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In Bi, there 3 unpaired electrons because on the periodic table it is in the third column of the p block which has 3 orbitals (according to Hund's rule each orbital must have one electron before you can add a second). In Si, there are 2 unpaired electrons because it is in the second column of the p block. Ni would have 2 unpaired electrons because it has 8 electrons and it is in the d block. This would result in 6 electrons being paired and the last two remaining unpaired.
It helps to draw this out visually for me. So, for the Bi it is in the p block and has 5 valence electrons. You know that the p block has 3 subshells, so draw out the subshells and begin to evenly distribute the electrons to each subshell (Hund's Rule). After doing that you will be able to see that there are two full subshells and one subshell is not full. That one subshell that is not full is the only one that is an unpaired electron which is why the answer is 1. You can follow these steps for the other elements. Hope this helped!
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