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Consider the isoelectronic species, \(\mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{F}^{-}\)and \(\mathrm{O}^{2-}\). The correct order of increasing length of their radii is ____.
(c) Amongst isoelectronic ions, ionic radii decrease with increase in nuclear charge:
\(
\mathrm{Mg}^{2+}(12)<\mathrm{Na}^{+}(11)<\mathrm{F^-}(10)<O^{2-}(8)
\)
Which of the following is not an actinoid?
(d) Elements with the atomic number \(\mathbf{Z}=90\) to 103 are called actinoids. Thus, terbium ( \(Z=65\) ) is not an actinoid. Terbium belongs to lanthanoids. \(
\text { Tb: }[X e] 4 f^9 5 d^0 6 s^2
\)
The order of screening effect of electrons of \(s, p, d\) and \(f\) orbitals of a given shell of an atom on its outer shell electrons is:
(a) HINT: The screening effect depends on the size of the orbital.
Explanation:
The screening effect or shielding effect is the phenomenon of reduction of nuclear force of attraction due to inner shell electrons towards valence electrons. The inner shell electrons protect the valence shell electrons from the nuclear force i.e. they shield them.
The order of screening effect of electrons of \(s, p, d\) and \(f\) orbitals of a given shell of an atom on its outer shell electrons is \(s>p>d>f\).
The screening effect decreases from s-orbital to f-orbital in an atom on account of the shape of the orbital.
The first ionisation enthalpies of \(\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}\) and \(\mathrm{Si}\) are in the order:
(a) The electronic configurations of \(\mathrm{Na}\) and \(\mathrm{Mg}\) are:
\(\mathrm{Na}\) (11): [Ne] \(3 s^1\) and \(\mathrm{Mg}(12):[\mathrm{Ne}] 3 \mathrm{~s}^2\). In both the atoms, the electron is to be removed from \(3 \mathrm{~s}\)-orbital but nuclear charge in \(\mathrm{Na}\) is less than \(\mathrm{Mg}\). Thus, ionisation energy of \(\mathrm{Na}\) is less than \(\mathrm{Mg}(\mathrm{Na}<\mathrm{Mg})\). The electronic configurations of \(\mathrm{Mg}\) and \(\mathrm{Al}\) are:
\(
\mathrm{Mg}:[\mathrm{Ne}] 3 \mathrm{~s}^2 ; \mathrm{Al}:[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^{1}
\)
In Mg, the electron is to be removed from 3s-orbital while in Al, it is to be removed from 3p-orbital. Since it is easier to remove an electron from 3p-orbital in comparison to 3s-orbital, the ionization enthalpy of \(\mathrm{Mg}\) is higher than \(\mathrm{Al}(\mathrm{Mg}>\mathrm{Al})\).
The electronic configurations of \(\mathrm{Al}\) and \(\mathrm{Si}\) are:
Al (13): [Ne] \(3 s^2 3 p^1; \mathrm{Si}(14):[\mathrm{Ne}] 3 s^2 3 p^2\)
In both the atoms, the electron is to be removed from 3p-orbital but nuclear charge in \(\mathrm{Si}\) is more than \(\mathrm{Al}\).
Thus, ionisation enthalpy of \(\mathrm{Al}\) is less than \(\mathrm{Si}\)
The electronic configuration of gadolinium (Atomic number 64) is
(c) HINT: Gadolium is a f block element.
Explanation:
The electronic configuration of \(\mathrm{La}(\mathrm{Z}=57)\) is \([\mathrm{Xe}] 5 \mathrm{~d}^1 6 \mathrm{~s}^2\). Therefore, further addition of electrons occurs in the lower energy \(4 \mathrm{f}\)-orbital till it is exactly halffilled at \(\mathrm{Eu}(\mathrm{Z}=64)\). Thus, the electronic configuration of Eu is \([\mathrm{Xe}] 4 \mathrm{f}^7 6 \mathrm{~s}^2\).
Thereafter, the addition of the next electron does not occur in the more stable exactly half-filled \(4 \mathrm{f}^7\) shell but occurs in the little higher energy \(5 \mathrm{~d}\)-orbital. Thus, the electronic configuration of \(\mathrm{Gd}(\mathrm{Z}=64)\) is \([\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2\).
The statement that is not correct for periodic classification of elements is:
(c) Hint: The energy of \(4 s\) is lower than \(3 d\)
In the case of transition elements (or any elements), the order of filling of electrons in various orbital is \(3 p<4 s<3 d\). Thus, \(3 d\) orbital is filled when \(4 s\) orbital gets completely filled.
For transition elements, the \(3 \mathrm{~d}\)-orbitals are filled with electrons after \(3 p\) and \(4 \mathrm{~s}\)-orbitals and before \(4 \mathrm{p}\)-orbitals. The order of filling the orbitals is: Is \(2 s 2 p 3 s 3 p 4 s 3 d 4 p 5 s 4 d 5 p 6 s\)…
Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is:
(c) Chlorine has higher electron gain enthalpy than fluorine. This is due to small size of fluorine atom, i.e., the electron density is high which resists the addition of an electron \((\mathrm{F}<\mathrm{Cl})\). In general, electron gain enthalpy decreases as atomic size increases. Thus, electron gain enthalpy follows the order:
\(
\mathrm{Cl}>\mathrm{Br}>\mathrm{I}
\)
\(
\text { Thus, the negative electron gain enthalpy follows the order: } \mathrm{F}<\mathrm{Cl}>\mathrm{Br}>\mathrm{I}
\)
The period number in the long form of the periodic table is equal to
(c) Since each period starts with the filling of electrons in a new principal quantum number, therefore, the period number in the long form of the periodic table refers to the maximum principal quantum number of any element in the period.
Period number \(=\) maximum \(n\) of any element (where, \(\mathrm{n}=\) principal quantum number).
The elements in which electrons are progressively filled in \(4 f\)-orbital are called
(c) HINT : \(f\) – block elements are lanthanoids and actinoids.
Explanation:
The elements in which electrons are progressively filled in \(4 \mathrm{f}\)-orbital are called lanthanoids.
Lanthanoids consist of elements from Z=58 (cerium) to 71 (lutetium).
Which of the following is the correct order of size of the given species:
(d) HINT: Size of anion is greater than size of cation of same element.
Explanation:
Anion formed after the gain of the electron to the neutral atom and cation formed after the loss of electron from outer shell.
Hence, cation has smaller size but anion has bigger size than its neutral atom (Anion is bigger than the parent atom and cation is smaller than the parent atom.)
Thus the order is \(
I^{-}>I>I^{+}
\)
The formation of the oxide ion, \(\mathrm{O}^{2-}(\mathrm{g})\), from the oxygen atom requires first an exothermic and then an endothermic step as shown below:
\(
\begin{aligned}
& \mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta H^{\ominus}=-141 \mathrm{~kJ} \mathrm{~mol}{ }^{-1} \\
& \mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta H^{\ominus}=+780 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\)
Thus process of formation of \(\mathrm{O}^{2-}\) in gas phase is unfavourable even though \(\mathrm{O}^{2-}\) is isoelectronic with neon. It is due to the fact that,
(c) Hint: Use the concept of electronic repulsion
Although \(\mathrm{O}^{2-}\) has noble gas configuration isoelectronic with neon but its formation is unfavourable due to the strong electronic repulsion (interelectronic repulsion) between the negatively charged \(\mathrm{O}^{-}\)ion and the second electron being added.
Hence, the electron repulsion outweighs the stability gained by achieving noble gas configuration.
Alternate:
There is a lot of repulsion when similar charges approach each other as \(\mathrm{O}^{-}(\mathrm{g})\), and electron are both negatively charged. To add an electron under such situation, the force of repulsion is to be overcome by applying external energy.
Comprehension given below is followed by some multiple-choice questions. Each question has one correct option. Choose the correct option.
In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, \(s, p, d\) and \(f\). The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Arfbau principle, the seven periods ( 1 to 7 ) have \(2,8,8,18,18,32\) and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of \(f\)-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table. The element with atomic number 57 belongs to
\(
\text { (c) The configuration of }{ }_{57} \mathrm{X} \text { is } 5 \mathrm{~d}^1 6 \mathrm{~s}^2 \text { hence it belongs to } d \text {-block. }
\)
Explanation: (c) The element with atomic number 57 belongs to \(d\)-block element as the last electron enters the \(5 d\)-orbital against the Aufbau principle. This anomalous behaviour can be explained on the basis of greater stability of the xenon (inert gas) core.
After barium ( \(\mathrm{Z}=56\) ), the addition of the next electron (i.e., \(57 \mathrm{th}\) ) should occur in \(4 \mathrm{f}\)-orbital in accordance with Aufbau principle. This will however, tend to destabilize the xenon core \((\mathrm{Z}=54),[\mathrm{Kr}]\left(4 \mathrm{~d}^{10} 4 \mathrm{f}^0 5 \mathrm{~s}^2 5 \mathrm{p}^6 5 \mathrm{~d}^0\right)\) since the \(4 \mathrm{f}\)-orbitals lie inside the core.
Therefore, the 57 th electron prefers to enter \(5 \mathrm{~d}\)-orbital which lies outside the xenon core and whose energy is only slightly higher than that of \(4 \mathrm{f}\)-orbital. In doing so, the stability caused by the addition of one electron to the higher energy \(5 \mathrm{~d}\)-orbital instead of the lower energy \(4 \mathrm{f}\)-orbital. Thus, the outer electronic configuration of \(\mathrm{La}(\mathrm{Z}=57)\) is \(5 \mathrm{~d}^1 6 \mathrm{~s}^2\) rather than the expected \(4 \mathrm{f}^1 6 \mathrm{~s}^2\).
Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.
In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, \(s, p, d\) and \(f\). The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Arfbau principle, the seven periods ( 1 to 7 ) have \(2,8,8,18,18,32\) and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of \(f\)-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table. The last element of the \(p\)-block in \(6^{\text {th }}\) period is represented by the outermost electronic configuration.
(c) The last element of \(p\)-block in \(6^{\text {th }}\) period is \({ }_{86} \mathrm{Rn}\) with configuration \(4 f^{14} 5 d^{10} 6 s^2 6 p^6\)
Explanation: Each period starts with the filling of electrons in a new principal energy shell. Therefore, the 6 th period starts with the filling of the \(6 \mathrm{~s}\)-orbital and ends when \(6 \mathrm{p}\)-orbitals are completely filled.
In between \(4 \mathrm{f}\) and \(5 \mathrm{~d}\)-orbitals are filled in accordance with the Aufbau principle. Thus, the outmost electronic configuration of the last element of the p-block in the 6 th period is \(6 \mathrm{~s}^2 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{p}^6\) or \(4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^2 6 \mathrm{p}^6\).
Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.
In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, \(s, p, d\) and \(f\). The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Arfbau principle, the seven periods ( 1 to 7 ) have \(2,8,8,18,18,32\) and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of \(f\)-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table. Which of the elements whose atomic numbers are given below, cannot be accommodated in the present set up of the long form of the periodic table?
(c) In present form of periodic table, elements till atomic number 118 are accommodated. So, element whose atomic number is 126 is not possible.
Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.
In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, \(s, p, d\) and \(f\). The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Arfbau principle, the seven periods ( 1 to 7 ) have \(2,8,8,18,18,32\) and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of \(f\)-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table. The electronic configuration of the element which is just above the element with atomic number 43 in the same group is ____.
\(
\text { (a) }{ }_{25} \mathrm{Mn} \text { has electronic configuration } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^2 \text {. }
\)
Explanation: Hint: Just substract 18 in atomic number 43 and find the next element which is just above the element with atomic number 43 The configuration of element which has 43 atomic number is as follows \(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^6 4 \mathrm{~d}^5 5 \mathrm{~s}^2\)
The next element which is just above the element with atomic number 43 has atomic number 25 . The electtronic configuration is as follows:
\(
1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^2
\)
Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.
In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, \(s, p, d\) and \(f\). The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods ( 1 to 7 ) have \(2,8,8,18,18,32\) and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of \(f\)-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table. The elements with atomic numbers 35, 53 and 85 are all ____.
\(
\text { (b) Halogens have atomic numbers } 9,17,35,53 \text { and } 85 \text {. }
\)
HINT: The group contains most electronegative element.
Explanation:
Each period ends with a noble gas. The atomic number of noble gases (i.e., group 18 elements) are 2, 10, \(18,36,54\) and 86 .
Therefore, elements with atomic numbers \(35(36-1), 53(54-1)\) and \(85(86-1)\), lie in a group before noble gases, i.e., halogens (group 17 ) elements.
Thus, the elements with atomic number 35,53 , and 85 all belong to halogens.
Electronic configurations of four elements A, B, C and D are given below :
(A) \(1 s^2 2 s^2 2 p^6\)
(B) \(1 s^2 2 s^2 2 p^4\)
(C) \(1 s^2 2 s^2 2 p^6 3 s^1\)
(D) \(1 s^2 2 s^2 2 p^5\)
Which of the following is the correct order of increasing tendency to gain electron :
(a) A \(1 s^2 2 s^2 2 p^6\): Noble gas configuration
B \(1 s^2 2 s^2 2 p^4\): 2 electrons short of stable configuration
C \(1 s^2 2 s^2 2 p^6 3 s^1\): Requires one electron to complete 5-orbital
D \(1 s^2 2 s^2 2 p^5\): Requires one electron to attain noble gas configuration
– Noble gases have no tendency to gain electrons since all their orbitals are completely filled. Thus, element A has the least electron gain enthalpy.
– Since element \(D\) has one electron less and element \(B\) has two electrons less than the corresponding noble gas configuration, hence, element \(D\) has the highest electron gain enthalpy followed by element B.
– Since, element \(\mathrm{C}\) has one electron in the 5-orbital and hence needs one more electron to complete it, therefore, electron gain enthalpy of \(\mathrm{C}\) is less than that of element \(\mathrm{B}\). Combining all the facts given above, the electron gain enthalpies of the four elements increase in the order \(A<C<B<D\).
HINT: Left to right across the period electron gain enthalpy increases.
Which of the following elements can show covalency greater than 4?
(b, c)
Elements Be and B lie in the 2 nd period. They can have a maximum of 8 electrons in the valence shell. In other words, they can have a maximum covalency of \(8 / 2=4\).
However, elements \(\mathrm{P}\) and \(\mathrm{S}\) have vacant \(\mathrm{d}\)-orbitals in their respective valence shells and hence can accommodate more than 8 electrons in their respective valence shell. In other words, they can show a covalency of more than 4.
Those elements impart colour to the flame on heating in it, the atoms of which require low energy for the ionisation (i.e., absorb energy in the visible region of spectrum). The elements of which of the following groups will impart colour to the flame?
(a, c) The elements of group 1 (alkali metals) and group 2 (alkaline earth metals) have low ionization enthalpies. Therefore, they impart colour to flame.
\(
\begin{array}{llll}
\text { Group 1 } & \text { Colour } & \text { Group 2 } & \text { Colour } \\
\mathrm{Li} & \text { Crimson } & \mathrm{Ca} & \text { Brick-red } \\
\mathrm{Na} & \text { Yellow } & \mathrm{Sr} & \text { Crimson red } \\
\mathrm{K} & \text { Pale violet } & – & – \\
\mathrm{Rb} & \text { Red violet } & \mathrm{Ba} & \text { Apple green } \\
\mathrm{Cs} & \text { Blue } & \mathrm{Ra} & \text { Crimson }
\end{array}
\)
Which of the following sequences contain atomic numbers of only representative elements?
\((\mathrm{a}, \mathrm{d})\) Elements of s and \(\mathrm{p}\)-block elements are called representative elements. Elements of \(\mathrm{f}\)-block ( \(\mathrm{Z}=21\) \(-30 ; 39-48 ; 57\) and \(72-80 ; 89\) and \(104-112\) ) are called transition elements while those of f-block (with \(Z=58-71\) and \(Z=90-103)\) are called inner transition elements.
(a) 3 – Group 1, 33 – group 15, 53 – group 17 and 87 – group 1 .
(d) 9 – Group 17,35 – Group 17, 51 – Group 15,88 – Group 2.
Which of the following elements will gain one electron more readily in comparison to other elements of their group?
\((a, d)\) Chlorine has the highest tendency to gain an electron and its electron gain enthalpy (-ve) is high. O and S belong to group 16 but \(S\) has larger tendency to accept electron.
Explanation:Â
Chlorine has the highest tendency to gain one electron because by doing so, it acquires the stable electronic configuration of teh nearest nobel gas, i.e., argo. Sulphur and oxygen belong to group 16 but the size of oxygen is much smaller than that of sulphur.
As a result, when an electron is added to them, the electron-electron repulsions in the smaller \(2 \mathrm{p}\)-subshell of oxygen are comparatively stronger than those present in the bigger \(3 p\)-subshell of sulphur. Therefore, \(\mathrm{S}\) has a higher tendency to gain an electron than \(\mathrm{O}\). \(\mathrm{Na}\), on the other hand, has only one electron in the valence shell and hence has a strong tendency to lose rather than gain one electron.
Which of the following statements are correct?
\((a, c, d)\) Chlorine has more negative electron gain enthalpy than fluorine due to bigger size and lesser electronic repulsion. Therefore, all other given statements are correct.
Which of the following sets contain only isoelectronic ions?
(a) \(\mathrm{Zn}^{2+}(30-2=28), \mathrm{Ca}^{2+}=(20-2=18), \mathrm{Ga}^{3+}(31-3=28), \mathrm{Al}^{3+}(13-3=10)\). These species have different number of electrons and hence are not isoelectronic ions.
(b) \(\mathrm{K}^{+}(19-1=18), \mathrm{Ca}^{2+}(20-2=18), \mathrm{Sc}^{3+}(21-3=18), \mathrm{Cl}^{-}(17+1=18)\) These are all isoelectronic ions since each one of them has 18 electrons.
(c) \(\mathrm{P}^{3-}(15+3=18), \mathrm{S}^{2-}(16+2=18), \mathrm{Cl}^{-}(17+1=18), \mathrm{K}^{+}(19-1=18)\).
These are all isoelectronic ions since each one of them has 18 electrons.
\(
\text { (d) } \mathrm{Ti}^{4+}(22-4=18), \operatorname{Ar}(18), \mathrm{Cr}^{3+}(24-3=21), \mathrm{V}^{5+}(23-5=18)
\)
These species have different number of electrons and hence are not isoelectronic ions.
In which of the following options order of arrangement does not agree with the variation of property indicated against it?
(b,c)
Due to the greater stability of the half-filled electronic configuration of nitrogen, its ionisation enthalpy is higher than that of oxygen. Thus, option (b) is incorrect. Due to stronger electron-electron repulsions in the small size of fluorine, the negative electron gain enthalpy of fluorine is lower than that of chlorine. Hence, option (c) is incorrect.
For increasing first ionization enthalpy, the order should be:
\(
\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}
\)
For increasing electron gain enthalpy, the order should be:
\(
\mathrm{I}<\mathrm{Br}<\mathrm{F}<\mathrm{Cl}
\)
Which of the following have no unit?
\(
\text { (a, d) Electron gain enthalpy and ionization enthalpy have units of enthalpy. }
\)
Ionic radii vary in
(a, c) Ionic radii ‘decreases as the effective nuclear charge increases.
\(
\text { Ionic radius } \propto \frac{1}{\text { effective nuclear charge }}
\)
Further, ionic radius increases as the screening effect increases.
Ionic radius \(\propto\) screening effect
An element belongs to \(3^{\text {rd }}\) period and group- 13 of the periodic table. Which of the following properties will be shown by the element?
(a & c) Except boron, all elements of groups 13 are metallic. These exists as solid. Being metallic in nature, aluminium is good conductor of electricity.
\((a, c)\) The element belonging to \(3^{\text {rd }}\) period and \(13^{\text {th }}\) group is aluminium which is a metal. Hence, it is solid, metallic and good conductor of electricity.
Match the correct atomic radius with the element.
\(
\begin{array}{|c|c|}
\hline \text { Element } & \text { Atomic radius (pm) } \\
\hline \text { (a) } \mathrm{Be} & \text { (p) } 74 \\
\hline \text { (b) } \mathrm{C} & \text { (q) } 88 \\
\hline \text { (c) } \mathrm{O} & \text { (r) } 111 \\
\hline \text { (d) } \mathrm{B} & \text { (s) } 77 \\
\hline \text { (e) } \mathrm{N} & \text { (t) } 66 \\
\hline
\end{array}
\)
(a) Atomic radius decreases from left to right within a period. This is caused by the increase in the effective nuclear charge.
\(
\mathrm{Be}>\mathrm{B}>\mathrm{C}>\mathrm{N}>\mathrm{O}
\)
a. \(\mathrm{Be} \rightarrow 111(\mathrm{r})\)
b. \(\mathrm{C} \rightarrow 77\) (s)
c. \(\mathrm{O} \rightarrow 66(\mathrm{t})\)
d. \(\mathrm{B} \rightarrow 88\) (q)
e. \(\mathrm{N} \rightarrow 74(\mathrm{p})\)
Match the correct ionisation enthalpies and electron gain enthalpies of the following elements.
\(
\begin{array}{|l|l|l|l|l|}
\hline \text { Elements } & & \boldsymbol{\Delta H}_{\mathbf{1}} & \mathbf{\Delta} \mathbf{H}_{\mathbf{2}} & \boldsymbol{\Delta}_{\mathbf{e g}} \mathbf{H} \\
\hline \text { (i) Most reactive non-metal } & \text { A. } & 419 & 3051 & -48 \\
\hline \text { (ii) Most reactive metal } & \text { B. } & 1681 & 3374 & -328 \\
\hline \text { (iii) Least reactive element e } & \text { C. } & 738 & 1451 & -40 \\
\hline \text { (iv) Metal forming binary halide } & \text { D. } & 2372 & 5251 & +48 \\
\hline
\end{array}
\)
\(
\begin{array}{|l|l|l|l|l|}
\hline & \text { A } & \text { B } & \text { C } & \text { D } \\
\hline a . & \text { ii } & \text { i } & \text { iv } & \text { iii } \\
\hline b . & \text { i } & \text { ii } & \text { iii } & \text { iv } \\
\hline c . & \text { i } & \text { iv } & \text { iii } & \text { ii } \\
\hline d . & \text { iv } & \text { i } & \text { iii } & \text { ii } \\
\hline
\end{array}
\)
(a) Hint: Most electronegative element has most negative electron gain enthalpy
(i) Most reactive non-metal has high \(\Delta \mathrm{H}_1\) and \(\Delta \mathrm{H}_2\) and most negative \(\Delta_{\mathrm{eg}} \mathrm{H}\). Therefore, the element is \(\mathrm{B}\).
(ii) Most reactive metal has low \(\Delta \mathrm{H}_1\) and high \(\Delta \mathrm{H}_2\) (because the second electron has to be lost from noble gas configuration) and has small negative \(\Delta_{\text {eg }} \mathrm{H}\). Therefore, the element is A.
(iii) Noble gases are the least reactive elements. They have very high \(\Delta \mathrm{H}_1\) and \(\Delta \mathrm{H}_2\) and have positive \(\Delta_{\mathrm{eg}} H\) values. Thus, the element is \(\mathrm{D}\).
(iv) Metal-forming binary halides are alkaline earth metals. They have \(\Delta \mathrm{H}_1\) and \(\Delta \mathrm{H}_2\) values little higher than those of most reactive metals (such as \(\mathrm{A}\) ) and have comparatively slightly less negative \(\Delta_{\mathrm{eg}} \mathrm{H}\) values. Thus, the element is \(\mathrm{C}\).
\(
\begin{array}{|c|c|c|c|c|c|}
\hline & \text { Elements } & & \Delta \mathrm{H}_1 & \Delta \mathrm{H}_2 & \Delta_{\text {eg }} \mathrm{H} \\
\hline \text { (i) } & \text { Most reactive non metal } & \mathrm{B} & 1681 & 3374 & -328 \\
\text { (ii) } & \text { Most reactive metal } & \text { A } & 419 & 3051 & -48 \\
\text { (iii) } & \text { Least reactive element } & \text { D. } & 2372 & 5251 & +48 \\
\text { (iv) } & \text { Metal forming binary halide } & \mathrm{C} & 738 & 1451 & -40 \\
\hline
\end{array}
\)
The electronic configuration of some elements is given in Column I and their electron gain enthalpies are given in Column II. Match the electronic configuration with electron gain enthalpy.
\(
\begin{array}{|l|l|}
\hline \begin{array}{l}
\text { Column (I) } \\
\text { Electronic configuration }
\end{array} & \begin{array}{l}
\text { Column (II) } \\
\text { Electron gain enthalpy/kJ mol } \mathbf{}^{-1}
\end{array} \\
\hline \text { (i) } 1 s^2 2 s^2 2 p^6 & \text { (A) }-53 \\
\hline \text { (ii) } 1 s^2 2 s^2 2 p^6 3 s^1 & \text { (B) }-328 \\
\hline \text { (iii) } 1 s^2 2 s^2 2 p^5 & \text { (C) }-141 \\
\hline \text { (iv) } 1 s^2 2 s^2 2 p^4 & \text { (D) }+48 \\
\hline
\end{array}
\)
(a)
\(
\text { (i) } \rightarrow \text { (D); (ii) } \rightarrow \text { (A) (iii) } \rightarrow \text { (B) (iv) } \rightarrow \text { (C) }
\)
(i) This electronic configuration corresponds to the noble gas i.e., neon. Since, noble gases have \(+\Delta_{\mathrm{eg}} \mathrm{H}\) values, therefore, electronic configuration (i) corresponds to the \(\Delta_{\text {eg }} \mathrm{H}=+48 \mathrm{~kJ} \mathrm{~mol}^{-1}\).
(ii) This electronic configuration corresponds to the alkali metal i.e., potassium. Alkali metals have small negative \(\Delta_{\mathrm{eg}} \mathrm{H}\) values, hence, electronic configuration (ii) correspongds to \(\Delta_{\text {eg }} \mathrm{H}=-53 \mathrm{~kJ} \mathrm{~mol}^{-1}\).
(iii) This electronic configuration corresponds to the halogen i.e., fluorine. Since, halogens have high negative \(\Delta_{\mathrm{eg}} \mathrm{H}\) values, therefore, electronic configuration (iii) corresponds to \(\Delta_{\mathrm{eg}} \mathrm{H}=328 \mathrm{~kJ} \mathrm{~mol}^{-1}\).
(iv) This electronic configuration corresponds to the chalcogen i.e., oxygen. Since, chalcogens have \(\Delta_{\mathrm{eg}} \mathrm{H}\) values less negative than those of halogens, therefore, electronic configuration (iv) corresponds to \(\Delta_{\mathrm{eg}} \mathrm{H}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1}\).
In the following questions, a statement of Assertion (A) followed by a statement of reason ( R ) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Generally, ionisation enthalpy increases from left to right in a period.
Reason (R): When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.
(b) Assertion and reason both are correct statements and reason is correct explanation of assertion. Ionisation enthalpy increases along a period because effective nuclear charge increases and atomic size decreases.
In the following questions, a statement of Assertion (A) followed by a statement of reason ( R ) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Boron has a smaller first ionisation enthalpy than beryllium.
Reason (R): The penetration of a 2 s electron to the nucleus is more than the \(2 p\) electron hence \(2 p\) electron is more shielded by the inner core of electrons than the 2 s electrons.
(c) Hint: Beryllium has a more stable configuration than Boron
Boron has a smaller first ionisation enthalpy than beryllium because the penetration of a \(2 \mathrm{~s}\) electron to the nucleus is more than the \(2 \mathrm{p}\) electron. Hence, the \(2 p\) electron is more shielded by the inner core of the electron than the 2 s electron.
The ionization enthalpy values are as follows:
\(
\begin{aligned}
& \mathrm{Be}=899 \mathrm{~kJ} / \mathrm{mol} \\
& \mathrm{B}=801 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
\)
Hence, Both assertion and reason are true and the reason is the correct explanation of assertion.
In the following questions, a statement of Assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.
Reason (R): The size of the atom increases on going down the group and the added electron would be farther from the nucleus.
(d) An assertion is not correct.
Explanation: Electron gain enthalpy does not always become less negative as we go down a group in Modern periodic table.
\(
\text { Ex. } \Delta \mathrm{H}_{\mathrm{eg}}(2 \mathrm{p} \text { series })<\Delta \mathrm{H}_{\mathrm{eg}}(3 \mathrm{p} \text { series }) \text {. }
\)
The reason i.e The size of the atom increases on going down the group in the Modern periodic table and added electron would be farther from the nucleus. is correct.
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