Question

The ${\mathbf{sp}}^{\mathbf{2}}$hybrid atomic orbitals have the following general form:

role="math" localid="1663746599823" $$\begin{gathered} {\mathit{\varphi }}_{\mathbf{1}}\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{3}}}{\mathit{\varphi }}_{\mathbf{s}}\mathbf{+}\mathbf{2}{\mathbf{A\varphi }}_{\mathbf{px}} \\ {\mathbf{\varphi }}_{\mathbf{2}}\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{3}}}{\mathbf{\varphi }}_{\mathbf{s}}\mathbf{-}\mathbf{2}{\mathbf{A\varphi }}_{\mathbf{px}}\mathbf{+}{\mathbf{B\varphi }}_{\mathbf{py}} \\ {\mathbf{\varphi }}_{\mathbf{3}}\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{3}}}{\mathbf{\varphi }}_{\mathbf{s}}\mathbf{-}\mathbf{2}{\mathbf{A\varphi }}_{\mathbf{px}}\mathbf{-}{\mathbf{B\varphi }}_{\mathbf{py}} \end{gathered}$$

where, ${\mathbf{\Phi }}_{\mathbf{s}}\mathbf{,}{\mathbf{\Phi }}_{\mathbf{px}}$androle="math" localid="1663746751426" ${\mathbf{\Phi }}_{\mathbf{py}}$represent orthonormal (normalized and orthogonalized) atomic orbitals. Calculate the values of Aand B.

Short answer

$A=\frac{1}{\sqrt{6}}$and$B=\frac{1}{2}$

The probability of finding a particle is one.

Step-by-step solution

Find A

Calculate the value of A as follows.

$\begin{array}{c}{\left(\frac{1}{\sqrt{3}}\right)}^2+4A^2=1\\ A^2=\frac{2}{12}\\ A=\frac{1}{\sqrt{6}}\end{array}$

Find B

Calculate the value of B as follows.

$\begin{array}{c}{\left(\frac{1}{\sqrt{3}}\right)}^2+A^2+B^2=1\\ B=1-\frac{1}{6}-\frac{1}{3}\\ B=\frac{1}{2}\end{array}$