40 electrons
Explanation:
The N for silicon tetrachloride, SiCl₄ is 40 electrons. The needed electrons that would be used to complete the lewis structure of the compound is actually 40 electrons. This number of electron will help the compound attain a noble configuration.
- The compound SiCl₄ is a covalent one. Here, there is sharing of electrons between two atoms.
- In drawing the electron dot formula, one must take into account the Available electrons and the Needed electrons.
- The Available electrons are sum of the valence electrons that can be accessed for the bonding. Si has 4 valence electrons, Cl has 7 valence electrons this makes a total of 4 + 7(4 atoms of chlorine), 32 available electrons.
- But to make a complete octet like that of noble gases, each atom most have 8 complete outer most electrons. This is the needed number of electrons. Since there are 5 atoms i.e 4 atoms of chlorine and 1 atom of Silicon, the needed electrons will be 5x8 = 40 electrons.
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Answer:
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Answer:
Rank in increasing order of effective nuclear charge:
Explanation:
This explains the meaning of effective nuclear charge, Zeff, how to determine it, and the calculations for a valence electron of each of the five given elements: F, Li, Be, B, and N.
<u>1) Effective nuclear charge definitions</u>
- While the total positive charge of the atom nucleus (Z) is equal to the number of protons, the electrons farther away from the nucleus experience an effective nuclear charge (Zeff) less than the total nuclear charge, due to the fact that electrons in between the nucleus and the outer electrons partially cancel the atraction from the nucleus.
- Such effect on on a valence electron is estimated as the atomic number less the number of electrons closer to the nucleus than the electron whose effective nuclear charge is being determined: Zeff = Z - S.
<u><em>2) Z eff for a F valence electron:</em></u>
- F's atomic number: Z = 9
- Total number of electrons: 9 (same numer of protons)
- Period: 17 (search in the periodic table or do the electron configuration)
- Number of valence electrons: 7 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 9 - 7 = 2
- Zeff = Z - S = 9 - 2 = 7
<u><em>3) Z eff for a Li valence eletron:</em></u>
- Li's atomic number: Z = 3
- Total number of electrons: 3 (same number of protons)
- Period: 1 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 1 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 3 - 1 = 2
- Z eff = Z - S = 3 - 2 = 1.
<em>4) Z eff for a Be valence eletron:</em>
- Be's atomic number: Z = 4
- Total number of electrons: 4 (same number of protons)
- Period: 2 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 2 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 4 - 2 = 2
- Z eff = Z - S = 4 - 2 = 2
<u><em>5) Z eff for a B valence eletron:</em></u>
- B's atomic number: Z = 5
- Total number of electrons: 5 (same number of protons)
- Period: 13 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 3 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 5 - 3 = 2
- Z eff = Z - S = 5 - 2 = 3
<u><em>6) Z eff for a N valence eletron:</em></u>
- N's atomic number: Z = 7
- Total number of electrons: 7 (same number of protons)
- Period: 15 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 5 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 7 - 5 = 2
- Z eff = Z - S = 7 - 2 = 5
<u><em>7) Summary (order):</em></u>
Atom Zeff for a valence electron
- <u>Conclusion</u>: the order is Li < Be < B < N < F
Answer:
Having a specific set of possible values
Answer: 19.4 g/cm3
Explanation: density is the relationship between mass over volume.
So density of gold is 15.7g/0.81cm3 = 19.4 g/cm3
