Question 1. What will be the name (IUPAC) and symbol if the element with atomic number 119 is discovered? Write its electronic configuration. Also, write the formulas of the stable chloride and oxide of this element.
Answer:
IUPAC name: Ununennium, Symbol: Uue
- Atomic number ofthe element =119 = 87 + 32
- It is known that the element with atomic number 87 is francium (Fr). Fr belongs to group-1 in the 7th period of the periodic table.
- So, the element with atomic number 119 will take its position in group-1 and 8th period just below francium(Fr).
- The electronic configuration of this element will be [UuojBs1, (where Uuo = Ununoctium, Z = 118). It will be an alkali metal with valency=1
- If the symbol ofthe element is ‘M’ then the formulas of its stable chloride and oxide will be MCI and M20 respectively.
periodic table question answer
Question 2. Formulas of oxide and chloride of an element A are A205 and AClg respectively. Which group of Mendeleev’s periodic table will the element belong to? State whether it is a metal or a non-metal.
Answer:
Hiking oxygen as standard, the valency ofthe element A is 5, and taking chlorine as standard, the valency of A is 3. Since oxygen-based valency (8-chlorine-based valency) is the same as that ofthe group number of the element, the element will be of group Mendeleev’s periodic table. Thus, itis a non-metal.
Question 3. A and B are two elements with atomic numbers 9 and 17 respectively. Explain why the element A is a more powerful oxidizing agent than the element B.
Answer:
Given that A and B are two elements with atomic numbers 9 and 17 respectively.
Electronic configuration of gA: ls22s22p5. Electronic configuration of B: ls²2s²2p63s23p5. Both A and B can accept electrons to form the anions of A¯ and B¯ having gas electronic configuration. However, the anions A¯ is more stable than B¯ because of the smaller atomic size of A. So A is a stronger oxidizing agent than B.
Classification of Elements and Periodicity in Properties Class 11
Question 4. Elements A, B, and C have atomic numbers (Z- 2), Z, and (Z +1) respectively. Of these, B is an inert gas. Which one of these has the highest electronegativity? Which one of these has the highest ionization potential? What is the formula of the compound formed by the combination of A and C? What is the nature of the bond in this compound?
Answer:
Since element B (atomic no =Z) is an inert gas, the element ‘A’ with atomic number (Z- 2) is included in group VIA. On the other hand, the element C, having an atomic number (Z + 1) must belong to groups (alkali metal). Hence, the electronegativity of the element A is maximum.
The element B, being an inert gas, has the highest value of ionization potential.
The valency of the element A, belonging to group (8- 6) = 2, and that ofthe element C, being an element of group IA, is 1. Therefore, the formula of the compound formed by A and C will be C2A.
A being a strongly electronegative element and C being a strongly electropositive element complete their octet through gain and loss of electrons respectively. So, the nature of the bond formed between C and A in C2A isionic or electrovalent bond.
Periodicity in Properties Class 11 Chemistry
Question 5. The atomic radius of 10Ne is more than that of gF —why?
Answer:
The atomic radius of 10Ne is more than that of gF
Fluorine forms diatomic molecules, thus the atomic radius of fluorine is a measure of half of the internuclear distance in its molecule (i.e., half of covalent bond length of F2 molecule) but neon being an inert gas, its atoms are incapable of forming covalent bonds by mutual combination amongst themselves.
The only force that comes into play between the atoms is the weak van der Waals force. So a measure of the atomic radius of Ne is equal to its van der Waals radius but the van der Waals radius is always greater than the covalent radius.
Thus, the atomic radius of neon is larger than that of fluorine. Furthermore, due to an increase in the number of electrons in the outermost 2p -orbital of Ne, there occurs an increase in electron-electron repulsion. So 2p-orbital of Ne suffers expansion leading to its increased atomic radius.
Question 6. The first electron affinity ofoxygen is negative but the second electron affinity is positive—explain.
Answer:
The first electron affinity ofoxygen is negative but the second electron affinity is positive
When an electron is added to the valence shell of an isolated gaseous O-atom in its ground state to form a negative ion, energy is released.
Because a neutral oxygen atom tends to complete its octet with electrons. So, the electron affinity of oxygen is an exothermic process and its value is negative.
When an extra electron is added to the O¯ ion, that second electron experiences a force of repulsion exerted by the negative charge ofthe anion. So, first, this process requires a supply of energy from an external source. This accounts for the endothermic nature of second electron affinity and has a positive value.
Periodic Classification of Elements Class 11 Notes
Question 7. The electron affinity of sodium is negative but magnesium has a positive value—why?
Answer:
The electron affinity of sodium is negative but magnesium has a positive value
Electronic configuration of 11Na: ls22s22p63s1 Electronic configuration of 12Mg: ls22s22p63s2. The addition of one electron to the 3s -orbital of Na leads to a comparatively stable electronic configuration with a fulfilled orbital.
So, this process of the addition of electrons to Na is an exothermic, process. So, the electron affinity of Na is negative. On the other hand, Mg has fulfilled 3s orbital and has a stable electronic configuration. So the addition of an electron to the 3p -orbital destabilizes the electronic configuration of Mg.
Additional energy is required for addition of electrons to the outermost shell of Mg i.e., this process is endothermic and thus the value of electron affinity of Mg is positive.
Trends in Periodic Table Class 11
Question 8. If the electron affinity of chlorine is 350 kJ.moI-1, then what is the amount of energy liberated to convert 1.775 g of chlorine (existing at atomic state) to chloride ions completely (in a gaseous state)
Answer:
Atomic mass of chlorine = 35.5 g .mol 1
The energy liberated in the conversion of 35.5 g of Cl- to Clion =350 kJ
∴ The energy liberated in the conversion of 1.775 g of Cl to Cl-
\(\text { ion }=\frac{350}{35.5} \times 1.775=17.5 \mathrm{~kJ}\)Modern Periodic Table and Its Properties Class 11
Question 9. The second ionization enthalpy of Mg is sufficiently high and the second electron-gain enthalpy of O has a positive value. How do you explain the existence of Mg2+02- rather than Mg+O-?
Answer:
The lattice energy of an ionic crystal depends on the force of attraction between the cations and anions \(\left(F \propto \frac{q_1 q_2}{r^2}\right)\)
So, the magnitude of lattice energy increases as the charges on the cation and anion increase. Consequently, the lattice energy of Mg²+O² is very much greater than that of Mg+O¯.
The lattice energy of Mg²+O²¯ is so high that it exceeds the unfavorable effects of the second ionization enthalpy of Mg and the second electron-gain enthalpy of O.
So, Mg²+O²¯ is a stable ionic compound and its formation is favored over Mg+O¯.
class 11 chemistry periodic classification solutions
Question 10. The atomic numbers of some elements are given below. Classify them into three groups so that the two elements in each group exhibit identical chemical behavior: 9, 12, 16, 34, 53, 56.
Answer:
Elements with atomic numbers 9 and 53 belong to the -block and they have similar outer electronic configurations (ns2np5). So they will exhibit similar chemical properties. Their group number = (10 + 2 + 5) = 17. Elements with atomic numbers 12 and 56 belong to s -block and theyhave similar outer electronic configurations (ns2).
class 11 chemistry chapter 3 exercise
So they will exhibit similar chemical properties. Their group number = 2. Elements with atomic numbers 16 and 34 belong to the -block and they have similar outer electronic configurations (ns2npA).
So, they will exhibit similar chemical properties. Their group number = (10 + 2 + 4) = 16. So, based on the similarity in chemical properties the given elements are divided into three groups :