Classification of Elements and Periodicity in Properties MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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231.
Which of the following statements is correct on the basis of the periodic table?
A
The most electronegative element in the periodic table is fluorine.
B
Scandium is the first transition element and belongs to fourth period.
C
Along a period halogens have maximum negative electron gain enthalpy.
D
All of these.
Answer :
All of these.
No explanation is given for this question. Let's discuss the answer together.
232.
Which of the following has the maximum number of unpaired electrons?
A
$$M{g^{2 + }}$$
B
$$T{i^{3 + }}\,$$
C
$${V^{3 + }}$$
D
$$F{e^{2 + }}$$
Answer :
$$F{e^{2 + }}$$
The electronic configuration of the given ions are as follows.
$${}_{12}M{g^{2 + }} = 1{s^2},2{s^2}2{p^6}$$ (No unpaired electron)
$$\,{}_{22}T{i^{3 + }} = 1{s^2},2{s^2}2{p^6}\,,3{s^2}3{p^6}3{d^1}$$ (One unpaired electron)
$${}_{23}{V^{3 + }} = 1{s^2},2{s^2}2{p^6}\,,3{s^2}3{p^6}3{d^2}$$ (Two unpaired electrons)
$${}_{26}F{e^{2 + }} = 1{s^2},2{s^2}2{p^6}\,,3{s^2}3{p^6}3{d^6}$$ (Four unpaired electrons)
233.
The first $$\left( {{\Delta _i}{H_1}} \right)$$ and second $$\left( {{\Delta _i}{H_2}} \right)$$ ionisation 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 :
Element
$${\Delta _l}{H_1}$$
$${\Delta _l}{H_2}$$
$${\Delta _{eg}}H$$
I
520
7300
− 60
II
419
3051
− 48
III
1681
3374
− 328
IV
1008
1846
− 295
V
2372
5251
+ 48
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 $$F$$, IV represents $$I$$, V represents $$He.$$
So, amongst these, II represents most reactive metal and V represents least reactive non-metal.
234.
Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides?
A
$$A{l_2}{O_3} < MgO < N{a_2}O < {K_2}O$$
B
$$MgO < {K_2}O < A{l_2}{O_3} < N{a_2}O$$
C
$$N{a_2}O < {K_2}O < MgO < A{l_2}{O_3}$$
D
$${K_2}O < N{a_2}O < A{l_2}{O_3} < MgO$$
Answer :
$$A{l_2}{O_3} < MgO < N{a_2}O < {K_2}O$$
On moving across a period ionisation energy increases hence the electropositive nature of metals decreases therefore the ease of formation of ion also decreases and hence the basic character decreases. Further basic
character ofalkali metals oxides increases from $$L{i_2}O$$ to
$$C{s_2}O.$$ Hence the correct order will be $$A{l_2}{O_3} < MgO < N{a_2}O < {K_2}O$$
235.
The statement that is not correct for periodic classification of elements is
A
the properties of elements are periodic function of their atomic numbers
B
non-metallic elements are less in number than metallic elements
C
for transition elements, the $$3d$$ orbitals are filled with electrons after $$3p$$ orbitals and before $$4s$$ orbitals
D
the first ionization enthalpies of elements generally increase with increase in atomic number as we go along a period
Answer :
for transition elements, the $$3d$$ orbitals are filled with electrons after $$3p$$ orbitals and before $$4s$$ orbitals
According to Aufbau principle, "in the ground state of the atoms, the orbitals are filled in order of their increasing energies." The increasing order of energy according to $$\left( {n + l} \right)$$ rule is : $$1s,2s,2p,3s,3p,4s,3d,4p,5s...$$ etc.
Therefore, the $$3d$$ orbitals are filled with electrons after $$4s$$ orbitals and before $$4p$$ orbitals.
236.
The electronic configuration of four elements are given below. Which element does not belong to the same family as others?
A
$$\left[ {Xe} \right]4{f^{14}},5{d^{10}},6{s^2}$$
B
$$\left[ {Kr} \right]4{d^{10}},5{s^2}$$
C
$$\left[ {Ne} \right]3{s^2},3{p^5}$$
D
$$\left[ {Ar} \right]3{d^{10}},4{s^2}$$
Answer :
$$\left[ {Ne} \right]3{s^2},3{p^5}$$
In a family, all elements have same outermost electronic configuration. Since $$\left[ {Ne} \right]3{s^2}3{p^5},$$ chlorine belongs to halogen family while the remaining three are in same group i.e. group 12.
$${}_{80}Hg = \left[ {Xe} \right]4{f^{14}}5{d^{10}}6{s^2}$$
$${}_{48}Cd = \left[ {Kr} \right]4{d^{10}}5{s^2}$$
$${}_{30}Zn = \left[ {Ar} \right]3{d^{10}}4{s^2}$$
237.
Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?
A
$$3{d^3},4{s^2}$$
B
$$3{d^5},4{s^1}$$
C
$$3{d^5},4{s^2}$$
D
$$3{d^2},4{s^2}$$
Answer :
$$3{d^5},4{s^2}$$
The sum of number of electrons (unpaired) in $$d$$-orbitals and number of electrons in $$s$$-orbital gives the number of oxidation states $$(os)$$ exhibited by a $$d$$-block element. Therefore,
$$\eqalign{
& \left( {\text{A}} \right)3{d^3},4{s^2} \Rightarrow OS = 3 + 2 = 5 \cr
& \left( {\text{B}} \right)3{d^5},4{s^1} \Rightarrow OS = 5 + 1 = 6 \cr
& \left( {\text{C}} \right)3{d^5},4{s^2} \Rightarrow OS = 5 + 2 = 7 \cr
& \left( {\text{D}} \right)3{d^2},4{s^2} \Rightarrow OS = 2 + 2 = 4 \cr} $$
Hence, element with $$3{d^5},4{s^2}$$ configuration exhibits largest number of oxidation states.
238.
Predict the formula of stable compound formed by an element $$'A'$$ with atomic number 114 and fluorine.
A
$$A{F_3}$$
B
$$A{F_2}$$
C
$$AF$$
D
$$A{F_4}$$
Answer :
$$A{F_4}$$
Atomic number 114 will fall into carbon family, hence it will be tetravalent. So, formula with $$F$$ will be $$A{F_4}.$$
239.
Given below are the names of few elements based on their position in the periodic table. Identify the element which is not correctly placed.
A
An element which tends to lose three electrons - Aluminium
B
An element which tends to gain two electrons - Iodine
C
An element with valency four - Silicon
D
A transuranium element - Plutonium
Answer :
An element which tends to gain two electrons - Iodine
Iodine can gain only one electron.
$$I + {e^ - } \to {I^ - }$$
240.
According to the Periodic Law of elements, the variation in properties of elements is related to their
A
nuclear masses
B
atomic numbers
C
nuclear neutron-proton number ratios
D
atomic masses
Answer :
atomic numbers
According to modern periodic law, the properties of the elements are repeated after certain regular intervals when these elements are arranged in order of their increasing stomic numbers.