Oxidation state of I is (-1) and for CO it is zero. Let's assume that the oxidation state of Fe in Fe(CO)₄I₂<span> (s) is x. For whole compound, the charge is zero.
Sum of oxidation numbers in all elements = Charge of the compound.
Here we have 1Fe , 4CO and 2I
hence we can find the oxidation state as;
x + 4*0 + 2*(-1) = 0
x + 0 - 2 = 0
x = +2
Hence the oxidation state of Fe in product </span>Fe(CO)₄I₂ (s) is +2.
Same as we can find the oxidation state (y) of Fe in Fe(CO)₅(s).
y + 5*0 = 0
y = 0
Since oxidation state of Fe increased from 0 to +2, the oxidized element is Fe in the given reaction.
Answer: 1.9 x 10²⁴ molecules Na
Explanation: To solve for the molecules of Na, we will use the Avogadro's number.
3.2 moles Na x 6.022 x10²³ molecules Na/ 1 mole Nà
= 1.9 x 10²⁴ molecules Na
Answer:
Molecules move freely around since they don't have a definite shape. This is the reason water stays liquid and couldn't be gripped.
An orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy.
Explanation:
The only true statement from the given options is that "an orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy." Inner orbitals which are also known to contain core electrons feels the bulk of the nuclear pull on them compared to the outermost orbitals containing the valence electrons.
- The nuclear pull is the effect of the nucleus pulling and attracting the electrons in orbitals.
- This pull is stronger for inner orbitals and weak on the outer ones.
- The outer orbitals are said to be well shielded from the pull of the nuclear charge.
- Also, based on the quantum theory, electrons in the outer orbitals have higher energies because they occupy orbitals at having higher energy value.
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