Chapter 3 Classification of Elements and Periodicity in Properties Solutions
Question - 21 : - Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer
Answer - 21 : -
When an electron is added to O atom to form O– ion, energy is released. Thus, the firstelectron gain enthalpy of O is negative.
On the other hand, when an electron is added to O– ion to form O2– ion,energy has to be given out in order to overcome the strong electronicrepulsions. Thus, the second electron gain enthalpy of O is positive.
Question - 22 : - What is the basic difference between the terms electron gain enthalpy and electronegativity?
Answer - 22 : -
| Electron gain enthalpy | Electronegativity |
| Tendency to gain electrons for an isolated gaseous atom is its electron gain enthalpy. | Tendency to attract the shared pairs of electrons for an atom which is in chemical compound is its electronegativity. |
Question - 23 : - How would you react to the statement that the electronegativity of N on the Pauling scale is 3.0 in all the nitrogen compounds?
Answer - 23 : - Electronegativity is a variable property of any element. Electronegativity is different for different compounds. Hence, the given statement “The electronegativity of Nitrogen on Pauling scale measures 3.0 for all nitrogen compounds” is incorrect. This is because the electronegativity of Nitrogen is different in NO2 and NH3.
Question - 24 : - Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
Answer - 24 : -
(a) As an atom expels a single electron, so the quantity of electron decreases by 1 but the nuclear charge does not change. Thus, the electron-electron repulsion decreases in an atom. So, there is an increase in the effective nuclear charge. Thus, there is a decrease in the radius of an atom.
(b) There is an increase in the size of an atom when it gains an electron. As it gains an electron then the quantity of electrons raises by 1. Thus, the electron-electron repulsion increases in an atom. There is an increase in effective nuclear charge as the quantity of proton remains unchanged. Thus, there is an increase in the radius of an atom.
Question - 25 : - Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer
Answer - 25 : - Ionization enthalpy of any atom relies on its number of protons and electrons. But, the isotopes of any element have an equal number of electrons and protons. Thus, the 1st ionization enthalpy of two isotopes of a single element is the same.
Question - 26 : - What are the major differences between metals and non-metals?
Answer - 26 : -
| Metals | Non- metals |
| a | They can easily expel an electron | a | They cannot easily expel an electron. |
| b | They cannot easily receive an electron. | b | They can easily receive an electron. |
| c | They form ionic compounds. | c | They form covalent compounds. |
| d | Their oxides are having basic nature. | d | Their oxides are having acidic nature. |
| e | Their ionization enthalpies are low. | e | Their ionization enthalpies are high. |
| f | Their electron affinity is less negative. | f | Their electron affinity is highly negative. |
| i | Their electronegativity is less. | i | Their electronegativity is more. |
| j | Their reducing power is high. | j | Their reducing power is low. |
Question - 27 : - Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room
temperature
Answer - 27 : -
(a) Element with 5 electrons in outer subshell is havingelectronic configuration ns2np5. Halogen group is having the sameelectronic configuration. So, the elements can be At, I, Br, Cl and F.
(b) The elements that tend to gain 2 electrons are those who canachieve stable inert gas configuration by gaining these electrons. They arehaving electronic configuration ns2np4. Oxygen containing group is having thesame electronic configuration.
(c) The elements that tend to expel 2 electrons are those whocan achieve stable inert gas configuration by expelling these electrons. Theyare having electronic configuration ns2.2nd Group elements are havingthe same electronic configuration. So, the elements can be Ba, Sr, Ca, Mg, Be.
(d) Elements of group 17 are having liquid, gas, metal andnon-metals as well at room temperature.
Question - 28 : - The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Answer - 28 : -
For group 1
There is only 1 valance electron in the elements of group 1. So, their tendency is to expel this electron in order to achieve a stable inert gas configuration. As we move down in group 1 the ionization enthalpies of the elements decreases. Thus, less energy will be required to expel the valence electron. As a result reactivity of elements increases as we move down in group 1.
For group 17
In the elements of group 17, there is a requirement of only 1 electron in order to achieve stable inert gas configuration. So, their tendency is to gain this 1 electron. As we move down in group 17 the ionization enthalpies of the elements increases. Thus, more energy will be required to expel the valence electron. As a result reactivity of elements decreases as we move down in group 17.
Question - 29 : - Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Answer - 29 : -
| Element | General outer electronic configuration |
| s–block | ns1–2, where n = 2 – 7 |
| p–block | ns2np1–6, where n = 2 – 6 |
| d–block | (n–1) d1–10 ns0–2, where n = 4 – 7 |
| f–block | (n–2)f1–14(n–1)d0–10ns2, where n = 6 – 7 |
Question - 30 : - Assign the position of the element having outer electronicconfiguration
(i) ns2 np4 for n = 3 (ii) (n –1)d2 ns2 for n = 4, and (iii) (n –2) f7 (n –1)d1 ns2 for n = 6, in the periodictable.
Answer - 30 : -
(i) Since n =3, the element belongs to the 3rd period.It is a p–blockelement since the last electron occupies the p–orbital.
There are four electrons in the p–orbital. Thus, thecorresponding group of the element
= Number of s–block groups + number of d–blockgroups + number of p–electrons
= 2 + 10 +4
= 16
Therefore, the element belongs to the 3rd period and 16th groupof the periodic table. Hence, the element is Sulphur.
(ii) Since n =4, the element belongs to the 4th period.It is a d–blockelement as d–orbitalsare incompletely filled.
There are 2 electrons in the d–orbital.
Thus, thecorresponding group of the element
= Number of s–block groups + number of d–blockgroups
= 2 + 2
= 4
Therefore, it is a 4th periodand 4th group element. Hence, the element is Titanium.
(iii) Since n =6, the element is present in the 6th period.It is an f –blockelement as the last electron occupies the f–orbital. It belongs to group 3of the periodic table since all f-block elements belong to group3. Its electronic configuration is [Xe] 4f7 5d1 6s2. Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence,the element is Gadolinium.