Question

What is the hybridization of the positively charged carbon in (ch3)3c+?

Answer 1

Answer : The hybridization of the positively charged carbon is $sp^{2}$.

Explanation :  Formula used for Hybridization

Z=1/2{number of valence electrons of central atom + number of monovalent atom attached to it + negative charge - positive charge}

where, Z = Hybridization

Now for the compound $(CH_{3} )_3C^+$

Number of valence electrons of central atom (C) = 4

Number of monovalent atom attached = 3

Number of negative charge = 0

Number of positive charge = 1

Now putting all the values in Z,

             Z = $\frac{1}{2} [4+3-1]$

                =  3

The corresponding Hybridization is $sp^{2}$ i.e Trigonal planar .

Answer 2

The positively charged of hybridization found in carbon from the compound (CH3)3C+ is 3, which is a Trigonal planar $(sp^2)$

• We can get the hybridization with this formula

• $Z=0.5$ [ (The attached monovalent atom) + (the valence electrons found in central atom) - ( positive charge) +( negative charge) ]

(valence electrons found in central atom) = 4

(Negative charge) =0

( positive charge) =1

The given compound is $(CH _{3} )3C+$

the valence electrons found in central atom=3

And Z= hybridization

• If we substitute into the formula we have,

$Z= 0.5 [ ( 3+4+0-1)$

=3

Therefore, it is a Trigonal planar.

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