Unless you're able to provide a diagram representing the problem, there's not much I can do so solve this problem.
The Correct Answer Is 3.2
Answer:
a) Mo the electron configuration: 42Mo: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d4
Mo3+ - is Paramagnetic
b) Au - [Xe] 4f14 5d10 6s1
For Au+ is not paramagnetic
c) Mn - [Ar] 3d5 4s2
Mn2+ is paramagnetic
d)Hf -[Xe] 4f¹⁴ 5d² 6s²
Hf2+ is not paramagnetic
Explanation:
An atom becomes positively charged when it looses an electron.
Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.
Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.
Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.
Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.
<u>Given:</u>
Change in internal energy = ΔU = -5084.1 kJ
Change in enthalpy = ΔH = -5074.3 kJ
<u>To determine:</u>
The work done, W
<u>Explanation:</u>
Based on the first law of thermodynamics,
ΔH = ΔU + PΔV
the work done by a gas is given as:
W = -PΔV
Therefore:
ΔH = ΔU - W
W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ
Ans: Work done is -9.8 kJ
