Chemistry, 07.11.2020 01:40 hcpscyruscm

Indicate the charge the following elements as they achieve the noble gas configuration. Ga O Br P Rb As
S Mg Al Se Li I

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Answer from: 91miketaylor

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Explanation:

Ga is in group 13 hence it must loose three electrons to form Ga^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.

O is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form O^2- since oxygen has six electrons on its outermost shell.

Br in group 17 has seven electrons in its outermost shell hence it must form Br^- (gain one electron) in order to attain the noble gas configuration.

P in group 15 must accept three electrons and form P^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.

S is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form S^2- since sulphur has six electrons on its outermost shell.

Mg in group 2 has two electrons on its outermost shell and must loose both to attain the noble gas configuration forming Mg^2+.

Al is in group 13 hence it must loose three electrons to form Al^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.

Se is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form Se^2- since selenium has six electrons on its outermost shell.

Lithium is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Li^+.

Rb is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Rb^+.

As in group 15 must accept three electrons and form As^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.

I in group 17 has seven electrons in its outermost shell hence it must form I^- (gain one electron) in order to attain the noble gas configuration.

Answer from: Quest

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.

Answer from: Quest

Nonmetal are moldable and easily shaped

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Indicate the charge the following elements as they achieve the noble gas configuration. Ga O Br P R...

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