Why is the electron gain enthalpy of chlorine more negative than fluorine?

The negative electron gain enthalpy of fluorine is less than that of chlorine. It is due to small size of fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2p orbitals of fluorine and thus, the incoming electron does not experience much attraction.

The ionization energy of nitrogen is more than that of oxygen because :

Ionization energy is the minimum energy required to remove the most loosely held electron from one mole of gaseous atoms to produce one mole of gaseous ions each with a charge of \\(+1\\). An atom is highly stable when its valence orbital is completely filled or half-filled and in nitrogen, the valence p orbital is exactly half filled. So, p orbital in nitrogen is more stable than in oxygen, which has one electron more than the half-filled configuration. So, the ionization energy of nitrogen is more than that of oxygen.

If \\(1^{st}\\) ionisation energy of Helium \\((He)\\) atom is \\(25.6\ eV/atom\\) then sum of \\(IE_{1}\\) and \\(IE_{2}\\) per atom of Helium \\((He)\\) would be:

The first \\(IE\\) of \\(He\\) is much higher than that of its' second \\(IE\\) as the first \\(e^-\\) removed is fully filled shell, i.e; \\(1s^2.\\) So, \\(IE_1\\) of \\(He=25.6eV\\)Then \\(He^+\\) is isoelectronic to hydrogen atom with \\(1s^1\\) configuration.\\(\triangle H_{E_2}=\left|E_1 (He^+)\right|\\)\\(=\left|-2^2\times \cfrac {13.6eV}{1^2}\right|\\)\\(=54.44eV\\)\\(\therefore IE_1+IE_2=25.6+54.4=80.04eV\\)

The first ionization enthalpies of \\(Na,\ Mg,\ Al\\) and \\(Si\\) are in the order:

As we move across the period, nuclear charge increases, atomic size decreases hence, ionization enthalpy increases. For \\(Al\\) (\\(3s^{2} 3p^{1}\\) ), electron has to be removed from partially filled \\(3p\\) orbital whereas in \\(Mg(3s^{2}\\)), electron has to be removed from stable fully filled \\(3s\\) orbital. Removal of an electron from stable, fully filled orbital requires more energy than removal of electron from partially filled orbital. Thus, ionisation enthalpy for \\(Mg\\) is greater than ionisation enthalpy for \\(Al\\). So, the correct order of first ionization enthalpies is: \\(Na Al<Si\\)

Which of the following will have the most negative electron gain enthalpy and which the least negative?P, S, Cl, F

Halogens have the highest negative value of electron gain enthalpy. Here Cl and F both are halogens but electron gain enthalpy of Cl is more negative than F because of the very small size of F which causes interelectronic repulsion among its electrons. Cl has the most negative one. P has the least negative one due to the larger size.

How do atomic radius vary in a period and in a group? How do you explain the variation?

In a period, from left to right, with increase in the atomic number, the atomic radius decreases due to increase in the nuclear charge which increases the attraction of the nucleus for the valence electrons. In a group, on moving from top to bottom, the ionic radius increases as a new energy level is added at each succeeding element but the number of valence electrons remains same.

The first \\(\displaystyle \left ( \Delta _{i}H _{1}\right )\\) and the second \\(\displaystyle \left ( \Delta _{i}H _{2}\right )\\) ionization enthalpies (in kJ mol\\(\displaystyle ^{-1}\\)) and the \\(\displaystyle \left ( \Delta _{eg}H \right )\\) electron gain enthalpy (in kJ mol\\(\displaystyle ^{-1}\\)) of a few elements are given below:Elements\\(\displaystyle \Delta H_{1}\\)\\(\displaystyle \Delta H_{2}\\)\\(\displaystyle \Delta _{eg}\\)HI5207300-60II4193051-48III16813374-328IV10081846-295V23725251+48VI7381451-40Which of the above elements is likely to be :(a) the least reactive element,(b) the most reactive metal,(c) the most reactive non-metal,(d) the least reactive non-metal,(e) the metal which can form a stable binary halide of the formula MX\\(\displaystyle _{2}\\) (X=halogen) and(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?

(a) The element V is the least reactive element as it has highest value of first ionization energy and positive electron gain enthalpy. (b) The metal II is the most reactive metal. It has very low value of first ionization energy.(c) The element III is the most reactive non metal. Its electron gain enthalpy is most negative.(d) The element IV is the least reactive non metal as it has least negative electron gain enthalpy among the non metals.(e) The metal VI can form a stable binary halide of the formula \\(MX_2\\) as it has lowest value of second ionization enthalpy.(f) The metal I can form a stable \\(MX\\) halide

The first ionization enthalpy values ( in kJ mol\\( \displaystyle ^{-1} \\) ) of group 13 element are :B Al Ga In Tl801 577 579 558 589How would you explain this deviation from the general trend?

On moving down the group, the ionization enthalpy decreases. This is true for B and Al. The ionization enthalpy of Ga is unexpectedly higher than that of Al. Ga contains 10d electrons in inner shell which are less penetrating. Their shielding is less effective than that of s and p electrons. The outer electron is held fairly strongly by the nucleus. The ionization enthalpy increases slightly. A similar increase is observed from In to Tl due to presence of 14f electrons in the inner shell of Tl which have poor shielding effect.

Explain the following:(i) Gallium has higher ionization enthalpy than aluminum. (ii) Boron does not exist as \\(B^{3+}\\)ion.(iii) Aluminum forms \\([AlF_6]^{3-}\\) ion but boron does not form \\([BF_6]^{3-}\\) ion.(iv) \\(PbX_2\\) is more stable than \\(PbX_4\\)(v) \\(Pb^{4+}\\) acts as an oxidizing agent, but \\(Sn^{2+}\\) acts as a reducing agent.(vi) Electron gain enthalpy of chlorine is more negative as compared to fluorine.(vii) \\(TI(NO_3)_3\\) acts as an oxidizing agent.(viii) Carbon shows catenation property but lead does not.(ix) \\(BF_3\\) does not hydrolyze.(x) Why does the element silicon, not form a graphite-like structure whereas carbon does?

(i) The ionization enthalpy value of Ga is higher than Al due to the inability of d- and f- electrons, which have a screening effect to compensate for the increase in nuclear charge.(ii) Boron has small size and sum of \\(\Delta_iH_1+\Delta_iH_2+\Delta_iH_3\\) is very high. Boron does not form \\(B^{3+}\\) ion therefore, give covalent compounds.(iii) \\(Al\\) has vacant 3d-orbitals and can expand its co-ordination no and forms \\([AlF_6]^{3-}\\). On the other hand, Boron does not have orbitals and can not form \\([BF_6]^{3-}\\) and cannot expand its covalence beyond \\(4\\) and thus, gives\\([BF_4].\\)(iv) Due to the inert pair effect, a \\(+2\\) oxidation state is more stable than a \\(+4\\) oxidation state.(v) \\(Pb^{4+}\\) by gaining \\(2\\) electrons change into \\(Pb^{2+}\\) which is more stable due to inert pair effect. \\(Sn^{2+}\\) is less stable than \\(Sn^{4+}\\) by losing electrons. Therefore, \\(Pb^{4+}\\) acts as an oxidizing agent while \\(Sn^{2+}\\) acts as a reducing agent.(vi) The size of F atom is very small and incoming electron feels interelectronic repulsion and electron gain enthalpy of F is less negative as compared to Cl.(vii) Due to the inert pair effect, Tl is more stable in \\(+1\\) oxidation state than that of \\(3+\\) oxidation state. Therefore, \\(Tl(NO_3)_3\\) acts as a strong oxidizing agent.(viii) Carbon atoms have the tendency to link with another through covalent bonds to form chains and rings. This property is called catenation. This is because \\(C-C\\) bonds are very strong. Down the group, the size increases, and electronegativity decreases, and thereby, the tendency to show catenation decreases. This can be clearly seen from bond enthalpy values. The order of catenation is \\(C>>Si>GE\approx Sn.\\) Lead does not show catenation.(ix) \\(BF_3\\) does not hydrolyze completely. It forms boric acid and fluroboric acid this is because the HF formed reacts with \\(H_3BO_3\\).\\(BF_3+3H_2O\rightarrow H_3BO_3+3HF \times 4\\)\\(H_3BO_3+3HF\rightarrow H^+[BF_4]^- +3H_2O\times 3\\)\\(4BF_3+3H_2O\rightarrow H_3BO_3+3[BF_4]^- +3H^+\\)(x) In graphite, carbon is \\(sp^3\\) hybridized. Carbon has a tendency to form multiple \\(p\pi-p\pi\\) bonds due to its small size and highest electronegativity in group \\(14\\). Silicon due to its large size and less electronegativity cannot form multiple bonds. Thus, silicon cannot form a graphite-like structure.

Trends in Atomic Radius and Ionic Radius | Definition, Examples, Diagrams

Q: Which of the following will have the most negative electron gain enthalpy and which the least negative?P, S, Cl, F

Halogens have the highest negative value of electron gain enthalpy. Here Cl and F both are halogens but electron gain enthalpy of Cl is more negative than F because of the very small size of F which causes interelectronic repulsion among its electrons. Cl has the most negative one. P has the least negative one due to the larger size.

Q: How do atomic radius vary in a period and in a group? How do you explain the variation?

In a period, from left to right, with increase in the atomic number, the atomic radius decreases due to increase in the nuclear charge which increases the attraction of the nucleus for the valence electrons. In a group, on moving from top to bottom, the ionic radius increases as a new energy level is added at each succeeding element but the number of valence electrons remains same.

Q: The first \\(\displaystyle \left ( \Delta _{i}H _{1}\right )\\) and the second \\(\displaystyle \left ( \Delta _{i}H _{2}\right )\\) ionization enthalpies (in kJ mol\\(\displaystyle ^{-1}\\)) and the \\(\displaystyle \left ( \Delta _{eg}H \right )\\) electron gain enthalpy (in kJ mol\\(\displaystyle ^{-1}\\)) of a few elements are given below:Elements\\(\displaystyle \Delta H_{1}\\)\\(\displaystyle \Delta H_{2}\\)\\(\displaystyle \Delta _{eg}\\)HI5207300-60II4193051-48III16813374-328IV10081846-295V23725251+48VI7381451-40Which of the above elements is likely to be :(a) the least reactive element,(b) the most reactive metal,(c) the most reactive non-metal,(d) the least reactive non-metal,(e) the metal which can form a stable binary halide of the formula MX\\(\displaystyle _{2}\\) (X=halogen) and(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?

(a) The element V is the least reactive element as it has highest value of first ionization energy and positive electron gain enthalpy. (b) The metal II is the most reactive metal. It has very low value of first ionization energy.(c) The element III is the most reactive non metal. Its electron gain enthalpy is most negative.(d) The element IV is the least reactive non metal as it has least negative electron gain enthalpy among the non metals.(e) The metal VI can form a stable binary halide of the formula \\(MX_2\\) as it has lowest value of second ionization enthalpy.(f) The metal I can form a stable \\(MX\\) halide

Q: The first ionization enthalpy values ( in kJ mol\\( \displaystyle ^{-1} \\) ) of group 13 element are :B Al Ga In Tl801 577 579 558 589How would you explain this deviation from the general trend?

On moving down the group, the ionization enthalpy decreases. This is true for B and Al. The ionization enthalpy of Ga is unexpectedly higher than that of Al. Ga contains 10d electrons in inner shell which are less penetrating. Their shielding is less effective than that of s and p electrons. The outer electron is held fairly strongly by the nucleus. The ionization enthalpy increases slightly. A similar increase is observed from In to Tl due to presence of 14f electrons in the inner shell of Tl which have poor shielding effect.

definition

Define Metallic Radius

The metallic radius is the radius of an atom joined by metallic bond. The metallic radius is half of the distance between the nuclei of two adjacent atoms in the metallic cluster.

Metallic radius for the given diagram is 180 pm.

definition

Atomic Size Trend Across the Period

Atomic radius decreases from left to right within a period. This is caused by the increase in the number of protons and electrons across a period. One proton has a greater effect than one electron cause of the coulombic law. Thus, electrons are pulled towards the nucleus, resulting in a smaller radius.

definition

Atomic Size Trend Down the Group

The number of energy levels increases moving down a group as the number of electrons increases. As a result, the distance between the nucleus and the outermost orbital increases and so the atomic radius increases.

example

Trend of atomic size in periodic table

The atomic radius of an element is a measure of the size of its atoms. It represents the mean distance from the nucleus to the outermost boundary of the surrounding cloud of electrons. Atomic radii vary in a predictable manner across the periodic table.

Trends in Atomic Radius and Ionic Radius

definition

Define Metallic Radius

definition

Atomic Size Trend Across the Period

definition

Atomic Size Trend Down the Group

example

Trend of atomic size in periodic table

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Trends in Ionisation Enthalpy

Trends in Electron Gain Enthalpy

Trends in Electronegativity

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Trends in Atomic and Ionic Radius

$6.02 \times 1 0^{20}$ molecules of urea are present in 100 mL of its solution. The concentration of solution is :

Why is the electron gain enthalpy of chlorine more negative than fluorine?

The ionization energy of nitrogen is more than that of oxygen because :

If $1^{s t}$ ionisation energy of Helium $(H e)$ atom is $25.6 e V / a t o m$ then sum of $I E_{1}$ and $I E_{2}$ per atom of Helium $(H e)$ would be:

The first ionization enthalpies of $N a, M g, A l$ and $S i$ are in the order:

Which of the following will have the most negative electron gain enthalpy and which the least negative?
P, S, Cl, F

How do atomic radius vary in a period and in a group? How do you explain the variation?

The first ionization enthalpy values ( in kJ mol $^{- 1}$ ) of group 13 element are :

B Al Ga In Tl
801 577 579 558 589

How would you explain this deviation from the general trend?

Elements	$Δ H_{1}$	$Δ H_{2}$	$Δ_{e g}$ H
I	520	7300	-60
II	419	3051	-48
III	1681	3374	-328
IV	1008	1846	-295
V	2372	5251	+48
VI	738	1451	-40

definition

Define Metallic Radius

definition

Atomic Size Trend Across the Period

definition

Atomic Size Trend Down the Group

example

Trend of atomic size in periodic table

REVISE WITH CONCEPTS

Trends in Ionisation Enthalpy

Trends in Electron Gain Enthalpy

Trends in Electronegativity

Quick Summary With Stories

Trends in Atomic and Ionic Radius

6.02×1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is :

Why is the electron gain enthalpy of chlorine more negative than fluorine?

The ionization energy of nitrogen is more than that of oxygen because :

If 1st ionisation energy of Helium (He) atom is 25.6 eV/atom then sum of IE1​ and IE2​ per atom of Helium (He) would be:

The first ionization enthalpies of Na, Mg, Al and Si are in the order:

Which of the following will have the most negative electron gain enthalpy and which the least negative?P, S, Cl, F

How do atomic radius vary in a period and in a group? How do you explain the variation?

The first ionization enthalpy values ( in kJ mol−1 ) of group 13 element are : B Al Ga In Tl 801 577 579 558 589 How would you explain this deviation from the general trend?

$6.02 \times 1 0^{20}$ molecules of urea are present in 100 mL of its solution. The concentration of solution is :

If $1^{s t}$ ionisation energy of Helium $(H e)$ atom is $25.6 e V / a t o m$ then sum of $I E_{1}$ and $I E_{2}$ per atom of Helium $(H e)$ would be:

The first ionization enthalpies of $N a, M g, A l$ and $S i$ are in the order:

Which of the following will have the most negative electron gain enthalpy and which the least negative?
P, S, Cl, F

The first ionization enthalpy values ( in kJ mol $^{- 1}$ ) of group 13 element are :

B Al Ga In Tl
801 577 579 558 589

How would you explain this deviation from the general trend?