Chemistry `color{red} ✎` PROBLEM SOLVING TECHNIQUES

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

 
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