First slide

Hybridization

Question

Number of $dπ - pπ$ bonds in Carbonate ion is ‘X’. Number of lone pairs in Carbonate ion is ‘Y’. Difference between X and Y is

Moderate

A

5
B

3
C

6
D

4

Solution

$Number of electron pairs (EP) = \frac{V + M - C + A}{2}; V = Group number of central atom; M = - H, - X, - OH . . . . monovalent groups attached to central atom; C = Charge of cation; A = Charge of anion . . . 1 in {NO}_{3}^{-}, 1 in {NO}_{2}^{-}, 1 in {ClO}_{4}^{-}, 1 in {ClO}_{3}^{-}, 1 in {ClO}_{2}^{-}, 1 in {ClO}^{-}, 2 in {CO}_{3}^{2 -}; . . . . 2 in {SO}_{3}^{2 -}, 2 in {SO}_{4}^{2 -}, 3 in {PO}_{4}^{3 -},; V = 2, 3.4, 5, 6, 7, 8 respectively for 2^{nd}, 13^{th}, 14^{th}, 15^{th}, 16^{th}, 17^{th}, 18^{th} group elements; Number bond pairs, BP (according to VSEPR theory) = Number of atoms directly attached to central atom; = 3 for {CO}_{3}^{2 -} * * * According to VSEPR theory, double bond and triple bond are counted as one bond pair only; If the bonded atom is Oxygen . . . . . Magnitude of Oxidation state of central atom = Total bonds; Number of atoms directly attached to central atom = sigma bonds; Total bonds - sigma bonds = pi bonds; Number lone pairs on central atom (LP) = EP - BP; = 0 in {CO}_{3}^{2 -}; {CO}_{3}^{2 -} -' 3' bp +' 0' lp on central atom; {sp}^{2} - Trigonal planar; . . . . Orientation of electron pairs . . . . Trigonal; . . . . Total 8 lone pairs / molecule; Note; Each double bonded oxygen atom has' 2' lone pairs on it; Each single bonded oxygen atom with a negative charge has three lone pairs on it; {CO}_{3}^{2 -} : Magnitude of Oxidation state of central atom = Total bonds = 4; Number of atoms directly attached to central atom = sigma bonds = 3; Total bonds - sigma bonds = pi bonds =' 1' . . . . . 1' pπ - pπ;$

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