Chemistry color{red} ✎ PROBLEM SOLVING TECHNIQUES
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### Topics covered

• Probelms based on predicting period,group and block of elements
• Problems based on ionization enthaply
• Problems based on electron gain enthalpy
•Problems based on valence and chemical reactivity

### PST -1 Question based on predicting period , group and block of elements.

The group of an element is predicted from the number of electrons in the valence shell or /and penultimate shell (last but one i.e. n-1 ) as follows :

color{red}☛ color{green} (" For s- block elements") , group number is equal to the  color{green} ("number of valence electrons.")

color{red}☛ color{green} (" For p- block elements") , group number is equal to  color{green} ("10+ number of electrons in the valence shell.")

color{red}☛ color{green} (" For d- block elements") : group number is equal to the number of electrons in  color{green} ("(n-1) d subshell + number of electrons in valence shell (nth shell) ")

Q 3157101084

Predict the position of the element in the periodic table satisfying the electronic configuration (n-1) d^1 ns^2 for n = 4.

Solution:

For n = 4 , the electronic configuration = (4-1)d^1 4s^2 = 3d^1 4s^2. The element corresponding to this configuration is scandium which is a d- block element .
Group number = No. of electrons in (n-1)d sub shell + No. of electrons in nth shell = 1+2 = 3
Q 3157201184

Elements A , B , C , D and E have the following electronic configurations.

A : 1s^2 , 2s^2 , 2p^1

B : 1 s^2 , 2s^2 , 2p^6 , 3s^2 , 3p^1

C : 1S^2 , 2s^2 , 2p^6 , 3s^2 , 3p^3

D : 1s^2 , 2s^2 , 2p^6 , 3s^2 , 3p^5

E : 1s^2 , 2s^2 , 2p^6 , 3s^2 , 3p^6 , 4s^2

Which among these will belong to the same group in the periodic table ?

Solution:

We know that elements having similar valence electronic configuration belong to the same group of the periodic table. Therefore , elements A and B having three electrons in the valence shell , i.e. 2s^2 , 2p^1 and 3s^2 , 3p^1 respectively belong to the same group , i.e. group 13 of the periodic table.

### PST- 2 Question based ionization enthalpy

Q 3117501480

The first Delta_iH_1 and the second (Delta_i H_2) ionization enthalpies (kj mol^(-1)) of a few elements designated by roman numericals are shown below :

tt (( "Element" , IE_1 , IE_2) , (I , 2372 , 5251) , (II, 520 , 7300) , (III , 900 , 1760) , (IV , 1680 , 3380))

Which of the above elements is likely to be
(a) a reactive metal . (b) a reactive non- metal.
(c) a noble gas. (d) a metal that forms a stable binary halide of the formula AX_2 = ( X = text(halogen)).

Solution:

(a) Since element II has a very low Delta_iH_1 but a very high Delta_iH_2 , therefore , it has only one electron in the valence shell and hence is likely to be a reactive metal ( i.e. an alkali metal).

(b) Since the Delta_iH_1 of element IV is very high and its Delta_iH_2 is not so high ( actually almost double ) , it is likely to be a reactive non-metal ( i.e. a halogen).

(c) Among the elements listed , Delta_iH_1 of element I is the highest and its Delta_iH_2 is also not so high therefore it must be a noble gas.

(d) The Delta_iH_1 of element III is higher than that of element II , but unlike element II its Delta_iH_2 is only about twice its Delta_iH_1 , therefore it is likely that element III has two electrons in the valence shell ( i.e. alkaline earth metal) . As such it will form a stable binary halide of the formula AX_2 where A is the metal and X is the halogen .

### PST-3 Question based on electron gain enthalpy

Q 3147501483

Which of the following pairs of elements would have more negative electron gain enthalpy ? explain.

(i) N or O
(ii) F or Cl
(iii) S or O
(iv) C or Si

Solution:

(a) The electron gain enthalpy of O is highly negative while that of N is slightly positive .

(b) Cl has more negative electron gain enthalpy than F

(c) S has more negative electron gain enthalpy than O

(d) C has more negative electron gain enthalpy than Si.

### PST-4 Question based on valence and chemical reactivity

Q 3187701687

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

(a) Silicon and oxygen. (b) aluminium and bromine . (c) calcium and iodine . (d) element 114 and fluorine. (e) element 120 and oxygen.

Solution:

(a) Silicon belongs to group 14. It has four valence electrons and hence its valence = 4 . Oxygen belongs to group 16. It has six electrons in the valence shell and hence its valence = 8-6 = 2 . Therefore , the formula of silicon oxide = SiO_2.

(b) Aluminium belongs to group 13 . It has three electrons in the valence shell and hence its valence = 3.
Bromine belongs to group 17. It has seven electrons in the valence shell and hence its valence = 8-7 = 1
Therefore , formula of aluminium bromide = AlBr_3.

(c) Calcium belongs to group 2. It has two electrons in the valence shell and hence its valence = 2. Iodine belongs to group 17. It has seven valence electrons and its valence = 8-7 = 1.
Therefore formula of calcium iodide = CaI_2
(d) In the 7th period , the 4th transition series ends at element with Z = 112 , therefore , filling of 7p- subshell will begin with element with Z = 113. Hence element with Z = 114 . will have two electrons in 7p - orbitals and two electrons in 7 s- orbital. In other words , the element Z = 114. will lie in group 14 and hence its valence = 4.
Fluorine belongs to group 17 with seven valence electrons and hence its valence = 8-7 = 1
Therefore , formula of compound formed by element 114 and fluorine is MF_4 where M represents the element.

(e) The 7th period will end at element with Z = 118. Therefore , filling of 8th shell i.e. 8s - orbital will begin with element Z = 119. and consequently element with Z = 120 will belongs to group 2 and its valence will be = 2.
Therefore , formula of compound formed by element Z = 120 and oxygen will be MO where M represents the element