(solved)Question 3.32 of NCERT Class XI Chemistry Chapter 3

Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements. (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorus and fluorine (f) Element 71 and fluorine

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Parveen,

Question 3.32
NCERT Class XI Chemistry

Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements. (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorus and fluorine (f) Element 71 and fluorine

Solution in Detail
(video solution below this)

$\displaystyle \underline{\underline{\text{Part (a) - Li and O}}}$

Li is $\displaystyle \text{[He]}2s^1$ G1 valence of 1

O is $\displaystyle \text{[He]}2s^22p^4$ G16 valence of 2

$\displaystyle Li_2O \text{ (ionic)}\:\underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (b) - Mg and N}}}$

Mg is $\displaystyle \text{[Ne]}2s^2$ G1 valence of 2

N is $\displaystyle \text{[He]}2s^22p^3$ G15 valence of 3

$\displaystyle Mg_3N_2 \text{ (covalent)}\:\underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (c) - Al and I}}}$

Aluminum has valency of 3, and iodine is a halogen of valency 1

$\displaystyle AlI_3 \text{ (covalent)}\:\underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (d) - Si and O}}}$

Si valency 4, oxygen 2

$\displaystyle SiO_2 \text{ (covalent)}\:\underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (e) - P and F}}}$

P valency 3, F has 1

$\displaystyle PF_3 \text{ (covalent)}\:\underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (f) - 71 and F}}}$

Z = 71 config is $\displaystyle [Xe]6s^24f^{14}5d^{1}$ lutetium

hence, valence of Z = 71 is ($\displaystyle 6s^2$ + $\displaystyle 5d^1$) = 3

$\displaystyle LuF_3 \text{ (covalent)}\:\underline{Ans}$

Video Explanation

Please watch this youtube video for a quick explanation of the solution:

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