Write the condensed ground-state electron configurations of these transition metal ions, and state which are paramagnetic:
1 answer:
Answer and Explanation :
Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.
(a) V³⁺
The electronic configuration is -
The electrons in 3d orbital = 1 (Unpaired)
Thus, the ion is paramagnetic as the electrons are unpaired.
(b) Cd²⁺
The electronic configuration is -
The electrons in 4d orbital = 10 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
(c) Co³⁺
The electronic configuration is -
The electrons in 3d orbital = 6 (Unpaired)
Thus, the ion is paramagnetic as the electrons are unpaired.
(d) Ag⁺
The electronic configuration is -
The electrons in 4d orbital = 10 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
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Answer:
64 molecules
Explanation:
we have the reaction
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
That is a balanced equation
We can see here that we need 5 molecules of O2 for every 4 molecules of NH3
we have in the beginning 10 molecules of O2 and 52 molecules of NH3
we can say:
if 4 molecules of NH3 need 5 molecules of O2 how many molecules are needed for 52 molecules of NH3
?
4 molecules of NH3 →5 molecules of O2
52 molecules of NH3 →X molecules of O2
Since we just have 10 molecules of O2 it is not enough so this is the limiting reagent and all the calculations have to be done base on it.
We use simple rule of three to get all the molecules.
5 molecules of O_2 → 4 molecules of NH_3
10 molecules of O_2 →X molecules of NH_3
Which means that we use all the O2 but just 8 molecules of NH3 and we have 44 less.
NO produced
5 molecules of O_2 → 4 molecules of NO
10 molecules of O_2 → X molecules of NO
H2O produced
5 molecules of O_2 → 6 molecules of H_2 O
10 molecules of O_2 → X molecules ofH_2 O
Total molecules
44 molecules of NH3 + 8 molecules of NO + 12 molecules of H2O = 64 molecules.