Question

Identify the hybridization of the c atom in ch2br2.

Answer 1

The hybridization of ${\text{C}}{{\text{H}}_{\text{2}}}{\text{B}}{{\text{r}}_{\text{2}}}$ is $\boxed{{\mathbf{s}}{{\mathbf{p}}^{\mathbf{3}}}}$ .

Further Explanation:

Prediction of hybridization:

The hybridization can be determined by calculating the number of hybrid orbitals (X) which is to be formed by the atom. The formula to calculate the number of hybrid orbitals (X) as follows:

$\boxed{{\text{X = }}\frac{1}{2}\left[{{\text{VE}}+{\text{MA}} - c + a}\right]}$

Where,

•VE is a total number of valence electrons of the central atom.

•MA is total number of monovalent atoms/groups surrounding the central atom.

•c is the charge on the cation if the given species is a polyatomic cation.

•a is the charge on the anion if the given species is a polyatomic anion.

Note: In MA only consider monovalent species should be considered and for divalent atoms or groups, MA is equal to zero.

Generally, the least electronegative atom is considered as the central atom. Calculate hybridization as follows:

1. If the value of X is 2 then it means two hybrid orbitals are to be formed and thusthe hybridization is sp.

2. If the value of X is 3 then it means three hybrid orbitals are to be formed and thusthe hybridization is $s{p^2}$.

3. If the value of X is 4 then it means four hybrid orbitals are to be formed and thus the hybridization is $s{p^3}$.

4. If the value of X is 5 then it means five hybrid orbitals are to be formed and thus the hybridization is $s{p^3}d$.

5. If the value of X is 6 then it means six hybrid orbitals are to be formed and thus the hybridization is $s{p^3}{d^2}$.

The ground state electronic configuration for carbon is,

$\left[{{\text{He}}}\right]{\text{ }}2{s^2}2{p^2}$

Therefore, the valence electrons of carbon (C) atom are 4.

Since bromine and hydrogen are monovalent atoms and thus the total number of monovalent atoms surrounding the central atom (MA) is 4.

Since the molecule is a neutral species thus the values of a and c are zero.

Substitute these values in the above formula as follows:

$\begin{aligned}&{\text{X=}}\frac{1}{2}\left[{{\text{VE}}+{\text{MA}} - c + a}\right]\\&=\frac{1}{2}\left[{{\text{4}}+{\text{4}} - 0 + 0}\right]\\&=\frac{1}{2}\left[ 8 \right]\\&=\boxed4\\\end{aligned}$

Since the value of X is 4, it means 4 hybrid orbitals are to be formed and therefore the hybridization of ${\mathbf{C}}{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{B}}{{\mathbf{r}}_{\mathbf{2}}}$ is ${\mathbf{s}}{{\mathbf{p}}^{\mathbf{3}}}$.

Learn more:

1. Calculation of mass of CO: brainly.com/question/6367198

2. Calculation of mass of oxygen in grams: brainly.com/question/8175791

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Covalent bonding and molecular structure

Keywords: hybridization, carbon, geometry, sp3, steric number, 4, p orbital, ch2br2-, Lewis structure, hybridization of ch2br2-, central atom, bromine, shape, total valence electron.

Answer 2

Hello!

The hybridization of the C atom in CH₂Br₂ is sp3

When bonding, the orbitals "s" and "p" from C atoms interact to form hybridized orbitals. If the C atom has 4 sigma bonds, as is the case in CH₂Br₂, there are 4 hybridized orbitals required, so 1 "s" orbital and 3 "p" orbitals hybridize to form an sp3 hybrid orbital. This orbital has tetrahedral geometry and the bond angle is 109,5 °.

Have a nice day!