(a) The `EN` of cesium `(Cs)` is `0.7` and that of chlorine `(Cl)` is `3.5`. Predict tha bond formed between them.
(b) The `X-X` bond length is `100 pm` and `C-C` bond length is `154 pm`. If `EN` of `'X'` and `'C'` are `3.0` and `2.0` respectively, calculate the `C-X` bond length.
(i) `127 pm` , (ii) `118 pm`
(iii) `108 pm` , (iv) `128 pm`
( c) Which of the properties can be predicted by `EN` values?
(d) If `a,b` and `c` are `EN`, `IE` and `EA` respectively. What is the formula of `EA ( c)` in the terms of `EN` (a) and `IE (b)` ?

Explanation

(a) `EN` difference between `Cs` and `Cl = (3.5 - 0.7) = 2.8`
Thus the bond between the two is ionic or electrovalent sice `(chi_(Cs)-chi_(Cl) gt 1.7`
(b) ii. Atomic radius of `X = ("Bond length")/(2) = (100)/(2) = 50 pm`
Atomic radius of `C = ("Bond length")/(2) = (154)/(2) = 77pm`
`C-X` bond length `= 50 + 77 = 127 pm`
As there is `EN` difference `(chi_(x)-chi_(c ) = 3.0 - 2.0 = 1.0)`, the extent of overlapping of orbitals will be more.
Therefore, the bond length will be slightly less than `127pm` i.e., `118 pm`.
( c) i. Bond energy of a molecule
ii. Polarity of a molecule
iii. Nature of an oxide.
iv. Metallic and non-metallic character of elements.
v. Percentage of ionic or covalent character of a molecule.
(d) `EN = a, IE = b, EA = c`
`EN = (IE+EA)/(2) rArr EA = 2EA - IE`
`rArr c = 2a-b`