Important Question for class 12 chemistry Academic team of entrancei prepared selected questions for class 12 chemistry download your free copy of class 12 chemistry questions from entrancei . do find NCERT solutions for class 12 chemistry
Prob 1.What is the significance of the terms isolated gaseous atom and ground state, while defining the ionization enthalpy and electron gain enthalpy? Sol:
The term isolated gaseous atom means that the atom should be free from other atoms in the gaseous state. No energy should be required to separate it further from other atoms. Whereas ground state means that the lowest energy state is possible for that atom. These two terms are used for comparison purpose.
Prob 2.Among the elements with atomic number 9,12 and 36, identify by atomic number of an element which is. a) highly electronegative b) highly electro positive c) an inert gas and give reasons for your choice. Sol:
The electronic configuration of the elements with atomic numbrs 9,12,36 are i) Atomic number 9 Þ 1s 2s 2p Þ Electro negative because this element can accept one more electron to aquire stable configuration. ii) Atomic number 12 Þ 1s 2 2s 2 2p6 3s 2 Þ Electro positive, because this element loses two electrons to have inert gas configuration. iii) Atomic number 36 Þ 1s 2s 2p 3s 3 p 3d 4s 4 p Þ Inert gas because this element has 8 electron in the outer most shell, hence it has no tendency either to lose or accept electrons. 2
2
5
2
2
6
Prob 3.The formation of F (g ) from (g) from O(g) is endothermic. Why? -
Sol:
i) ii)
exothermic F + e - ¾¾¾¾¾¾¾¾¾¾ ®F(g ) (g ) Energy is released
F (g )
2
6
10
2
6
is exothermic where as that of
O 2-
(monovalent anion)
exothermic +e O( g ) + e - ¾¾¾¾¾¾¾¾¾¾ ® O - ¾¾¾¾¾¾¾¾¾¾ ¾ ® O2 (g ) (g ) Energy is released , endothermic (energy is required )
Prob 4.Give the formula of species that will be isoelectronic with the following atoms or ions : i) F ii) Ar iii) K d) Sr -
Entrancei
+
2+
Sol:
i) Species iso electronic to F (10 es) are N 3- ,O 2- , Ne, Na + , Mg 2+ , Al 3+ ,etc. ii) Species isoelectronic to Ar (18 es) are P 3- ,S 2- ,Cl - , K + ,Ca2+ , etc. iii) Species isoelectronic to K (18 es) are P 3 - , s 2 - ,Cl - , K + ,Ca2+ ,etc. iv) Species isoelectronic to Sr (36 es) are Br - , Kr , Rb + ,etc. -
+
2+
Prob 5.Calculate the electronegativity of carbon from the following data, EN = 104.2 k cal, EN = 83 .1 kcal EN = 98.8 k Cal ; c = 2.1 H -H
C -H
Sol:
C -C
H
According to Pauling equation 1 é ù êE AB - 2 [E AA + EBB ]ú ë û
c A - c B = 0.208
1/ 2
1
é
cC - c H = 0.208 êEC - H ë
2 1 ( EC -C + EH -H )ùú 2 û
1
é
cC - 2.1 = 0.208 ê98.8, ë
2 1 ( 83.1 + 104.2 )ùú 2 û
cC - 2.1 = 0.208 ( 5.15 )
1/ 2
cC - 2.1 = 0.208 ´ 2.269
cC - 2.1 = 0.4719 \ cC = 2.5719
Thus, electronegativity of carbon is 2.5719 Prob 6.On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements. Sol:
The sixth period (n=6) of periodic table involves the filling of 6s, 4f, 5d and 6p subshell in the increasing order of energy. The total number of orbitals available are 16 and therefore, the maximum numbe of electrons that can be accommodated is 32. Thus, the 6th period of periodic table should have 32 elements.
Entrancei