Long chains of carbon molecules.
The quantum numbers for the last valence electron in nickel are (3, 2, 0, -½).
The electron configuration of nickel is [Ar]4s²3d⁸
The last electron added is a 3d electron, so <em>n</em> = 3 and <em>l</em> = 2.
We construct a table of quantum numbers.
<u>Element </u><em><u>n</u></em><u> </u><em><u>l</u></em><u> </u><em><u>m</u></em><u>ₗ mₛ </u>
Sc 3 2 2 +½
Ti 3 2 1 +½
V 3 2 0 +½
Cr 3 2 -1 +½
Mn 3 2 -2 +½
Fe 3 2 2 -½
Co 3 2 1 -½
Ni 3 2 0 -½
The quantum numbers for the last electron in nickel are (3, 2, 0, -½).
we are asked to multiply 5.4 mm by 6.02 mm
5.4 mm has 2 significant figures
6.02 mm has 3 significant figures
When multiplying these 2 numbers the answer should have the least number of significant figures
From the 2 numbers the least number of significant figures is 2 , therefore answer should be rounded off to 2 significant numbers
5.4 x 6.02 = 32.5 rounded off is 33
Answer is b. 33 m^2
