12th Grade > Chemistry
PERIODIC CLASSIFICATION OF ELEMENTS MCQs
Total Questions : 30
| Page 3 of 3 pages
Answer: Option A. -> F < Cl > Br > I
:
A
In a group of non-metals, electrongain enthalpy of the 2nd element is maximum. Among halogens, chlorine has the highest election affinity.
We would expect fluorine would have the highest electron affinity because of being a halogen and very small size, but its so small that it leads to interelectronic repulsions and the overall energy released when an electron is added is reduced.
:
A
In a group of non-metals, electrongain enthalpy of the 2nd element is maximum. Among halogens, chlorine has the highest election affinity.
We would expect fluorine would have the highest electron affinity because of being a halogen and very small size, but its so small that it leads to interelectronic repulsions and the overall energy released when an electron is added is reduced.
Answer: Option C. -> ns2 np3
:
C
Elements with either half-filled or completely filled sub shells have astable electronic configuration. The elements with stable configurations have less electron affinity or electron gain enthalpy.
ns2np3 ⟶ 15th group (V A group) has half filled p-subshell.
:
C
Elements with either half-filled or completely filled sub shells have astable electronic configuration. The elements with stable configurations have less electron affinity or electron gain enthalpy.
ns2np3 ⟶ 15th group (V A group) has half filled p-subshell.
Answer: Option B. -> Ionization enthalpy: Li
:
B
Beryllium has more ionization enthalpy than boron as it has a completely filled 's' orbital.
Be -1s22s2
B - 1s22s22p1
:
B
Beryllium has more ionization enthalpy than boron as it has a completely filled 's' orbital.
Be -1s22s2
B - 1s22s22p1
Answer: Option A. -> -320 kJ/mole
:
A
Ionization enthalpy is exactly opposite to electron gain enthalpy in this case. If you are talking about ionization enthalpy of F and electron gain enthalpy of F, then we can't say the same thing.
F−⟶ F + e− ΔH=320kJ/mole(I.E)
F+e−⟶ F− ΔH=−320kJ/mole(I.E)
:
A
Ionization enthalpy is exactly opposite to electron gain enthalpy in this case. If you are talking about ionization enthalpy of F and electron gain enthalpy of F, then we can't say the same thing.
F−⟶ F + e− ΔH=320kJ/mole(I.E)
F+e−⟶ F− ΔH=−320kJ/mole(I.E)
Answer: Option D. -> The recurrence of similar outer electronic configuration
:
D
The recurrence of similar outer electronic configuration is the reason for the periodicity of properties.
Although Mendeleev discovered that properties were periodic and created a table enlisting the elements, he didn't know the reason behind this occurrence.
:
D
The recurrence of similar outer electronic configuration is the reason for the periodicity of properties.
Although Mendeleev discovered that properties were periodic and created a table enlisting the elements, he didn't know the reason behind this occurrence.
Answer: Option A. -> 2.53
:
A
△ = EC−H - √EH−H×EC−C
= 98.8 - [(104.2)×(83.1)]12 = 5.75 kcal.
(XC−H) E.N of C-H bond = 0.218 √△ = 0.218 (5.75)12
= 0.43
XC = 0.43 +XH = 0.43 + 2.1 = 2.53
:
A
△ = EC−H - √EH−H×EC−C
= 98.8 - [(104.2)×(83.1)]12 = 5.75 kcal.
(XC−H) E.N of C-H bond = 0.218 √△ = 0.218 (5.75)12
= 0.43
XC = 0.43 +XH = 0.43 + 2.1 = 2.53
Answer: Option B. -> Representative or normal elements
:
B
In the modern periodic table,elements are classified into representative elements, transition elements, inner transition, inner gases based on electronic configuration.
When the group numberof the elements represents the number of valence electrons they arecalledrepresentative elements.
Group - 1 (1A) - 1 valence electrons
2 (IIA) - 2 valence electrons
3 (IIIA) - 3 valence electrons
4 (IVA) - 4 valence electrons
15 (VA) - 5 valence electrons
16 (VIA) - 6 valance electrons
17 (VIIA) - 7 valence electrons
:
B
In the modern periodic table,elements are classified into representative elements, transition elements, inner transition, inner gases based on electronic configuration.
When the group numberof the elements represents the number of valence electrons they arecalledrepresentative elements.
Group - 1 (1A) - 1 valence electrons
2 (IIA) - 2 valence electrons
3 (IIIA) - 3 valence electrons
4 (IVA) - 4 valence electrons
15 (VA) - 5 valence electrons
16 (VIA) - 6 valance electrons
17 (VIIA) - 7 valence electrons
Answer: Option C. -> sp > sp2 > sp3
:
C
With the increase in % of s-character in hybrid orbitals the electronegativity of the hybrid orbitals increases.
sp - 50% s character,
sp2- 33.3 % s character
sp3 - 25% s character.
:
C
With the increase in % of s-character in hybrid orbitals the electronegativity of the hybrid orbitals increases.
sp - 50% s character,
sp2- 33.3 % s character
sp3 - 25% s character.
Answer: Option C. -> Cl + e−= Cl−
:
C
Generally, to determine whether energy is released or absorbed when an electron is added to a neutral gaseous atom, you need to look at the stability of the species.
If we add an electron to Cl atom, energy will be released because Cl− is more stable than Cl atom.
If we add an electron to N atom, energy will be absorbed because N is more stable than N−
In the case of anions, energy is always absorbed when an electron is added.
So, if an electron is added to O− ion a lot of energy is needed to make itO2−
:
C
Generally, to determine whether energy is released or absorbed when an electron is added to a neutral gaseous atom, you need to look at the stability of the species.
If we add an electron to Cl atom, energy will be released because Cl− is more stable than Cl atom.
If we add an electron to N atom, energy will be absorbed because N is more stable than N−
In the case of anions, energy is always absorbed when an electron is added.
So, if an electron is added to O− ion a lot of energy is needed to make itO2−
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