Classification of Elements and Periodicity in Properties MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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91.
Aqueous solutions of two compounds $${M_1} - O - H$$ and $${M_2} - O - H$$ are prepared in two different beakers. If, the electronegativity of $${M_1} = 3.4,{M_2} = 1.2,O = 3.5$$ and $$H = 2.1$$ then the nature of two solutions will be respectively :
A
acidic, basic
B
acidic, acidic
C
basic, acidic
D
basic, basic
Answer :
acidic, basic
The electronegativity difference between $${M_1}$$ and $$O$$ is $$0.1,$$ which indicates $${M_1} - O$$ bond will be covalent, since $$O - H$$ bond having more ionic character thus bond will break and $${H^ + }\,ions$$ gets release and acidic solution is formed and whereas difference between electronegativity of $${M_2}O$$ is $$2.3,$$ thus, $${M_2} - O$$ bond will break. Hence, solution will be basic in nature.
92.
First and second ionisation enthalpies $$\left( {{\text{in}}\,\,kJ/mol} \right)$$ of few elements are given below :
Element
$$I{E_1}$$
$$I{E_2}$$
1.
520
7300
2.
900
1760
3.
1680
3380
4.
2080
3963
Which of the above elements will form halides with formula $$M{X_2}?$$
A
(i) and (ii)
B
(i) and (iii)
C
(ii) and (iii)
D
(i) and (iv)
Answer :
(ii) and (iii)
(ii) is a reactive metal from second group, and (iii) is a reactive non-metal from 17th group.
For second group elements, the difference in successive $$IE$$ is less. The compound formed will be $$M{X_2}.$$
93.
The electronic configuration of gadolinium ( atomic number 64 ) is
No explanation is given for this question. Let's discuss the answer together.
94.
Which of the following will have lowest electron affinity?
A
Nitrogen
B
Oxygen
C
Argon
D
Boron
Answer :
Argon
Electron affinity of noble gases is zero.
95.
A sudden large jump between the values of second and third ionisation energies of an element would be associated with which of the following electronic configuration?
A
$$1{s^2},2{s^2}2{p^6},3{s^1}3{p^2}$$
B
$$1{s^2},2{s^2}2{p^6},3{s^2}3{p^1}$$
C
$$1{s^2},2{s^2}2{p^6},3{s^1}$$
D
$$1{s^2},2{s^2}2{p^6},3{s^2}$$
Answer :
$$1{s^2},2{s^2}2{p^6},3{s^2}$$
A sudden large jump between the values of 2nd and 3rd ionisation energies of an element indicates, that the element has two electrons in its valence shell. Then, the possible electronic configuration will be $$1{s^2},2{s^2},2{p^6},3{s^2}.$$
96.
In the periodic table, the maximum chemical reactivity is at the extreme left ( alkali metals ) and extreme right (halogens). Which properties of these two groups are responsible for this?
A
Least ionisation enthalpy on the left and highest negative electron gain enthalpy on the right.
B
Non-metallic character on the left and metallic character on the right.
C
High atomic radii on the left and small atomic radii on the right.
D
Highest electronegativity on the left and least electronegativity on the right.
Answer :
Least ionisation enthalpy on the left and highest negative electron gain enthalpy on the right.
Elements on the left side have lowest ionisation enthalpy due to which they can very easily lose electrons while the elements on the right can accept electrons easily as they show highest negative electron gain enthalpy.
97.
Identify the wrong statement in the following.
A
Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius
B
Amongst isoelectronic species, greater the negative charge on the anion, larger is the ionic radius
C
Atomic radius of the elements increases as one moves down the first group of the periodic table
D
Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table
Answer :
Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius
Atomic radius of the elements decreases across a period from left to right due to increase in effective nuclear charge. On moving down a group, since, number of shells increases, so atomic radius
increases.
Amongst isoelectronic species, ionic radius increases with increase in negative charge or decrease in positive charge.
98.
Aqueous solutions of two compounds $$M - O - H$$ and $$M' - O - H$$ have been prepared in two different beakers. If the electronegativity of $$M = 3.5,M' = 1.72,O = 3.0$$ and $$H = 2.1,$$ then the solutions respectively are
A
acidic, acidic
B
acidic, basic
C
basic, basic
D
basic, acidic
Answer :
acidic, basic
In the compound $$M - O - H,$$ if $$IE$$ or $$E.N$$ of $$M$$ is low, the compound will act as a base and if the $$IE$$ or $$E.N.$$ of $$M$$ is high, the compound will behave as an acid. Therefore, $$M - O - H$$ will act as an acid as $$E.N.$$ of $$M$$ is high (3.5) and $$M' - O - H$$ will act as a base as $$E.N.$$ of $$M'$$ is low (1.72).
99.
The increasing order of the ionic radii of the given isoelectronic species is :
Among isoelectronic species ionic radii increases as the charge increases.
Order of ionic radii $$C{a^{2 + }} < {K^ + } < \,C{l^ - } < {S^{2 - }}$$
The number of electrons remains the same but nuclear charge increases with increase in the atomic number causing decrease in size.
100.
The formation of the oxide ion, $$O_{\left( g \right)}^{2 - },$$ from oxygen atom requires first an exothermic and then an endothermic step as shown below :
$$\eqalign{
& {O_{\left( g \right)}} + {e^ - } \to O_{\left( g \right)}^ - ;\Delta {H^ \circ } = - 141\,kJ\,mo{l^{ - 1}} \cr
& O_{\left( g \right)}^ - + {e^ - } \to O_{\left( g \right)}^{2 - };\Delta {H^ \circ } = + 780\,kJ\,mo{l^{ - 1}} \cr} $$
Thus, process of formation of $${O^{2 - }}$$ in gas phase is unfavourable even though $${O^{2 - }}$$ is isoelectronic with neon. It is due to the fact that,
A
oxygen is more electronegative
B
addition of electron in oxygen results in larger size of the ion
C
electron repulsion outweighs the stability gained by achieving noble gas configuration
D
$${O^ - }$$ ion has comparatively smaller size than oxygen atom
Answer :
electron repulsion outweighs the stability gained by achieving noble gas configuration
Due to small size of $$O$$ atom, electron repulsion outweighs the stability gained by achieving noble gas configuration.