*An element has three naturally occuring isotopes with the following masses and abundances Isotopic mass(amu):
27.977
28.976
29. 974
Fractional abundance:
0.9221
0.0470
0.0309

a. calculate the atomic weight of this element
b. What is the identity of the element?​

a. 28.08 amu

b. Silicon

Explanation:

Step 1: Given data

Isotope         Isotopic mass (mi)         Fractional abundance (abi)

 ²⁸X               27.977 amu                   0.9221

 ²⁹X               28.976 amu                  0.0470

 ³⁰X               29. 974 amu                 0.0309

Step 2: Calculate the atomic weight (aw) of this element

We will use the following expression.

aw = ∑mi × abi

aw = 27.977 amu × 0.9221 + 28.976 amu × 0.0470 + 29. 974 amu × 0.0309

aw = 28.08 amu

Looking at the periodic table, the element with atomic weight = 28.08 amu is silicon.

"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."