Chapter 3 Classification of Elements and Periodicity in Properties Solutions
Question - 31 : - The first (ΔiH1) and the second (ΔiH) ionization enthalpies(in kJ mol–1) and the (ΔegH) 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 |
Which of the aboveelements is likely to be :
(a) the least reactiveelement.
(b) the most reactivemetal.
(c) the most reactivenon-metal.
(d) the least reactivenon-metal.
(e) the metal which can form a stable binaryhalide of the formula MX2, (X=halogen).
(f)the metal which can form a predominantly stable covalent halide of the formulaMX (X=halogen)?
Answer - 31 : -
(a) ElementV is likely to be the least reactive element. This is because it has thehighest first ionization enthalpy (ΔiH1) and a positive electron gain enthalpy (ΔegH).
(b) Element II islikely to be the most reactive metal as it has the lowest first ionizationenthalpy (ΔiH1) and a low negative electron gain enthalpy (ΔegH).
(c) Element III islikely to be the most reactive non–metal as it has a high first ionizationenthalpy (ΔiH1) and the highest negative electron gain e(a) ElementV is likely to be the least reactive element. This is because it has thehighest first ionization enthalpy (ΔiH1) and a positive electron gain enthalpy (ΔegH).
(b) Element II islikely to be the most reactive metal as it has the lowest first ionizationenthalpy (ΔiH1) and a low negative electron gain enthalpy (ΔegH).
(c) Element III islikely to be the most reactive non–metal as it has a high first ionizationenthalpy (ΔiH1) and the highest negative electron gain enthalpy (ΔegH).
(d) Element V islikely to be the least reactive non–metal since it has a very high firstionization enthalpy (ΔiH2) and a positive electron gain enthalpy (ΔegH).
(e) Element VI has alow negative electron gain enthalpy (ΔegH). Thus, it is ametal. Further, it has the lowest second ionization enthalpy (ΔiH2). Hence, it can form a stable binary halide of theformula MX2 (X=halogen).
(f) Element V has the highest first ionization energyand high second ionization energy. Therefore, it can form a predominantlystable covalent halide of the formula MX (X=halogen).
nthalpy (ΔegH).
(d) Element V islikely to be the least reactive non–metal since it has a very high firstionization enthalpy (ΔiH2) and a positive electron gain enthalpy (ΔegH).
(e) Element VI has alow negative electron gain enthalpy (ΔegH). Thus, it is ametal. Further, it has the lowest second ionization enthalpy (ΔiH2). Hence, it can form a stable binary halide of theformula MX2 (X=halogen).
(f) Element V has the highest first ionization energyand high second ionization energy. Therefore, it can form a predominantlystable covalent halide of the formula MX (X=halogen).
Question - 32 : - Predict the formula of the stable binary compounds thatwould be formed by the combination of the following pairs of elements.
(a) Lithiumandoxygen (b) Magnesium and nitrogen
(c)Aluminium and iodine (d) Silicon and oxygen
(e) Phosphorus and fluorine (f) Element 71 and fluorine
Answer - 32 : -
(a) Li2O
(b) Mg3N2
(c) AlI3
(d) SiO2
(e) PF3 orPF5
(f) The element with the atomic number 71 is Lutetium(Lu). It has valency 3. Hence, the formula of the compound is LuF3.
Question - 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 - 33 : -
(c) The period in Modern periodic table indicates the value of ‘n’ i.e. a principal quantum number.
Question - 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 the value of an azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
Answer - 34 : -
(b) The d-block has 8 columns because a maximum of 8 electrons can occupy all the orbitals in a d-subshell
Question - 35 : - “ Anything that influences the valence electrons will affect the chemistry of the element”. Which of the factors given below is not affecting the valence shell?
(a) Valence Principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons
Answer - 35 : -
(c) Nuclear mass
Question - 36 : - The size of isoelectronic species — F–, Ne and Na+ is affected by
(a) nuclear charge (Z )
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitals
(d) none of the factors because their size is the same.
Answer - 36 : -
The size of an isoelectronic speciesincreases with a decrease in the nuclear charge (Z). For example, the order of theincreasing nuclear charge of F–,Ne, and Na+ is as follows:
F– < Ne < Na+
Z 9 10 11
Therefore, the order of the increasing size of F–, Ne and Na+ is as follows:
Na+ < Ne < F–
Question - 37 : - Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on the removal of an electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of an electron from orbitals bearing lower n value is easier than from orbital having higher n value
Answer - 37 : -
(d) is the incorrect statement
For orbitals having a lower value of ‘n’ the removal of an electron is easy compared to the orbitals having a higher value of ‘n’.
Because the electrons of orbitals having a lower value of ‘n’ are highly attracted to the nucleus than that of the electrons of orbitals having a higher value of ‘n’
Question - 38 : - Considering the elements B, Al, Mg, and K, the correct order of their metallic
character is :
(a) B > Al > Mg > K (b) Al > Mg > B > K
(c) Mg > Al > K > B (d) K > Mg > Al > B
Answer - 38 : -
(d) K > Mg > Al > B
Reason:
As we move from left and right in a period the metallic character of the elements decreases. Thus, Mg > Al.
As we move down from a group the metallic character of the elements decreases. Thus, Al > B.
From the above two statements, it can be stated that K > Mg.
Thus, K > Mg > Al > B
Question - 39 : - Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :
- B > C > Si > N > F
- Si > C > B > N > F
- F > N > C > B > Si
- F > N > C > Si > B
Answer - 39 : -
(3) F > N > C > B > Si
Reason:
As we move from left and right in a period the non-metallic characteristic of the elements decreases. Thus, F > N > C > B.
As we move down from a group the metallic character of the elements decreases. Thus, C > Si.
From the above two statements, it can be stated that B > Si.
Thus, F > N > C > B > Si
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Answer - 40 : -
(b) F > O > Cl > N
Reason:
As we move from left and right in a period the non-metallic characteristic of the elements increases. Thus, F > O > N.