Answer:
There are no unpaired electrons.
Explanation:
There are no unpaired electrons in the Lewis symbol for a nitride ion(
).The nitride ion has a charge of -3. The negative charge on the Nitride ion indicates a gain in electrons . Nitrogen has 5 valence electrons that is the number of electrons that are in its outer shell .The total number of electrons that the nitride ion has is equal to 5+3 = 8 electrons . Electrons usually appear in pairs and obey the octet rule therefore the nitride ion has four electron pairs no unpaired electrons.
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Explanation:
The given cation:
(Rf₂Al₂F₃)³⁺
The oxidation number gives the extent to which a specie is oxidized in a reaction.
This number is assigned based on some rules:
- Elements in combined state whose atoms combines with themselves have an oxidation number of zero.
- The charge carried on simple ions gives their oxidation number.
- Algebraic sum of all the oxidation numbers of atoms in neutral compound is zero. In an ion with more than one kind of atom, the charge on it is the oxidation number.
for the specie given;
Known:
oxidation number of Al = +3
F = -1
charge = +3
let the oxidation number of Rf = k
2k + 2(3) + 3(-1) = +3
2k + 6 - 3 = 3
2k = 0
k = 0
The oxidation state of rutherfodium is 0
learn more:
Oxidation state brainly.com/question/10017129
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Lyman Series Working Formula:
1/λ = RH (1-(1/n^2))
Given:
n = 6
RH = Rydberg's constant = 1.0968x10^7 m^-1
c = speed of light = 3x10^8 m/s
Required:
Frequency (Hertz or cycles per second)
Solution:
To solve for the wavelength λ, we substitute the given in the working formula
1/λ = RH (1-(1/n^2))
1/λ = 1.0968x10^7 m^-1 (1-(1/6^2))
λ = 0.0000000938 m or 93.8 nm
To get the frequency, we will use the formula below.
f = c/λ
We then substitute c or the speed of light,
f = (3x10^8 m/s) / 0.0000000938 m
Therefore,
f = 3.2x10^15 s^-1
<em>ANSWER: Frequency = </em><em /><em>3.2x10^15 s^-1</em>
Answer:
<h2>
The equilibrium constant Kc for this reaction is 19.4760</h2>
Explanation:
The volume of vessel used= 
ml
Initial moles of NO= 
moles
Initial moles of H2= 
moles
Concentration of NO at equilibrium= 
M
Moles of NO at equilibrium= 
=
moles
2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)
<u>Initial</u> :1.3*10^-2 2.6*10^-2 0 0 moles
<u>Equilibrium</u>:1.3*10^-2 - x 2.6*10^-2-x x x/2 moles
∴
⇒
![Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH2O%5D%5E2%5BN2%5D%7D%7B%5BH2%5D%5E2%5BNO%5D%5E2%7D%20%28volume%20of%20vesselin%20litre%29)
<u>Equilibrium</u>:0.31*10^-2 1.61*10^-2 0.99*10^-2 0.495*10^-2 moles
⇒
⇒
First let us determine the electronic configuration of
Bromine (Br). This is written as:
Br = [Ar] 3d10 4s2 4p5
Then we must recall that the greatest effective nuclear
charge (also referred to as shielding) greatly increases as distance of the
orbital to the nucleus also increases. So therefore the electron in the
farthest shell will experience the greatest nuclear charge hence the answer is:
<span>4p orbital</span>
