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QUESTION IMAGE

here is a sketch of a 1s orbital: image of 1s orbital with x, y, z axes…

Question

here is a sketch of a 1s orbital:

image of 1s orbital with x, y, z axes and a slider

this sketch is about 200 pm wide.
the coordinate (x, y, and z) axes are also shown.
you can rotate the sketch for a better view of the orbital by dragging the slider with your mouse.

suppose an atom with its nucleus at the origin has an electron in a 1s orbital. complete each row of the table below by deciding how ( p_a ), the probability of finding the electron at point ( a ), compares to ( p_b ), the probability of finding the electron at point ( b ).

point ( a )point ( b )compare ( p_a ) to ( p_b )
10 pm below the nucleus along the (-z) axis.10 pm in front of the nucleus, along the (+y) axis.( circ ) ( p_a < p_b ) <br> ( circ ) ( p_a = p_b ) <br> ( circ ) ( p_a > p_b )
10 pm to the right of the nucleus, along the (+x) axis.10 pm behind the nucleus, along the (-y) axis.( circ ) ( p_a < p_b ) <br> ( circ ) ( p_a = p_b ) <br> ( circ ) ( p_a > p_b )

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Explanation:

A 1s orbital is spherically symmetric, meaning the probability of finding an electron at a given distance from the nucleus depends only on the radial distance (the distance from the nucleus), not on the direction (x, y, or z axis).

Row 1:

Step 1: Determine radial distances

Point A: 10 pm from nucleus (along -x axis).
Point B: 10 pm from nucleus (along +z axis).

Step 2: Compare probabilities

Since both points are 10 pm from the nucleus, and 1s orbital is spherically symmetric, \( P_A = P_B \).

Row 2:

Step 1: Determine radial distances

Point A: 10 pm from nucleus (along -z axis).
Point B: 10 pm from nucleus (along -y axis).

Step 2: Compare probabilities

Both points are 10 pm from the nucleus. Spherical symmetry implies \( P_A = P_B \).

Row 3:

Step 1: Determine radial distances

Point A: 10 pm from nucleus (along +x axis).
Point B: 10 pm from nucleus (along +y axis).

Step 2: Compare probabilities

Both points are 10 pm from the nucleus. Spherical symmetry gives \( P_A = P_B \).

Answer:

point Apoint Bcompare \( P_A \) to \( P_B \)
10 pm below the nucleus along the -z axis.10 pm in front of the nucleus, along the -y axis.\( P_A = P_B \)
10 pm to the right of the nucleus, along the +x axis.10 pm behind the nucleus, along the +y axis.\( P_A = P_B \)