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Solve this question ~

Find the Hybridization of N in compound " NH3 " and determine its shape ~ ​

Answer 1

Let's check Electronic configuration of N in ground state

$\\ \sf\longmapsto 1s^22s^22p^3$

• In exited state

$\\ \sf\longmapsto 1s^22s^22px^12py^12pz^2$

• 3 p orbitals

In hydrogen case

$\\ \sf\longmapsto 1s^1$

• 1 s orbital

Hence Hybridization

$\\ \sf\longmapsto s-p-p-p$

$\\ \sf\longmapsto sp^3$

Structure is tetrahedral .

Spare way:-

• Bond pairs==3=>Bonding electrons=6
• Lone pair=1=>Anti bonding electrons=2

Hybridization

$\\ \sf\longmapsto \dfrac{1}{2}[\sigma+\sigma *]$

$\\ \sf\longmapsto \dfrac{1}{2}[6+2]$

$\\ \sf\longmapsto \dfrac{1}{2}[8]$

$\\ \sf\longmapsto 4$

• sp3 Hybridization
• Shape-Tetrahedral .