What is the electron configuration the following?
3. Arsenic
4. Lead

3. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3

4. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 5d10 4f14 6p2

Explanation:

Divide the periodic tables into blocks to find electron configurations.

Groups 1-2 are the s block.

Groups 13-18 are the p block.

Groups 3-12 are the d block (excluding Lanthanides and Actinides).

The Lanthanides and Actinides are the f block.

To find the electron configuration of arsenic, move across the periodic table starting from the top left corner of the periodic table. H and He (which is actually part of the s block) are 1s2 (period, followed by block, followed by how many elements over). Move to the next period (2). Li and Be are 2s1 and 2s2 respectively, but we would only write 2s2 in our configuration (the last/rightmost of the block). Then we have 2p6, which passes B, C, N, O, F, and Ne. We continue this process: 3s2, 3p6, and 4s2. Now we reach the d block (Groups 3-12). The period of these elements will be written 1 less (for example, Scandium would not be ...4d1; instead, it would be ...3d1). So we continue: 3d10, and then back onto the 4th period in the p block. Since arsenic is 3 elements over in the p block, we will finish the configuration with 4p3.

As for lead, you will have to pass the lanthanides, which are f block. They are similar to the d block, except you write the period as 2 less (for example, Cerium would be ...4f1, rather than 6f1). The elements lanthanum and actinium are not in the f block; they are in the d block.

One more thing to note are noble gas configurations. Noble gas configurations use the Group 18 gases (noble gases) to simplify our configurations. Let's take arsenic as an example again: since the noble gas before arsenic is argon, Ar (the element in group 18 in the previous period), we can write argon in brackets, [Ar], and then write only the configuration of the period arsenic is in: 4s2 3d10 4p3. As a result, our noble gas configuration for arsenic would look like this: [Ar] 4s2 3d10 4p3.

The noble gas configuration for lead is [Xe] 6s2 5d10 4f14 6p2.

Sorry for the long response, but I wanted to clarify the method to approach these types of problems for the future. I hope you at least get a general idea of how to solve these problems. :)

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o