Page No 94: - Chapter 3 - Classifications Of Elements & Periodicity In Properties - Ncert Solutions class 11 - Chemistry

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Question 3.24:

Describe the theory associated with the radius of an atom as it

(a) gains an electron

(b) loses an electron

Answer:

(a) When an atom gains an electron, its size increases. When an electron is added, the number of electrons goes up by one. This results in an increase in repulsion among the electrons. However, the number of protons remains the same. As a result, the effective nuclear charge of the atom decreases and the radius of the atom increases.

(b) When an atom loses an electron, the number of electrons decreases by one while the nuclear charge remains the same. Therefore, the interelectronic repulsions in the atom decrease. As a result, the effective nuclear charge increases. Hence, the radius of the atom decreases.

Question 3.25:

Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your Answer.

Answer:

The ionization enthalpy of an atom depends on the number of electrons and protons (nuclear charge) of that atom. Now, the isotopes of an element have the same number of protons and electrons. Therefore, the first ionization enthalpy for two isotopes of the same element should be the same.

Question 3.26:

What are the major differences between metals and non-metals?

Answer:

	Metals		Non–metals
1.	Metals can lose electrons easily.	1.	Non-metals cannot lose electrons easily.
2.	Metals cannot gain electrons easily.	2.	Non-metals can gain electrons easily.
3.	Metals generally form ionic compounds.	3.	Non–metals generally form covalent compounds.
4.	Metals oxides are basic in nature.	4.	Non–metallic oxides are acidic in nature.
5.	Metals have low ionization enthalpies.	5.	Non–metals have high ionization enthalpies.
6.	Metals have less negative electron gain enthalpies.	6.	Non­–metals have high negative electron gain enthalpies.
7.	Metals are less electronegative. They are rather electropositive elements.	7.	Non–metals are electronegative.
8.	Metals have a high reducing power.	8.	Non–metals have a low reducing power.

Use the periodic table to Answer the following Questions.Question 3.27:

(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:

(a) The electronic configuration of an element having 5 electrons in its outermost subshell should be ns² np⁵. This is the electronic configuration of the halogen group. Thus, the element can be F, CL, Br, I, or At.

(b) An element having two valence electrons will lose two electrons easily to attain the stable noble gas configuration. The general electronic configuration of such an element will be ns². This is the electronic configuration of group 2 elements. The elements present in group 2 are Be, Mg, Ca, Sr, Ba.

(c) An element is likely to gain two electrons if it needs only two electrons to attain the stable noble gas configuration. Thus, the general electronic configuration of such an element should be ns² np⁴. This is the electronic configuration of the oxygen family.

(d) Group 17 has metal, non–metal, liquid as well as gas at room temperature.

Question 3.28:

The increasing order of reactivity among group 1 elements is Li < Na < K < Rb CI > Br > I. Explain.

Answer:

The elements present in group 1 have only 1 valence electron, which they tend to lose. Group 17 elements, on the other hand, need only one electron to attain the noble gas configuration. On moving down group 1, the ionization enthalpies decrease. This means that the energy required to lose the valence electron decreases. Thus, reactivity increases on moving down a group. Thus, the increasing order of reactivity among group 1 elements is as follows:

Li < Na < K < Rb < Cs

In group 17, as we move down the group from Cl to I, the electron gain enthalpy becomes less negative i.e., its tendency to gain electrons decreases down group 17. Thus, reactivity decreases down a group. The electron gain enthalpy of F is less negative than Cl. Still, it is the most reactive halogen. This is because of its low bond dissociation energy. Thus, the decreasing order of reactivity among group 17 elements is as follows:

F > Cl > Br > I

Question 3.29:

Write the general outer electronic configuration of s-, p-, d- and f- block elements.

Answer:

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

Assign the position of the element having outer electronic configurationQuestion 3.30:

(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 periodic table.

Answer:

(i) Since n = 3, the element belongs to the 3^rd period. It is a p–block element since the last electron occupies the p–orbital.

There are four electrons in the p–orbital. Thus, the corresponding group of the element

= Number of s–block groups + number of d–block groups + number of p–electrons

= 2 + 10 + 4

= 16

Therefore, the element belongs to the 3^rd period and 16^th group of the periodic table. Hence, the element is Sulphur.

(ii) Since n = 4, the element belongs to the 4^th period. It is a d–block element as d–orbitals are incompletely filled.

There are 2 electrons in the d–orbital.

Thus, the corresponding group of the element

= Number of s–block groups + number of d–block groups

= 2 + 2

= 4

Therefore, it is a 4^th period and 4^th group element. Hence, the element is Titanium.

(iii) Since n = 6, the element is present in the 6^th period. It is an f –block element as the last electron occupies the f–orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe] 4f⁷ 5d¹ 6s². Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium.

Question 3.31:

The first (ΔiH1) and the second (ΔiH) ionization enthalpies (in kJ mol^–1) and the (Δ_egH) electron gain enthalpy (in kJ mol^–1) of a few elements are given below:

Elements	ΔiH	ΔiH	ΔegH
I	520	7300	–60
II	419	3051	–48
III	1681	3374	–328
IV	1008	1846	–295
V	2372	5251	+48
VI	738	1451	–40

(a) the least reactive element.Which of the above elements is likely to be :

(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 MX₂, (X=halogen).

(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?

Answer:

(a) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy (ΔiH₁) and a positive electron gain enthalpy (ΔegH).

(b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (ΔiH₁) and a low negative electron gain enthalpy (ΔegH).

(c) Element III is likely to be the most reactive non–metal as it has a high first ionization enthalpy (ΔiH₁) and the highest negative electron gain enthalpy (ΔegH).

(d) Element V is likely to be the least reactive non–metal since it has a very high first ionization enthalpy (ΔiH₂) and a positive electron gain enthalpy (ΔegH).

(e) Element VI has a low negative electron gain enthalpy (ΔegH). Thus, it is a metal. Further, it has the lowest second ionization enthalpy (ΔiH₂). Hence, it can form a stable binary halide of the formula MX₂ (X=halogen).

(f) Element V has the highest first ionization energy and high second ionization energy. Therefore, it can form a predominantly stable covalent halide of the formula MX (X=halogen).

Question 3.32:

Predict the formula of the stable binary compounds that would be formed by the combination of the following pairs of elements.

(a) Lithium and oxygen                     (b) Magnesium and nitrogen

(c) Aluminium and iodine                   (d) Silicon and oxygen

(e) Phosphorus and fluorine              (f) Element 71 and fluorine

Answer:

(a) Li₂O

(b) Mg₃N₂

(c) AlI₃

(d) SiO₂

(e) PF₃ or PF₅

(f) The element with the atomic number 71 is Lutetium (Lu). It has valency 3. Hence, the formula of the compound is LuF_3.

Question 3.33:

In the modern periodic table, the period indicates the value of:

(a) Atomic number

(b) Atomic mass

(c) Principal quantum number

(d) Azimuthal quantum number.

Answer:

The value of the principal quantum number (n) for the outermost shell or the valence shell indicates a period in the Modern periodic table.

Question 3.34:

Which of the following statements related to the modern periodic table is incorrect?

(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.

(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.

(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.

(d) The block indicates value of azimuthal quantum number (l ) for the last subshell that received electrons in building up the electronic configuration.

Answer:

The d-block has 10 columns because a maximum of 10 electrons can occupy all the orbitals in a d subshell.