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Chemistry 2 Marks Questions

PART - 1 CH - 3 Classification of Elements and Periodicity in Properties · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 16 questions.

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Question 12 Marks
The electronic configuration of some elements are as :
(i) $1 s^2 2 s^2 2 p^5$
(ii) $1 s^2 2 s^2 2 p^6 3 s^2 3 p^2$
(iii) $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^2 4 s^2$
Find out the halogen, transition element and carbon family element from them.
Answer
(i) Halogens (ii) Carbon family element
(iii) Transition element.
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Question 22 Marks
In elements of second period give reason for difference in properties of Li and Be .
Answer
The properties of Li and Be are different from other members of their group and they show more similarity in their properties with the diagonally located elements i.e. Mg and Al , this is called diagonal relationship. The reasons for this are as follows :
(i) Small size of elements
(ii) High electronegativity
(iii) High charge/radius ratio
(iv) Absence of $d$-orbitals in the valence shell.
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Question 32 Marks
Calculate the percentage ionic character of bond in CsF. If electronegativity of Cs and F is 0.7 and 4.0 respectively.
Answer
$\begin{array}{l} \text {  % ionic character of bond in CsF } \\
=16 \Delta+3.5 \Delta^2 \\ \Delta=\text { Difference in electronegativity } \\
=4.0-0.7=3.3 \\ \text { Hence % ionic of bond }=(16 \times 3.3)+\left[3.5(3.3)^2\right] \\
=52.8+3.5 \times 10.89 \\ \text { Character of bond }=52.8+38.1 \\
=90.9 \%\end{array}$
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Question 42 Marks
(i) From $B$ to $F$ in the second period and in third period from Al to Cl in increasing order of electron gain enthalpy.
(ii) Arrange elements of halogen family in decreasing order of electron gain enthalpy.
Answer
(i) In second period the arrangement electron gain enthalpy of B to F is as : $N < B < C < O < F$.
In third period the sequence of electron gain enthalpy from Al to Cl : $Al < P < Si < S < Cl$.
(ii) The order of electron gain enthalpy of elements of halogen family: $Cl > F > Br > I$.
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Question 52 Marks
Why the group 17 (halogen) elements of periodic table have more negative value of electron gain enthalpy?
Answer
The following reasons are responsible for the more negative value of electron gain enthalpy of group 17 elements : (i) Small atomic size (ii) More effective nuclear charge (iii) $ns ^2 np ^5$ configuration.
Hence their electron gaining tendency is more and they accept an electron and gain the stable electronic configuration $\left( ns ^2 np ^6\right)$ like noble gases.
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Question 62 Marks
Answer the following questions.
(i) Why group 1 elements form $M ^{+1}$ instead of $M ^{+2} ?$
(ii) Why second group elements form $M ^{+2}$ instead of $M ^{+1}$ ?
Answer
(i) From first group elements (alkali metals) the difference between II and I ionization enthalpy is more than 16 eV . Hence, their lower oxidation state is more stable. Hence, they form $M ^{+1}$ not $M ^{+2}$.
(ii) The difference between II and I ionization enthalpy of second group is less than 11 eV . Hence their higher oxidation state is more stable and they form $M ^{+2}$ not $M ^{+1}$
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Question 72 Marks
(i) The value of ionization enthalpy of phosphorus is greater than the ionization energy of sulphur whereas the atomic size of sulphur is less than that of phosphorus, why?
(ii) Why is the ionization enthalpy of Mn greater than the ionization enthalpy of Cr ?
Answer
(i) The outermost electronic configuration of phosphorus is $3 p ^3$ which is half filled and hence it is more stable while the outermost electronic configuration of sulphur is $3 p ^4$ which is less stable than $3 p ^3$. Hence it requires more energy to remove electron from stable configuration of phosphorus. Hence ionization enthalpy of phosphorus is more than that of sulphur.
(ii) The outermost electronic configuration of Mn is $3 d^5 4 s^2$ while that of Cr is $3 d^5 4 s^1$. In comparison to $4 s^1$ (unpaired), $4 s^2$ (paired) configuration is more stable, which requires more energy to remove electron. Hence Mn have more ionization enthalpy in comparison to that of Cr .
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Question 82 Marks
It is easy to form $Cs ^{+}$from Cs in comparison to $Na ^{+}$from Na . Why?
Answer
On moving from up to down in group atomic size increases. Hence ionization enthalpy decreases. In comparison to Na , the size of Cs is more. Hence its ionization enthalpy is less in comparison to Na . Hence from this electrons are easily removed. Therefore it is easy to form $Cs ^{+}$from Cs in comparison to $Na ^{+}$from Na .
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Question 92 Marks
Which of the following element have second highest ionization enthalpy, and why?
$Be , Li , B$ and C
Answer
Out of $Be , Li , B$ and C , the second ionization enthalpy of Li is highest because when one electron is removed from it, Li attain the inert gas configuration [He] which is stable, and it require more energy to remove an electron from it.
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Question 102 Marks
Arrange $C , N , O$ and F in increasing order of second ionization enthalpy, with reason.
Answer
The outermost electronic configuration of C , $N , O$ and F is $2 p ^2, 2 p ^3, 2 p ^4$ and $2 p ^5$ respectively, from which when one electron is removed following electronic configuration is obtained :
$
\begin{array}{llll}
C^{+} & N^{+} & O^{+} & F^{+} \\
2 p^1 & 2 p^2 & 2 p^3 & 2 p^4
\end{array}
$
As in comparison to $2 p^4$, from $2 p^3$ the removal of electron is typical. Hence, the increasing order of second ionization potential of these elements are as :
$
C < N < F< O
$
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Question 112 Marks
Why is the ionization enthalpy of B less than the ionization enthalpy of Be?
Answer
In $Be \left(2 s^2\right)$ last electron is ejected from $s -$ orbital which is completely filled and in comparison to $p$-orbital it is more closed to nucleus (more penetrating power). Hence to remove it out, it requires more energy. While the $p$-electron of Boron ( $2 p ^1$ ) can be easily removed out. That's why the ionization enthalpy of B is less than that of Be.
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Question 122 Marks
In periodic table how reactivity of elements get changed ?
Answer
In a period, the chemical reactivity of the elements of the first group is very high, it decreases after reaching the middle and after reaching group 17 , it increases again and becomes very high. Similarly, in a group of common elements (e.g. alkali metals) the reactivity increases from top to bottom, whereas in a group of non-metals (e.g. halogen family) the reactivity decreases from top to bottom.
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Question 132 Marks
Write the properties of $d$-block element.
Answer
The properties of $d$-block elements are as follows: The elements belongs to group $3^{\text {rd }}$ to $12^{\text {th }}$ of periodic table are known as d-block elements. Their last electron enters in ( $n -1$ ) d orbitals. The outermost electronic configuration of these elements is $(n-1) d^{1-10}$ $ns ^{0-2}$ (where $n =4$ to 7 ). All of these elements are metals. The ions of these elements are generally coloured and represent the oxidation state and paramagnetism. They are also used as catalyst. $Zn , Cd$ and Hg (group 12) are non-transition elements although they have general electronic configuration $( n -1) d ^{10} ns^2$.
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Question 142 Marks
Inner-transition elements have been kept separately below the main periodic table. Why?
Answer
Inner-transition elements are the elements in which lanthanoids and actinoids are present. Lanthanoid series starts after $La _{51}$ and actinoid series starts after ${ }_{89} Ac$. La and Ac are the elements of third group. The element after Lanthanoids is ${ }_{72} Hf$ and element after actinoid is $104 U _{ nq }$. Hence in reality all lanthanoids and actinoids are $d$-block elements but if we put in periodic table that the structure of periodic table get disturbed. Therefore, to maintain its structure and to follow the principle of classification by placing elements with similar properties in the same class these elements have been kept separately at the bottom of the periodic table.
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Question 152 Marks
In comparison to the atomic mass of an element, the atomic number is more capable of indicating the properties of that element. Confirm the statement.
Answer
Moseley observed that there is regularity in the characteristic x -ray spectra of elements and the graph between atomic number $( z )$ and $\sqrt{V}$ is a straight line as $Z \propto \sqrt{V}( V =$ frequency of x -rays $)$ but between atomic mass and $\sqrt{V}$ a straight line do not formed and properties of elements depend on atomic number not on atomic mass. It proves the above statement.
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Question 162 Marks
Write the contribution of Lothar-Meyer in classification of elements.
Answer
Lothar-Meyer made graphs between the physical properties of elements, such as atomic volume, melting point, boiling point and atomic weight and found that elements with similar properties are at the same position. Hence they show similarity among the elements of a certain group, they are called Lothar-Meyer's Curve. They helped in classification of elements.
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