Pi3cl2 is a nonpolar molecule. based on this information, determine the i−p−i bond angle, the cl−p−cl bond angle, and the i−p−cl bond angle. enter the number of degrees of the i−p−i, cl−p−cl, and i−p−cl bond angles, separated by commas (e. g., 30,45,90)

120, 180,90

Explanation:

According to VSEPR theory with a central atom with 5 gropus, we will expect a trigonal bipyramidal structure. This structure look like a trgonal plane containing the three iodine atoms, and above and below this plane we have the two bromine atoms 180º apart. Why can we say this? The key is that the problem is telling us the PIₐCl₂ is a nonpolar molecule and this shape is the only one that will result in a nonpolar structure. The Chlorine atoms will be in the axial positions 180 degrees apart and the bond angles will be :

I-P- I       120º

Cl-P-Cl   180º

I-P-Cl      90 º

See attachment.

The two Cl atoms are in the axial positions, and the three I atoms are in what we call equatorial position in the trigonal plane.

Any other arrangement will result in a polar molecule. Place a Cl atom in the trigonal plane and I in the axial position, and you can see that the molecule will be polar because the positions will no longer be equivalent and the dipoles will not cancel each other.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally