Classification of Elements and Periodicity in Properties MCQ Questions & Answers in Inorganic Chemistry | Chemistry
Learn Classification of Elements and Periodicity in Properties MCQ questions & answers in Inorganic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
221.
Which of the following electronic configuration of an atom has the lowest ionisation enthalpy?
A
$$1{s^2},2{s^2}2{p^5}$$
B
$$1{s^2},2{s^2}2{p^3}$$
C
$$1{s^2},2{s^2}2{p^5},3{s^1}$$
D
$$1{s^2},2{s^2}2{p^6}$$
Answer :
$$1{s^2},2{s^2}2{p^5},3{s^1}$$
The electronic configuration $$1{s^2},2{s^2}2{p^5},3{s^1}$$ shows lowest ionisation energy because this configuration is unstable due to the presence of one electron in $$s$$-orbital. Hence, less energy is required to remove the electron.
222.
Which of the following properties of isotopes of an element is different?
A
First ionisation enthalpy
B
Effective nuclear charge
C
Electron affinity
D
Melting point and boiling point
Answer :
Melting point and boiling point
Isotopes differ in physical properties.
223.
What is the order of ionisation energies of the coinage metal
A
$$Cu > Ag < Au$$
B
$$Cu > Ag > Au$$
C
$$Cu < Ag < Au$$
D
$$Au > Ag < Cu$$
Answer :
$$Cu > Ag < Au$$
$$Cu = 3{d^{10}}4{s^1},Ag = 4{d^{10}}5{s^1},Au = 4{f^{14}}5{d^{10}}6{s^1}$$ In all the above given cases, unpaired $$s$$ - electron has to be removed. In the case of $$Cu,$$ a $$4s$$ electron is to be removed which is closer to the nucleus than the $$5s$$ electron of $$Ag .$$
So, $$I{E_1}$$ of $$Cu > I{E_1}$$ of $$Ag.$$
However, in case of $$Au,$$ due to imperfect screening effect of $$14{e^ - }s$$ of $$4f$$ orbitals, the nuclear charge increases and therefore $$5s\,{e^ - }$$ of $$Au$$ is more tightly held.
Thus, the order of $$I{E_1}$$ is $$Cu > Ag < Au.$$
224.
The element with the highest first ionization potential is
A
boron
B
carbon
C
nitrogen
D
oxygen
Answer :
nitrogen
Amongst $$B, C, N$$ and $$O;N$$ has the highest first ionization energy, because of its half filled $$2p$$ orbital which is more stable.
225.
The basic character of $$MgO,BaO,N{a_2}O$$ and $$FeO$$ follow the order
A
$$MgO < FeO < BaO < N{a_2}O$$
B
$$FeO < MgO < N{a_2}O < BaO$$
C
$$FeO < MgO < BaO < N{a_2}O$$
D
$$N{a_2}O < MgO < FeO < BaO$$
Answer :
$$FeO < MgO < BaO < N{a_2}O$$
The basic character of metal oxides decreases from left to right in a period and
increases down the group.
226.
Which of the following sets of oxides is amphoteric in nature?
A
$$A{l_2}{O_3},A{s_2}{O_3},ZnO$$
B
$$CO,NO,{N_2}O$$
C
$$S{O_3},S{O_2},C{l_2}{O_7}$$
D
$$N{a_2}O,MgO,BaO$$
Answer :
$$A{l_2}{O_3},A{s_2}{O_3},ZnO$$
$$A{l_2}{O_3},A{s_2}{O_3}$$ and $$ZnO$$ are amphoteric in nature.
227.
Which of the following arrangements represents the correct order of electron gain enthalpy?
A
$$O < S < F < Cl$$
B
$$Cl < F < S < O$$
C
$$S < O < Cl < F$$
D
$$F < Cl < O < S$$
Answer :
$$O < S < F < Cl$$
Electron gain enthalpy becomes more negative across a period while it becomes less negative in a group. However, electron gain enthalpy of $$O$$ or $$F$$ is less than that of the succeeding element. When electron is added to $$n =2,$$ the repulsion is more than when it is added to $$n=3$$ in case of $$Cl$$ or $$S.$$
228.
The screening effect of inner electrons of the nucleus causes
A
decrease in the ionization energy
B
increase in the ionization energy
C
no effect on the ionization energy
D
increases the attraction of the nucleus for the electrons
Answer :
decrease in the ionization energy
Higher the screening effect, lower is the $$I.E.$$
229.
Match the atomic numbers of the elements given in column I with the periods given in column II and mark the appropriate choice.
Column I (Atomic number)
Column (period)
a.
31
1.
5
b.
50
2.
3
c.
56
3.
4
d.
14
4.
6
A
a - 1, b - 2, c - 3, d - 4
B
a - 2, b - 1, c - 4, d - 3
C
a - 3, b - 4, c - 1, d - 2
D
a - 3, b - 1, c - 4, d - 2
Answer :
a - 3, b - 1, c - 4, d - 2
No explanation is given for this question. Let's discuss the answer together.
230.
The first $$\left( {{\Delta _i}{H_1}} \right)$$ and second $$\left( {{\Delta _i}{H_2}} \right)$$ ionization enthalpies $$\left( {{\text{in}}\,kJ\,mo{l^{ - 1}}} \right)$$ and the electron gain enthalpy $$\left( {{\Delta _{eg}}H} \right)\left( {{\text{in}}\,kJ\,mo{l^{ - 1}}} \right)$$ of the elements (i), (ii), (iii),(iv) and (v) are given below.
$$\eqalign{
& {\text{Element}}\,\,\,\,{\Delta _i}{H_1}\,\,\,{\Delta _i}{H_2}\,\,\,\,\,\,{\Delta _{eg}}H \cr
& \left( {\text{i}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,520\,\,\,\,\,\,\,\,\,7300\,\,\,\,\,\, - 60 \cr
& \left( {{\text{ii}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,419\,\,\,\,\,\,\,\,\,3051\,\,\,\,\,\, - 48 \cr
& \left( {{\text{iii}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1681\,\,\,\,\,\,\,3374\,\,\,\,\, - 328 \cr
& \left( {{\text{iv}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1008\,\,\,\,\,\,\,1846\,\,\,\,\, - 295 \cr
& \left( {\text{v}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2372\,\,\,\,\,\,\,5251\,\,\,\,\,\, + 48 \cr} $$
The most reactive metal and the least reactive non-metal of these are respectively
A
(i) and (v)
B
(v) and (ii)
C
(ii) and (v)
D
(iv) and (v)
Answer :
(ii) and (v)
(i) represents $$Li,$$ (ii) represents $$K$$
(iii) represents $$Br,$$ (iv) represents $$I$$
(v) represents $$He$$
So, amongst these, (ii) represents most reactive metal and (v) represents least reactive non-metal.