2)

$AlCl_3$ ,

$Al$ atom bonds with

$3$

$Cl$ atoms. But here it violates octet rule. There are no bonds or lone pairs, which makes the surrounding electron count

$8$ . It has only

$3$ bonds which makes

$6$ electrons. So, it is an electron deficient compound. So, both

$AlCl_3$ and

$B_2H_6$ are electron deficient compounds.

1185718_1120553_ans_4c93171577bf4e56bb22579926016461.png

Correct order of the tendency of metals to loss of electrons.

0%
$Zn > Cu > Ag$
0%
$Zn < Cu < Ag$
0%
$Zn > Cu < Ag$
0%
$Cu > Zn > Ag$

Explanation

The tendency of metals to lose electrons can be determined by their position in electrochemical series.

Elements that lose electrons more easily have lower reduction potential. If we consider the reduction potentials of

$Zn,Cu$ &

$Ag$ , their order of tendency to lose electrons is given by

$Zn >Cu >Ag$

$E^o=-0.763,+0.153,+0.799$

Which of the following match is correct?

0%
last natural element- Uub
0%
General electronic configuration of IA group - ${ ns }^{ 2 }$
0%
inert gas lies in ${ 2 }^{ nd }$ - ${ 6 }^{ th }$ period
0%
Typical elements - ${ 3 }^{ rd }$ period elements

Which of the following group bear maximum number of elements?

0%
$3^{rd}$ group
0%
$2^{nd}$ group
0%
$18^{th}$ group
0%
$13^{th}$ group

Explanation

Since the third group contains

$d$ -block as well as

$f$ -block elements also which are much more than any group elements .

Which of the following sequence regarding ionisation potential of coinage metal is correct?

0%
$Cu>Ag>Au$
0%
$Cu<Ag<Au$
0%
$Cu>Ag<Au$
0%
$Ag>Cu<Au$

Explanation

The ionisation energy of the elements is expected to decrease down the group with an increase of distance between the valence electron and the nucleus and as well as because of the decrease in the effective nuclear charge.

So, the amount of energy required to ionize electron will be least for gold (atomic radius max). therefore the sequence regarding ionisation potential of coinage metal is:

$Cu$

$>$

$Ag$

$>$

$Au$

Hence, the correct option is

$\text{A}$

The one electron species having ionization energy of 54.4 eV is

0%
$H$
0%
$He^+$
0%
$B^{4+}$
0%
$Li^{2+}$
0%
$Be^{2+}$

Explanation

$I.E = \dfrac{13.6 Z^2}{n^2} eV = 13.6 Z^2$ for one electron species.

$13.6 Z^{2} = 54.4$ or

$Z^{2} = 4$ or

$Z = 2$ i.e.,

$He^{+}$ .

Strongest reducing agent is:

0%
K
0%
Mg
0%
Both
0%
None

Explanation

Reducing nature of alkali metal increase down the group due to ease of loosing electrons down the group

So,

$Na<K;Mg<Ca$

Alkali metals are strong reducing agents that alkaline earthmetals.

so,

$K>Mg$

The elements with atomic number 117 and 120 are yet to be discovered. In which group would you place these elements when discovered?

0%
17, 2
0%
16, 4
0%
15, 3
0%
18, 2

Which of the following metal oxides is most basic?

0%
$ZnO$
0%
$Al_2O_3$
0%
$As_2O_3$
0%
$K_2O$

$1s^2,2s^2,2p^6,3s^1$ shows configuration of:

0%
$Al^{3+}$ in ground state
0%
$Ne$ in excited state
0%
$Mg^+$ in excited state
0%
None of these

Name the element which has a total of three shells with five electrons in its valence shells:

0%
Nitogen
0%
Phosphorus
0%
Arsenic
0%
Anitmony
0%
Bismuth

Which of the following elements never show positive oxidation number?

0%
O
0%
Fe
0%
Ga
0%
F

The electronic configuration of an element is

$1s^22s^22p^63s^23p^63d^{10}4s^1$ . What is the atomic number of the next element of the same group?

0%
$20$
0%
$119$
0%
$30$
0%
$47$

Nuclides have:

0%
specific atomic number
0%
same numbers of proton
0%
same numbers of neutrons
0%
specific atomic number and mass number

Explanation

A nuclide is a species of an atom with a specific number of protons and neutrons in the nucleus leads a specific mass number.
The nuclide concept (referring to individual nuclear species) emphasizes nuclear properties over chemical properties.
for example, $carbon-13$ with $6$ protons and $7$ neutrons.

Element having highest I.P. value is:

0%
Ne
0%
He
0%
Be
0%
N

Explanation

The electronic configuration of the given element is:

Ne :

$[He]2s^22p^6$

He :

$1s^2$

Be:

$1s^22s^2$

N :

$[He]2s^22p^3$

Although Ne, He and N all have fully filled stable configuration but the electron is to be removed from a 1s shell in case of He. 1s shell being closest to the nucleus makes quite tough to remove an electron from it. Thus, the element having the highest I.P. value is Helium.

In general, the properties that decrease and increase down a group in the periodic table, respectively, are:

0%
electronegativity and electron gain enthalpy.
0%
electronegativity and atomic radius.
0%
atomic radius and electronegativity
0%
electron gain enthalpy and electronegativity

The element with

$Z = 120$ (not yet discovered) will be an/ a:

0%
transition metal
0%
inner-transition metal
0%
alkaline earth metal
0%
alkali metal

Explanation

$Z = 120$
Its general electronic configuration may be represented as

$[Noble\ gas]ns^{2}$ , like other alkaline earth metals.

Thus the given element will be an alkaline earth metal.

Which of the following is affected by the stable electron configuration of an atom ?

I. Electronegativity
II. Ionisation enthalpy
III. Electron gain enthalpy

0%
Only I
0%
Only II
0%
Both II & III
0%
All the above

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 9 - 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 9 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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