Answer:
p orbitals only
Explanation:
Carbon has an atomic number of 6 so its electron configuration will be 1s² 2s² 2p². It has two orbitals as indicated with the 2 as its period number with the outer orbital have 4 valence electrons. So carbon is in the p-orbital, period 2 and in group 4.
a) Alkali metals
=> group 1
=> Li: 1s2 2s => 1s
Na: [Ne] 3s => 3s
K: [Ar] 4s => 4s
Rb: [Kr] 5s => 5s
Cs: [Xe] 6s => 6s
Fr: [Rn] 7s => 7s
=> outer electron configuration is ns, where n is the main energy level: 1, 2, 3, 4, 5, 6,7.
b) Alkaline earth metals
=> group 2 => you have to add 1 electron to the alkaly metal of the same row.
=> Be: [He] 2s2 => 2s2
Mg: [Ne] 3s2 => 3s2
Ca: [Ar] 4s2 => 4s2
Sr: [Kr] 5s2 => 5s2
Ba: [Xe] 6s2 => 6s2
Ra: [Rn[ 7s2 => 7s2
=>outer electron configuration is n s2, where n is the main energy level: 1, 2, 3, 4, 5, 6, 7
c) halogens
=> group 7
=> F: [He] 2s2 2p5 => 2s2 2p5
Cl: [Ne] 3s2 3p5 => 3s2 3p5
Br: [Ar] 3d10 4s2 4p5 => 4s2 4p5
I: [Kr] 4d10 5s2 5p6 => 5s2 5p5
At: [Xe] 4f14 5d10 6s2 6p5
=> outer electron configuration is ns2 np5, where n is the main energy level 1, 2, 3, 4, 5, 6, 7
d) Noble gases
=> group 8
I will show only the outer shell which is what is requested
=> He: 1s2
Ne: ... 2s2 2p6
Ar: ... 3s2 3p6
Kr: ... 4s2 4p6
Xe: ... 5s2 5p6
Rn: ... 6s2 6p6
=> the outer electron configuration is ns2 np6, except for He for which it is 1s2
Answer: 3 moles Na
Explanation: To find the number of moles of Na, divide the number of atoms of Na with the Avogadro's Number.
1.806x10²⁴ atoms Na x 1 mole Na / 6.022x10²³ atoms Na
= 2.99 or 3 moles Na
It’s probably Mercury I’m not sure
Answer:
All of the above.
Explanation:
Hello there!
In this case, according to the definition of the Lewis structures, which are represented by the valence electrons, we first identify that the N atom has five valence electrons and each fluorine has seven valence electrons.
In such a way, we cans say that N is the central atom due to its lower electronegativity, the molecule has 7+7+7+5=26 valence electrons and the three F-N bonds are covalent, therefore the answer is all of the above.
Regards!
