S and S²⁻ do not have the outer subshell fully filled with electrons.
Explanation:
We look at electronic configurations:
Ca 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² - the outer subshell 4s² is fully-filled with electrons
S 1s² 2s² 2p⁶ 3s² 3p⁴ - the outer subshell 3p⁴ is not fully-filled with electrons
Zn²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s⁰ - here the 4s subshell is higher in energy than 3d subshell so will consider 3d¹⁰ the out subshell which is fully-filled with electrons
S²⁻ 1s² 2s² 2p⁶ 3s² 3p² - the outer subshell 3p² is not fully-filled with electrons
Ca²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ - the outer subshell 3p⁶ is fully-filled with electrons
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electron configurations
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Answer:
C. Lose three electrons to have a full outer shell
Explanation:
Al is in Group 13 of the Periodic Table, so it has three valence electrons.
It must either lose three electrons or gain five to achieve a stable octet.
It is easier to lose three electrons than it is to gain five, so Al loses three electrons.
D. is wrong, for the same reason.
A. is wrong. If Al lost three electrons, it would be breaking into a stable inner shell.
C. is wrong. Al is a metal, so it will lose electrons in a reaction.
We observe that heat capacity of salted water we will find that it is less than pure water. We now that it takes less energy to increase the temperature of the salt water 1°C than pure water. Which means that the salted water heats up faster and eventually reaches to its boiling point first.
hope it helps
Answer:
The answer to your question is: letter c
Explanation:
Data
V1 = 612 ml n1 = 9.11 mol
V2 = 123 ml n2 = ?
Formula
n2 = 1.83 mol
Explanation:
96.485 columbs=1 faraday will
deposit 64/2g= 32 g cu ion
therfore it will require
96,485 ×2/32 =? coulombs or 1/16 of
Faraday= 1 / 16 mole of electrons .
